# Math Help - Proving the theorem (if a numbers digits is divisible by 3, so is the number)

1. ## Proving the theorem (if a numbers digits is divisible by 3, so is the number)

Use the theorem below to answer the question that follows.

If the sum of a number's digits is divisible by three, the number is divisible by three.

Which of the following ways of expressing a three digit number, n, best demonstrates why this theorem is true?

A. $n = a(10^2) + b(10) + c$
B. $n = a(100) + b(10) + c$
C. $n = a(99 + 1) + b(9 + 1) + c$
D. $n = a(300-200) + b(30-20) + c(9-8)$

2. ## Re: Proving the theorem (if a numbers digits is divisible by 3, so is the number)

Originally Posted by jbwtucker
Use the theorem below to answer the question that follows.

If the sum of a number's digits is divisible by three, the number is divisible by three.

Which of the following ways of expressing a three digit number, n, best demonstrates why this theorem is true?

A. $n = a(10^2) + b(10) + c$
B. $n = a(100) + b(10) + c$
C. $n = a(99 + 1) + b(9 + 1) + c$
D. $n = a(300-200) + b(30-20) + c(9-8)$
C.

Tell me why?

3. ## Re: Proving the theorem (if a numbers digits is divisible by 3, so is the number)

I'll do my best. I'm thinking...

99 is divisible by 3. So if a(99 + 1) is divisible by 3, then a must be divisible by 3. Same goes for the b term.

...am I heading the right direction? Because c could be anything. So I don't see what that proves.

4. ## Re: Proving the theorem (if a numbers digits is divisible by 3, so is the number)

Originally Posted by jbwtucker
I'll do my best. I'm thinking...

99 is divisible by 3. So if a(99 + 1) is divisible by 3, then a must be divisible by 3. Same goes for the b term.

...am I heading the right direction? Because c could be anything. So I don't see what that proves.
No.

a(99+1)+b(9+1)+c

=99a+a+9b+b+c

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

=9(11a+b)+(a+b+c)

Now, 9(11a+b) divisible by 3, so... What can you tell me...?

5. ## Re: Proving the theorem (if a numbers digits is divisible by 3, so is the number)

I see. 9(11a+b) divided by 3 is 3(11a+b). And the original theorem states that the sum of a number's digits (a, b and c) is divisible by 3.

Thanks!