Problem 4

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.)

Step 1 of 6: Bound the number of digits of

4444^{4444}

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In plain words

Chasing the exact value of $4444^{4444}$ is hopeless (it has tens of thousands of digits), so the trick is to never compute it: bound how large its digit sum can possibly be, then bound the digit sum of that, and finally pin down the last digit sum using arithmetic modulo 9, which digit sums always respect.

4444^{4444} < 10000^{4444} = 10^{17776}

Detailed analysis

Since $4444 < 10000 = 10^4$ , raising to the $4444$ th power gives $4444^{4444} < (10^4)^{4444} = 10^{17776}$ . A positive integer less than $10^{17776}$ has at most $17776$ digits, and each digit is at most $9$ , so its digit sum $A$ satisfies $A \le 9 \cdot 17776 = 159984$ .