I was both hoping and fearing that someone would correct my simplification. (That is more nested logs than I would’ve guessed however.)

Yes. A number like

123456 = 1 * 100000 + 2 * 10000 + 3 * 1000 + 4 * 100 + 5 * 10 + 6 = 1 * (99999+1) + 2 * (9999+1) + 3 * (999+1) + 4 * (99+1) + 5 * (9+1) + 6

When checking whether it is a multiple of some k, you can add/subtract multiples of k without changing the result, and those 99...9 are multiples of both 3 and 9.

So 123456 is a multiple of 3 (or 9) iff

1 * 1 + 2 * 1 + 3 * 1 + 4 * 1 + 5 * 1 + 6 * 1 = 1 + 2 + 3 + 4 + 5 + 6

is. Apply the same rule as often as you want -- that is, until you only have one digit left, because then it won't get simpler anymore.

It is basically because $10 mod 3 == 1$ (as 10 = 3*3 + 1). So if you are in the ring modulo 3, where every number is equal to the remainder of its division by 3, the sum of the digits of the number in its decimal representation equals the number itself (modulo 3), because in that ring 10 is actually 1, so the 10s in the decimal sum become 1s. Ie if n_k is the kth digit of n, you have

    (mod 3) n == n_0 + n_1*10 + n_2*10^2 + ... == n_0 + n_1 + n_2 + ...

Hence, n is divisible by 3 iff $n mod 3 == 0$ iff $(n_0 + n_1 + n_2 + ...) mod 3 == 0$.

Of course, summing up the digits may not give you a 1-digit number, but it gives you a number that you know is divisible by 3 (if the original number is divisible by 3). So you can apply the same idea/process again, summing up the digits of that number, and get another number that is divisible by 3. Repeat until you end up with one digit (hence the recursion mentioned).