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If A is the sum of the digits of $4444^{4444}$ and B is the sum of the digits of A then find, the sum of the digits of B.

I tried using mod(10), Fermat's Theorem alongside Euler's Totient Function but I am not able to make it through, Please help with this question.

Shridhar Sharma
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    Hint. $4444^{4444}\equiv A\equiv B\pmod9$. – MH.Lee Feb 12 '22 at 03:13
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    The punchline is that $4444^{4444}$ has fewer than $5000\cdot \log_{10}(10000)=25000$ digits (we could be more specific if it mattered, but it doesn't matter, and it is obvious that $4444\cdot \log_{10}(4444)$ is smaller than even this) and so has digit sum fewer than $9\cdot 25000$. The sum of the digits of this result is capped by a certain amount. The sum of the digits of that is capped by a certain result.... to the point of us having only one possibility that is the correct value mod9 – JMoravitz Feb 12 '22 at 03:31

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