# 1.16 Factors, Multiples, Prime Numbers and Divisibility

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2 Square numbers have an odd number of factors because they have a repeated factor. What kind of numbers have n factors? How can you decide how many factors a number will have without working them all out? You can set a challenge such as finding a number with exactly 13 factors. 1 One answer would be p (where p is a prime number), so choosing p 3 gives the number The factors are 1, 3, 9, 7, 81, 43, 79, 187, 6 561, , , and , and there are thirteen of them. So it s to do with the number of factors that n has (and n is how many factors the original number has)! For prime factorisation, it s possible to draw tree diagrams going down the page. Stop and put a ring around it when you reach a prime number. Try each time to split the number into two factors that are as nearly equal as possible, because that leads to fewer steps. Could discuss why we don t count 1 as a prime number Tests for divisibility. Build up a big table of all the tests on the board (see sheet). Then make 3 columns headed by 3 random numbers (e.g., 016, 3465 and 400) and put the numbers 1-1 down the side. Place either a tick or a cross in each column to say whether the column number is divisible by the row number I come into the classroom and ask the class to get into pairs but 1 person is left over. We have to have groups of equal size, so I say never mind, instead get into groups of 3. Again 1 person is left over. We try groups of 4. Again, 1 person is left over. Finally groups of 5 works. How many people do you think there are in the class? Suppose that p, q, r, are prime numbers, and a, b, c, are integers 0. Any power of a prime a number can be written as p, and will have a 1 3 a factors ( 1, p, p, p,... p ). Hence p has factors and p has 3 factors, as above. Since every number can be factorised into primes, every number x can a b c be written as p q r... and will have ( a 1)( b 1)( c 1)... factors, since any of the a 1 powers of p can be multiplied by any of the powers of the others. So the possibilities are as below. no. of factors 1 1 p e.g., for 4 prime factorisations of numbers that have that many factors p 3 p or pq 4 p 5 p or 6 p so 4 = p or 8 p or 9 p or p q 3 p q or pqr p q 4 p q 4 / \ 6 4 / \ / \ 3 One reason is that prime factorisation would go on for ever! It s useful to have one unique way (apart from the order you write it in) of prime factorising every integer. It s productive to look for patterns in the answers (e.g., if a number is divisible by 10 then it s necessarily divisible by and by 5, etc.). Note for example that divisible by 4 divisible by, but not the other way round. Answer: 5 pupils; 85, 145, 05, etc. also work but hopefully would not be real class sizes! One strategy is to say that n 1 must be a multiple of, 3 and 4; i.e., a multiple of 1, and go through the multiples of 1 until you find one where the number that is one higher is a multiple of I m thinking of a number or I m thinking of Lots of possibilities.

6 Then begin somewhere in the middle of a sheet of A4 0.5 cm 0.5 cm squared paper. Mark a dot. Choose a column in the table, say the 3 s. Draw a line 3 squares long straight up the paper, turn right, draw a line 6 squares long, turn right, then 9, then 3, and so on down the column, always turning right (see drawing on the right S is the starting point). When you get to the bottom of a column of numbers, just start again at the top. Keep going until you get back to the place where you started. The easiest columns to do are 3, 6, 9 (but 9 is boring a square, obviously) Diagonals in Rectangles. If an x y rectangle is drawn on squared paper, how many squares does the diagonal line pass through? x and y are integers. e.g., for a 5 3 rectangle, the line goes through 7 squares. All of them will fit on A4 0.5 cm 0.5 cm squared paper, but you need to start in a sensible place. Teacher may need to assist. The tricky thing is seeing that turning right when coming down the page towards you means actually going left. Some people find it helpful to turn the paper as you go (like turning a map), so you re always drawing away from you, and some people just find that confusing. It s also useful to mention the name spirolateral so we expect it to spiral in, not meander further and further towards the edge of the paper! Answer: If x and y are co-prime (HCF=1) then the line must go through x y 1 squares. This is the smallest number of squares between opposite corners. (Imagine a curly line see below). We count it as going through a square even if it only just clips the square. Only if it goes exactly through a crossing-point do we not count the square Snooker investigation. A snooker ball is projected from the near left corner (A, below) of a rectangular snooker table at 45 to the sides. If there are pockets at all four corners, and the table has dimensions x y ( x is the horizontal width), which pocket will the ball end up in? x and y are integers. B A C Assume that every time the ball hits a side it rebounds at 45, and that the ball never runs out of kinetic energy. Above for a 5 3 table, the answer is pocket C. D Try some examples on squared paper. 1 If HCF (, x y) 1 then every of the way HCF ( x, y ) along the diagonal line will be a crossing point, x since and HCF ( x, y ) HCF ( x, y) y will be integers. Each of these HCF (, x y) 1 crossing points means one fewer square for the diagonal to go through. So that means the total number of squares will be x y 1 ( HCF( x, y) 1) x y HCF( x, y) Answer: Every diagonal step forward moves the ball 1 square horizontally and 1 square vertically. Since x is the horizontal distance, after x,3 x,5 x,... steps the ball will be at the right wall. After,4,6,... x x x steps the ball will be at the left wall. Similarly, after y,3 y,5 y,... steps the ball will be at the top side. After y,4 y,6 y,... steps the ball will be at the bottom side. Therefore, the first time that a multiple of x is equal to a multiple of y (i.e., after LCM (, x y ) steps), the ball will be in one of the corners. It will never be corner A, because the ball only reaches there if it travels an even number of xs ' and an even number of ys. ' That will never be the LCM of x and y because half that many xs ' would match half that many ys. ' If LCM ( x, y) x is odd, then the pocket will be either

7 Hint: How many diagonal steps will the ball move before it lands in a pocket? What is the significance of the digit sum of an integer when the integer isn t a multiple of 9? Try working some out. Find out what casting out 9 s refers to. (Suitable for a homework: ask grandparents.) You can prove that the digit sum of a 3-digit number, say, is the remainder when dividing by 9 by writing " abc ", as n 100a 10b c. (This process would work just as well however many digits the number had.) n 100a 10b c 99a 9b a b c 9(11 a b) ( a b c) so a b c is the remainder when n is divided by 9. (This assumes that a b c 9. If it s equal to 9, then n is a multiple of 9;; if it s more than 9, then we can just start again and find the digit sum of this number, because the remainder of this number when divided by 9 will be the same as the remainder of n when divided by 9.) Why not define highest common multiple and lowest common factor, as well? C or D. Otherwise, it will be A or B. LCM ( x, y) If is odd, then the pocket will be either y B or C. Otherwise, it will be A or D. Taken together, this means you can always predict which pocket the ball will end up in. e.g., for a 10 4 table, LCM (10, 4) is even so AB side; 0 4 pocket B. is odd so BC side. Hence Answer: It s the remainder when you divide the number by 9. E.g., , remainder 4. And the digit sum of 38 is 4 (actually 13, but the digit sum of 13 is 4). (The digit sum of a multiple of 9 is itself a multiple of 9.) Hence the method of casting out 9 s in which every integer in the calculation is replaced by its digit sum. When the calculation is done with these numbers, the answer should be the digit sum of the answer to the original question. This provides a way of checking. e.g., , replacing 946 and 36 by their digit sums gives 1 3, and 3 is the digit sum of 17. This doesn t guarantee that the sum is correct, but if this test doesn t work then the sum is definitely wrong. This is a bit like the modern check-sums method used on bar-codes to make sure the machine has read the numbers accurately. Here a single mistake can always be identified. Answer: If you think about it, LCF would always be 1 and HCM would always be infinite!

8

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10 Prime Numbers

11 Tests for Divisibility integer test divides into the number if 1 no test any integer look at units digit 0,, 4, 6 or 8 3 digit sum 3, 6, 9 4 look at last digits divisible by 4 5 look at units digit 0 or 5 6 test for and test for 3 passes both tests 7 double the units digit and subtract from the rest of the number divisible by 7 8 divide it by divisible by 4 9 digit sum 9 10 look at units digit 0 11 alternating digit sum ( + + +) 0 Notes means do the same thing to the answer and keep going until you have only 1 digit left digit sum on 73 gives (so passes the test for 3) alternating digit sum ( + + +) on gives (working from right to left and beginning with +) = 0 (so passes the test for 11) double the units digit and subtract from the rest of the number on gives = = = 1, divisible by 7 (so passes the test for 7) You can use a calculator to generate multiples to use for practice. (E.g., type in a 4-digit random integer, multiply by 7 and you have a random multiple of 7 for trying out the test for divisibility by 7.) A test for divisibility by 1 could be to pass the tests for divisibility by 3 and by 4. It wouldn t be any good to use the tests for divisibility by and by 6 because passing the test for 6 means that the number must be even so the test for adds nothing. 3 and 4 are co-prime (HCF = 1), but 6 and aren t. These tests are worth memorising and practising.

12 Stamps Investigation (Proof) If the stamps are x pence and y pence ( x and y co-prime), then the highest impossible value (with unlimited quantities of each) is xy x y pence. This is easy to show for a particular pair of stamp values. For example, for x 3 and y 5 we write down all the positive integers using three columns (see right imagine the columns going down for ever). We re going to shade in all the numbers that are possible. Clearly the right column will all be possible by using just the 3p stamps, because the numbers are all multiples of 3. In the second row, 5 will be possible (one 5p stamp) so we shade that in. Also everything under 5 in the second column will be possible by using one 5p stamp and different numbers of 3p stamps Similarly, 10 (two 5p stamps) and everything below it will be possible. So 7 is the highest impossible amount (and this is ). Now we try the same thing generally, with x pence and y pence stamps ( x and y co-prime). The diagram is shown on the right. We will assume that x y. As before, the right column will all be possible (multiples of x pence), so we shade it in. 1 3 x 1 x 1 x x 3 x 1 x x The column containing y will all be possible from y downwards ( y, y x, y x, ) and this cannot be the same column as the x column ( x, x, 3x, ) because y is not a multiple of x (they re co-prime). x 1 x x 3 3x 1 3x For the same reason, y cannot be in either of the two columns dealt with so far (unless there are only two columns because x ). (It can t be in the x column because y cannot be a multiple of x, and the extra y places moved on from y can t equal a multiple of x, which would be necessary if it were in the same column as y ). So we keep locating the next multiple of y, and we always find it in a previously unvisited column, and we shade it in and we shade in the rest of that column below it. There are x columns altogether, so when we reach ( x 1) y and shade that in (and all the numbers beneath it) the highest impossible amount will be the number directly above ( x 1) y (since all the other numbers in that row will already be shaded in). This number will be ( x 1) y x xy x y, and so this will be the highest impossible value.

13 Number of ways n c of making up a cost c pence out of stamps c 5p, 7p 3p, 7p 5p, 8p 5p, 9p 7p, 10p continued

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