# 1975 IMO Problems/Problem 4

## Problem

When $4444^{4444}$ is written in decimal notation, the sum of its digits is $A$. Let $B$ be the sum of the digits of $A$. Find the sum of the digits of $B$. ( $A$ and $B$ are written in decimal notation.)

## Solution

Note that $$4444^{4444}<10000^{4444}=\left(10^4\right)^{4444}=10^{17776}$$

Therefore $4444^{4444}$ has fewer than 17776 digits. This shows that $A<9\cdot 17775=159975$. The sum of the digits of $A$ is then maximized when $A=99999$, so $B\leq 45$. Note that out of all of the positive integers less than or equal to 45, the maximal sum of the digits is 12.

It's not hard to prove that any base-10 number is equivalent to the sum of its digits modulo 9. Therefore $4444^{4444}\equiv A\equiv B(\bmod{9})$. This motivates us to compute $X$, where $1\leq X \leq 12$, such that $4444^{4444}\equiv X(\bmod{9})$. The easiest way to do this is by searching for a pattern. Note that $$4444^1\equiv 7(\bmod 9)\\4444^2\equiv 4(\bmod 9)\\4444^3\equiv 1(\bmod 9)$$

and since $4444=3\times 1481+1$, $$4444^{4444}\equiv 4444^{3\times1481+1}\equiv \left(4444^3\right)^{1481}\times 4444\equiv 1\times 4444\equiv 7(\bmod{9})$$

Thus, $X=7$, which means that the sum of the digits of $B$ is $\boxed{7}$.