Thread: 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.
B.
C.
D.

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.
B.
C.
D.

C.

Tell me why?

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.

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...?

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!