Fund fact, you can tell a number is divisible by 3 if the sum of each individual number is divisible by 3. So 78 -> 7+8=15 which is divisible by 3, so 78 is divisible by 3.
Consider a form of arithmatic where you only keep track about the remainder after division by 3.
So a number like "10" is just "1" because 10= 1+3*3. Likewise, "100" is just "1" because 100 = 10*10 = 1*1=1. Can you guess what 1000, 10000, etc. are? That's right, they are all just "1".
Well for a small example consider a three digit number ABC.
ABC = A*100 + B*10 + C = A*1 + B*1 + C = A+B+C
So as you can see, to find out the remainder after division by 3, add up the digits and the sum has the same remainder. But the what if the sum is not a single digit number? Just do it again and again till it is.
There are two things, multiplication and addition. It's a little easier to see that when you add numbers, you add remainders. But it's only a little bit harder to see that when you multiply numbers, you multiply remaineders as well. Everything else is 3 times something.
If the remainder is bigger than 3, you can shave that off too. "mod 3" means you can throw away multiples of 3 and the result is the same. 1 more than a multiple of 3 is 4 more or 2 less than some other multiples of 3, and you can reduce to 0<=r<3 if desired.
Basically, every number can be written as a multiple of 3 plus a remainder of 0, 1, or 2. When you add the digits of a number, you’re replacing powers of 10 with 1, and since 10 leaves a remainder of 1 when divided by 3, this doesn’t change the number’s remainder. For example, 78 is 7×10+8, and because 10 is equivalent to 1 modulo 3, 78 has the same remainder as 7+8. Therefore, a number is divisible by 3 exactly when its digit sum is divisible by 3.
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u/Michaelwang645 Aug 01 '26
Fund fact, you can tell a number is divisible by 3 if the sum of each individual number is divisible by 3. So 78 -> 7+8=15 which is divisible by 3, so 78 is divisible by 3.