What is Divisibility ?

1. What is Divisibility? • Lets say, we have 15 books to be given out to 5 pupils. • Can we evenly distribute all the books? • 15 ÷ 5 = 3 • Each pupil will gets 3 books and • we don’t have any extra left. (No Remainder) YES! 15 is divisible by 5

2. What is Divisibility? • How if, we have 17 books to be given out to 5 pupils. • Can we evenly distribute all the books (without any extra left)? • 17 ÷ 5 = 3 (R)2 • Each pupil will gets 3 books and • We HAVE2 extra left. (Remainder is 2) NO! 17 is NOT divisible by 5

3. 25 Divisible By 5 NO remainder • 25 toys given out to 5 children? • 45 toys given out to 5 children? • 75 toys given out to 5 children? • 68 toys given out to 5 children? 45 Divisible By 5 NO remainder 75 Divisible By 5 NO remainder 68 NOT Divisible By 5 Remainder 3

4. Divisible by 2, 5; 4, 25; 8, 125 • Numbers that are divisible by 2 or 5: • Check their last digit • Numbers that are divisible by 4 or 25: • Check their last two digits • Numbers that are divisible by 8 or 125: • Check their last three digits

5. Divisible by 3 or 9 • Sum up the all the digits of the number. • E.g.: 123456, • The sum of its digits is: 1+2+3+4+5+6 = 21 • 21 is divisible by 3. So, 123456 is divisible by 3 • 21 is NOT divisible by 9. So, 123456 is NOT divisible by 9 • E.g.: 234567, • The sum of its digits is: 2+3+4+5+6+7 = 27 • 27 is divisible by both 3 and 9 respectively. • So, 234567 is also divisible by both 3 and 9.

6. Divisible by 7, 11 or 13 • Find the difference between: • “the number formed by last three digits”with • “the number formed by the digits of the rest” • E.g.: 420321: • Number formed by last three digits: 321 • Number formed by the digit of the rest: 420 • Their difference: 420 – 321 = 99 (divisible by 11) • So, 420321 is also divisible by 11.

7. Divisible by 7 or 21 • Subtract 2 times the last digit from the rest. • E.g.: 714 • 2X4 = 8; 71-8=63 (divisible by 7 and 21) • 714 is also divisible by 7 and 21 respectively • E.g.: 476 • 2X6 = 12; 47-12=35 (divisible by 7 but NOT 21) • So, 476 is divisible by 7 but NOT by 21

8. Divisible by 11 • The alternating sum*1 of its digits is divisible by 11 • E.g.: 781 • 7-8+1 =0 or • (7+1) – (8) =0 • So, 781 is divisible by 11 • E.g.: 34859 • 3-4+8-5+9= ? • (3+8+9) – (4+5) = 11 • So, 34859 is divisible by 11

9. Summary of divisibility facts Fact 1 8 is divisible by 4 12 is divisible by 4 12 + 8 = 20 is divisible by 4 12 - 8 = 4 is divisible by 4 Fact 2 36 is divisible by 3 36 is divisible by 4 (3 and 4 are coprime*2) 36 is divisible by 12 (= 3X4)

10. Example 1 • Given a 4-digit number 83Y2, which is divisible by 9. What is the possible digit for Y? • 83Y2 is divisible by 9 means that: Sum of digits : 8+3+Y+2 is divisible by 9 • 13+ Y is divisible by 9 （8+3+2=13） • Numbers that are divisible by 9 after 13 are: 18, 27, 36, …. • 18= 13 +5; 27 = 13+ 14; 36= 13 + 23;…… • Y is 1-digit number. So, Y = 5. • 8352 is divisible by 9

11. Example 2 • Given two numbers, M and N, where both of them are divisible by 7. Their sum, M+N is a 5-digit number, 3802_. What is the last digit of the sum M+N? • Both M and N are divisible by 7, means that their sum, (M+N= )3802_ also divisible by 7 • difference between 38 and 02_ is divisible by 7 • 38 – 2_ = multiple of 7 (list out some possible ans) • 38 – 24 = 14; the last digit is 4 • 38024 is divisible by 7

12. Example 3 • Find the smallest 4-digit number which is divisible by 2, 5, and 11. • An integer which is divisible by 2, 5 and 11 is also divisible by 2x5x11 = 110 (since 2, 5 and 11 are coprime) • 9x110=990; 10x110=1100; 11x110=1210 • Hence, the smallest 4-digit integer required is 1100.

13. Exercise 1 • Which of the following numbers are divisible by 8? • A)34 342 • B)23 624 • C)120 458 • D)348 624

14. Solution 1 • A)1 234 342 • B)3 273 624 • C)5 120 458 • D)7 348 000 • (B) and (D) since 342 and 000 is divisible by 8 . • (A) and (C) is not divisible by 8 since 342 and 458 is not divisible by 8

15. Exercise 2 • Is 10-digit number, 987654321Y a multiple of 45? If ‘yes’, what is the digit for ‘Y’?

16. Solution 2 • 45=5×9, • If “987654321Y” is a multiple of 45, then it is divisible by 5 and 9 respectively. • For integer that is divisible by 5, the last digit must be either 0 or 5. So, Y is either 0 or 5. • For integer that is divisible by 9: the sum of its digits, that is 9+8+7+6+5+4+3+2+1+Y, must be 0 or multiple of 9. • 9+8+7+6+5+4+3+2+1=45, a multiple of 9. So, Y must be 0 or multiple of 9. • Combine the 2 conditions, Y must be 0.

17. Exercise 3 • Given five numbers: 0, 1, 3, 7, 9. Without repetition, pick any 4 numbers from the list and form a 4-digit number that is divisible by 55.

18. Solution 3 • 55 = 5 X 11 • Hence, the number formed must be divisible by both 5 and 11. • Numbers that divisible by 5 ends with digit 0 or 5. • AXB0 ; the last digit must be 0. • (A+B) – (X+0) is 0 or multiple of 11 (11, 22, 33….) • Take X as 1, then A and B could be 3 or 9. • Ans: 3190 or 9130.

19. *1) Alternating Sum • The alternating sum of the digits for a number, let’s say 72853 is as followed: • -+ -+ or • The difference between: • The sum of digits in odd places and • The sum of digits in even places. • (++) – (+) 7 7 2 2 8 8 5 5 3 3 Back

20. *2) Coprime • Two numbers are coprime, if • They are in the simplest form when we put them into fraction form. • E.g.: 6 and 9 are NOTcoprime, as is NOT in the simplest fraction form. • E.g.: 2 and 5 are coprime, as is in the simplest fraction form. • No integers can divide them simultaneously, besides 1. Back