Let $A$ be the sum of the digits of the number $4444^{4444}$, and $B$ be the sum of the digits of the number $A$. Find the sum of the digits of the number $B$.

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The sum of the digits is comparable to $4444^{4444} \equiv (-2)^{4444} = (2^6)^{740} \cdot 2^4 \equiv 1^{740} \cdot 7 \pmod{9}$. Moreover, $4444^{4444} < 10^{4 \cdot 5000}$, meaning this number has no more than 20000 digits. Therefore, $A \leq 180000, B \leq 45$, and the sum of the digits of the number $B$ is no more than 14.

## Answer

7.