\section*{Problem 4 - 330944}

Someone finds the statement

\[
22!=11240007277 * * 607680000
\]

In this, the two digits indicated by \(*\) are illegible. He wants to determine these digits without performing the multiplications that correspond to the definition of 22!.

Conduct such a determination and justify it! It may be used that the given digits are correct.

Hint: For every positive integer \(n\), \(n\)! is defined as the product of all positive integers from 1 to \(n\).
}

Let \(a\) be the front and \(b\) be the back of the two illegible digits. Since 22! is divisible by 9, their sum of digits must be divisible by 9. The sum of digits is \(1+1+2+4+7+2+7+7+a+b+6+7+6+8=58+a+b\), so given \(0 \leq a+b \leq 18\), then \(a+b \in\{5,14\}\) holds.

Furthermore, 22! is divisible by 11, so its alternating sum of digits must be divisible by 11. This is \(1-1+2-4+0-0+0-7+2-7+7-a+b-6+0-7+6-8=-22-a+b\), so \(a-b\) must be divisible by 11, which, given \(-9 \leq a-b \leq 9\), leads to \(a-b=0\) and thus \(a=b\).

Since there is no solution in natural numbers for \(a+b=5\) and \(a=b\), it must be that \(a+b=14\) and thus \(a=b=7\). These are the sought-after digits.