Determine the iterated sum of the digits of $88^{88}$?

Question

First I use the fact that if the sum of the digits of a number is divided by $9$ then the number is divided by $9$.

So : $88^0\equiv 1\ [9]$

$88^1\equiv 7\ [9]$

$88^2\equiv 4\ [9]$

$88^3\equiv 1\ [9]$

Then $88^{88}=\big(88^{29}\big)^3\times88^{1}\equiv 7 \ [9]$

I try to see the possibilities : we have $88^{88}\le 10^{176}$ which means that the sum of the digits is bounded by $9\times176=1584$. Then the sum of the sum is bounded by $1+45+81+36=163$. Then the sum of the sum of the sum is bounded by $1+54+27=82$. Then the sum of the sum of the sum of the sum is bounded by $8+9=17$.

So it could be $7$ or $16$ but $1+6=7$. That's why it's $7$ the final answer.

Thanks in advance !

Answer 1

Eventually the iterated sum of the digits will be a single digit number. A number is congruent to the sum of its decimal digits mod $9$, so every iterated sum will be congruent to $7$ mod $9$ (assuming your calculation is correct). Therefore the final iterated sum of the digits is $7$.