The problems in this booklet are organized into strands. A problem often appears in multiple strands. The problems are suitable for most students in


 Morris West
 2 years ago
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1 The problems in this booklet are organized into strands. A problem often appears in multiple strands. The problems are suitable for most students in Grade 11 or higher.
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3 Problem E Mathletes in Action Four mathletes, Jen, John, JP and Judith, competed in the finals of the CEMC Mental Math Bowl. Prizes were awarded for the top three competitors as follows: first place received an anniversary Math Faculty Pink Tie, second place received a book of Brain Teasers, and third place received an I Love Math Tshirt. Three staff members of the CEMC predicted how the prizes would be awarded. Dean predicted that Judith would win the pink tie, John would win the book and JP would win the Tshirt. Ian predicted that Jen would win the pink tie, Judith would win the book and JP would win the Tshirt. Sandy predicted that John would win the pink tie, Jen would win the book and Judith would win the Tshirt. It turns out that each staff member predicted exactly one prize winner correctly. Determine precisely how the prizes were awarded.
4 Problem Problem E and Solution Mathletes in Action Four mathletes, Jen, John, JP and Judith, competed in the finals of the CEMC Mental Math Bowl. Prizes were awarded for the top three competitors as follows: first place received an anniversary Math Faculty Pink Tie, second place received a book of Brain Teasers, and third place received an I Love Math Tshirt. Three staff members of the CEMC predicted how the prizes would be awarded. Dean predicted that Judith would win the pink tie, John would win the book and JP would win the Tshirt. Ian predicted that Jen would win the pink tie, Judith would win the book and JP would win the Tshirt. Sandy predicted that John would win the pink tie, Jen would win the book and Judith would win the Tshirt. It turns out that each staff member predicted exactly one prize winner correctly. Determine precisely how the prizes were awarded. Solution Let s look at Dean s prize predictions. Assume he is correct that Judith won the pink tie. This leads to the following six possibilities. First  Pink Tie Second  Brain Book Third  Tshirt Fourth  No Prize Judith Jen John JP (1) Judith Jen JP John () Judith John Jen JP (3) Judith John JP Jen (4) Judith JP Jen John (5) Judith JP John Jen (6) Since the other two parts of Dean s prize prediction are not correct, John cannot win the book prize so we can rule out (3) and (4), and JP cannot win the T  shirt so we can also rule out (). This leaves (1), (5) and (6) as the only possibilities for Dean. First  Pink Tie Second  Brain Book Third  Tshirt Fourth  No Prize Judith Jen John JP (1) Judith JP Jen John (5) Judith JP John Jen (6) Now look at Ian s prize predictions in light of Dean s three valid possibilities. Ian makes exactly one true prize prediction. None of Dean s possibilities would ever make one of Ian s prize predictions true. Therefore, our original assumption that Dean correctly predicted Judith to win the pink tie was incorrect. So now we assume that Dean correctly predicts that John would win the book prize. This leads to the following six possibilities. First  Pink Tie Second  Brain Book Third  Tshirt Fourth  No Prize Jen John JP Judith (7) Jen John Judith JP (8) JP John Jen Judith (9) JP John Judith Jen (10) Judith John Jen JP (11) Judith John JP Jen (1)
5 Since the other two parts of Dean s prize prediction are not correct, Judith cannot win the pink tie so we can rule out (11) and (1), and JP cannot win the T  shirt so we can also rule out (7). This leaves (8), (9) and (10) as the only possibilities for Dean. First  Pink Tie Second  Brain Book Third  Tshirt Fourth  No Prize Jen John Judith JP (8) JP John Jen Judith (9) JP John Judith Jen (10) Now look at Ian s prize predictions in light of Dean s three valid possibilities. Ian makes exactly one true prize prediction. Possibility (8) is the only one of the possibilities that works for Ian. Now we must check Sandy s prize prediction to see if it is still valid with (8). Sandy predicted that John would win the pink tie, Jen would win the book and Judith would win the Tshirt. This prediction works since exactly one of Sandy s prize predictions, that Judith wins the Tshirt, is true. The other two predictions are false. We should check to see that this is the only correct solution. We do this by assuming Dean s third prediction, JP wins the Tshirt, was his only correct solution. This leads to the following six possibilities. First  Pink Tie Jen Second  Brain Book John Third  Tshirt JP Fourth  No Prize Judith (13) Jen Judith JP John (14) John Jen JP Judith (15) John Judith JP Jen (16) Judith Jen JP John (17) Judith John JP Jen (18) Since the other two parts of Dean s prize prediction are not correct, Judith cannot win the pink tie so we can rule out (17) and (18), and John cannot win the Brain Teaser Book so we can also rule out (13). This leaves (14), (15) and (16) as the only possibilities for Dean. First  Pink Tie Second  Brain Book Third  Tshirt Fourth  No Prize Jen Judith JP John (14) John Jen JP Judith (15) John Judith JP Jen (16) Now look at Ian s prize predictions in light of Dean s three valid possibilities. Ian makes exactly one true prize prediction. Possibility (15) is the only one of the possibilities that works for Ian. In both (14) and (16) Ian would make at least two correct prize predictions. Now we must check Sandy s prize prediction to see if it is still valid with (15). Sandy predicted that John would win the pink tie, Jen would win the book and Judith would win the Tshirt. Since two of Sandy s prize predictions would be true in (15) but Sandy only made one correct prize prediction, (15) does not work for Sandy. the only possibility is that Jen finishes first and wins the pink tie, John finishes second and wins the Brain Teaser book, and Judith finishes third and gets the Tshirt. JP finishes fourth and leaves with the satisfaction of making it to the finals of the Mental Math Bowl. By the way, the correct answer to the Sample Question from the CEMC Mental Math Bowl was Remember, no calculators are used to perform mental math.
6 Problem E What s Your Angle Anyway III? AB is a diameter of a circle with centre C. D is a point on the circumference of the circle other than A and B. Determine the measure of ADB. If you need a hint, consider solving this week s Problem D, What s Your Angle Anyway II?
7 Problem E and Solution What s Your Angle Anyway III? Problem AB is a diameter of a circle with centre C. D is a point on the circumference of the circle other than A and B. Determine the measure of ADB. Solution Join D to the centre C. Since CA, CB and CD are radii of the circle, CA = CB = CD. Since CA = CD, CAD is isosceles and CAD = CDA = x. Since CB = CD, CBD is isosceles and CBD = CDB = y. This new information is marked on the following diagram. The angles in a triangle add to 180 so in ABD ADB + DAB + DBA = 180 (x + y ) + x + y = 180 (x + y ) = 180 x + y = 90 But ADB = x + y so ADB = 90. This result is often expressed as a theorem for circles: An angle ( ADB) inscribed in a circle by the diameter (AB) of a circle is 90.
8 Problem E Partly Salted When one kilogram of salt is added to a solution of salt and water, the solution becomes 1 salt by mass. One kilogram of water is then added to the new 3 solution resulting in a solution that is 3 10 salt by mass. What fraction of the original solution was salt? +
9 Problem E and Solution Partly Salted Problem When one kilogram of salt is added to a solution of salt and water, the solution becomes 1 3 salt by mass. One kilogram of water is then added to the new solution resulting in a solution that is 3 10 Solution salt by mass. What fraction of the original solution was salt? Let S represent the mass of salt in the original solution. Let W represent the mass of water in the original solution. Then S + W represents the total mass of the original solution. After adding 1 kg of salt to the original solution, there is (S + 1) kg of salt and (S + W + 1) kg of solution. Since this new solution is one third salt by mass, S + 1 S + W + 1 = 1 3 3(S + 1) = 1(S + W + 1) 3S + 3 = S + W + 1 S + = W (1) After adding 1 kg of water to the new solution, there is still (S + 1) kg of salt in (S + W + ) kg of solution. Since this new solution is three tenths salt by mass, S + 1 S + W + = (S + 1) = 3(S + W + ) Substituting (1) into (), 10S + 10 = 3S + 3W + 6 7S 3W = 4 () 7S 3(S + ) = 4 7S 6S 6 = 4 S = kg Substituting for S in (1), W = 6 kg. Then the original solution was S + W = 8 kg. The fraction of the original solution that was salt was 8 = 1 4. one quarter of the original solution was salt by mass.
10 Problem E Pair Possibilities Determine all possible ordered pairs of positive integers (a, b) that satisfy 1 a + b = 8 a + b and a + 5b 54.
11 Problem E and Solution Pair Possibilities Problem Determine all possible ordered pairs of positive integers (a, b) that satisfy 1 a + b = 8 and a + 5b 54. a + b Solution Since a and b are positive integers satisfying a + 5b 54, we could write out all ordered pairs that satisfy this inequality and then determine which ones also satisfy the first equation. There will be a large number of possibilities to check so we need to find a way to reduce the number of possibilities. Working with the first equation: Find a common denominator: b + a ab = 8 a + b Crossmultiply : (b + a)(a + b) = 8ab Expand and simplify: 4a + 4ab + b = 8ab It follows that a b = 0 and b = a. Rearrange: 4a 4ab + b = 0 Factor: (a b) = 0 Each of the ordered pairs (a, b) will look like (a, a). We substitute a for b in the inequality obtaining a + 5(a) 54 or 1a 54 or a 4.5. Since a is a positive integer, a can only take on integer values 1,, 3 and 4. The ordered pairs follow quickly: {(1, ), (, 4), (3, 6), (4, 8)}. the only ordered pairs of positive integers (a, b) that satisfy 1 a + b = 8 a + b and a + 5b 54 are (1, ), (, 4), (3, 6), and (4, 8).
12 Problem E Get Your Analytic Skills On OAB is an isosceles right triangle with vertex O at the origin (0, 0), vertices A and B on the line x + 3y 13 = 0 and AOB = 90. Determine the area of OAB.
13 Problem OAB is an isosceles right triangle with vertex O at the origin (0, 0), vertices A and B on the line x + 3y 13 = 0 and AOB = 90. Determine the area of OAB. Solution Problem E and Solution Get Your Analytic Skills On In analytic geometry problems, a representative diagram is important and often provides clues for the solution of the problem. The diagram above has the given information plus a couple of pieces of information that will be justified now. Let B have coordinates (p, q). The slope of OB = q. Since AOB = p 90, OB OA and the slope of OA is the negative reciprocal of OB. Therefore the slope of OA = p. Since the q triangle is isosceles, OA = OB and it follows that the coordinates of A are ( q, p). (We can verify this by finding the length of OA and the length of OB and showing that both are equal to p + q.) Since B(p, q) is on the line x + 3y 13 = 0, it satisfies the equation of the line. p + 3q 13 = 0 (1) Since A( q, p) is on the line x + 3y 13 = 0, it satisfies the equation of the line. q + 3p 13 = 0 or 3p q 13 = 0 () Since we have two equations and two unknowns, we can use elimination to solve for p and q. (1) 4p + 6q 6 = 0 () 3 9p 6q 39 = 0 Adding, we obtain 13p 65 = 0 p = 5 Substituting in (1) q 13 = 0 3q = 3 q = 1 The point B is (5, 1) and the length of OB = = 6. Since OA = OB, OA = 6. AOB is a right triangle so we can use OB as the base and OA as the height in the formula for the area of a triangle. Then the area of AOB = OA OB = 6 6 = 13. the area of AOB is 13 units.
14 Problem E Enough Information? In P QR, P Q = 7, P R = 9 and median P M = 7. Determine the length of QR.
15 Problem Problem E and Solutions Enough Information? In P QR, P Q = 7, P R = 9 and median P M = 7. Determine the length of QR. Solution 1 Since P Q = P M = 7, P QM is isosceles. In P QM, draw an altitude from P to QM, intersecting at N. In an isosceles triangle, the altitude drawn to the base bisects the base. Therefore QN = NM = x. Since P M is a median in P QR, MR = QM = x. Let P N = h. P NM is a right triangle. Using Pythagoras Theorem, P N = P M NM h = 7 x h = 49 x (1) P NR is a right triangle. Using Pythagoras Theorem, P N = P R NR h = 9 (x + x) h = 81 (3x) h = 81 9x () In equations (1) and (), the left side of each equation is h. Therefore, the right side of equation (1) must equal the right side of equation (). So 49 x = 81 9x x + 9x = x = 3 x = 4 x =, x > 0 QR = QN + NM + MR = x + x + x = 4x = 8 units.
16 Problem In P QR, P Q = 7, P R = 9 and median P M = 7. Determine the length of QR. Solution This solution is presented for students who have done some trigonometry and know the law of cosines. Since P M is a median, QM = MR = x. Then QR = x. Using the law of cosines in P QM, P M = P Q + QM (P Q)(QM)cos(Q) 7 = 7 + x (7)(x)cos(Q) 49 = 49 + x 14xcos(Q) 14xcos(Q) = x (1) Using the law of cosines in P QR, P R = P Q + QR (P Q)(QR)cos(Q) 9 = 7 + (x) (7)(x)cos(Q) 81 = x 8xcos(Q) 8xcos(Q) = 4x 3 () Using elimination to solve for x, (1) 8xcos(Q) = x () 8xcos(Q) = 4x 3 Subtracting 0 = x + 3 x = 3 x = 16 x = 4, x > 0 QR = x = 8 the length of QR is 8 units.
17 Problem E A Simple System of Equations? If 3x = 16 y+1 and x = 5y 17, determine the value of x + y. Try to see if doing this problem without a calculator affects how you think about the problem.
18 Problem E and Solutions A Simple System of Equations? Problem If 3x = 16 y+1 and x = 5y 17, determine the value of x + y. Solution 1 For this solution we will attempt to answer the question only. We will not determine the values of x and y that generate the sum. 3x = 16 y+1 3x = ( 4) y+1 3x = 4y+4 3x = 4y + 4 (1) But x = 5y 17 () (1) () x = y + 1 Rearranging x + y = 1 x + y = 1. Notice that the problem only asks for x + y, it is not necessary to find values for x and y. Solution This solution carries on from equations (1) and () in solution 1 to find the values of x and y, and then determines the sum. (1) 5 15x = 0y + 0 (3) () 4 8x = 0y 68 (4) (3) (4) 7x = 88 x = 88 ( ) 7 88 Substitue in () = 5y 17 7 Multiply by = 35y 119 As before (but with much more work) x + y = = 35y y = = 59 7 x + y = = = 1
19 Problem E You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. The following information may be helpful in the solution of the problem. An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3,5,7,9 is an arithmetic sequence with four terms and constant difference. The general term of an arithmetic sequence is t n = a + (n 1)d, where a is the first term, d is the constant difference and n is the number of terms. The sum, S n, of the first n terms of an arithmetic sequence can be found using either S n = n ( ) [a + (n 1)d] or S t1 + t n n = n, where t 1 is the first term of the sequence and t n is the n th term of the sequence. The following example is provided to verify the accuracy of the formulas and to illustrate their use. For the arithmetic sequence 3,5,7,9, a = t 1 = 3, d =, n = 4 and t n = t 4 = 9. S n = = 4 S n = n [a + (n 1)d] = 4 [(3) + (3)] = [1] = 4 ( ) ( ) t1 + t n S n = n = 4 = 4(6) = 4
20 Problem Problem E and Solutions You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Three solutions are provided for this problem. Solution 1 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 5 = 3 and t = a 3 a = 1 5 = 7. Since the new sequence is arithmetic, the difference between consecutive terms is constant. Therefore the constant difference is d = t t 1 = 7 3 = 4. We can now generate a few more terms in the sequence of differences by adding the constant to the previous term. t 3 = t + 4 = = 11 and t 4 = t = = 15. Using the information we can present the information in tabular form. Term Number First Sequence First Difference Second Difference n a n t n (Difference of the differences) Since the second difference is constant we can represent the general term of the first sequence with a quadratic function. Let a n = pn + qn + r. For n = 1, a 1 = = p(1) + q(1) + r. p + q + r =. (1) For n =, a = 5 = p() + q() + r. 4p + q + r = 5. () For n = 3, a 3 = 1 = p(3) + q(3) + r. 9p + 3q + r = 1. (3) Subtracting (1) from (), 3p + q = 3. (4) Subtracting () from (3), 5p + q = 7. (5) Subtracting (4) from (5), p = 4 and p = follows. Substituting p = into (4), 3() + q = 3 and q = 3 follows. Substituting p =, q = 3 into (1), 3 + r = and r = 3 follows. a n = n 3n + 3 is the general term of the given sequence in terms of n.
21 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. We can generate more terms: t 3 = 11, t 4 = 15, t 5 = 19,. Each term in the new sequence is the difference between consecutive terms of the original sequence. a a 1 = t 1 a 3 a = t a 4 a 3 = t 3 a n a n 3 = t n 3 a n 1 a n = t n a n a n 1 = t n 1 Adding, a n a 1 = t 1 + t + t t n + t n 1 = S n 1. (1) To find the sum S n 1 = t 1 + t + t t n + t n 1 we can use the formula S n = n [a + (n 1)d] with a = 3 and d = 4 for (n 1) terms. So S n 1 = t 1 + t + t t n + t n 1 = n 1 [a + ((n 1) 1)d] = n 1 [(3) + ((n 1) 1)(4)] = n 1 [6 + (n )(4)] = n 1 [6 + 4n 8] = n 1 [4n ] = (n 1)(n 1) From (1) above, a n a 1 = S n 1 so a n = S n 1 + a 1 = (n 1)(n 1) + = n 3n + 3. a n = n 3n + 3 is the general term of the given sequence in terms of n.
22 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution 3 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. Using the formula for the general term of an arithmetic sequence, t n = a + (n 1)d, the general term of the sequence of differences is t n = (3) + (n 1)(4) = 3 + 4n 4 = 4n 1. To generate the original sequence we start with the first term and add more terms from the arithmetic sequence of differences. For example, a 1 = 3, a = + t 1 = + [4(1) 1] = + 3 = 5, and a 3 = + t 1 + t = + [4(1) 1] + [4() 1] = = 1 So, a n = + t 1 + t + t t n 1 = + [4(1) 1] + [4() 1] + [4(3) 1] + + [4(n 1) 1] = + [4(1) + 4() + 4(3) + + 4(n 1)] + (n 1)( 1) = + 4[ (n 1)] n + 1 [ ] (n 1)(n) = 3 n + 4 (1) = 3 n + (n n) = 3 n + n n = n 3n + 3 In (1) above, (n 1) is an arithmetic sequence with t 1 = 1, t n 1 = (n 1) and the number of terms (n ( 1) terms. ) Using the formula for the sum of the terms of an t1 + t n arithmetic sequence, S n = n, we obtain ( ) 1 + (n 1) ( n ) (n 1)(n) S n 1 = (n 1) = (n 1) = a n = n 3n + 3 is the general term of the given sequence in terms of n.
23 Problem E Love is Blind Valentine A heart is constructed by attaching two white semicircles, each with radius cm, onto the hypotenuse of an isosceles right triangle. The heart is then pasted onto a rectangular sheet of red construction paper as illustrated below. (The dashed lines, the side measurement and the right angle symbol will not actually be on the finished card.) Cupid shoots an arrow that lands somewhere on the card. Determine the probability that Cupid s arrow lands on the heart. cm
24 Problem E and Solution Love is Blind Valentine Problem A heart is constructed by attaching two white semicircles, each with radius cm, onto the hypotenuse of an isosceles right triangle. The heart is then pasted onto a rectangular sheet of red construction paper. Cupid shoots an arrow that lands somewhere on the card. Determine the probability that Cupid s arrow lands on the heart. Solution The semicircles are placed along the hypotenuse of the triangle. The hypotenuse is equal in length to four radii and is therefore 4 = 8 cm long. Let x represent the length of the sides of equal length in the triangle and let h represent the height of the triangle drawn from the vertex between the two equal sides to the opposite side. Place the given information on the diagrams below. cm x h x 8 cm h x x Since the triangle is a right triangle we can use Pythagoras Theorem to find the value of x. x + x = 8 x = 64 x = 3 x = 4, since x > 0 The altitude of an isosceles, right triangle bisects the hypotenuse so we can determine the height of the triangle, h. This statement is justified at the end of the solution. h + 4 = x h + 16 = 3 h = 16 h = 4, since h > 0 We now have enough information to find the area of the heart and the entire shape.
25 The total height of the rectangle is equal to the height of the triangle plus the radius of the semicircle. The height of the rectangle is 4 + = 6 cm. The width of the rectangle is the same as the length of the hypotenuse and is therefore equal to 8 cm. The area of the rectangle is 8 6 = 48 cm. The area of the heart is equal to the area of the triangle plus the area of two congruent semicircles of radius cm. Since the triangle is isosceles right we can use one of the equal sides as the base and the other as the height. The area of the triangle is bh = (4 ) (4 ) = 3 = 16 cm. The area of the two semicircles is the same as the area of one circle with radius cm. The area of the two semicircles is πr = π() = 4π cm. The total area of the heart is (4π + 16) cm. The probability that Cupid s arrow lands on the heart is equal to the area of the heart divided by the area of the rectangle. The probability equals 4π+16 = In other words, Cupid has 48 about a 60% chance of landing his arrow on the heart. Happy Valentine s Day. Justification of the statement: The altitude of an isosceles right triangle bisects the hypotenuse. There are many ways to justify this. Since the triangle is an isosceles right triangle, the two angles opposite the sides of equal length are equal. In a triangle, the sum of the angles is 180 so the two equal angles must add to the remaining 90 and each angle is 45. Construct the altitude, h, as shown in the following diagram, splitting the opposite sides into two parts, labelled a and b. x a h b x The altitude hits the opposite side at 90 so each of the two smaller triangles contains a right angle and a 45 angle. The missing angle in each of the two smaller triangles is then 45. Each of the smaller triangles is then isosceles and it follows that h = a and h = b. a = b and the altitude bisects the hypotenuse.
26 Problem E Seeing Red A cube has edges of length n, where n is a positive integer. Three faces, meeting at a corner, are painted red. The cube is then cut into n 3 smaller cubes of unit length. That is, the side lengths of the new cubes are 1 unit. If exactly 15 of these cubes have no faces painted red, determine the value of n. n n n
27 Problem E and Solutions Seeing Red Problem A cube has edges of length n, where n is a positive integer. Three faces, meeting at a corner, are painted red. The cube is then cut into n 3 smaller cubes of unit length. That is, the side lengths of the new cubes are 1 unit. If exactly 15 of these cubes have no faces painted red, determine the value of n. n 1 n1 A n B 1 1 n1 n1 C Solution 1 n This solution requires no content beyond grade ten. The second solution will use the factor theorem which is generally taught in grade twelve. In the diagram above, the sides painted red are labelled A, B, and C. We know that there are n 3 unit cubes. To determine the number of unpainted cubes we can subtract the number of cubes with some red from the total number of cubes. Side A has dimensions n by n by 1 and so contains n unit cubes with some red. Side B has dimensions n by (n 1) by 1 and so contains n (n 1) unit cubes with some red. Side C has dimensions (n 1) by (n 1) by 1 and so contains (n 1)(n 1) unit cubes with some red. The number of unpainted cubes is n 3 n n(n 1) (n 1)(n 1). We can simplify this as follows: n 3 n n(n 1) (n 1)(n 1) = n (n 1) n(n 1) (n 1)(n 1) Each term contains a common factor of (n 1) so the expression simplifies to (n 1)(n n (n 1)) = (n 1)(n n + 1). This further simplifies to (n 1) 3. If the solver pauses here to think about this, if the unit cubes on side A then side B and finally side C are removed we are left with a cube whose side lengths are (n 1) and (n 1) 3 unit cubes. But (n 1) 3 = 15, the actual number of unpainted cubes. Taking the cube root, n 1 = 5 and n = 6 follows.
28 Solution The second solution will use the factor theorem which is generally taught in grade twelve. This solution picks up from the expression giving us the number of unpainted cubes, setting it equal to 15, the number of unpainted cubes. n 3 n n(n 1) (n 1)(n 1) = 15 n 3 n n + n n + n 1 = 15 n 3 3n + 3n 16 = 0 Let f(n) = n 3 3n + 3n 16. When n = 6, f(6) = 6 3 3(6 ) + 3(6) 16 = = 4 4 = 0. Since f(6) = 0, (n 6) is a factor of f(n). After long division (or synthetic division), f(n) = (n 6)(n + 3n + 1). So (n 6)(n + 3n + 1) = 0. n + 3n + 1 = 0 has no real roots so n = 6 is the only root. Therefore the original cube has edges of length 6.
29 Problem E This Triangle is All Right A rightangled triangle has sides whose lengths are twodigit integers. The digits of the length of the hypotenuse are the reverse of the digits of the length of one of the other sides. Determine the smallest possible length of the hypotenuse. a c b
30 Problem E and Solution Problem This Triangle is All Right A rightangled triangle has sides whose lengths are twodigit integers. The digits of the length of the hypotenuse are the reverse of the digits of the length of one of the other sides. Determine the smallest possible length of the hypotenuse. Solution Let x represent the tens digit of the hypotenuse such that x is an integer from 1 to 9. Let y represent the units digit of the hypotenuse such that y is an integer from 1 to 9. Then the length of the hypotenuse is 10x + y. Since one of the sides has the same digits as the hypotenuse in reverse order, the length of this side is 10y + x. Let the third side be z such that z is a two digit integer. Using the Pythagorean Theorem: (10y + x) + z = (10x + y) Expanding: 100y + 0xy + x + z = 100x + 0xy + y Rearranging: z = 99x 99y Factoring: z = 99(x y)(x + y) Since z is a perfect square, 99(x + y)(x y) must also be a perfect square. But 99(x + y)(x y) = 9(11)(x + y)(x y). To be a perfect square (x + y)(x y) must be a multiple of 11 and contain a factor which is a perfect square. Since x and y are each integers from 1 to 9, x y cannot be 11, and so we must have x + y = 11. Again, since x and y are integers from 1 to 9, there are three possibilities for x y that give a perfect square; x y = 1, x y = 4 or x y = 9. These are the three possibilities: (x + y)(x y) = 11(1) or (x + y)(x y) = 11(4) or (x + y)(x y) = 11(9). If x + y = 11 and x y = 1, we solve the system of equations obtaining x = 6 and y = 5. This gives a hypotenuse of 10x + y = 65 and second side 10y + x = 56. Then solving for z, z = 99(x + y)(x y) = 99(11)(1) = 1089 and z = 33. This solution is easily confirmed but is it the only solution? If x + y = 11 and x y = 4, we solve the system of equations obtaining x = 7.5 and y = 3.5. But x and y are both integers so this solution is inadmissible. If x + y = 11 and x y = 9, we solve the system of equations obtaining x = 10 and y = 1. But x must be an integer from 1 to 9 so this solution is inadmissible. Therefore the only solution is a right triangle with hypotenuse of length 65.
31 Problem E Advanced Training A blue train is 1000 km north of a red train. The blue train is travelling south, and the red train is travelling east. Both trains travel at the same speed of 100 km/h. How much time elapses until the total distance travelled by the two trains is equal to the distance between the trains.
32 Problem E and Solution Advanced Training Problem A blue train is 1000 km north of a red train. The blue train is travelling south, and the red train is travelling east. Both trains travel at the same speed of 100 km/h. How much time elapses until the total distance travelled by the two trains is equal to the distance between the trains. Solution Let t be the time in hours until the total distance travelled by the two trains is equal to the distance between the trains. Since each train is travelling at 100 km/h, the distance travelled by each train in t hours is 100t km. The total distance travelled by the two trains is 100t = 00t km. The following diagram represents the positions of the two trains after t hours. The blue train starts at B and moves to C. The red train starts at R and moves to S. Then BC = RS = 100t and CR = t. We want the time t when CS = BC + RS = 100t + 100t = 00t. B 1000 km C 100t km R 100t km S But CRS is right angled so CS = CR + RS 0000t t = 0 (00t) = ( t) + (100t) 40000t = t t t t + 10t 50 = 0 Using the Quadratic Formula, t = 10 ± 100 4( 50) t = 10 ± 10 3 t = 5 ± 5 3 Since t > 0, is inadmissible. t = = 3.66 hours. The distance between the two trains will be equal to their total distance travelled in ( ) hours which is approximately 3 hours and 40 minutes after they leave their initial positions.
33 Problem E Median Madness A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. median In ABC, ABC = 90. A median is drawn from A meeting BC at M such that AM = 5. A second median is drawn from C meeting AB at N such that CN = 10. Determine the length of the longest side of ABC.
34 Problem E and Solution Median Madness Problem A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. In ABC, ABC = 90. A median is drawn from A meeting BC at M such that AM = 5. A second median is drawn from C meeting AB at N such that CN = 10. Determine the length of the longest side of ABC. Solution A x N x B y M y C Since AM is a median, M is the midpoint of BC. Then BM = MC = y. Since CN is a median, N is the midpoint of AB. Then AN = NB = x. NBC is right angled since B = 90. Using the Pythagorean Theorem, NB + BC = CN x + (y) = ( 10) x + 4y = 40 (1) ABM is right angled since B = 90. Using the Pythagorean Theorem, AB + BM = AM (x) + y = 5 4x + y = 5 () Adding (1) and (), 5x + 5y = 65 Dividing by 5, x + y = 13 (3) The longest side of ABC is the hypotenuse AC. Using the Pythagorean Theorem, AC = AB + BC = (x) + (y) = 4x + 4y = 4(x + y ) Substituting from (3) above, AC = 4(13) Taking the square root, AC = 13 the length of the longest side is 13 units. Note: The solver could actually solve a system of equations to find x = and y = 3 and then proceed to solve for the longest side. The above approach was provided to expose the solver to alternate thinking about the solution of this problem.
35 Problem E A NotSoAverage Task Four numbers are selected such that when each number is added to the average of the other three the following sums are obtained: 5, 37, 43, and 51. Determine the average of the four numbers.
36 Problem Problem E and Solution A NotSoAverage Task Four numbers are selected such that when each number is added to the average of the other three the following sums are obtained: 5, 37, 43, and 51. Determine the average of the four numbers. Solution It is possible to precisely determine the four numbers but the problem only asks for their average. Let a, b, c, d represent each of the four numbers. We are looking for a+b+c+d 4. When the first number is added to the average of the other three numbers the result is 5. a + b + c + d = 5 which simplifies to 3a + b + c + d = 75 (1) 3 When the second number is added to the average of the other three numbers the result is 37. b + a + c + d = 37 which simplifies to a + 3b + c + d = 111 () 3 When the third number is added to the average of the other three numbers the result is 43. c + a + b + d = 43 which simplifies to a + b + 3c + d = 19 (3) 3 When the fourth number is added to the average of the other three numbers the result is 51. d + a + b + c = 51 which simplifies to a + b + c + 3d = 153 (4) 3 If we add (1), (), (3), and (4) we get 6a + 6b + 6c + 6d = = 468. Dividing by 6, a + b + c + d = 78. The average of the four numbers is 78 4 = Therefore the average of the four numbers is (By solving the system of equations we can actually determine that the numbers are 1.5, 16.5, 5.5 and 37.5.)
37 Problem E Five Solutions  Really? There are five values of x that satisfy the equation (x 5x + 5) x +4x 60 = 1. Determine these five values of x. x 1 = x = x 3 = x 4 = x 5 =
38 Problem E and Solution Five Solutions  Really? Problem There are five values of x that satisfy the equation (x 5x + 5) x +4x 60 = 1. Determine these five values of x. Solution Let s consider the ways that an expression of the form a b can be 1: The base, a, is 1. In this case, the exponent can be any value and we need to solve x 5x + 5 = 1. So x = 4 or x = 1. x 5x + 5 = 1 x 5x + 4 = 0 (x 4)(x 1) = 0 The exponent, b, is 0. In this case, the base can be any number other than 0 and we need to solve x + 4x 60 = 0. So x = 6 or x = 10. x + 4x 60 = 0 (x 6)(x + 10) = 0 When x = 6, the base is 6 5(6) + 5 = When x = 10, the base is ( 10) 5( 10) + 5 = The base, a, is 1 and the exponent, b, is even. We first need to solve x 5x + 5 = 1. So x = or x = 3. x 5x + 5 = 1 x 5x + 6 = 0 (x )(x 3) = 0 When x =, the exponent is + 4() 60 = 48, which is even. Therefore, when x =, (x 5x + 5) x +4x 60 = 1. When x = 3, the exponent is 3 + 4(3) 60 = 39, which is odd. Therefore, when x = 3, (x 5x + 5) x +4x 60 = 1. So x = 3 is not a solution. Therefore, the values of x that satisfy (x 5x + 5) x +4x 60 = 1 are x = 1, x =, x = 4, x = 6 and x = 10.
39 Problem E A Skinny Quadrilateral In the diagram, QP 1 R 1 is rightangled with QP 1 = and QR 1 = 5. Lines QP 1 and QR 1 are extended and many more points are labelled at intervals of 1 unit, so that P 1 P = P P 3 = P 3 P 4 = P 4 P 5 = = 1, and R 1 R = R R 3 = R 3 R 4 = R 4 R 5 = = 1. In fact, P 1 P j = j 1 and R 1 R k = k 1 for any positive integers j and k. For example, P 1 P 5 = 5 1 = 4 and R 1 R 4 = 4 1 = 3. Determine the value of n so that the area of quadrilateral P n P n+1 R n+1 R n is 01. That is, determine the value of n so that the area of the quadrilateral with vertices P n, P n+1, R n+1, and R n is 01.
40 Problem E and Solution A Skinny Quadrilateral Problem In the diagram, QP 1 R 1 is rightangled with QP 1 = and QR 1 = 5. Lines QP 1 and QR 1 are extended and many more points are labelled at intervals of 1 unit, so that P 1 P = P P 3 = P 3 P 4 = P 4 P 5 = = 1, and R 1 R = R R 3 = R 3 R 4 = R 4 R 5 = = 1. In fact, P 1 P j = j 1 and R 1 R k = k 1 for any positive integers j and k. For example, P 1 P 5 = 5 1 = 4 and R 1 R 4 = 4 1 = 3. Determine the value of n so that the area of quadrilateral P n P n+1 R n+1 R n is 01. Solution In order to solve the problem, looking at the calculation of a specific area may prove helpful. So let s determine the area of quadrilateral P 4 P 5 R 5 R 4. Area of quadrilateral P 4 P 5 R 5 R 4 = Area P 5 QR 5 Area P 4 QR 4 = 1 (QP 5)(QR 5 ) 1 (QP 4)(QR 4 ) = 1 ( + (5 1))(5 + (5 1)) 1 ( + (4 1))(5 + (4 1)) = 1 (6)(9) 1 (5)(8) = 7 0 = 7 units We will pattern the more general solution off the above example. Area of quad. P n P n+1 R n+1 R n = Area P n+1 QR n+1 Area P n QR n 01 = 1 (QP n+1)(qr n+1 ) 1 (QP n)(qr n ) 01 = 1 ( + ((n + 1) 1))(5 + ((n + 1) 1)) 1 ( + (n 1))(5 + (n 1)) 01 = 1 ( + n)(5 + n) 1 (1 + n)(4 + n) 404 = ( + n)(5 + n) (1 + n)(4 + n) multiplying by 404 = n + 7n + 10 (n + 5n + 4) 404 = n + 7n + 10 n 5n = n = n 009 = n Therefore the value of n is 009. Using the method of the specific example, this result is easily confirmed.
41
42 Problem E Powerful Factors at Work The product of the integers 1 to 64 can be written in an abbreviated form as 64! and we say 64 factorial. So 64! = In general, the product of the positive integers 1 to n is n! = n (n 1) (n ) 3 1. Determine the largest positive integer value of n so that 64! is divisible by 1 n.
43 Problem E and Solution Powerful Factors at Work Problem The product of the integers 1 to 64 can be written in an abbreviated form as 64! and we say 64 factorial. So 64! = In general, the product of the positive integers 1 to n is n! = n (n 1) (n ) 3 1. Determine the largest positive integer value of n so that 64! is divisible by 1 n. Solution The prime factorization of 1 is 3. We must determine how many times the factors 3 are repeated in the factorization of P. First we will count the number of factors of in P by looking at the 3 even numbers. Each of the numbers {, 4, 6,, 60, 6, 64} contains a factor of. That is a total of 3 factors of. Dividing the even numbers by, we obtain the numbers {1,, 3,, 30, 31, 3}. This list contains 16 even numbers so we gain another 16 factors of bringing the total to = 48. Dividing the even numbers by, we obtain the numbers {1,, 3,, 14, 15, 16}. This list contains 8 even numbers so we gain another 8 factors of bringing the total to = 56. Dividing the even numbers by, we obtain the numbers {1,, 3,, 6, 7, 8}. This list contains 4 even numbers so we gain another 4 factors of bringing the total to = 60. Dividing the even numbers by, we obtain the numbers {1,, 3, 4}. This list contains even numbers so we gain another factors of bringing the total to 60 + = 6. Finally, dividing the even numbers by, we obtain the numbers {1, }. This list contains 1 even number so we gain another factor of bringing the total to = 63. So when P is factored there are 63 factors of. In fact, the largest power of that P is divisible by is 63. Next we will count the number of factors of 3 in P by looking at the 1 multiples of 3, namely {3, 6, 9,, 57, 60, 63}. Each of these numbers 1 numbers contains a factor of 3. Dividing the numbers by 3, we obtain the numbers {1,, 3,, 19, 0, 1}. This list contains 7 multiples of 3 so we gain another 7 factors of 3 bringing the total to = 8. Dividing the multiples of 3 by 3, we obtain the numbers {1,, 3, 4, 5, 6, 7}. This list contains multiples of 3 so we gain another factors of 3 bringing the total to 8 + = 30. Dividing the multiples of 3 by 3, we obtain the numbers {1, }. This list contains no multiples of 3 so when P is factored there is a total of 30 factors of 3. In fact, the largest power of 3 that P is divisible by is Combining the results P is divisible by = = 3 ( ) = 3 (( ) 3) 30 = 3 (1) 30 P is divisble by 1 30 and the largest value of n is 30. (Since all of the powers of 3 have been used, none remain to combine with any of the remaining s to form an additional factor of 1.)
44 Problem E You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. The following information may be helpful in the solution of the problem. An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3,5,7,9 is an arithmetic sequence with four terms and constant difference. The general term of an arithmetic sequence is t n = a + (n 1)d, where a is the first term, d is the constant difference and n is the number of terms. The sum, S n, of the first n terms of an arithmetic sequence can be found using either S n = n ( ) [a + (n 1)d] or S t1 + t n n = n, where t 1 is the first term of the sequence and t n is the n th term of the sequence. The following example is provided to verify the accuracy of the formulas and to illustrate their use. For the arithmetic sequence 3,5,7,9, a = t 1 = 3, d =, n = 4 and t n = t 4 = 9. S n = = 4 S n = n [a + (n 1)d] = 4 [(3) + (3)] = [1] = 4 ( ) ( ) t1 + t n S n = n = 4 = 4(6) = 4
45 Problem Problem E and Solutions You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Three solutions are provided for this problem. Solution 1 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 5 = 3 and t = a 3 a = 1 5 = 7. Since the new sequence is arithmetic, the difference between consecutive terms is constant. Therefore the constant difference is d = t t 1 = 7 3 = 4. We can now generate a few more terms in the sequence of differences by adding the constant to the previous term. t 3 = t + 4 = = 11 and t 4 = t = = 15. Using the information we can present the information in tabular form. Term Number First Sequence First Difference Second Difference n a n t n (Difference of the differences) Since the second difference is constant we can represent the general term of the first sequence with a quadratic function. Let a n = pn + qn + r. For n = 1, a 1 = = p(1) + q(1) + r. p + q + r =. (1) For n =, a = 5 = p() + q() + r. 4p + q + r = 5. () For n = 3, a 3 = 1 = p(3) + q(3) + r. 9p + 3q + r = 1. (3) Subtracting (1) from (), 3p + q = 3. (4) Subtracting () from (3), 5p + q = 7. (5) Subtracting (4) from (5), p = 4 and p = follows. Substituting p = into (4), 3() + q = 3 and q = 3 follows. Substituting p =, q = 3 into (1), 3 + r = and r = 3 follows. a n = n 3n + 3 is the general term of the given sequence in terms of n.
46 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. We can generate more terms: t 3 = 11, t 4 = 15, t 5 = 19,. Each term in the new sequence is the difference between consecutive terms of the original sequence. a a 1 = t 1 a 3 a = t a 4 a 3 = t 3 a n a n 3 = t n 3 a n 1 a n = t n a n a n 1 = t n 1 Adding, a n a 1 = t 1 + t + t t n + t n 1 = S n 1. (1) To find the sum S n 1 = t 1 + t + t t n + t n 1 we can use the formula S n = n [a + (n 1)d] with a = 3 and d = 4 for (n 1) terms. So S n 1 = t 1 + t + t t n + t n 1 = n 1 [a + ((n 1) 1)d] = n 1 [(3) + ((n 1) 1)(4)] = n 1 [6 + (n )(4)] = n 1 [6 + 4n 8] = n 1 [4n ] = (n 1)(n 1) From (1) above, a n a 1 = S n 1 so a n = S n 1 + a 1 = (n 1)(n 1) + = n 3n + 3. a n = n 3n + 3 is the general term of the given sequence in terms of n.
47 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution 3 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. Using the formula for the general term of an arithmetic sequence, t n = a + (n 1)d, the general term of the sequence of differences is t n = (3) + (n 1)(4) = 3 + 4n 4 = 4n 1. To generate the original sequence we start with the first term and add more terms from the arithmetic sequence of differences. For example, a 1 = 3, a = + t 1 = + [4(1) 1] = + 3 = 5, and a 3 = + t 1 + t = + [4(1) 1] + [4() 1] = = 1 So, a n = + t 1 + t + t t n 1 = + [4(1) 1] + [4() 1] + [4(3) 1] + + [4(n 1) 1] = + [4(1) + 4() + 4(3) + + 4(n 1)] + (n 1)( 1) = + 4[ (n 1)] n + 1 [ ] (n 1)(n) = 3 n + 4 (1) = 3 n + (n n) = 3 n + n n = n 3n + 3 In (1) above, (n 1) is an arithmetic sequence with t 1 = 1, t n 1 = (n 1) and the number of terms (n ( 1) terms. ) Using the formula for the sum of the terms of an t1 + t n arithmetic sequence, S n = n, we obtain ( ) 1 + (n 1) ( n ) (n 1)(n) S n 1 = (n 1) = (n 1) = a n = n 3n + 3 is the general term of the given sequence in terms of n.
48 Problem E Pick a Card  Any Card Luke has a unique deck of cards. Each card in the deck has a positive three digit number on it. There is exactly one card in the deck for every three digit positive integer. The deck is shuffled. Luke selects a card at random from the deck and calculates the sum of the digits. If the sum of the digits is 15, the card is a winner. Determine the probability that Luke selects a winning card.
49 Problem E and Solution Pick a Card  Any Card Problem Luke has a unique deck of cards. Each card in the deck has a positive three digit number on it. There is exactly one card in the deck for every three digit positive integer. The deck is shuffled. Luke selects a card at random from the deck and calculates the sum of the digits. If the sum of the digits is 15, the card is a winner. Determine the probability that Luke selects a winning card. Solution To begin, we need to determine the number of cards in the deck. Since there is a card for each three digit positive integer there are 900 cards in the deck. We must be careful calculating this number. There are 999 positive integers less than Of this set, 99 are two digit numbers. Therefore there are = 900 three digit positive integers. Next we need to determine the digit combinations on a card that have a sum of 15. We will determine the possibilities using cases. Then we will look at the specific groups of numbers that sum to 15 to count the number of cards produced from each group. 1. One of the digits on the card is a zero. The other two digits on the card must add to 15. This leads to two groups of numbers: (0,6,9) and (0,7,8).. One of the digits on the card is a one but the number does not contain a zero. The other two digits on the card must add to 14. This leads to three groups of numbers: (1,5,9), (1,6,8) and (1,7,7). 3. One of the digits on the card is a two but the number does not contain a zero or one. The other two digits on the card must add to 13. This leads to three groups of numbers: (,4,9), (,5,8) and (,6,7). 4. One of the digits on the card is a three but the number does not contain a zero, one or two. The other two digits on the card must add to 1. This leads to four groups of numbers: (3,3,9), (3,4,8), (3,5,7) and (3,6,6). 5. One of the digits on the card is a four but the number does not contain a zero, one, two or three. The other two digits on the card must add to 11. This leads to two groups of numbers: (4,4,7) and (4,5,6). 6. One of the digits on the card is a five but the number does not contain a zero, one, two, three or four. The other two digits on the card must add to 10. This leads to only one group of numbers: (5,5,5).
50 Now that we have the groups of numbers, we can determine the number of cards that can be created from each group of three numbers. We will do this again with cases: groups containing a zero, groups containing three distinct numbers but not zero, groups containing exactly two numbers the same but not zero, and groups containing three numbers the same but not zero. 1. One of the numbers in the group is zero. Earlier we found that there were two such groups: (0,6,9) and (0,7,8). This is a special case since zero cannot appear in the number as the hundreds digit for the number to be a three digit number. For each of the two groups of numbers, the zero can be placed in two ways, in the tens digit or the units digit. Once the zero is placed, the other two numbers can be placed in the remaining two spots in two ways. Each group can form = 4 three digit numbers. Since there are two groups, there are 4 = 8 cards in the deck that contain a zero and add to 15.. All three digits on the card are different but the number does not contain a zero. From the earlier cases there are eight groups in which all three numbers are different: (1,5,9), (1,6,8), (,4,9), (,5,8), (,6,7), (3,4,8), (3,5,7), and (4,5,6). For each of these groups, the hundreds digit can be filled three ways. For each of these three choices for hundreds digit, the tens digit can be filled two ways. Once the hundreds digit and tens digit are selected the units digit must get the third digit. So each group can form 3 = 6 different numbers. Since there are eight groups, there are 8 6 = 48 cards in the deck that contain three different digits other than zero and add to Two of the digits on the card are the same and the number does not contain a zero. From the earlier cases, there are four groups of numbers in which exactly two of the numbers in the group are the same: (1,7,7), (3,3,9), (3,6,6), and (4,4,7). For each of these groups, the unique number can be placed in one of three spots. Once the unique number is placed the other two numbers must go in the remaining two spots. So each group can form 3 different numbers. Since there are four groups, there are 4 3 = 1 cards in the deck that do not contain a zero but contain two digits the same and add to The three digits on the card are the same. From the earlier cases we discovered only one such group: (5,5,5). Only one card can be produced using the numbers from this group. Combining the counts from the above four cases, there are = 69 cards in the deck with a digit sum of 15. Therefore the probability that Luke selects a winning card is. This translates to approximately a 7.7% chance of winning. 69 = A game is considered fair if there is close to a 50% chance of winning. This game is definitely not fair. If you changed the game to if the card chosen has a sum that is divisible by 5, there is a 0% chance of winning. This is better but still not fair. If you changed the game to if the card chosen has a sum less than 14, there is a 46% chance of winning. This game is much fairer but not really fun. Can you create a game using this specific deck of cards that is reasonably fair and fun to play?
51 Problem E A Product for the Ages A mother has four children, each with a different age. The product of their ages is The sum of the ages of the three oldest children is 40 and the sum of the ages of the three youngest children is 3. Determine the ages of the four children.
52 Problem Problem E and Solution A Product for the Ages A mother has four children, each with a different age. The product of their ages is The sum of the ages of the three oldest children is 40 and the sum of the ages of the three youngest children is 3. Determine the ages of the four children. Solution Let the ages of the children from youngest to oldest be a, b, c, d. Since the ages of the three oldest children sum to 40, b + c + d = 40. (1) Since the ages of the three youngest children sum to 3, a + b + c = 3. () Subtracting () from (1), we obtain d a = 8. This means that the difference between the age of the oldest child and the age of the youngest child is 8. Now = = ( 3) ( 3) ( 3) 5 = Since all of the ages are different, this statement tells us that at least one child is over 1 and one child is under 10. There is a limited number of possibilities such that the difference between the oldest and youngest is 8 which also satisfy the condition that the youngest is under 10 and the oldest is over 1. The possibilities for youngest and oldest are (5,13), (6,14), (7,15), (8,16), and (9,17). No other combination would be possible since the oldest child must be over 1 and the youngest child must be under 10. The numbers 7, 13, and 17 are primes and are not factors of Therefore we can eliminate the possibilities where an age is one of 7, 13, or 17, leaving (6,14) and (8,16). But 14 contains a prime factor of 7 which is not a factor of Since there is only one possibility left, (8,16), we can conclude that the youngest is 8 and the oldest is 16. Now = Using the remaining factors 3 3 and 5, we need to create two numbers between 8 and 16. The only possibilities are 3 = 9 and 3 5 = 15. Therefore the ages of the children are 8, 9, 15, and 16. It is easy to verify that this is the correct solution.
53 Problem E Show Me The Money As a cashier, Igotta Job needs to give change to people when they buy merchandise. She knows that the Canadian coin system has the following greedy property: the best way to give change is to use the fewest number of coins possible. For example, if Igotta needed to provide 30 change, she would use one quarter (5 ) and one nickel (5 ), a total of two coins. Igotta would not give three dimes (3 10 ) since this option uses three coins instead of two. She would not use two dimes ( 10 ) and two nickels ( 5 ) since this option uses four coins instead of two. And she certainly would not give thirty pennies (30 1 ). Each combination would provide the correct amount of change but only the first option uses the fewest number of coins and therefore follows the greedy property. How many different sums of money, less than one dollar, can Igotta make using the greedy property and exactly four coins chosen from pennies (1 coins), nickels (5 coins), dimes (10 coins) and quarters (5 coins)? ($1 = 100 )
54 Problem Problem E and Solution Show Me The Money As a cashier, Igotta Job needs to give change to people when they buy merchandise. She knows that the Canadian coin system has the following greedy property: the best way to give change is to use the fewest number of coins possible. For example, if Igotta needed to provide 30 change, she would use one quarter (5 ) and one nickel (5 ), a total of two coins. Igotta would not give three dimes (3 10 ) since this option uses three coins instead of two. She would not use two dimes ( 10 ) and two nickels ( 5 ) since this option uses four coins instead of two. And she certainly would not give thirty pennies (30 1 ). Each combination would provide the correct amount of change but only the first option uses the fewest number of coins and therefore follows the greedy property. How many different sums of money, less than one dollar, can Igotta make using the greedy property and exactly four coins chosen from pennies (1 coins), nickels (5 coins), dimes (10 coins) and quarters (5 coins)? ($1 = 100 ) Solution For the solution, we will consider cases to carefully count the possibilities. 4 pennies There is only one way to use four pennies for change. This gives a sum of 4. 3 pennies There are three possibilities here: three pennies and one nickel giving a sum of 8, three pennies and one dime giving a sum of 13, and three pennies and one quarter giving a sum of 8. pennies and two of one other coin. This leads to two possibilities: two pennies and two dimes for a sum of, and two pennies and two quarters for a sum of 5. Note that two pennies and two nickels gives a sum of 1. This same sum can be made using fewer coins, namely two pennies and one dime, and is therefore not valid. and one of each of two other coins. This leads to three valid possibilities: two pennies, one nickel and one dime for a sum of 17, two pennies, one nickel and one quarter for a sum of 3, and two pennies, one dime and one quarter for a sum of penny and three of one other coin. This leads to one valid possibility: one penny and three quarters for a sum of 76. The possibilities one penny and three nickels, and one penny and three dimes are both invalid since these sums can be made up with fewer coins. two of a second type of coin and one of a third type of coin. This leads to three valid possibilities: one penny, one nickel and two quarters for a sum of 56, one penny, one dime and two quarters for a sum of 61, and one penny, one quarter and two dimes for a sum of 46. The sums obtained using one penny, one nickel and two dimes, or one penny, one dime and two nickels, or one penny, one quarter and two nickels can be obtained using fewer coins and hence are invalid.
55 1 penny (continued) and one of each of the other three coins. This leads to one valid possibility: one penny, one nickel, one dime and one quarter for a sum of pennies and four of one other coin. There are no valid possibilities here since each possible sum can be made up with fewer coins. Four nickels is 0. The same sum can be made using two dimes and hence fewer coins. Four dimes is 40. The same sum can be made using one quarter, one dime and one nickel, and hence fewer coins. Four quarters would give $1=100 but we are looking for sums less than one dollar so this is invalid. three of a second type of coin and one of a third type of coin. This leads to two valid possibilities: three quarters and one nickel for a sum of 80, and three quarters and one dime for a sum of 85. Using three nickels and any other coin would be invalid since three nickels could be replaced with two coins, a dime and a nickel, to produce the same sum. Using three dimes and any other coin would be invalid since three dimes could be replaced with two coins, a quarter and a nickel, to produce the same sum. two of one coin and two of another coin. This leads to one valid possibility: two dimes and two quarters for a sum of 70. Two nickels could never be paired with any other pair since two nickels can be replaced by one dime. two of one coin and one of each of the remaining two coins. This leads to one valid possibility: two quarters, one nickel, and one dime for a sum of 65. Two dimes, one nickel and one quarter gives a sum of 50. This sum can be made with fewer coins by using two quarters. Two nickels, one dime and one quarter gives a sum of 45. This sum can be made with fewer coins by using one quarter and two dimes. The total number of different sums of money is the obtained by adding the number of possibilities from each case, namely = 18 possible sums. The 18 possible sums under one dollar, using the greedy property and four coins, are 4, 8, 13, 17,, 8, 3, 37, 41, 46, 5, 56, 61, 65, 70, 76, 80, 85.
56 Problem E A Card Game To Flip Over Too E Z Dealer has a deck consisting of 100 cards numbered from 1 to 100. Each card has the same number printed on both sides. One side of the card is yellow and the other side of the card is red. E Z places all the cards, red side up, on the table. He first turns over every card that has a number which is a multiple of. He then examines all the cards, and turns over every card that has a number which is a multiple of 3. He again examines all the cards, and turns over every card that has a number which is a multiple of 4. Finally, he examines all the cards and turns over every card that has a number which is a multiple of 5. After E Z has finished, how many cards have the red side facing up?
57 Problem Problem E and Solution A Card Game To Flip Over Too E Z Dealer has a deck consisting of 100 cards numbered from 1 to 100. Each card has the same number printed on both sides. One side of the card is yellow and the other side of the card is red. E Z places all the cards, red side up, on the table. He first turns over every card that has a number which is a multiple of. He then examines all the cards, and turns over every card that has a number which is a multiple of 3. He again examines all the cards, and turns over every card that has a number which is a multiple of 4. Finally, he examines all the cards and turns over every card that has a number which is a multiple of 5. After E Z has finished, how many cards have the red side facing up? Solution If a card number is a multiple of, 3, 4 and 5, it will be flipped four times. This card will go from red to yellow to red to yellow to red again. So the card will still be red once E Z has finished. If a card number is a multiple of exactly three of, 3, 4 and 5, it will be flipped three times. This card will go from red to yellow to red to yellow. So the card will be yellow once E Z has finished. If a card number is a multiple of exactly two of, 3, 4 and 5, then it will be flipped twice. This card will go from red to yellow to red again. So the card will still be red once E Z has finished. If a card number is a multiple of exactly one of, 3, 4 and 5, it will be flipped once. This card will go from red to yellow. So the card will be yellow once E Z has finished. If a card number is a multiple of none of, 3, 4 and 5, then this card will not be flipped and so the card will still be red once E Z has finished. To determine how many cards have the red side facing up once E Z has finished, let s determine how many cards have the yellow side facing up once E Z has finished. To do so, we need to determine how many card numbers are multiples of exactly three of, 3, 4 and 5 and how many cards are multiples of exactly one of, 3, 4 and 5. Let s consider the cases: A card number is a multiple of, 3 and 4, but not 5 If a card number is a multiple of, 3 and 4, then it must be a multiple of 1, the lowest common multiple of, 3 and 4. So we are want card numbers that are multiples of 1 but not 5. If a card number is a multiple of 1 and 5, then it is a multiple of 1 5 = 60. So we want all multiples of 1 that are not multiples of 60. There are 8 multiples of 1 from 1 to 100, but one is 60. So there are 8 1 = 7 numbers that are multiples of, 3 and 4, but not 5. A card number is a multiple of, 3 and 5, but not 4 If a card number is a multiple of, 3 and 5, then it must be a multiple of 30, the lowest common multiple of, 3 and 5. So we want all multiples of 30 that are not multiples of 4. There are 3 multiples of 30 from 1 to 100, but one is 60, which is also a multiple of 4. So there are numbers from 1 to 100 that are multiples of, 3 and 5, but not 4.
58 A card number is a multiple of, 4 and 5, but not 3 If a card number is a multiple of, 4 and 5, then it must be a multiple of 0, the lowest common multiple of, 4 and 5. So we want all multiples of 0 that are not multiples of 3. There are 5 multiples of 0 from 1 to 100, but one is 60, which is a multiple of 3. So there are 4 numbers from 1 to 100 that are multiples of, 4 and 5, but not 3. A card number is a multiple of 3, 4 and 5, but not It is not possible for a card number to be a multiple of 4 but not. So there are no card numbers in this case. A card number is a multiple of but not 3, 4, or 5 There are 50 numbers from 1 to 100 which are multiples of and 5 numbers from 1 to 100 which are multiples of 4 (and thus ). So there are 50 5 = 5 numbers from 1 to 100 multiples of by but not 4. These are {, 6, 10, 14, 18,, 6, 30, 34, 38, 4, 46, 50, 54, 58, 6, 66, 70, 74, 78, 8, 86, 90, 94, 98} We need to remove numbers that are still multiples of 3 or 5. After doing so we are left with {, 14,, 6, 34, 38, 46, 58, 6, 74, 8, 86, 94, 98} So there are 14 numbers from 1 to 100 that are multiples of but not 3, 4 or 5. A card number is a multiple of 3 but not, 4, or 5 There are 33 multiples of 3 from 1 to 100, {3, 6, 9, 1, 15,, 87, 90, 93, 96, 99}. In this group of multiples, there are 17 numbers that are odd. So there are 17 numbers from 1 to 100 that are multiples of 3 but not. These numbers are {3, 9, 15, 1, 7, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99}. We still need to remove numbers that are multiples of 5. After doing so we are left with {3, 9, 1, 7, 33, 39, 51, 57, 63, 69, 81, 87, 93, 99}. So there are 14 numbers from 1 to 100 that are multiples of 3 but not, 4 or 5. A card number is a multiple of 4 but not, 3, or 5 It is not possible for a card number to be a multiple of 4 but not. So there are no card numbers in this case. A card number is a multiple of 5 but not, 3, or 4 There are 0 multiples of 5 from 1 to 100, but half of those are multiples of. The multiples of 5 which are not multiples of are {5, 15, 5, 35, 45, 55, 65, 75, 85, 95}. We still need to remove numbers that are multiples of 3. After doing so we are left with {5, 5, 35, 55, 65, 85, 95}. So there are 7 numbers from 1 to 100 that are multiples of 5 but not, 3 or 4. Therefore, once he has finished, EZ Dealer is left with 100 ( ) = = 5 cards with the red side facing up. Extension: Suppose E Z Dealer continues flipping cards in this manner. So, after he has flipped all cards whose number is a multiple of 5, he then flips all cards whose card number is a multiple of 6, then 7, then 8, and so on until he flips all cards whose number is a multiple of 100. Once E Z has finished, how many cards will have the red side facing up?
59
60 Problem E So Many Triangles P QR has side QP extended to X so that QP = P X, P R extended to Z so that P R = RZ, and RQ extended to Y so that RQ = QY. If the area of XY Z = 1176 cm, determine the area of P QR.
61 Problem E and Solution So Many Triangles Problem P QR has side QP extended to X so that QP = P X, P R extended to Z so that P R = RZ, and RQ extended to Y so that RQ = QY. If the area of XY Z = 1176 cm, determine the area of P QR. Solution On the above diagram the lengths of the equal sides, QP = P X, P R = RZ, and RQ = QY, have been marked. Join P to Y, Q to Z, and R to X. P QR and P QY have a common altitude drawn from vertex P to the line segment RY, meeting it at A. The triangles have equal base lengths, RQ = QY. area P QR = area P QY = x. At this point we can proceed to look at various other triangles with equal areas. P QY and P XY have the same height and equal base lengths. area P XY = area P QY =x. P XR and P QR have the same height and equal base lengths. area P XR = area P QR = x. P XR and RXZ have the same height and equal base lengths. area RXZ = area P XR = x. P QR and QRZ have the same height and equal base lengths. area QRZ = area P QR = x. QRZ and QY Z have the same height and equal base lengths. area QY Z = area QRZ = x. Then the area of XY Z = area P XY +area P QY +area P QR+area P XR+area RXZ +area QRZ +area QY Z = x + x + x + x + x + x + x 7x = 1176 and x = 168 cm. the area of P QR is 168 cm.
62 Problem E What s Your Angle Anyway III? AB is a diameter of a circle with centre C. D is a point on the circumference of the circle other than A and B. Determine the measure of ADB. If you need a hint, consider solving this week s Problem D, What s Your Angle Anyway II?
63 Problem E and Solution What s Your Angle Anyway III? Problem AB is a diameter of a circle with centre C. D is a point on the circumference of the circle other than A and B. Determine the measure of ADB. Solution Join D to the centre C. Since CA, CB and CD are radii of the circle, CA = CB = CD. Since CA = CD, CAD is isosceles and CAD = CDA = x. Since CB = CD, CBD is isosceles and CBD = CDB = y. This new information is marked on the following diagram. The angles in a triangle add to 180 so in ABD ADB + DAB + DBA = 180 (x + y ) + x + y = 180 (x + y ) = 180 x + y = 90 But ADB = x + y so ADB = 90. This result is often expressed as a theorem for circles: An angle ( ADB) inscribed in a circle by the diameter (AB) of a circle is 90.
64 Problem E Partly Salted When one kilogram of salt is added to a solution of salt and water, the solution becomes 1 salt by mass. One kilogram of water is then added to the new 3 solution resulting in a solution that is 3 10 salt by mass. What fraction of the original solution was salt? +
65 Problem E and Solution Partly Salted Problem When one kilogram of salt is added to a solution of salt and water, the solution becomes 1 3 salt by mass. One kilogram of water is then added to the new solution resulting in a solution that is 3 10 Solution salt by mass. What fraction of the original solution was salt? Let S represent the mass of salt in the original solution. Let W represent the mass of water in the original solution. Then S + W represents the total mass of the original solution. After adding 1 kg of salt to the original solution, there is (S + 1) kg of salt and (S + W + 1) kg of solution. Since this new solution is one third salt by mass, S + 1 S + W + 1 = 1 3 3(S + 1) = 1(S + W + 1) 3S + 3 = S + W + 1 S + = W (1) After adding 1 kg of water to the new solution, there is still (S + 1) kg of salt in (S + W + ) kg of solution. Since this new solution is three tenths salt by mass, S + 1 S + W + = (S + 1) = 3(S + W + ) Substituting (1) into (), 10S + 10 = 3S + 3W + 6 7S 3W = 4 () 7S 3(S + ) = 4 7S 6S 6 = 4 S = kg Substituting for S in (1), W = 6 kg. Then the original solution was S + W = 8 kg. The fraction of the original solution that was salt was 8 = 1 4. one quarter of the original solution was salt by mass.
66 Problem E Pair Possibilities Determine all possible ordered pairs of positive integers (a, b) that satisfy 1 a + b = 8 a + b and a + 5b 54.
67 Problem E and Solution Pair Possibilities Problem Determine all possible ordered pairs of positive integers (a, b) that satisfy 1 a + b = 8 and a + 5b 54. a + b Solution Since a and b are positive integers satisfying a + 5b 54, we could write out all ordered pairs that satisfy this inequality and then determine which ones also satisfy the first equation. There will be a large number of possibilities to check so we need to find a way to reduce the number of possibilities. Working with the first equation: Find a common denominator: b + a ab = 8 a + b Crossmultiply : (b + a)(a + b) = 8ab Expand and simplify: 4a + 4ab + b = 8ab It follows that a b = 0 and b = a. Rearrange: 4a 4ab + b = 0 Factor: (a b) = 0 Each of the ordered pairs (a, b) will look like (a, a). We substitute a for b in the inequality obtaining a + 5(a) 54 or 1a 54 or a 4.5. Since a is a positive integer, a can only take on integer values 1,, 3 and 4. The ordered pairs follow quickly: {(1, ), (, 4), (3, 6), (4, 8)}. the only ordered pairs of positive integers (a, b) that satisfy 1 a + b = 8 a + b and a + 5b 54 are (1, ), (, 4), (3, 6), and (4, 8).
68 Problem E Get Your Analytic Skills On OAB is an isosceles right triangle with vertex O at the origin (0, 0), vertices A and B on the line x + 3y 13 = 0 and AOB = 90. Determine the area of OAB.
69 Problem OAB is an isosceles right triangle with vertex O at the origin (0, 0), vertices A and B on the line x + 3y 13 = 0 and AOB = 90. Determine the area of OAB. Solution Problem E and Solution Get Your Analytic Skills On In analytic geometry problems, a representative diagram is important and often provides clues for the solution of the problem. The diagram above has the given information plus a couple of pieces of information that will be justified now. Let B have coordinates (p, q). The slope of OB = q. Since AOB = p 90, OB OA and the slope of OA is the negative reciprocal of OB. Therefore the slope of OA = p. Since the q triangle is isosceles, OA = OB and it follows that the coordinates of A are ( q, p). (We can verify this by finding the length of OA and the length of OB and showing that both are equal to p + q.) Since B(p, q) is on the line x + 3y 13 = 0, it satisfies the equation of the line. p + 3q 13 = 0 (1) Since A( q, p) is on the line x + 3y 13 = 0, it satisfies the equation of the line. q + 3p 13 = 0 or 3p q 13 = 0 () Since we have two equations and two unknowns, we can use elimination to solve for p and q. (1) 4p + 6q 6 = 0 () 3 9p 6q 39 = 0 Adding, we obtain 13p 65 = 0 p = 5 Substituting in (1) q 13 = 0 3q = 3 q = 1 The point B is (5, 1) and the length of OB = = 6. Since OA = OB, OA = 6. AOB is a right triangle so we can use OB as the base and OA as the height in the formula for the area of a triangle. Then the area of AOB = OA OB = 6 6 = 13. the area of AOB is 13 units.
70 Problem E Sphere Pressure A cube rests inside a sphere so that each vertex touches the sphere. The radius of the sphere is 6 cm. Determine the volume of the cube.
71 Problem E and Solution Sphere Pressure Problem A cube rests inside a sphere so that each vertex touches the sphere. The radius of the sphere is 6 cm. Determine the volume of the cube. Solution Label four vertices of the cube A, B, C, D as shown in the diagram. Let x represent the side length of the cube. Then AB = BC = CD = x. The diagonals of a cube intersect in a point such that the distance from the intersection point to each vertex is equal. Since each vertex of the cube touches the sphere, the diagonal of the cube, AD, is equal in length to the diameter of the sphere. Therefore AD = (6) = 1 cm. Each face of a cube is a square so ABC = 90. Using Pythagoras Theorem, AC = AB + BC = x + x = x. In a cube the sides are perpendicular to the base. In particular, DC is perpendicular to the base and it follows that DC AC. Therefore DCA is a right angled triangle. Using Pythagoras Theorem, AD = AC + CD = x + x = 3x. But AD = 1 so AD = 144. Then 3x = 144, x = 48 and x = 4 3 since x > 0. The volume of the cube is x 3 = (4 3) 3 = 19 3 cm 3. the volume of the cube is 19 3 cm 3.
72 Problem E Enough Information? In P QR, P Q = 7, P R = 9 and median P M = 7. Determine the length of QR.
73 Problem Problem E and Solutions Enough Information? In P QR, P Q = 7, P R = 9 and median P M = 7. Determine the length of QR. Solution 1 Since P Q = P M = 7, P QM is isosceles. In P QM, draw an altitude from P to QM, intersecting at N. In an isosceles triangle, the altitude drawn to the base bisects the base. Therefore QN = NM = x. Since P M is a median in P QR, MR = QM = x. Let P N = h. P NM is a right triangle. Using Pythagoras Theorem, P N = P M NM h = 7 x h = 49 x (1) P NR is a right triangle. Using Pythagoras Theorem, P N = P R NR h = 9 (x + x) h = 81 (3x) h = 81 9x () In equations (1) and (), the left side of each equation is h. Therefore, the right side of equation (1) must equal the right side of equation (). So 49 x = 81 9x x + 9x = x = 3 x = 4 x =, x > 0 QR = QN + NM + MR = x + x + x = 4x = 8 units.
74 Problem In P QR, P Q = 7, P R = 9 and median P M = 7. Determine the length of QR. Solution This solution is presented for students who have done some trigonometry and know the law of cosines. Since P M is a median, QM = MR = x. Then QR = x. Using the law of cosines in P QM, P M = P Q + QM (P Q)(QM)cos(Q) 7 = 7 + x (7)(x)cos(Q) 49 = 49 + x 14xcos(Q) 14xcos(Q) = x (1) Using the law of cosines in P QR, P R = P Q + QR (P Q)(QR)cos(Q) 9 = 7 + (x) (7)(x)cos(Q) 81 = x 8xcos(Q) 8xcos(Q) = 4x 3 () Using elimination to solve for x, (1) 8xcos(Q) = x () 8xcos(Q) = 4x 3 Subtracting 0 = x + 3 x = 3 x = 16 x = 4, x > 0 QR = x = 8 the length of QR is 8 units.
75 Problem E A Simple System of Equations? If 3x = 16 y+1 and x = 5y 17, determine the value of x + y. Try to see if doing this problem without a calculator affects how you think about the problem.
76 Problem E and Solutions A Simple System of Equations? Problem If 3x = 16 y+1 and x = 5y 17, determine the value of x + y. Solution 1 For this solution we will attempt to answer the question only. We will not determine the values of x and y that generate the sum. 3x = 16 y+1 3x = ( 4) y+1 3x = 4y+4 3x = 4y + 4 (1) But x = 5y 17 () (1) () x = y + 1 Rearranging x + y = 1 x + y = 1. Notice that the problem only asks for x + y, it is not necessary to find values for x and y. Solution This solution carries on from equations (1) and () in solution 1 to find the values of x and y, and then determines the sum. (1) 5 15x = 0y + 0 (3) () 4 8x = 0y 68 (4) (3) (4) 7x = 88 x = 88 ( ) 7 88 Substitue in () = 5y 17 7 Multiply by = 35y 119 As before (but with much more work) x + y = = 35y y = = 59 7 x + y = = = 1
77 Problem E You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. The following information may be helpful in the solution of the problem. An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3,5,7,9 is an arithmetic sequence with four terms and constant difference. The general term of an arithmetic sequence is t n = a + (n 1)d, where a is the first term, d is the constant difference and n is the number of terms. The sum, S n, of the first n terms of an arithmetic sequence can be found using either S n = n ( ) [a + (n 1)d] or S t1 + t n n = n, where t 1 is the first term of the sequence and t n is the n th term of the sequence. The following example is provided to verify the accuracy of the formulas and to illustrate their use. For the arithmetic sequence 3,5,7,9, a = t 1 = 3, d =, n = 4 and t n = t 4 = 9. S n = = 4 S n = n [a + (n 1)d] = 4 [(3) + (3)] = [1] = 4 ( ) ( ) t1 + t n S n = n = 4 = 4(6) = 4
78 Problem Problem E and Solutions You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Three solutions are provided for this problem. Solution 1 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 5 = 3 and t = a 3 a = 1 5 = 7. Since the new sequence is arithmetic, the difference between consecutive terms is constant. Therefore the constant difference is d = t t 1 = 7 3 = 4. We can now generate a few more terms in the sequence of differences by adding the constant to the previous term. t 3 = t + 4 = = 11 and t 4 = t = = 15. Using the information we can present the information in tabular form. Term Number First Sequence First Difference Second Difference n a n t n (Difference of the differences) Since the second difference is constant we can represent the general term of the first sequence with a quadratic function. Let a n = pn + qn + r. For n = 1, a 1 = = p(1) + q(1) + r. p + q + r =. (1) For n =, a = 5 = p() + q() + r. 4p + q + r = 5. () For n = 3, a 3 = 1 = p(3) + q(3) + r. 9p + 3q + r = 1. (3) Subtracting (1) from (), 3p + q = 3. (4) Subtracting () from (3), 5p + q = 7. (5) Subtracting (4) from (5), p = 4 and p = follows. Substituting p = into (4), 3() + q = 3 and q = 3 follows. Substituting p =, q = 3 into (1), 3 + r = and r = 3 follows. a n = n 3n + 3 is the general term of the given sequence in terms of n.
79 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. We can generate more terms: t 3 = 11, t 4 = 15, t 5 = 19,. Each term in the new sequence is the difference between consecutive terms of the original sequence. a a 1 = t 1 a 3 a = t a 4 a 3 = t 3 a n a n 3 = t n 3 a n 1 a n = t n a n a n 1 = t n 1 Adding, a n a 1 = t 1 + t + t t n + t n 1 = S n 1. (1) To find the sum S n 1 = t 1 + t + t t n + t n 1 we can use the formula S n = n [a + (n 1)d] with a = 3 and d = 4 for (n 1) terms. So S n 1 = t 1 + t + t t n + t n 1 = n 1 [a + ((n 1) 1)d] = n 1 [(3) + ((n 1) 1)(4)] = n 1 [6 + (n )(4)] = n 1 [6 + 4n 8] = n 1 [4n ] = (n 1)(n 1) From (1) above, a n a 1 = S n 1 so a n = S n 1 + a 1 = (n 1)(n 1) + = n 3n + 3. a n = n 3n + 3 is the general term of the given sequence in terms of n.
80 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution 3 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. Using the formula for the general term of an arithmetic sequence, t n = a + (n 1)d, the general term of the sequence of differences is t n = (3) + (n 1)(4) = 3 + 4n 4 = 4n 1. To generate the original sequence we start with the first term and add more terms from the arithmetic sequence of differences. For example, a 1 = 3, a = + t 1 = + [4(1) 1] = + 3 = 5, and a 3 = + t 1 + t = + [4(1) 1] + [4() 1] = = 1 So, a n = + t 1 + t + t t n 1 = + [4(1) 1] + [4() 1] + [4(3) 1] + + [4(n 1) 1] = + [4(1) + 4() + 4(3) + + 4(n 1)] + (n 1)( 1) = + 4[ (n 1)] n + 1 [ ] (n 1)(n) = 3 n + 4 (1) = 3 n + (n n) = 3 n + n n = n 3n + 3 In (1) above, (n 1) is an arithmetic sequence with t 1 = 1, t n 1 = (n 1) and the number of terms (n ( 1) terms. ) Using the formula for the sum of the terms of an t1 + t n arithmetic sequence, S n = n, we obtain ( ) 1 + (n 1) ( n ) (n 1)(n) S n 1 = (n 1) = (n 1) = a n = n 3n + 3 is the general term of the given sequence in terms of n.
81 Problem E Find The Area  No Parabolem! y B(0, 5) A(r, 0) x C(5, 0) D(p, q) In the diagram, D(p, q) is the vertex of the parabola. The parabola cuts the xaxis at A(r, 0) and C(5, 0). The area of ABC is 5. Determine the area of DBC.
82 Problem Problem E and Solutions Find The Area  No Parabolem! B(0, 5) y In the diagram, D(p, q) is the vertex of the parabola. The parabola cuts the xaxis at A(r, 0) and C(5, 0). The area of ABC is 5. Determine the area of DBC. Solution 1 In ABC, the height is the distance from the xaxis to B(0, 5). Therefore the height is 5 units. The base is AC = 5 r. Using the formula for the area of a triangle, Area ABC = (5 r)(5) = 5. Then 5 r = and r = 3 follows. The coordinates of A are (3, 0). The axis of symmetry is a vertical line through the midpoint of AC which is (4, 0). It follows that the xcoordinate of the vertex is p = 4. Therefore, the vertex is D(4, q). Since the two xintercepts of the parabola are 3 and 5, the equation of the parabola in factored form can be written as y = a(x 3)(x 5). Since B(0, 5) is on the parabola, we can solve for a by substituting x = 0 and y = 5 into y = a(x 3)(x 5). This leads to a = 1 3 and y = 1 3 (x 3)(x 5). To find q, the ycoordinate of D, substitute x = 4, y = q into y = 1 3 (x 3)(x 5). Then q = (4 3)(4 5) = 3. Construct a rectangle with sides parallel to the x and yaxes so that the points B, D, and C are on the rectangle. B is one vertex of the rectangle. We can determine the coordinates of the other three vertices of the rectangle. The results are shown on the diagram to the right. It is straight forward to then determine the lengths required for finding various areas: BE = = 16 3, ED = 4, DF = 1, F C = 1 3, CG = 5 and BG = 5. To find the area of DBC, subtract the areas of BED, DF C, and BCG from the area of rectangle BEF G. The area of rectangle BEF G = BE BG = = 80 3 units. The area of BED = BE ED = = 3 3 units. DF F C The area of DF C = = = 1 6 units. The area of BCG = CG BG = 5 5 = 5 units. y B(0, 5) A(r, 0) A(3, 0) D(p, q) x C(5, 0) x C(5, 0) E(0, 3) D(4, ) F(5, 3) Area DBC = Area Rectangle BEF G Area BED Area DF C Area BCG = = 10 3 units the area of DBC = 10 3 units. The second solution uses the equation of a line and is very different from the first solution. 3 G(5, 5)
83 y Problem In the diagram, D(p, q) is the vertex of the parabola. The parabola cuts the xaxis at A(r, 0) and C(5, 0). The area of ABC is 5. Determine the area of DBC. B(0, 5) Solution To begin this solution we will pick up a result from the first solution and do not repeat the work here. In solution 1, we found vertex D to be the point (4, 1 3 ). A(r, 0) D(p, q) x C(5, 0) Let P (t, 0) be the point where the line through B and D crosses the xaxis. We will determine the equation of the line containing B, P, and D. The slope of the line is The yintercept of the line is 5. Therefore the equation of the line through B, P, and D is y = 4 3 x = 16 4 = 4 To find t, the xcoordinate of P we substitute x = t and y = 0 into the equation of the line. y = 4 3 x = 4 3 t + 5 4t = 15 t = 15 4 B(0, 5) y A(3, 0) P(t, 0) 1 D(4, 3) x C(5, 0) To determine the area of BDC, we find the sum of the area of BP C and the area of DP C. We will use P C as the base in both triangles. Then P C = = 5 4. In BP C, the height is the perpendicular distance from the xaxis to point B. The height is 5. The area of BP C = = 5 8 units. In DP C, the height is the perpendicular distance from the xaxis to point D. The height is 1 3. The area of DP C = = 5 4 units. the area of DBC = 10 3 units. Area DBC = Area BP C + Area DP C = = = 80 4 = 10 3 units
84 Problem E Advanced Training A blue train is 1000 km north of a red train. The blue train is travelling south, and the red train is travelling east. Both trains travel at the same speed of 100 km/h. How much time elapses until the total distance travelled by the two trains is equal to the distance between the trains.
85 Problem E and Solution Advanced Training Problem A blue train is 1000 km north of a red train. The blue train is travelling south, and the red train is travelling east. Both trains travel at the same speed of 100 km/h. How much time elapses until the total distance travelled by the two trains is equal to the distance between the trains. Solution Let t be the time in hours until the total distance travelled by the two trains is equal to the distance between the trains. Since each train is travelling at 100 km/h, the distance travelled by each train in t hours is 100t km. The total distance travelled by the two trains is 100t = 00t km. The following diagram represents the positions of the two trains after t hours. The blue train starts at B and moves to C. The red train starts at R and moves to S. Then BC = RS = 100t and CR = t. We want the time t when CS = BC + RS = 100t + 100t = 00t. B 1000 km C 100t km R 100t km S But CRS is right angled so CS = CR + RS 0000t t = 0 (00t) = ( t) + (100t) 40000t = t t t t + 10t 50 = 0 Using the Quadratic Formula, t = 10 ± 100 4( 50) t = 10 ± 10 3 t = 5 ± 5 3 Since t > 0, is inadmissible. t = = 3.66 hours. The distance between the two trains will be equal to their total distance travelled in ( ) hours which is approximately 3 hours and 40 minutes after they leave their initial positions.
86 Problem E Five Solutions  Really? There are five values of x that satisfy the equation (x 5x + 5) x +4x 60 = 1. Determine these five values of x. x 1 = x = x 3 = x 4 = x 5 =
87 Problem E and Solution Five Solutions  Really? Problem There are five values of x that satisfy the equation (x 5x + 5) x +4x 60 = 1. Determine these five values of x. Solution Let s consider the ways that an expression of the form a b can be 1: The base, a, is 1. In this case, the exponent can be any value and we need to solve x 5x + 5 = 1. So x = 4 or x = 1. x 5x + 5 = 1 x 5x + 4 = 0 (x 4)(x 1) = 0 The exponent, b, is 0. In this case, the base can be any number other than 0 and we need to solve x + 4x 60 = 0. So x = 6 or x = 10. x + 4x 60 = 0 (x 6)(x + 10) = 0 When x = 6, the base is 6 5(6) + 5 = When x = 10, the base is ( 10) 5( 10) + 5 = The base, a, is 1 and the exponent, b, is even. We first need to solve x 5x + 5 = 1. So x = or x = 3. x 5x + 5 = 1 x 5x + 6 = 0 (x )(x 3) = 0 When x =, the exponent is + 4() 60 = 48, which is even. Therefore, when x =, (x 5x + 5) x +4x 60 = 1. When x = 3, the exponent is 3 + 4(3) 60 = 39, which is odd. Therefore, when x = 3, (x 5x + 5) x +4x 60 = 1. So x = 3 is not a solution. Therefore, the values of x that satisfy (x 5x + 5) x +4x 60 = 1 are x = 1, x =, x = 4, x = 6 and x = 10.
88
89 Problem E So Many Triangles P QR has side QP extended to X so that QP = P X, P R extended to Z so that P R = RZ, and RQ extended to Y so that RQ = QY. If the area of XY Z = 1176 cm, determine the area of P QR.
90 Problem E and Solution So Many Triangles Problem P QR has side QP extended to X so that QP = P X, P R extended to Z so that P R = RZ, and RQ extended to Y so that RQ = QY. If the area of XY Z = 1176 cm, determine the area of P QR. Solution On the above diagram the lengths of the equal sides, QP = P X, P R = RZ, and RQ = QY, have been marked. Join P to Y, Q to Z, and R to X. P QR and P QY have a common altitude drawn from vertex P to the line segment RY, meeting it at A. The triangles have equal base lengths, RQ = QY. area P QR = area P QY = x. At this point we can proceed to look at various other triangles with equal areas. P QY and P XY have the same height and equal base lengths. area P XY = area P QY =x. P XR and P QR have the same height and equal base lengths. area P XR = area P QR = x. P XR and RXZ have the same height and equal base lengths. area RXZ = area P XR = x. P QR and QRZ have the same height and equal base lengths. area QRZ = area P QR = x. QRZ and QY Z have the same height and equal base lengths. area QY Z = area QRZ = x. Then the area of XY Z = area P XY +area P QY +area P QR+area P XR+area RXZ +area QRZ +area QY Z = x + x + x + x + x + x + x 7x = 1176 and x = 168 cm. the area of P QR is 168 cm.
91 Problem E Get Your Analytic Skills On OAB is an isosceles right triangle with vertex O at the origin (0, 0), vertices A and B on the line x + 3y 13 = 0 and AOB = 90. Determine the area of OAB.
92 Problem OAB is an isosceles right triangle with vertex O at the origin (0, 0), vertices A and B on the line x + 3y 13 = 0 and AOB = 90. Determine the area of OAB. Solution Problem E and Solution Get Your Analytic Skills On In analytic geometry problems, a representative diagram is important and often provides clues for the solution of the problem. The diagram above has the given information plus a couple of pieces of information that will be justified now. Let B have coordinates (p, q). The slope of OB = q. Since AOB = p 90, OB OA and the slope of OA is the negative reciprocal of OB. Therefore the slope of OA = p. Since the q triangle is isosceles, OA = OB and it follows that the coordinates of A are ( q, p). (We can verify this by finding the length of OA and the length of OB and showing that both are equal to p + q.) Since B(p, q) is on the line x + 3y 13 = 0, it satisfies the equation of the line. p + 3q 13 = 0 (1) Since A( q, p) is on the line x + 3y 13 = 0, it satisfies the equation of the line. q + 3p 13 = 0 or 3p q 13 = 0 () Since we have two equations and two unknowns, we can use elimination to solve for p and q. (1) 4p + 6q 6 = 0 () 3 9p 6q 39 = 0 Adding, we obtain 13p 65 = 0 p = 5 Substituting in (1) q 13 = 0 3q = 3 q = 1 The point B is (5, 1) and the length of OB = = 6. Since OA = OB, OA = 6. AOB is a right triangle so we can use OB as the base and OA as the height in the formula for the area of a triangle. Then the area of AOB = OA OB = 6 6 = 13. the area of AOB is 13 units.
93 Problem E Sphere Pressure A cube rests inside a sphere so that each vertex touches the sphere. The radius of the sphere is 6 cm. Determine the volume of the cube.
94 Problem E and Solution Sphere Pressure Problem A cube rests inside a sphere so that each vertex touches the sphere. The radius of the sphere is 6 cm. Determine the volume of the cube. Solution Label four vertices of the cube A, B, C, D as shown in the diagram. Let x represent the side length of the cube. Then AB = BC = CD = x. The diagonals of a cube intersect in a point such that the distance from the intersection point to each vertex is equal. Since each vertex of the cube touches the sphere, the diagonal of the cube, AD, is equal in length to the diameter of the sphere. Therefore AD = (6) = 1 cm. Each face of a cube is a square so ABC = 90. Using Pythagoras Theorem, AC = AB + BC = x + x = x. In a cube the sides are perpendicular to the base. In particular, DC is perpendicular to the base and it follows that DC AC. Therefore DCA is a right angled triangle. Using Pythagoras Theorem, AD = AC + CD = x + x = 3x. But AD = 1 so AD = 144. Then 3x = 144, x = 48 and x = 4 3 since x > 0. The volume of the cube is x 3 = (4 3) 3 = 19 3 cm 3. the volume of the cube is 19 3 cm 3.
95 Problem E Find The Area  No Parabolem! y B(0, 5) A(r, 0) x C(5, 0) D(p, q) In the diagram, D(p, q) is the vertex of the parabola. The parabola cuts the xaxis at A(r, 0) and C(5, 0). The area of ABC is 5. Determine the area of DBC.
96 Problem Problem E and Solutions Find The Area  No Parabolem! B(0, 5) y In the diagram, D(p, q) is the vertex of the parabola. The parabola cuts the xaxis at A(r, 0) and C(5, 0). The area of ABC is 5. Determine the area of DBC. Solution 1 In ABC, the height is the distance from the xaxis to B(0, 5). Therefore the height is 5 units. The base is AC = 5 r. Using the formula for the area of a triangle, Area ABC = (5 r)(5) = 5. Then 5 r = and r = 3 follows. The coordinates of A are (3, 0). The axis of symmetry is a vertical line through the midpoint of AC which is (4, 0). It follows that the xcoordinate of the vertex is p = 4. Therefore, the vertex is D(4, q). Since the two xintercepts of the parabola are 3 and 5, the equation of the parabola in factored form can be written as y = a(x 3)(x 5). Since B(0, 5) is on the parabola, we can solve for a by substituting x = 0 and y = 5 into y = a(x 3)(x 5). This leads to a = 1 3 and y = 1 3 (x 3)(x 5). To find q, the ycoordinate of D, substitute x = 4, y = q into y = 1 3 (x 3)(x 5). Then q = (4 3)(4 5) = 3. Construct a rectangle with sides parallel to the x and yaxes so that the points B, D, and C are on the rectangle. B is one vertex of the rectangle. We can determine the coordinates of the other three vertices of the rectangle. The results are shown on the diagram to the right. It is straight forward to then determine the lengths required for finding various areas: BE = = 16 3, ED = 4, DF = 1, F C = 1 3, CG = 5 and BG = 5. To find the area of DBC, subtract the areas of BED, DF C, and BCG from the area of rectangle BEF G. The area of rectangle BEF G = BE BG = = 80 3 units. The area of BED = BE ED = = 3 3 units. DF F C The area of DF C = = = 1 6 units. The area of BCG = CG BG = 5 5 = 5 units. y B(0, 5) A(r, 0) A(3, 0) D(p, q) x C(5, 0) x C(5, 0) E(0, 3) D(4, ) F(5, 3) Area DBC = Area Rectangle BEF G Area BED Area DF C Area BCG = = 10 3 units the area of DBC = 10 3 units. The second solution uses the equation of a line and is very different from the first solution. 3 G(5, 5)
97 y Problem In the diagram, D(p, q) is the vertex of the parabola. The parabola cuts the xaxis at A(r, 0) and C(5, 0). The area of ABC is 5. Determine the area of DBC. B(0, 5) Solution To begin this solution we will pick up a result from the first solution and do not repeat the work here. In solution 1, we found vertex D to be the point (4, 1 3 ). A(r, 0) D(p, q) x C(5, 0) Let P (t, 0) be the point where the line through B and D crosses the xaxis. We will determine the equation of the line containing B, P, and D. The slope of the line is The yintercept of the line is 5. Therefore the equation of the line through B, P, and D is y = 4 3 x = 16 4 = 4 To find t, the xcoordinate of P we substitute x = t and y = 0 into the equation of the line. y = 4 3 x = 4 3 t + 5 4t = 15 t = 15 4 B(0, 5) y A(3, 0) P(t, 0) 1 D(4, 3) x C(5, 0) To determine the area of BDC, we find the sum of the area of BP C and the area of DP C. We will use P C as the base in both triangles. Then P C = = 5 4. In BP C, the height is the perpendicular distance from the xaxis to point B. The height is 5. The area of BP C = = 5 8 units. In DP C, the height is the perpendicular distance from the xaxis to point D. The height is 1 3. The area of DP C = = 5 4 units. the area of DBC = 10 3 units. Area DBC = Area BP C + Area DP C = = = 80 4 = 10 3 units
98 Problem E Love is Blind Valentine A heart is constructed by attaching two white semicircles, each with radius cm, onto the hypotenuse of an isosceles right triangle. The heart is then pasted onto a rectangular sheet of red construction paper as illustrated below. (The dashed lines, the side measurement and the right angle symbol will not actually be on the finished card.) Cupid shoots an arrow that lands somewhere on the card. Determine the probability that Cupid s arrow lands on the heart. cm
99 Problem E and Solution Love is Blind Valentine Problem A heart is constructed by attaching two white semicircles, each with radius cm, onto the hypotenuse of an isosceles right triangle. The heart is then pasted onto a rectangular sheet of red construction paper. Cupid shoots an arrow that lands somewhere on the card. Determine the probability that Cupid s arrow lands on the heart. Solution The semicircles are placed along the hypotenuse of the triangle. The hypotenuse is equal in length to four radii and is therefore 4 = 8 cm long. Let x represent the length of the sides of equal length in the triangle and let h represent the height of the triangle drawn from the vertex between the two equal sides to the opposite side. Place the given information on the diagrams below. cm x h x 8 cm h x x Since the triangle is a right triangle we can use Pythagoras Theorem to find the value of x. x + x = 8 x = 64 x = 3 x = 4, since x > 0 The altitude of an isosceles, right triangle bisects the hypotenuse so we can determine the height of the triangle, h. This statement is justified at the end of the solution. h + 4 = x h + 16 = 3 h = 16 h = 4, since h > 0 We now have enough information to find the area of the heart and the entire shape.
100 The total height of the rectangle is equal to the height of the triangle plus the radius of the semicircle. The height of the rectangle is 4 + = 6 cm. The width of the rectangle is the same as the length of the hypotenuse and is therefore equal to 8 cm. The area of the rectangle is 8 6 = 48 cm. The area of the heart is equal to the area of the triangle plus the area of two congruent semicircles of radius cm. Since the triangle is isosceles right we can use one of the equal sides as the base and the other as the height. The area of the triangle is bh = (4 ) (4 ) = 3 = 16 cm. The area of the two semicircles is the same as the area of one circle with radius cm. The area of the two semicircles is πr = π() = 4π cm. The total area of the heart is (4π + 16) cm. The probability that Cupid s arrow lands on the heart is equal to the area of the heart divided by the area of the rectangle. The probability equals 4π+16 = In other words, Cupid has 48 about a 60% chance of landing his arrow on the heart. Happy Valentine s Day. Justification of the statement: The altitude of an isosceles right triangle bisects the hypotenuse. There are many ways to justify this. Since the triangle is an isosceles right triangle, the two angles opposite the sides of equal length are equal. In a triangle, the sum of the angles is 180 so the two equal angles must add to the remaining 90 and each angle is 45. Construct the altitude, h, as shown in the following diagram, splitting the opposite sides into two parts, labelled a and b. x a h b x The altitude hits the opposite side at 90 so each of the two smaller triangles contains a right angle and a 45 angle. The missing angle in each of the two smaller triangles is then 45. Each of the smaller triangles is then isosceles and it follows that h = a and h = b. a = b and the altitude bisects the hypotenuse.
101 Problem E Seeing Red A cube has edges of length n, where n is a positive integer. Three faces, meeting at a corner, are painted red. The cube is then cut into n 3 smaller cubes of unit length. That is, the side lengths of the new cubes are 1 unit. If exactly 15 of these cubes have no faces painted red, determine the value of n. n n n
102 Problem E and Solutions Seeing Red Problem A cube has edges of length n, where n is a positive integer. Three faces, meeting at a corner, are painted red. The cube is then cut into n 3 smaller cubes of unit length. That is, the side lengths of the new cubes are 1 unit. If exactly 15 of these cubes have no faces painted red, determine the value of n. n 1 n1 A n B 1 1 n1 n1 C Solution 1 n This solution requires no content beyond grade ten. The second solution will use the factor theorem which is generally taught in grade twelve. In the diagram above, the sides painted red are labelled A, B, and C. We know that there are n 3 unit cubes. To determine the number of unpainted cubes we can subtract the number of cubes with some red from the total number of cubes. Side A has dimensions n by n by 1 and so contains n unit cubes with some red. Side B has dimensions n by (n 1) by 1 and so contains n (n 1) unit cubes with some red. Side C has dimensions (n 1) by (n 1) by 1 and so contains (n 1)(n 1) unit cubes with some red. The number of unpainted cubes is n 3 n n(n 1) (n 1)(n 1). We can simplify this as follows: n 3 n n(n 1) (n 1)(n 1) = n (n 1) n(n 1) (n 1)(n 1) Each term contains a common factor of (n 1) so the expression simplifies to (n 1)(n n (n 1)) = (n 1)(n n + 1). This further simplifies to (n 1) 3. If the solver pauses here to think about this, if the unit cubes on side A then side B and finally side C are removed we are left with a cube whose side lengths are (n 1) and (n 1) 3 unit cubes. But (n 1) 3 = 15, the actual number of unpainted cubes. Taking the cube root, n 1 = 5 and n = 6 follows.
103 Solution The second solution will use the factor theorem which is generally taught in grade twelve. This solution picks up from the expression giving us the number of unpainted cubes, setting it equal to 15, the number of unpainted cubes. n 3 n n(n 1) (n 1)(n 1) = 15 n 3 n n + n n + n 1 = 15 n 3 3n + 3n 16 = 0 Let f(n) = n 3 3n + 3n 16. When n = 6, f(6) = 6 3 3(6 ) + 3(6) 16 = = 4 4 = 0. Since f(6) = 0, (n 6) is a factor of f(n). After long division (or synthetic division), f(n) = (n 6)(n + 3n + 1). So (n 6)(n + 3n + 1) = 0. n + 3n + 1 = 0 has no real roots so n = 6 is the only root. Therefore the original cube has edges of length 6.
104 Problem E This Triangle is All Right A rightangled triangle has sides whose lengths are twodigit integers. The digits of the length of the hypotenuse are the reverse of the digits of the length of one of the other sides. Determine the smallest possible length of the hypotenuse. a c b
105 Problem E and Solution Problem This Triangle is All Right A rightangled triangle has sides whose lengths are twodigit integers. The digits of the length of the hypotenuse are the reverse of the digits of the length of one of the other sides. Determine the smallest possible length of the hypotenuse. Solution Let x represent the tens digit of the hypotenuse such that x is an integer from 1 to 9. Let y represent the units digit of the hypotenuse such that y is an integer from 1 to 9. Then the length of the hypotenuse is 10x + y. Since one of the sides has the same digits as the hypotenuse in reverse order, the length of this side is 10y + x. Let the third side be z such that z is a two digit integer. Using the Pythagorean Theorem: (10y + x) + z = (10x + y) Expanding: 100y + 0xy + x + z = 100x + 0xy + y Rearranging: z = 99x 99y Factoring: z = 99(x y)(x + y) Since z is a perfect square, 99(x + y)(x y) must also be a perfect square. But 99(x + y)(x y) = 9(11)(x + y)(x y). To be a perfect square (x + y)(x y) must be a multiple of 11 and contain a factor which is a perfect square. Since x and y are each integers from 1 to 9, x y cannot be 11, and so we must have x + y = 11. Again, since x and y are integers from 1 to 9, there are three possibilities for x y that give a perfect square; x y = 1, x y = 4 or x y = 9. These are the three possibilities: (x + y)(x y) = 11(1) or (x + y)(x y) = 11(4) or (x + y)(x y) = 11(9). If x + y = 11 and x y = 1, we solve the system of equations obtaining x = 6 and y = 5. This gives a hypotenuse of 10x + y = 65 and second side 10y + x = 56. Then solving for z, z = 99(x + y)(x y) = 99(11)(1) = 1089 and z = 33. This solution is easily confirmed but is it the only solution? If x + y = 11 and x y = 4, we solve the system of equations obtaining x = 7.5 and y = 3.5. But x and y are both integers so this solution is inadmissible. If x + y = 11 and x y = 9, we solve the system of equations obtaining x = 10 and y = 1. But x must be an integer from 1 to 9 so this solution is inadmissible. Therefore the only solution is a right triangle with hypotenuse of length 65.
106 Problem E Median Madness A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. median In ABC, ABC = 90. A median is drawn from A meeting BC at M such that AM = 5. A second median is drawn from C meeting AB at N such that CN = 10. Determine the length of the longest side of ABC.
107 Problem E and Solution Median Madness Problem A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. In ABC, ABC = 90. A median is drawn from A meeting BC at M such that AM = 5. A second median is drawn from C meeting AB at N such that CN = 10. Determine the length of the longest side of ABC. Solution A x N x B y M y C Since AM is a median, M is the midpoint of BC. Then BM = MC = y. Since CN is a median, N is the midpoint of AB. Then AN = NB = x. NBC is right angled since B = 90. Using the Pythagorean Theorem, NB + BC = CN x + (y) = ( 10) x + 4y = 40 (1) ABM is right angled since B = 90. Using the Pythagorean Theorem, AB + BM = AM (x) + y = 5 4x + y = 5 () Adding (1) and (), 5x + 5y = 65 Dividing by 5, x + y = 13 (3) The longest side of ABC is the hypotenuse AC. Using the Pythagorean Theorem, AC = AB + BC = (x) + (y) = 4x + 4y = 4(x + y ) Substituting from (3) above, AC = 4(13) Taking the square root, AC = 13 the length of the longest side is 13 units. Note: The solver could actually solve a system of equations to find x = and y = 3 and then proceed to solve for the longest side. The above approach was provided to expose the solver to alternate thinking about the solution of this problem.
108 Problem E A Skinny Quadrilateral In the diagram, QP 1 R 1 is rightangled with QP 1 = and QR 1 = 5. Lines QP 1 and QR 1 are extended and many more points are labelled at intervals of 1 unit, so that P 1 P = P P 3 = P 3 P 4 = P 4 P 5 = = 1, and R 1 R = R R 3 = R 3 R 4 = R 4 R 5 = = 1. In fact, P 1 P j = j 1 and R 1 R k = k 1 for any positive integers j and k. For example, P 1 P 5 = 5 1 = 4 and R 1 R 4 = 4 1 = 3. Determine the value of n so that the area of quadrilateral P n P n+1 R n+1 R n is 01. That is, determine the value of n so that the area of the quadrilateral with vertices P n, P n+1, R n+1, and R n is 01.
109 Problem E and Solution A Skinny Quadrilateral Problem In the diagram, QP 1 R 1 is rightangled with QP 1 = and QR 1 = 5. Lines QP 1 and QR 1 are extended and many more points are labelled at intervals of 1 unit, so that P 1 P = P P 3 = P 3 P 4 = P 4 P 5 = = 1, and R 1 R = R R 3 = R 3 R 4 = R 4 R 5 = = 1. In fact, P 1 P j = j 1 and R 1 R k = k 1 for any positive integers j and k. For example, P 1 P 5 = 5 1 = 4 and R 1 R 4 = 4 1 = 3. Determine the value of n so that the area of quadrilateral P n P n+1 R n+1 R n is 01. Solution In order to solve the problem, looking at the calculation of a specific area may prove helpful. So let s determine the area of quadrilateral P 4 P 5 R 5 R 4. Area of quadrilateral P 4 P 5 R 5 R 4 = Area P 5 QR 5 Area P 4 QR 4 = 1 (QP 5)(QR 5 ) 1 (QP 4)(QR 4 ) = 1 ( + (5 1))(5 + (5 1)) 1 ( + (4 1))(5 + (4 1)) = 1 (6)(9) 1 (5)(8) = 7 0 = 7 units We will pattern the more general solution off the above example. Area of quad. P n P n+1 R n+1 R n = Area P n+1 QR n+1 Area P n QR n 01 = 1 (QP n+1)(qr n+1 ) 1 (QP n)(qr n ) 01 = 1 ( + ((n + 1) 1))(5 + ((n + 1) 1)) 1 ( + (n 1))(5 + (n 1)) 01 = 1 ( + n)(5 + n) 1 (1 + n)(4 + n) 404 = ( + n)(5 + n) (1 + n)(4 + n) multiplying by 404 = n + 7n + 10 (n + 5n + 4) 404 = n + 7n + 10 n 5n = n = n 009 = n Therefore the value of n is 009. Using the method of the specific example, this result is easily confirmed.
110
111 Problem E Hosers Nine large hoses can fill a swimming pool in four hours and six small hoses can fill the same swimming pool in eight hours. How long will it take four large hoses and eight small hoses working together to fill the swimming pool?
112 Problem Problem E and Solutions Hosers Nine large hoses can fill a swimming pool in four hours and six small hoses can fill the same swimming pool in eight hours. How long will it take four large hoses and eight small hoses working together to fill the swimming pool? Solution 1 9 large hoses can fill 1 swimming pool in 4 hours. 9 large hoses can fill 1 of the swimming pool in 1 hour. 4 1 large hose can fill 1 1 = 1 of the swimming pool in 1 hour small hoses can fill 1 swimming pool in 8 hours. 6 small hoses can fill 1 of the swimming pool in 1 hour. 8 1 small hose can fill 1 1 = 1 of the swimming pool in 1 hour In one hour, 4 large hoses and 8 small hoses fill 4 ( ) ( ) 48 = = + 3 = 5 of the pool Since it is 5 full in 1 hour, 18 the pool will be completely full in 18 5 Solution = large hoses can fill 1 swimming pool in 4 hours. 1 large hose can fill large hoses can fill 1 9 of the swimming pool in 4 hours. of the swimming pool in 1 hour. 6 small hoses can fill 1 swimming pool in 8 hours. 1 small hose can fill small hoses can fill 1 6 of the swimming pool in 8 hours. of the swimming pool in 1 hour. Together in one hour, 4 large hoses and 8 small hoses fill = + 3 = 5 of the pool Since it is 5 full in 1 hour, 18 the pool will be completely full in 18 5 = or 3 hours and 36 minutes. or 3 hours and 36 minutes. with four large hoses and eight small hoses the pool can be filled in 3 hours and 36 minutes.
113 Problem E Partly Salted When one kilogram of salt is added to a solution of salt and water, the solution becomes 1 salt by mass. One kilogram of water is then added to the new 3 solution resulting in a solution that is 3 10 salt by mass. What fraction of the original solution was salt? +
114 Problem E and Solution Partly Salted Problem When one kilogram of salt is added to a solution of salt and water, the solution becomes 1 3 salt by mass. One kilogram of water is then added to the new solution resulting in a solution that is 3 10 Solution salt by mass. What fraction of the original solution was salt? Let S represent the mass of salt in the original solution. Let W represent the mass of water in the original solution. Then S + W represents the total mass of the original solution. After adding 1 kg of salt to the original solution, there is (S + 1) kg of salt and (S + W + 1) kg of solution. Since this new solution is one third salt by mass, S + 1 S + W + 1 = 1 3 3(S + 1) = 1(S + W + 1) 3S + 3 = S + W + 1 S + = W (1) After adding 1 kg of water to the new solution, there is still (S + 1) kg of salt in (S + W + ) kg of solution. Since this new solution is three tenths salt by mass, S + 1 S + W + = (S + 1) = 3(S + W + ) Substituting (1) into (), 10S + 10 = 3S + 3W + 6 7S 3W = 4 () 7S 3(S + ) = 4 7S 6S 6 = 4 S = kg Substituting for S in (1), W = 6 kg. Then the original solution was S + W = 8 kg. The fraction of the original solution that was salt was 8 = 1 4. one quarter of the original solution was salt by mass.
115 Problem E Five Prime Mates The product of five different odd prime numbers is a fivedigit number of the form strst, where r = 0. Determine all possible numbers. See the next page for a summary of divisibility tests for the integers to 1.
116 Divisibility Tests Divisibility by : A number is divisible by if the last digit is even. Divisibility by 3: A number is divisible by 3 if the sum of the digits is divisible by 3. For example, 195 is not divisible by 3 since = 17 which is not divisible by 3. However, 196 is divisible by 3 since = 18 which is divisible by 3. Divisibility by 4: A number is divisible by 4 if the last two digits are divisible by 4. For example, 195 is not divisible by 4 since 95 is not divisible by 4. However, 196 is divisible by 4 since 96 is divisible by 4. Divisibility by 5: A number is divisible by 5 if the last digit is a 0 or 5. Divisibility by 6: A number is divisible by 6 if it is divisible by both and 3. The number 395 is not divisible by 6 since it is not even and hence is not divisible by. The number 86 is not divisible by 6 since it is not divisible by 3 ( = 16 which is not divisible by 3). The number 964 is divisible by 6. It is even and is therefore divisible by. It is divisible by 3 since = 1 which is divisible by 3. Since 964 is divisible by both and 3, it is divisible by 6. Divisibility by 7: We can follow an unusual algorithm to determine if an number is divisible by 7: Remove the unit s digit, double that digit and subtract it from the leftover number. If the difference is divisible by 7, the original number is divisible by seven. If unsure, repeat the algorithm with the new number. Is 1356 divisible by 7? Remove the 6, double the 6 to 1, subtract from 135 leaving 13. Is 13 divisible by 7? Remove the 3, double the 3 to 6, subtract from 1 leaving 6. 6 is not divisible by 7 and therefore 1356 is not divisible by 7. Is divisible by 7? Remove the 4, double to 8, subtract from 450 giving Repeat. Remove the 4, double to 8, subtract from 449 giving 441. Repeat. Remove the 1, double to, subtract from 44 giving 4 which is divisible by 7. Therefore, is divisible by 7. Divisibility by 8: A number is divisible by 8 if the last three digits are divisible by 8. For example, 195 is not divisible by 8 since 95 is not divisible by 8. However, 196 is divisible by 8 since 96 is divisible by 8. Divisibility by 9: A number is divisible by 9 if the sum of the digits is divisible by 9. For example, 195 is not divisible by 9 since = 17 which is not divisible by 9. However, 196 is divisible by 9 since = 18 which is divisible by 9. Divisibility by 10: A number is divisible by 10 if the last digit is a 0. Divisibility by 11: We can follow an unusual algorithm to determine if an number is divisible by 11: Add the numbers in the even positions. Add the numbers in the odd positions. Subtract the two sums. If this difference is divisible by 11, the original number is divisible by 11. Is divisible by 11? The sums are = 14 and = 13. The difference of the sums is 1, which is not divisible by 11. Therefore, the number is not divisible by 11. Is divisible by 11? The sums are = 5 and = 7. The difference of the sums is, which is divisible by 11. Therefore, the number is divisible by 11. Is divisible by 11? The sums are = 18 and = 18. The difference of the sums is 0, which is divisible by 11. Therefore, the number is divisible by 11. Divisibility by 1: A number is divisible by 1 if it is divisible by 4 and 3. The number 394 is not divisible by 1 since 94 is not divisible by 4. The number 964 is not divisible by 1 since it is not divisible by 3. (The sum of the digits is 19 which is not divisible by 3.) The number 964 is divisible by 1. The last two digits, 64, are divisible by 4 and therefore 964 is divisible by 4. The sum of the digits is 1 which is divisible by 3. Since 964 is divisible by 4 and 3, it is divisible by 1.
117 Problem E and Solutions Five Prime Mates Problem The product of five different odd prime numbers is a fivedigit number of the form strst, where r = 0. Determine all possible numbers. Solution 1 Be sure to read Solution after Solution 1 A number is divisible by 11 if the difference between the sum of the even position numbers and sum of the odd position numbers is a multiple of 11. (This problem can still be done without knowing this divisibility fact but the task is made simpler with it.) The sum of the digits in the even positions is strst is t + s. The sum of the digits in the odd positions is s + r + t but r = 0 so the sum is s + t. The difference of the two sums is (t + s) (s + t) = 0 which is a multiple of 11. Therefore, st0st is divisible by 11. Only odd factors are used so the product will be odd. This means that the product looks like 10 1, 30 3, 50 5, 70 7, or So we begin to systematically look at the possibilities. First, we will examine numbers that have 3 (and 11) as a factor. To be divisible by three, the sum of the digits will be divisible by three. To be divisible by nine, the sum of the digits will be divisible by nine. But if the number is divisible by nine, it is divisible by 3 and would have a repeated prime factor which is not allowed. So we want numbers divisible by 3 but not 9. The possibilities are as follows: 101, 51051, 33033, 93093, 15015, 75075, 57057, 87087, 39039, and The sum of the digits of these numbers is divisible by three so the numbers are divisible by three. The numbers 81081, 63063, 45045, 707 and are divisible by 9 and have therefore been eliminated. Now we examine the prime factorization of each of these numbers to see which numbers satisfy the conditions. 101 = = = Since the prime factor 7 is repeated, this is not a valid number = = = Since there are 5 different odd prime factors, is a valid number = = = Since the prime factor 11 is repeated, this is not a valid number = = = Since there are 5 different odd prime factors, is a valid number.
118 15015 = = = Since there are 5 different odd prime factors, is a valid number = = = = Since the prime factor 5 is repeated and there are six prime factors, this is not a valid number = = = Since there are 5 different odd prime factors, is a valid number = = = Since there are 5 different odd prime factors, is a valid number = = = Since the prime factor 13 is repeated, this is not a valid number = = = Since there are 5 different odd prime factors, is a valid number. Second, we will examine numbers that are divisible by 5 but not 3, since divisibility by three has been examined. If a number is divisible by 5 it ends in 5 or 0. Since the number is odd, we can exclude any number ending in 0 leaving 505, 35035, 55055, 65065, and as possible numbers. (15015, 45045, were examined above and have been excluded.) Now we examine the prime factorization of each of these numbers to see which numbers satisfy the conditions. 505 = = = Since the prime factor 5 is repeated, this is not a valid number = = = Since the prime factor 7 is repeated, this is not a valid number = = = Since the prime factor 11 is repeated, this is not a valid number = = = Since the prime factor 13 is repeated, this is not a valid number = = = Since there are 5 different odd prime factors, is a valid number = = = Since there are 5 different odd prime factors, is a valid number. Thirdly, we will look at numbers that are divisible by 7 but not 3 or 5. If we multiply 7 by the next four odd prime numbers we get = 3333, a six digit number so we are beyond all possible solutions. Therefore there are 8 numbers of the form st0st which are the product of five different odd prime numbers, namely 51051, 93093, 15015, 57057, 87087, 69069, and Look at Solution for a much more insightful approach to the problem.
119 Problem The product of five different odd prime numbers is a fivedigit number of the form strst, where r = 0. Determine the number of possible numbers. Solution Since r = 0, the number is of the form st0st. Then st0st = st(1000) + st = st( ) = st(1001) This means that the number st0st is divisible by 1001 which is the product of the three odd prime factors 7, 11, and 13. So st is a two digit number which is the product of two different odd prime factors none of which can be 7, 11 or 13. It is now a straight forward matter of generating all possible two digit products, a b say, using odd prime factors other than 7, 11 and 13. Prime Factor Prime Factor st Five Different Product a b = a b Odd Primes st0st , 5, 7, 11, , 7, 11, 13, , 7, 11, 13, , 7, 11, 13, , 7, 11, 13, , 7, 11, 13, , 7, 11, 13, , 7, 11, 13, No other two digit product of two different odd prime factors other than 7, 11 and 13 exists. Therefore there are 8 numbers of the form st0st which are the product of five different odd prime numbers, namely 15015, 51051, 57057, 69069, 87087, 93093, and Look on the following page for a summary of the divisibility tests for the numbers through 1.
120 Divisibility Tests Divisibility by : A number is divisible by if the last digit is even. Divisibility by 3: A number is divisible by 3 if the sum of the digits is divisible by 3. For example, 195 is not divisible by 3 since = 17 which is not divisible by 3. However, 196 is divisible by 3 since = 18 which is divisible by 3. Divisibility by 4: A number is divisible by 4 if the last two digits are divisible by 4. For example, 195 is not divisible by 4 since 95 is not divisible by 4. However, 196 is divisible by 4 since 96 is divisible by 4. Divisibility by 5: A number is divisible by 5 if the last digit is a 0 or 5. Divisibility by 6: A number is divisible by 6 if it is divisible by both and 3. The number 395 is not divisible by 6 since it is not even and hence is not divisible by. The number 86 is not divisible by 6 since it is not divisible by 3 ( = 16 which is not divisible by 3). The number 964 is divisible by 6. It is even and is therefore divisible by. It is divisible by 3 since = 1 which is divisible by 3. Since 964 is divisible by both and 3, it is divisible by 6. Divisibility by 7: We can follow an unusual algorithm to determine if an number is divisible by 7: Remove the unit s digit, double that digit and subtract it from the leftover number. If the difference is divisible by 7, the original number is divisible by seven. If unsure, repeat the algorithm with the new number. Is 1356 divisible by 7? Remove the 6, double the 6 to 1, subtract from 135 leaving 13. Is 13 divisible by 7? Remove the 3, double the 3 to 6, subtract from 1 leaving 6. 6 is not divisible by 7 and therefore 1356 is not divisible by 7. Is divisible by 7? Remove the 4, double to 8, subtract from 450 giving Repeat. Remove the 4, double to 8, subtract from 449 giving 441. Repeat. Remove the 1, double to, subtract from 44 giving 4 which is divisible by 7. Therefore, is divisible by 7. Divisibility by 8: A number is divisible by 8 if the last three digits are divisible by 8. For example, 195 is not divisible by 8 since 95 is not divisible by 8. However, 196 is divisible by 8 since 96 is divisible by 8. Divisibility by 9: A number is divisible by 9 if the sum of the digits is divisible by 9. For example, 195 is not divisible by 9 since = 17 which is not divisible by 9. However, 196 is divisible by 9 since = 18 which is divisible by 9. Divisibility by 10: A number is divisible by 10 if the last digit is a 0. Divisibility by 11: We can follow an unusual algorithm to determine if an number is divisible by 11: Add the numbers in the even positions. Add the numbers in the odd positions. Subtract the two sums. If this difference is divisible by 11, the original number is divisible by 11. Is divisible by 11? The sums are = 14 and = 13. The difference of the sums is 1, which is not divisible by 11. Therefore, the number is not divisible by 11. Is divisible by 11? The sums are = 5 and = 7. The difference of the sums is, which is divisible by 11. Therefore, the number is divisible by 11. Is divisible by 11? The sums are = 18 and = 18. The difference of the sums is 0, which is divisible by 11. Therefore, the number is divisible by 11. Divisibility by 1: A number is divisible by 1 if it is divisible by 4 and 3. The number 394 is not divisible by 1 since 94 is not divisible by 4. The number 964 is not divisible by 1 since it is not divisible by 3. (The sum of the digits is 19 which is not divisible by 3.) The number 964 is divisible by 1. The last two digits, 64, are divisible by 4 and therefore 964 is divisible by 4. The sum of the digits is 1 which is divisible by 3. Since 964 is divisible by 4 and 3, it is divisible by 1.
121 Problem E Powerfully Perfect Squares The prime factorization of 0 is 5. The divisors of 0 are: = 1, = 5, =, = 10, 5 0 = 4, and 5 1 = 0. The number 0 has 6 divisors. Two of the divisors, 1 and 4, are perfect squares. How many divisors of are perfect squares?
122 Problem E and Solution Powerfully Perfect Squares Problem The prime factorization of 0 is 5. The divisors of 0 are = 1, = 5, =, = 10, 5 0 = 4, and 5 1 = 0. The number 0 has 6 divisors. Two of the divisors, 1 and 4, are perfect squares. How many divisors of are perfect squares? Solution First, let s look at the prime factorization of some perfect squares. 9 = 3, 16 = 4, 36 = 3, and 144 = 4 3 are the prime factorizations of four perfect squares. Note that the exponent on each of the prime factors is even. For some integer a, if m is an even integer greater than or equal to zero then a m is a perfect square. Now = ( 503) 01 = ( ) = All divisors of will be of the form k 503 n, 0 k 404, 0 n 01 where k and n are both integers. For k to be a perfect square, k must be an even integer such that 0 k 404. There are 404 = 01 even numbers from 1 to 404. The number 0 is also even so there are 013 values of k such that k is a perfect square. For 503 n to be a perfect square, n must be an even integer such that 0 n 01. There are 1006 even numbers from 1 to 01. Zero is also even so there are 1007 values of n such that 503 n is a perfect square. For each of the 013 values of k, there are 1007 values of n so there are = perfect square divisors of has perfect square divisors. (For a smaller number like 1 6 we could also determine the number of perfect square divisors using the above approach. We could verify that the approach is valid by physically listing all of the divisors and then counting the number that are perfect squares. The number 1 6 has only 91 divisors, 8 of which are perfect squares. It is not practical to list all the divisors of and then count the perfect squares. But our approach allows us to still count the perfect square divisors.)
123 Problem E Powerful Factors at Work The product of the integers 1 to 64 can be written in an abbreviated form as 64! and we say 64 factorial. So 64! = In general, the product of the positive integers 1 to n is n! = n (n 1) (n ) 3 1. Determine the largest positive integer value of n so that 64! is divisible by 1 n.
124 Problem E and Solution Powerful Factors at Work Problem The product of the integers 1 to 64 can be written in an abbreviated form as 64! and we say 64 factorial. So 64! = In general, the product of the positive integers 1 to n is n! = n (n 1) (n ) 3 1. Determine the largest positive integer value of n so that 64! is divisible by 1 n. Solution The prime factorization of 1 is 3. We must determine how many times the factors 3 are repeated in the factorization of P. First we will count the number of factors of in P by looking at the 3 even numbers. Each of the numbers {, 4, 6,, 60, 6, 64} contains a factor of. That is a total of 3 factors of. Dividing the even numbers by, we obtain the numbers {1,, 3,, 30, 31, 3}. This list contains 16 even numbers so we gain another 16 factors of bringing the total to = 48. Dividing the even numbers by, we obtain the numbers {1,, 3,, 14, 15, 16}. This list contains 8 even numbers so we gain another 8 factors of bringing the total to = 56. Dividing the even numbers by, we obtain the numbers {1,, 3,, 6, 7, 8}. This list contains 4 even numbers so we gain another 4 factors of bringing the total to = 60. Dividing the even numbers by, we obtain the numbers {1,, 3, 4}. This list contains even numbers so we gain another factors of bringing the total to 60 + = 6. Finally, dividing the even numbers by, we obtain the numbers {1, }. This list contains 1 even number so we gain another factor of bringing the total to = 63. So when P is factored there are 63 factors of. In fact, the largest power of that P is divisible by is 63. Next we will count the number of factors of 3 in P by looking at the 1 multiples of 3, namely {3, 6, 9,, 57, 60, 63}. Each of these numbers 1 numbers contains a factor of 3. Dividing the numbers by 3, we obtain the numbers {1,, 3,, 19, 0, 1}. This list contains 7 multiples of 3 so we gain another 7 factors of 3 bringing the total to = 8. Dividing the multiples of 3 by 3, we obtain the numbers {1,, 3, 4, 5, 6, 7}. This list contains multiples of 3 so we gain another factors of 3 bringing the total to 8 + = 30. Dividing the multiples of 3 by 3, we obtain the numbers {1, }. This list contains no multiples of 3 so when P is factored there is a total of 30 factors of 3. In fact, the largest power of 3 that P is divisible by is Combining the results P is divisible by = = 3 ( ) = 3 (( ) 3) 30 = 3 (1) 30 P is divisble by 1 30 and the largest value of n is 30. (Since all of the powers of 3 have been used, none remain to combine with any of the remaining s to form an additional factor of 1.)
125 Problem E You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. The following information may be helpful in the solution of the problem. An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3,5,7,9 is an arithmetic sequence with four terms and constant difference. The general term of an arithmetic sequence is t n = a + (n 1)d, where a is the first term, d is the constant difference and n is the number of terms. The sum, S n, of the first n terms of an arithmetic sequence can be found using either S n = n ( ) [a + (n 1)d] or S t1 + t n n = n, where t 1 is the first term of the sequence and t n is the n th term of the sequence. The following example is provided to verify the accuracy of the formulas and to illustrate their use. For the arithmetic sequence 3,5,7,9, a = t 1 = 3, d =, n = 4 and t n = t 4 = 9. S n = = 4 S n = n [a + (n 1)d] = 4 [(3) + (3)] = [1] = 4 ( ) ( ) t1 + t n S n = n = 4 = 4(6) = 4
126 Problem Problem E and Solutions You Make a Difference The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Three solutions are provided for this problem. Solution 1 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 5 = 3 and t = a 3 a = 1 5 = 7. Since the new sequence is arithmetic, the difference between consecutive terms is constant. Therefore the constant difference is d = t t 1 = 7 3 = 4. We can now generate a few more terms in the sequence of differences by adding the constant to the previous term. t 3 = t + 4 = = 11 and t 4 = t = = 15. Using the information we can present the information in tabular form. Term Number First Sequence First Difference Second Difference n a n t n (Difference of the differences) Since the second difference is constant we can represent the general term of the first sequence with a quadratic function. Let a n = pn + qn + r. For n = 1, a 1 = = p(1) + q(1) + r. p + q + r =. (1) For n =, a = 5 = p() + q() + r. 4p + q + r = 5. () For n = 3, a 3 = 1 = p(3) + q(3) + r. 9p + 3q + r = 1. (3) Subtracting (1) from (), 3p + q = 3. (4) Subtracting () from (3), 5p + q = 7. (5) Subtracting (4) from (5), p = 4 and p = follows. Substituting p = into (4), 3() + q = 3 and q = 3 follows. Substituting p =, q = 3 into (1), 3 + r = and r = 3 follows. a n = n 3n + 3 is the general term of the given sequence in terms of n.
127 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. We can generate more terms: t 3 = 11, t 4 = 15, t 5 = 19,. Each term in the new sequence is the difference between consecutive terms of the original sequence. a a 1 = t 1 a 3 a = t a 4 a 3 = t 3 a n a n 3 = t n 3 a n 1 a n = t n a n a n 1 = t n 1 Adding, a n a 1 = t 1 + t + t t n + t n 1 = S n 1. (1) To find the sum S n 1 = t 1 + t + t t n + t n 1 we can use the formula S n = n [a + (n 1)d] with a = 3 and d = 4 for (n 1) terms. So S n 1 = t 1 + t + t t n + t n 1 = n 1 [a + ((n 1) 1)d] = n 1 [(3) + ((n 1) 1)(4)] = n 1 [6 + (n )(4)] = n 1 [6 + 4n 8] = n 1 [4n ] = (n 1)(n 1) From (1) above, a n a 1 = S n 1 so a n = S n 1 + a 1 = (n 1)(n 1) + = n 3n + 3. a n = n 3n + 3 is the general term of the given sequence in terms of n.
128 Problem The sequence a 1 =, a = 5, a 3 = 1,, a n,, where a n is the n th term in the sequence, has the property that the difference between consecutive terms forms an arithmetic sequence. Find a formula for a n in terms of n. Solution 3 Let the sequence of differences be t 1, t, t 3,, t n,. Then t 1 = a a 1 = 3 and t = a 3 a = 7. Since the new sequence is arithmetic, the constant difference is d = 7 3 = 4. Using the formula for the general term of an arithmetic sequence, t n = a + (n 1)d, the general term of the sequence of differences is t n = (3) + (n 1)(4) = 3 + 4n 4 = 4n 1. To generate the original sequence we start with the first term and add more terms from the arithmetic sequence of differences. For example, a 1 = 3, a = + t 1 = + [4(1) 1] = + 3 = 5, and a 3 = + t 1 + t = + [4(1) 1] + [4() 1] = = 1 So, a n = + t 1 + t + t t n 1 = + [4(1) 1] + [4() 1] + [4(3) 1] + + [4(n 1) 1] = + [4(1) + 4() + 4(3) + + 4(n 1)] + (n 1)( 1) = + 4[ (n 1)] n + 1 [ ] (n 1)(n) = 3 n + 4 (1) = 3 n + (n n) = 3 n + n n = n 3n + 3 In (1) above, (n 1) is an arithmetic sequence with t 1 = 1, t n 1 = (n 1) and the number of terms (n ( 1) terms. ) Using the formula for the sum of the terms of an t1 + t n arithmetic sequence, S n = n, we obtain ( ) 1 + (n 1) ( n ) (n 1)(n) S n 1 = (n 1) = (n 1) = a n = n 3n + 3 is the general term of the given sequence in terms of n.
129 Problem E Minimum Multiplier The number 1867 is multiplied by a positive integer k. The last four digits of the product are 199. Determine the minimum value of k ??????
130 Problem E and Solution Minimum Multiplier Problem The number 1867 is multiplied by a positive integer k. The last four digits of the product are 199. Determine the minimum value of k ?????? Solution To begin with we will show that k has four digits or less. A number with five digits, pqrst for example, can be written p qrst = p qrst. The digit p in the multiplier cannot affect the final four digits in the product. Therefore the minimum k is a number with four or fewer digits. Let the multiplier be abcd such that 1867 abcd is a number whose last four digits are 199. Then multiplying 7, the units digit of 1867, by d, the units digit in abcd, produces a number ending in. The only possible value for d is 6 since 7 6 = 4. (Note the possible last digits when 7 multiplies a single digit number: 7 0 = 0, 7 1 = 7, 7 = 14, 7 3 = 1, 7 4 = 8, 7 5 = 35, 7 6 = 4, 7 7 = 49, 7 8 = 56, 7 9 = 63.) Therefore the multiplier is abc6. The second last digit in the product 199 is 9. This digit is produced by multiplying 67 from 1867 with c6 from abc c c. 9 So 0 + 7c is a number that ends in 9. The only possible value for c is 7. (Refer back to the product list given above.) Therefore the multiplier is a number of the form ab76. Is k = 76? The product, = , does not end in 199. So k is at least a three digit number. The third last digit in the product 199 is 9. This digit is produced by multiplying 867 from 1867 with b76 from ab b b So b is a number that ends in 9 and it follows that 7b is a number that ends in 1. The only possible value for b is 3. (Refer back to the product list given above.) Therefore the multiplier is a number of the form a376. Is k = 376? The product = does end in 199. So 376 multiplied by 1867 produces a number ending in 199. the smallest value of k is 376.
131 Problem E Seeing Red A cube has edges of length n, where n is a positive integer. Three faces, meeting at a corner, are painted red. The cube is then cut into n 3 smaller cubes of unit length. That is, the side lengths of the new cubes are 1 unit. If exactly 15 of these cubes have no faces painted red, determine the value of n. n n n
132 Problem E and Solutions Seeing Red Problem A cube has edges of length n, where n is a positive integer. Three faces, meeting at a corner, are painted red. The cube is then cut into n 3 smaller cubes of unit length. That is, the side lengths of the new cubes are 1 unit. If exactly 15 of these cubes have no faces painted red, determine the value of n. n 1 n1 A n B 1 1 n1 n1 C Solution 1 n This solution requires no content beyond grade ten. The second solution will use the factor theorem which is generally taught in grade twelve. In the diagram above, the sides painted red are labelled A, B, and C. We know that there are n 3 unit cubes. To determine the number of unpainted cubes we can subtract the number of cubes with some red from the total number of cubes. Side A has dimensions n by n by 1 and so contains n unit cubes with some red. Side B has dimensions n by (n 1) by 1 and so contains n (n 1) unit cubes with some red. Side C has dimensions (n 1) by (n 1) by 1 and so contains (n 1)(n 1) unit cubes with some red. The number of unpainted cubes is n 3 n n(n 1) (n 1)(n 1). We can simplify this as follows: n 3 n n(n 1) (n 1)(n 1) = n (n 1) n(n 1) (n 1)(n 1) Each term contains a common factor of (n 1) so the expression simplifies to (n 1)(n n (n 1)) = (n 1)(n n + 1). This further simplifies to (n 1) 3. If the solver pauses here to think about this, if the unit cubes on side A then side B and finally side C are removed we are left with a cube whose side lengths are (n 1) and (n 1) 3 unit cubes. But (n 1) 3 = 15, the actual number of unpainted cubes. Taking the cube root, n 1 = 5 and n = 6 follows.
133 Solution The second solution will use the factor theorem which is generally taught in grade twelve. This solution picks up from the expression giving us the number of unpainted cubes, setting it equal to 15, the number of unpainted cubes. n 3 n n(n 1) (n 1)(n 1) = 15 n 3 n n + n n + n 1 = 15 n 3 3n + 3n 16 = 0 Let f(n) = n 3 3n + 3n 16. When n = 6, f(6) = 6 3 3(6 ) + 3(6) 16 = = 4 4 = 0. Since f(6) = 0, (n 6) is a factor of f(n). After long division (or synthetic division), f(n) = (n 6)(n + 3n + 1). So (n 6)(n + 3n + 1) = 0. n + 3n + 1 = 0 has no real roots so n = 6 is the only root. Therefore the original cube has edges of length 6.
134 Problem E This Triangle is All Right A rightangled triangle has sides whose lengths are twodigit integers. The digits of the length of the hypotenuse are the reverse of the digits of the length of one of the other sides. Determine the smallest possible length of the hypotenuse. a c b
135 Problem E and Solution Problem This Triangle is All Right A rightangled triangle has sides whose lengths are twodigit integers. The digits of the length of the hypotenuse are the reverse of the digits of the length of one of the other sides. Determine the smallest possible length of the hypotenuse. Solution Let x represent the tens digit of the hypotenuse such that x is an integer from 1 to 9. Let y represent the units digit of the hypotenuse such that y is an integer from 1 to 9. Then the length of the hypotenuse is 10x + y. Since one of the sides has the same digits as the hypotenuse in reverse order, the length of this side is 10y + x. Let the third side be z such that z is a two digit integer. Using the Pythagorean Theorem: (10y + x) + z = (10x + y) Expanding: 100y + 0xy + x + z = 100x + 0xy + y Rearranging: z = 99x 99y Factoring: z = 99(x y)(x + y) Since z is a perfect square, 99(x + y)(x y) must also be a perfect square. But 99(x + y)(x y) = 9(11)(x + y)(x y). To be a perfect square (x + y)(x y) must be a multiple of 11 and contain a factor which is a perfect square. Since x and y are each integers from 1 to 9, x y cannot be 11, and so we must have x + y = 11. Again, since x and y are integers from 1 to 9, there are three possibilities for x y that give a perfect square; x y = 1, x y = 4 or x y = 9. These are the three possibilities: (x + y)(x y) = 11(1) or (x + y)(x y) = 11(4) or (x + y)(x y) = 11(9). If x + y = 11 and x y = 1, we solve the system of equations obtaining x = 6 and y = 5. This gives a hypotenuse of 10x + y = 65 and second side 10y + x = 56. Then solving for z, z = 99(x + y)(x y) = 99(11)(1) = 1089 and z = 33. This solution is easily confirmed but is it the only solution? If x + y = 11 and x y = 4, we solve the system of equations obtaining x = 7.5 and y = 3.5. But x and y are both integers so this solution is inadmissible. If x + y = 11 and x y = 9, we solve the system of equations obtaining x = 10 and y = 1. But x must be an integer from 1 to 9 so this solution is inadmissible. Therefore the only solution is a right triangle with hypotenuse of length 65.
136 Problem E A Card Game To Flip Over Too E Z Dealer has a deck consisting of 100 cards numbered from 1 to 100. Each card has the same number printed on both sides. One side of the card is yellow and the other side of the card is red. E Z places all the cards, red side up, on the table. He first turns over every card that has a number which is a multiple of. He then examines all the cards, and turns over every card that has a number which is a multiple of 3. He again examines all the cards, and turns over every card that has a number which is a multiple of 4. Finally, he examines all the cards and turns over every card that has a number which is a multiple of 5. After E Z has finished, how many cards have the red side facing up?
137 Problem Problem E and Solution A Card Game To Flip Over Too E Z Dealer has a deck consisting of 100 cards numbered from 1 to 100. Each card has the same number printed on both sides. One side of the card is yellow and the other side of the card is red. E Z places all the cards, red side up, on the table. He first turns over every card that has a number which is a multiple of. He then examines all the cards, and turns over every card that has a number which is a multiple of 3. He again examines all the cards, and turns over every card that has a number which is a multiple of 4. Finally, he examines all the cards and turns over every card that has a number which is a multiple of 5. After E Z has finished, how many cards have the red side facing up? Solution If a card number is a multiple of, 3, 4 and 5, it will be flipped four times. This card will go from red to yellow to red to yellow to red again. So the card will still be red once E Z has finished. If a card number is a multiple of exactly three of, 3, 4 and 5, it will be flipped three times. This card will go from red to yellow to red to yellow. So the card will be yellow once E Z has finished. If a card number is a multiple of exactly two of, 3, 4 and 5, then it will be flipped twice. This card will go from red to yellow to red again. So the card will still be red once E Z has finished. If a card number is a multiple of exactly one of, 3, 4 and 5, it will be flipped once. This card will go from red to yellow. So the card will be yellow once E Z has finished. If a card number is a multiple of none of, 3, 4 and 5, then this card will not be flipped and so the card will still be red once E Z has finished. To determine how many cards have the red side facing up once E Z has finished, let s determine how many cards have the yellow side facing up once E Z has finished. To do so, we need to determine how many card numbers are multiples of exactly three of, 3, 4 and 5 and how many cards are multiples of exactly one of, 3, 4 and 5. Let s consider the cases: A card number is a multiple of, 3 and 4, but not 5 If a card number is a multiple of, 3 and 4, then it must be a multiple of 1, the lowest common multiple of, 3 and 4. So we are want card numbers that are multiples of 1 but not 5. If a card number is a multiple of 1 and 5, then it is a multiple of 1 5 = 60. So we want all multiples of 1 that are not multiples of 60. There are 8 multiples of 1 from 1 to 100, but one is 60. So there are 8 1 = 7 numbers that are multiples of, 3 and 4, but not 5. A card number is a multiple of, 3 and 5, but not 4 If a card number is a multiple of, 3 and 5, then it must be a multiple of 30, the lowest common multiple of, 3 and 5. So we want all multiples of 30 that are not multiples of 4. There are 3 multiples of 30 from 1 to 100, but one is 60, which is also a multiple of 4. So there are numbers from 1 to 100 that are multiples of, 3 and 5, but not 4.
138 A card number is a multiple of, 4 and 5, but not 3 If a card number is a multiple of, 4 and 5, then it must be a multiple of 0, the lowest common multiple of, 4 and 5. So we want all multiples of 0 that are not multiples of 3. There are 5 multiples of 0 from 1 to 100, but one is 60, which is a multiple of 3. So there are 4 numbers from 1 to 100 that are multiples of, 4 and 5, but not 3. A card number is a multiple of 3, 4 and 5, but not It is not possible for a card number to be a multiple of 4 but not. So there are no card numbers in this case. A card number is a multiple of but not 3, 4, or 5 There are 50 numbers from 1 to 100 which are multiples of and 5 numbers from 1 to 100 which are multiples of 4 (and thus ). So there are 50 5 = 5 numbers from 1 to 100 multiples of by but not 4. These are {, 6, 10, 14, 18,, 6, 30, 34, 38, 4, 46, 50, 54, 58, 6, 66, 70, 74, 78, 8, 86, 90, 94, 98} We need to remove numbers that are still multiples of 3 or 5. After doing so we are left with {, 14,, 6, 34, 38, 46, 58, 6, 74, 8, 86, 94, 98} So there are 14 numbers from 1 to 100 that are multiples of but not 3, 4 or 5. A card number is a multiple of 3 but not, 4, or 5 There are 33 multiples of 3 from 1 to 100, {3, 6, 9, 1, 15,, 87, 90, 93, 96, 99}. In this group of multiples, there are 17 numbers that are odd. So there are 17 numbers from 1 to 100 that are multiples of 3 but not. These numbers are {3, 9, 15, 1, 7, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99}. We still need to remove numbers that are multiples of 5. After doing so we are left with {3, 9, 1, 7, 33, 39, 51, 57, 63, 69, 81, 87, 93, 99}. So there are 14 numbers from 1 to 100 that are multiples of 3 but not, 4 or 5. A card number is a multiple of 4 but not, 3, or 5 It is not possible for a card number to be a multiple of 4 but not. So there are no card numbers in this case. A card number is a multiple of 5 but not, 3, or 4 There are 0 multiples of 5 from 1 to 100, but half of those are multiples of. The multiples of 5 which are not multiples of are {5, 15, 5, 35, 45, 55, 65, 75, 85, 95}. We still need to remove numbers that are multiples of 3. After doing so we are left with {5, 5, 35, 55, 65, 85, 95}. So there are 7 numbers from 1 to 100 that are multiples of 5 but not, 3 or 4. Therefore, once he has finished, EZ Dealer is left with 100 ( ) = = 5 cards with the red side facing up. Extension: Suppose E Z Dealer continues flipping cards in this manner. So, after he has flipped all cards whose number is a multiple of 5, he then flips all cards whose card number is a multiple of 6, then 7, then 8, and so on until he flips all cards whose number is a multiple of 100. Once E Z has finished, how many cards will have the red side facing up?
139 Problem E A NotSoAverage Task Four numbers are selected such that when each number is added to the average of the other three the following sums are obtained: 5, 37, 43, and 51. Determine the average of the four numbers.
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