\section*{Task 4 - 331044}

Someone finds the statement

\[
23!=2585201673 * 8849 * 6640000
\]

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 23!.

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 the two unknown digits be \(x\) and \(y\) (\(0 \leq x, y \leq 9\)), i.e., the digit representation

\[
23!=2585201673 x 8849 y 6640000
\]

23! is guaranteed to be divisible by 3, 9, and 11 due to its construction. A number that is divisible by 3 has a sum of digits that is divisible by 3. A number leaves the same remainder when divided by 9 as its sum of digits. Additionally, a number is divisible by 11 if its alternating sum of digits is divisible by 11.

The sum of the digits is \(Q=2+5+\ldots+3+x+8+\ldots+8+y+6+\ldots+0=84+x+y\). The alternating sum of the digits is \(A=2-5+6-5 \pm \ldots+x-8 \pm \ldots y+6 \mp \ldots 0=47+x-37-y=10+x-y\).

From \(84 \equiv 0(\bmod 3)\) it follows that \(x+y\) must be divisible by 3. Therefore, the remaining possibilities are

\begin{tabular}{cc|cc|cc|cc|cc}
\(x\) & \(y\) & \(x\) & \(y\) & \(x\) & \(y\) & \(x\) & \(y\) & \(x\) & \(y\) \\
0 & \(0,3,6,9\) & 1 & \(2,5,8\) & 2 & \(1,4,7\) & 3 & \(0,3,6,9\) & 4 & \(2,5,8\) \\
5 & \(1,4,7\) & 6 & \(0,3,6,9\) & 7 & \(2,5,8\) & 8 & \(1,4,7\) & 9 & \(0,3,6,9\)
\end{tabular}

From \(84 \equiv 3(\bmod 9)\) it follows that \(x+y \equiv 6(\bmod 9)\) must hold. Of the possibilities listed in the table, only the following remain:

\begin{tabular}{ll|cc|cc|cc|cc}
\(x\) & \(y\) & \(x\) & \(y\) & \(x\) & \(y\) & \(x\) & \(y\) & \(x\) & \(y\) \\
0 & 6 & 1 & 5 & 2 & 4 & 3 & 3 & 4 & 2 \\
5 & 1 & 6 & 0,6 & 7 & 5 & 8 & 4 & 9 & 3
\end{tabular}

For divisibility by 11, \(10+x-y \equiv 0(\bmod 11)\) must hold. Testing the remaining digit pairs \((x, y)\), only \(x=8, y=7\) satisfies the last requirement. The unreadable digits are 8 and 7, so it holds that

\[
23!=25852016738884976640000
\]

as a control multiplication shows.