Prove that an integer is divisible by 3 if and only if the sum of its digits is divisible by 3.?

Answer 1

Suppose the digits of n are an,an-1...a1,a0, that is,

n = 10nan + 10n-1an-1 + 10a1 + a0
= (an + an-1 + a1 + a0) + ((10n - 1)an + (10n-1 - 1)an-1 + … + (10 - 1)a1).

Since 10r - 1 = 99…9 (r digits) is divisible by 3, the second (large) bracket is divisible by 3, and the result follows.

Answer 2

As a simple example, consider a 2 digit number in base 10: ab. You can write it as

10a + b =9a + (a+b).

We know that 9a is divisible by 3, so 10a+b will be divisible by 3 if and only if a+b is.

Similarly for a 3 digit number abc: 100a+10b+c
you can write it as:

99a+9b+(a+b+c),

and 99a+9b is divisible by 3, so the total will be iff a+b+c is.

Do you see how you can apply this pattern to any length base 10 number?

Answer 3

12

Answer 4

You can also prove this using ring theory. See any graduate algebra book for help.