1. KS3 Divisibility Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 30 th November 2014

2. 2Last number is even. 3Digits add up to multiple of 3. e.g: 1692: 1+6+9+2 = 18 4Last two digits are divisible by 4. e.g. 143328 5Last digit is 0 or 5. 6Number is divisible by 2 and 3 (so use tests for 2 and 3). 7 There isn’t really any trick that would save time. You could double the last digit and subtract it from the remaining digits, and see if the result is divisible by 7. e.g: 2464 -> 246 – 8 = 238 -> 23 – 16 = 7. But you’re only removing a digit each time, so you might as well long divide! 8 Last three digits divisible by 8. 9Digits add up to multiple of 9. 10Last digit 0. 11When you sum odd-positioned digits and subtract even-positioned digits, the result is divisible by 11. e.g. 47949: (4 + 9 + 9) – (7 + 4) = 22 – 11 = 11, which is divisible by 11. 12Number divisible by 3 and by 4. ? ? ? ? ? ? ? ? ? ? ? Divisibility Rules How can we tell if a number is divisible by: 

3. Quickfire Divisibility 467911 726 168 9196 252 1001 91 216 87912 ????? ????? ????? ? ? ??? ????? ????? ????? ?????

4. Breaking Down Divisibility Rules Are these statements true or false? If we want to show that a number is divisible by 15, we can show it is divisible by 3 and 5. If we want to show that a number is divisible by 24, we can show it is divisible by 6 and 4.  FalseTrue  FalseTrue The problem is that 12 is divisible by 6 and 4, but it is not divisible by 24! We need to pick two numbers which are coprime, i.e. do not share any factors. How can we therefore test if a number is divisible by 24?

5. Quickfire What divisibility rules would we use if we wanted to test divisibility by: 182 and 9 rules 455 and 9 364 and 9 405 and 8 An easy Year 10 Maths Olympiad problem: Find the smallest positive integer which consists only of 0s and 1s, and which is divisible by 12. Since in must be divisible by 4, the only possibility for the last two digits is 00. It must have at least three 1s to be divisible by 3 (as we can’t have zero 0s). Therefore 11100 is the answer. ?

6. Exercises Problem sheet of Junior and Intermediate Olympiad problems. Work in pairs/groups if you wish. Answers on next slides. (File Ref: KS3_DivisibilityQuestions.docx)

7. Question 1 [J31] Every digit of a given positive integer is either a 3 or a 4 with each occurring at least once. The integer is divisible by both 3 and 4. What is the smallest such integer? ?

8. Question 2 [J50] The eight-digit number “ppppqqqq”, where p and q are digits, is a multiple of 45. What are the possible values of p? ?

9. Question 3 [M07] (a) A positive integer N is written using only the digits 2 and 3, with each appearing at least once. If N is divisible by 2 and by 3, what is the smallest possible integer N? (b) A positive integer M is written using only the digits 8 and 9, with each appearing at least once. If M is divisible by 8 and by 9, what is the smallest possible integer M? ? ?

10. Question 4 [M55] A palindromic number is one which reads the same when its digits are reversed, for example 23832. What is the largest six-digit palindromic number which is exactly divisible by 15? ?

11. Question 5 [J16] Find a rule which predicts exactly when five consecutive integers have sum divisible by 15. ?

12. Question 6 [M96] Find the possible values of the digits p and q, given that the five-digit number ‘p543q’ is a multiple of 36. ?

13. Question 7 ?

14. Question 8 [M31] Find the smallest positive multiple of 35 whose digits are all the same as each other. ?

15. Question 9 ?

16. Question 10 [Based on NRich] If the digits 5, 6, 7 and 8 are inserted at random in 3_1_4_0_92 (one in each space), what is the probability that the number created will be a multiple of 396 if: a)Each of 5, 6, 7, 8 is used exactly once in each of the four gaps. b)Each of 5, 6, 7, 8 can be used multiple times. ?