Exercise 6 (Divisibility criteria). Let a 2 1 be an integer. We may write for so

Question

Exercise 6 (Divisibility criteria). Let a 2 1 be an integer. We may write for some d > 0 so that ao-...ad are integers in the set {0 integers ad,... ao are the digits of the integer a. Show that: 9), with ad 0. The 1. The integer a is even if and only if its last digit ao is even. 2. The integer a is divisible by 5 if and only if its last digit ao is either 0 or 5 3. The integer a is divisible by 4 if and only if the number 10ai + ao given by its last two digits is divisible by 4 divisible by 3 divisible by 9 4. The integer a is divisible by 3 if and only if the sum ad + + ao of its digits is 5. The integer a is divisible by 9 if and only if the sum ad + +ao of its digits is

Explanation / Answer

a = 10d * ad + 10d-1 * ad-1 +...+ 10a1 + a0 (Given)

a = 10 * (10d-1 * ad + 10d-2 * ad-1 +...+ a1) + a0

Let y = (10d-1 * ad + 10d-2 * ad-1 +...+ a1)

a = 10 * y + a0 ..................... (1)

1. Criteria for divisibility by 2 is number should be even

if a0 is even that is multiple of 2 then R.H.S. is multiple of 2 (10 is multiple of 2, so 10 * y is divisible by 2 and a0 is also divisible by 2)

so whole expression on RHS is multiple of 2 that means RHS is even.

   Hence a will be even only if a0 is even is proved

2. Criteria for divisibility by 5 is number should be end with 0 or 5

On RHS. in eq (1), 10 * y is multiple of 5 meaning it is perfectly divisible by 5.

   Now. if last digit a0 is either 0 or 5 so it will also be divisible by 5

   So whole RHS is divisible by 5 and so a is divisible by 5 is proved.

3. Criteria for divisibilty by 4 is Number formed by Last 2 digits should be divisible by 4.

Number formed by Last two digits in given expression is (10a1 + a0 ).

If this number is divisible by 4 then RHS. will be divisible by 4 and so a will be divisible by 4. It is proved.

4. Criteria for multiple of 3 is sum of all digits in given number should be divisible by 3.

In given expression sum of digits is ad + ad-1 + ... + a0. If this sum is divisible by 3 then whole RHS will be divisible by 3.

   So a will be divisible by 3 is proved

5. Criteria for multiple of 9 is sum of all digits in given number should be divisible by 9.

In given expression sum of digits is ad + ad-1 + ... + a0. If this sum is divisible by 9 then whole RHS will be divisible by 3.

   So a will be divisible by 9 is proved

6. Criteria for multiple of 11 is Subtraction of sum of the even digits from the sum of the odd digits should be divisible by 11.

so in given expression- this subtraction is given by

(-1)kak = (-1)dad + (-1)d-1ad-1 + .... + (-1)a1 + a0

   If above expression is divisible by 11 then only a will be divisible by 11 is proved.

7. 152460 = 105(1) + 104(5) + 103(2) + 102(4) + 101(6) + 0

Last digit is 0 so number is divisible by 5.

152460 = 5 * 30492

Now 30492 = 104(3) + 103(0) + 102(4) + 101(9) + 2

Last digit is 2 which is even so number is divisible by 2.

30492 = 2 * 15246

so 152460 = 2 * 5 * 15246

Now in 15246 sum of even digits (1+2+6=9) and sum of odd digits (5+4=9). Diiference in then is 0 which is divisible by 11. so number is divisible by 11.

so 152460 = 2 * 5 * 11 * 1386

  Now in 1386 sum of even digits (1+8=9) and sum of odd digits (6+3=9). Diiference in then is 0 which is divisible by 11. so number is divisible by 11.

so 152460 = 2 * 5 * 11 * 11 * 126

Now in 126 sum of digits is 1+2+6 = 9 which is divisible by 3. so 126 is divisible by 3.

so 152460 = 2 * 5 * 11 * 11 * 3 * 42

Now in 42 sum of digits is 4+2 = 6 which is divisible by 3. so 42 is divisible by 3.

so 152460 = 2 * 5 * 11 * 11 * 3 * 3 * 14

In 14 last digit is 4 which is even. so 14 is multiple of 2.

  so 152460 = 2 * 5 * 11 * 11 * 3 * 3 * 2 * 7

152460 = 2 * 2 * 3 * 3 * 5 * 7 * 11 * 11

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