Challenge: Explain why the divisibility rule for 3 works

Here’s what I tell kids (6th – 8th grades) when I explain this:

10 is three 3’s with one left over.
100 is thirty-three 3’s with one left over.
It’s easy to see that every power of 10 leaves a remainder of 1 (one left over) when divided by 3.

Now, if we have two 10’s, we’re going to have a bunch of 3’s and two left-overs. Same if we have two 100’s and so on. The number of 10’s or 100’s or 1000’s and so on tells us how many leftovers we are going to have if we set aside our three, or thirty-three, or three hundred thirty-three, etc. 3’s from each set of 10’s or 100’s or 1000’s.

Now if we take a number (written in base 10) each of the digits tells us how many of a certain power of 10 we have. So each digit is going to also contribute that number of “leftovers” when we take out the original 3’s.

So if we have a number like 357, and we want to know if it’s divisible by 3, what we really want to know is, could we split it up into piles of 3, with nothing leftover. So we know our 300 gives us a bunch of piles of 3, with 3 leftover. Our 50 gives us another bunch of piles of 3, with 5 leftover. And then we throw our 7 ones in with the other leftovers. That gives us 3 + 5 + 7 items left to sort into piles of 3, and see if there’s anything left over. Since 3 + 5 + 7 = 15 which is divisible by 3, we can combine all the leftovers into even piles of 3 with nothing left over, so 357 is divisible by 3.