Divisibility by Three: Proof

Date: 07/22/2003 at 01:08:12
From: Emily Cheshire
Subject: Sum of the digits of multiples of 3.- PROOF

Why is the sum of the digits of a multiple of 3 divisible by 3?

Date: 07/22/2003 at 20:46:40
From: Doctor Barrus
Subject: Re: Sum of the digits of multiples of 3.- PROOF

Hey, Emily!

I'm going to start with an example, and then I'll give a proof.

First the example: 3 * 98 = 294, so 294 is divisible by 3. Now also

2 + 9 + 4 = 15, 

and 15 is also divisible by 3. Why is this? Well, looking at what 
each digit means in the place it's in, we see that

294 = 2*100 + 9*10 + 4*1
    = 2*(99+1) + 9*(9+1) + 4*(0+1)
    = (2*99 + 9*9 + 4*0) + (2*1 + 9*1 + 4*1)

(((*)))   294 - (2*99 + 9*9 + 4*0) = 2 + 9 + 4

So the sum of the digits, which is found on the right, is equal to 
294 (which is divisible by 3) minus (2*99 + 9*9 + 4*0). Now note that 
99, 9, and 0 are all divisible by 3, so their sum is, too:

2*99 + 9*9 + 4*0 = 3 * (2*33 + 9*3 + 4*0)

(see how I factored a 3 out of everything?)

So now the left-hand side of the (((*))) equation is one number 
divisible by 3 minus another number divisible by three. The answer 
has to be divisible by 3 too. Then the sum of the digits (the right-
hand side) is equal to a number divisible by 3.

PROOF: Let abc be a 3-digit number with a as the hundreds digit, b as 
the tens digit, and c as the units digit. (So we're not multiplying 
a*b*c - abc are the actual digits of the number.) Suppose that abc is 
divisible by 3. Then 3 is a factor of

100*a + 10*b + c, which is the same as

(99+1)*a + (9+1)*b + (0+1)*c, or

(99*a + 9*b + 0*c) + (1*a + 1*b + 1*c)

So now

abc - (99*a + 9*b + 0*c) = a + b + c

And every number on the left-hand side is divisible by 3, so the sum 
a + b + c on the right-hand side must be divisible by 3 also. So 
every number that's divisible by 3 has a digit sum also divisible by 
3, and by working backwards (and changing things slightly) you can 
show that vice versa is true, as well.

This proof was for 3-digit numbers, but you can do it for any number 
of digits, using these same steps.

I hope this helps! If you have more questions, feel free to ask them. 
Good luck!

- Doctor Barrus, The Math Forum
  http://mathforum.org/dr.math/