2000 Solutions Fermat Contest (Grade 11)

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1 anadian Mathematics ompetition n activity of The entre for Education in Mathematics and omputing, University of Waterloo, Waterloo, Ontario 000 s Fermat ontest (Grade ) for The ENTRE for EDUTION in MTHEMTIS and OMPUTING wards 000 Waterloo Mathematics Foundation

2 000 Fermat s Part :. The sum is equal to () 6 () 4 4 () 8 8 (D) 64 0 (E) = 64 = 6 NSWER: (E). If the following sequence of five arrows repeats itself continuously, what arrow will be in the 48th position?,,,, () () () (D) (E) Since this sequence repeats itself, once it has completed nine cycles it will be the same as starting at the beginning. Thus the 48th arrow will be the same as the third one. NSWER: () 3. farmer has 7 cows, 8 sheep and 6 goats. How many more goats should be bought so that half of her animals will be goats? () 8 () () (D) 9 (E) 6 If the cows and sheep were themselves goats we would have goats. This means that she would need nine extra goats.

3 000 Fermat s 3 Let the number of goats added be x. Therefore, 6 + x + x = ross multiplying gives, ( 6+ x)= + x + x = + x x = 9. s in solution, she would add 9 goats. NSWER: (D) 4. The square of 9 is divided by the cube root of. What is the remainder? () 6 () 3 () 6 (D) (E) The square of 9 is 8 and the cube root of is. When 8 is divided by, the quotient is 6 and the remainder is. Since 8 = 6 +, the remainder is. NSWER: (E). The product of, 3,, and y is equal to its sum. What is the value of y? () 3 () 0 3 () 0 9 (D) 3 0 (E) 0 3 Since ( )( 3)( )( y)= y. 30y= 0 + y 9y = 0 and y = 0. 9 NSWER: () 6. student uses a calculator to find an answer but instead of pressing the x key presses the x key by mistake. The student s answer was 9. What should the answer have been? () 43 () 8 () 79 (D) 3 (E) 66 Since x = 9, x = 8. Thus, x = 8 = 66.

4 000 Fermat s 4 Since the square root of the number entered is 9, the number must have been 8. The answer desired is 8 = 66. NSWER: (E) 7. The sum of the arithmetic series ( 300)+ ( 97)+ ( 94) is () 309 () 97 () 6 (D) 98 (E) 8 The given series is ( 300)+ ( 97)+ ( 94) ( 3) The sum of the terms from 300 to 300 is 0. The sum of all the terms is thus = 98. NSWER: (D) 8. In a school referendum, 3 of a student body voted yes and 8% voted no. If there were no spoiled ballots, what percentage of the students did not vote? () 7% () 40% () 3% (D) % (E) 88% Since 3 or 60% of the students voted yes and 8% voted no, 88% of the student body voted. Hence, 00% 88% or %, of the students did not vote. NSWER: (D) 9. The numbers 6, 4, x, 7, 9, y, 0 have a mean of 3. What is the value of x+ y? () 0 () () 3 (D) (E) 3 If the 7 numbers have a mean of 3, this implies that these numbers would have a sum of 7 3 or 9. We now can calculate x+ y since, x y + 0 = 9 or, x + y+ 6 = 9 Therefore, x+ y=3. NSWER: (E) 0. 3 If xxx ( ( + )+ )+ 3= x + x + x 6 then x is equal to () () 9 () 4 or 3 (D) or 0 (E)

5 000 Fermat s xxx ( ( + )+ )+ 3= xx ( + x+ )+ 3 3 = x + x + x Since x + x + x+ 3= x + x + x 6 x+ 3= x 6 x = 9 NSWER: () Part :. When the regular pentagon is reflected in the line PQ, and then rotated clockwise 44 about the centre of the pentagon, its position is P Q () P Q () P Q () (D) P Q (E) P Q fter the reflection, the new pentagon is Since each of the five central angles equals 7 o o, a clockwise rotation of 44 7 will place the pentagon in the shown position. o ( ) about its centre NSWER: () 6 7. If the expression 8 was evaluated, it would end with a string of consecutive zeros. How many zeros are in this string? () 0 () 8 () 6 (D) 3 (E) zero at the end of a number results from the product of and. The number of zeros at the end of a number equals the number of product pairs of and that can be formed from the prime factorization of that number Since 8 = ( ) ( ) ( )

6 000 Fermat s = = There will be ten zeros in the string at the end of the number. NSWER: () 3. Rectangle D is divided into five congruent rectangles as shown. The ratio : is () 3: () : () : (D) :3 (E) 4:3 D If we let the width of each rectangle be x units then the length of each rectangle is 3x units. (This is illustrated in the diagram.) The length,, is now 3x+ x+ x or x units and = 3 x. Thus : = x: 3x = 3 :, x 0. D 3x x x x x x 3x NSWER: (D) 4. In the regular hexagon DEF, two of the diagonals, F and D, intersect at G. The ratio of the area of quadrilateral FEDG to the area of G is F G () 3 3: () 4 : () 6 : (D) 3: (E) : E D Join E to and D to as shown. lso join E to and draw a line parallel to E through the point of intersection of E and D. Quadrilateral FEDG is now made up of five triangles each of which has the same area as G. The required ratio is :. F E G D

7 000 Fermat s 7 For convenience, assume that each side of the hexagon has a length of units. Each angle in the hexagon equals 0 o so G = ( 0 )= 60. Now label G as shown. Using the standard ratios for a triangle we have G = 3 and G =. 3 G 60 The area of G = () 3 =. Dividing the quadrilateral FGDE as illustrated, it will have an area of ( 3 )+ ()( 3)= 3. The required ratio is 3 3 : or :, as in solution. 3 F E G 3 3 D NSWER: (E). In a sequence, every term after the second term is twice the sum of the two preceding terms. The seventh term of the sequence is 8, and the ninth term is 4. What is the eleventh term of the sequence? () 60 () 304 () 8 (D) 6 (E) 64 Let the seventh through eleventh terms of the sequence be t7, t8, t9, t0, and t. Since t = t + t = 8 + t = 8 + t ( ) ( ) t8 = 4. Hence t0 = ( t8 + t9)= ( 4+ 4)= 6 and t = 4 ( + 6) = , NSWER: () 6. The digits,, 3, and are randomly arranged to form a four digit number. What is the probability that the sum of the first and last digits is even? () 4 () 3 () 6 (D) (E) 3

8 000 Fermat s 8 The numbers that can be formed from the digits,, 3, and, in ascending order, are 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, and 3. These are twelve possibilities. For the sum of two digits to be even, both must be even or both must be odd. From the above list, the numbers for which the sum of the first and last digits is even, are 3, 3, 3, and 3. These are four possibilities. Thus, the probability of getting one of these numbers is 4 or. 3 NSWER: () 7. Three circles have centres, and with radii, 4 and 6 respectively. The circles are tangent to each other as shown. Triangle has () obtuse () = 90 () = 90 (D) all angles acute (E) = Since the circles are mutually tangent, the lines joining their centres pass through the points of tangency. Thus, the sides of have lengths 6, 8 and 0 if we write in the radii as shown. Since 0 = 6 + 8, the triangle is right-angled at. NSWER: () If P = and Q = 3 3 then the value of P Q is () () () 0 (D) (E) 4 P Q = ( P+ Q) ( P Q)= [( )+( 3 3 )]( ) [ ( )] = ( )( ) = 43 = 4. o P Q = ( ) ( )

9 000 Fermat s 9 = = = o o 4000 NSWER: (E) 9. n ant walks inside a 8 cm by 0 cm rectangle. The ant s path follows straight lines which always make angles of 4 to the sides of the rectangle. The ant starts from a point X on one of the shorter sides. The first time the ant reaches the opposite side, it arrives at the mid-point. What is the distance, in centimetres, from X to the nearest corner of the rectangle? () 3 () 4 () 6 (D) 8 (E) 9 If we took a movie of the ant s path and then played it backwards, the ant would now start at the point E and would then end up at point X. Since the ant now starts at a point nine cm from the corner, the first part of his journey is from E to. This amounts to nine cm along the length of the rectangle since E is an isosceles right-angled triangle. This process continues as illustrated, until the ant reaches point. y the time the ant has reached, it has travelled or 3 cm along the length of the rectangle. To travel from to X, the ant must travel cm along the length of the rectangle which puts the ant 3 cm from the closest vertex. 9 cm E 9 cm 36 cm 8 cm 8 cm 8 cm 8 cm X 3 cm cm 9 cm 8 cm 8 cm 8 cm 36 cm 36 cm 3 cm cm NSWER: () 0. Given a+ b+ 3c+ 4d+ e= k and a= 4b= 3c= d = e find the smallest positive integer value for k so that a, b, c, d, and e are all positive integers. () 87 () () 0 (D) 0 (E) 60 From the equalities a= 4b= 3c= d = e, we conclude that e is the largest and a is the smallest of the integers. Since e is divisible by, 4, 3, and, the smallest possible value for e is 60. The corresponding values for a, b, c, and d, are,, 0, and 30, respectively. Thus, the smallest positive integer value for k is ( )+ ( )+ ( )+ ( )=.

10 000 Fermat s 0 NSWER: () Part :. Two circles of radius 0 are tangent to each other. tangent is drawn from the centre of one of the circles to the second circle. To the nearest integer, what is the area of the shaded region? () 6 () 7 () 8 (D) 9 (E) 0 Let the centres of the circles be and and the point of contact of the tangent line from to the circle with centre be. Thus, = 90 o using properties of tangents to a circle. Since the sides of right-angled triangle are in the ratio of : : 3, = 30 o and = 60 o. The area of the shaded region is equal to the area of minus the area of the two sectors of the circles. The area of the two sectors is equivalent to the area of a circle with radius 0. 4 The area of the shaded region = ( 0)( 0 3) π( 0) = 0 3 π To the nearest integer, the area is 8. 4 NSWER: (). The left most digit of an integer of length 000 digits is 3. In this integer, any two consecutive digits must be divisible by 7 or 3. The 000th digit may be either a or b. What is the value of a+ b? () 3 () 7 () 4 (D) 0 (E) 7 We start by noting that the two-digit multiples of 7 are 7, 34,, 68, and 8. Similarly we note that the two-digit multiples of 3 are 3, 46, 69, and 9. The first digit is 3 and since the only two-digit number in the two lists starting with 3 is 34, the second digit is 4. Similarly the third digit must be 6. The fourth digit, however, can be either 8 or 9. Let s consider this in two cases.

11 000 Fermat s ase If the fourth digit is 8, the number would be and would stop here since there isn t a number in the two lists starting with 7. ase If the fourth digit is 9, the number would be and the five digits 3469 would continue repeating indefinitely as long as we choose 9 to follow 6. If we consider a 000 digit number, its first 99 digits must contain 399 groups of The last groups of five digits could be either 3469 or 3468 which means that the 000th digit may be either or so that a+ b= + = 7. NSWER: () 3. circle is tangent to three sides of a rectangle having side lengths and 4 as shown. diagonal of the rectangle intersects the circle at points and. The length of is () () 4 () (D) (E) Of the many ways to solve this problem perhaps the easiest is to use a artesian co-ordinate system. There are two choices for the origin, the centre of the circle and the bottom left vertex of the rectangle. The first solution presented uses the centre of the circle as the origin where the axes are lines drawn parallel to the sides of the rectangle through ( 0, 0). The equation of the circle is now x + y = and the equation of the line containing the diagonal is y+ = x+ or x y=. ( ) y (0, ) (, 0) x 4 Equation of Line: y+ = ( x+ ) x y= Equation of ircle: x + y = ( ) + = Solving to find the intersection points of the line and circle gives, y+ y y + 4y= 0 y( y+ 4)= 0 y= 0 or y= If we substitute these values of y into x y= we find, 3 4 The points of intersection are, and, 0. 4 ( ) ( )

12 000 Fermat s The length of is = =. Note: The other suggested choice for the origin would involve using the equation of the line y x and the circle x y. = ( ) + ( ) = The diameter, G, is parallel to the sides of the rectangle and has its endpoint at which is the centre of the rectangle. Draw O where O is the centre of the circle and by chord properties of a circle, =. Since O DE, DE = O =α. Thus, O is similar to FDE. Since DF =, = and = 4. The length of is or 4. NSWER: () 4 4. For the system of equations x + x y + x y = and x+ xy+ xy = 3, the sum of the real y values that satisfy the equations is () 0 () () (D) (E) onsider the system of equations 4 x + x y + x y = () and x+ xy+ xy = 3 () The expression on the left side of equation () can be rewritten as, ( )( + + ) Thus, x + xy xy x xy xy Substituting from () gives, x+ xy xy Now subtracting (3) from (), xy = 0, x Substituting for x in (3) gives, ( ) ( x xy xy) ( x + xy + xy) 4 x + x y + x y = x+ xy x y ( )( )= = + = 3 or, x+ xy xy = (3) 0 y = 0 y + 0y 0 =

13 000 Fermat s 3 0y y+ 0 = 0 y y+ = 0 ( y ) ( y )= 0 y = or y = The sum of the real y values satisfying the system is. NSWER: (E). The given cube is cut into four pieces by two planes. The first plane is parallel to face D and passes through the midpoint of edge G. The second plane passes through the midpoints of edges, D, HE, and GH. Determine the ratio of the volumes of the smallest and largest of the four pieces. F D E () 3:8 () 7:4 () 7: (D) 7:7 (E) : G H T For convenience, let each edge of the cube have length. The plane PQRS through the mid-point of G and parallel N D to face D bisects the volume of the cube. The plane K through K, L, M, and N also bisects the volume of the S cube and contains the line segment QS. Hence the two P planes divide the cube into four pieces, two equal Q E F R smaller pieces and two equal larger pieces. Extend M QK and SN to meet at T. G L H We note that TN is similar to TPS. Since T N = and N =, PS = and P = it is easy TP PS to calculate to find T =. The volume of the upper smallest piece is equal to the volume of the tetradedron TQSP - volume of the tetradedron TKN. This volume is, 3 ( )( )( ) 3 ()()() 4 7 = = The volume of a largest piece is, 7 7 =. 6 6 The ratio of the volume of the smallest piece to that of the largest piece is 7:7. NSWER: (D)

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