Olympiad digits!

Define $f(n)$ , for $n\in \mathbb{N}$ , to be $f(n)=n$ , when $n\in A=\{1,\dots, 9\}$ , and to be the sum of the digits of $n$ , if $n \not\in A$ .

It can be shown that, If $n \not\in A$ ,then $f(n)<n$ . Therefore, for some $m \in \mathbb{N}$ , the $m$ -fold application of $f_m(n)$ would be a member of $A$ . To determine which member of $A$ , $n \ mod \ 9$ can be calculated. For the case $4444^{4444}$ , the answer is $7$ .

Then we need to prove that $f_3(4444^{4444})$ is actually $7$ . In other words, applying $f$ , three times, would give us a single digit number.

$f(4444^{4444}) \leq 9 \times \lfloor \log_{10} 4444^{4444} \rfloor =145890$

$f(4444^{4444})$ would have a maximum of $6$ digits. Therefore

$f_2(4444^{4444}) \leq 9 \times 5 = 45$

finally,

$f_3(4444^{4444}) \leq 3+9=12$

The only natural number, less than or equal to $12$ ,that is congruent to $7$ ,mod $9$ , is $7$ itself.

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 * 17775=159975. The sum of the digits of A is then maximized when A = 99999, so B is less than or equal to 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 $4444^{4444}$ is equivalent to A which is equivalent to B (modulo 9). This motivates us to compute X, where 1 is less than or equal to X which is less than or equal to 12, such that $4444^{4444}$ is equivalent to X (modulo 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 * 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 7.