1975 IMO Problems/Problem 4

Problem

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.)

Solution

Let $N=4444^{4444}$ . We now take the base-10 logarithm of $N$ :

$\log N=4444\log 4444<4444\cdot 4=17776$

Therefore $N$ has less than 17776 digits. This shows that $A<9\cdot 17775=159975$ . The sum of the digits of $A$ is then maximized when $A=99999$ , so $B\leq 45$ . 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 provethat any base-10 number is equivalent to the sum of its digits modulo 9. Therefore $N\equiv A\equiv B\bmod{9}$ . We now compute $N\bmod{9}$ :

$N\equiv 4444^{4444}\equiv (4437+7)^{4444}\equiv (9\cdot 493 +7)^{4444}\bmod{9}$

After expanding, every term except $7^{4444}$ is divisible by 9, so they all cancel out. This shows that $N\equiv 7^{4444}\bmod{9}$ . Note that $7^3=343=9\cdot 36+1$ . Therefore

$N\equiv 7^{4444}\equiv 7^{3\cdot 1481 +1}\equiv 7\cdot (7^3)^{1481}\equiv 7\cdot (9\cdot 36+1)^{1481}\bmod{9}$

After expanding, every term except $7$ is divisible by 9, so they all cancel out. This shows that $N\equiv 7\bmod{9}$ . Therefore $A\equiv B\equiv 7\bmod{9}$ , and the sum of the digits of $B$ is also $7\bmod{9}$ . However, we established that the sum of the digits of $B$ is at most 12. This proves that the sum of the digits of $B$ is $\boxed{7}$ .

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1975 IMO (Problems) • Resources
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1975 IMO Problems/Problem 4

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