Let $A$ be the sum of the digits of the number $4444^{4444}$ written in the decimal system, and $B$ be the sum of the digits of $A$! Calculate the sum of the digits of $B$!

( $A$ and $B$ are also written in the decimal system!)
Let's denote the sum of the digits of $B$ by $C$, and our task is to calculate $C$. First, let's give an upper estimate for the values of $A$, $B$, and $C$! Since $4444^{4444} < 10000^{4444} < 10000^{5000} = 10^{20000}$, our number has at most 20000 digits, and thus the sum of its digits, which is $A$, is at most $9 \cdot 20000 < 200000$; therefore, $A$ is at most a six-digit number. This means that the sum of the digits of $B$ is not more than $6 \cdot 9 = 54$. This, in turn, means that $C$ can be at most 13, since the sum of the digits of numbers not greater than 54 is at most 49, and then it is exactly 13. Thus, we have:

\[
C \leq 13.
\]

On the other hand, we know that a number divided by nine gives the same remainder as the sum of its digits. According to this, $4444^{4444}$, $A$, $B$, and $C$ all give the same remainder when divided by nine. From the following transformation, we immediately get that this remainder is 7: $4444^{4444} = (4444^{4444} - 7^{4444}) + 7(7^{3 \cdot 1481} - 1) + 7$. Indeed, $a^n - b^n$ is divisible by $(a - b)$ for any natural number $n$, and here $4444 - 7 = 9 \cdot 493$ and $7^3 - 1 = 342 = 9 \cdot 38$ are divisible by nine.

Thus, $C$ is at most 13 and gives a remainder of 7 when divided by nine, so the value of $C$, which is the sum of the digits of $B$, is 7.

Zalán Bodó (Budapest, I. István Gymnasium, 3rd grade)