Give a simple proof that an integer can be divided by 3 if the sum of its digits can be divided by 3.?

Question

The question came across my head one day and I tried to solve it myself but failed. I knew there was some simple proofs when I was in school but forget now. Help me to pick up my mathematics.

The question came across my head one day and I tried to solve it myself but failed. I knew there were some simple proofs when I was in school but forget now. Help me to pick up my mathematics.

Answer 1

Say you have a two digit number:
10x + y = (9+1)x + y = 9x + (x + y)
Thus 10x + y is divisible by 3 if and only if (x + y) is divisible by 3.
since 9x is already divisible by 3.

The rule can be extended to any n-digit number.
Note: 100 = 99 + 1...


d:

Answer 2

d_c has a good proof for a 2 digit number... here's a general proof for a number of any length.
Take a number of the written form y1y2y3y4...yn where y1 is the first digit, y2 is the second, etc. and yn is the last. It'll look a little confusing because I don't know how to put subscripts into YA.
So, for example, the number 3694 would have y1 = 3, y2 = 6, y3 = 9, y4 = 4.

Suppose that (y1 + y2 + ... + yn) is divisible by 3. We now need to show that the original number is divisible by three.
We can rewrite the original number as:

10^(n-1)*y1 + 10^(n-2)*y2 + ... + 10*y(n-1) + yn,
We can rewrite this as:
(y1 + y2 + ... + yn) + 9999...(n times)*y1 + 9999... (n-1 times)y2 + .... + 9*y(n-1)
In the above number, 9999....(n times) means there are n 9s in the number.

we know that (y1 + y2 + ... + yn) is divisible by three, and everything else has a coefficient made up of just nines. It is easy to check that all of these are divisible by three, so the whole thing is divisible by three.

Answer 3

333
3+3+3=9
9/3 =3

so 333 is divisible by three also ---- 333/3 =111

Answer 4

9, 18, 27, 36, 45, 54, 63, 81, 90.