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

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.

• To find the digital sum, you will have to compute $88^{88}$ explicitly. The reason being that taking remainder modulo $9$ does not at all simplify the calculation of the sum,it only provides a bound. Aug 31, 2016 at 11:56
• 88 doesn't look very compatible with the decimal system.. So I suspect there is no easy way round it.
– Vim
Aug 31, 2016 at 11:57
• Are you sure you don't mean the iterated digit sum? That's $7$, for the reason you mentioned. The answer to your question is $745$ by computer, but I don't see any short cut to seeing that.
– lulu
Aug 31, 2016 at 11:58
• @lulu by iterated you mean "the sum of the sum of the sum" ? Aug 31, 2016 at 12:00
• @Maman Indeed. This is the kind of sum (along with the mod-11 sum) that we use to check hand-written calculations. Aug 31, 2016 at 12:01

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 $7$.
• Why do you exclude $16$ ? Aug 31, 2016 at 12:22
• @Maman The sum of the digits of $16$ is $7$. Aug 31, 2016 at 12:22