Also, any number whose digits sum to a multiple of 3 is divisible by 3. For 51, 5+1=6, and 6 is a multiple of 3, so 51 can be cleanly divided by 3.
Does this also work the other way round, i.e. do all multiples of three have digits that sum to a multiple of 3? All the ones I’ve checked so far do, but is it proven?
Indeed, an integer is divisible by 3 if and only if the sum of its digits is divisible by 3.
For proof, take the polynomial representation of an integer n = a_0 * 10^k + a_1 * 10^{k-1} + … + a_k * 1. Note that 10 mod 3 = 1, which means that 10^i mod 3 = (10 mod 3)^i = 1. This makes all powers of 10 = 1 and you’re left with n = a_0 + a_1 + … + a_k. Thus, n is divisible by 3 iff a_0 + a_1 + … + a_k is. Also note that iff answers your question then; all multiples of 3 have to, by definition, have digits whose sum is a multiple of 3
I’d forgotten this trick. It works for large numbers too.
122,300,223÷3 = 40,766, 741
1+2+2+3+2+2+3 = 15
threw up and died while reading this
I wish I could read 😞
Just squint and wing it.
