## Task B-4.3.

If $26!=403291 x 6112660563558 x y 00000$, determine the digits $x$ and $y$.
## Solution.

First, let's determine how many zeros the number 26! ends with. We need to determine the power of the number $5$ in the prime factorization of this number (since the factor 2 appears more frequently), which is equal to $5^{6}$.

The number 26! ends with 6 zeros. Therefore, $y=0$.

Let's determine the value of the digit $x$. The given number must be divisible by 9, so the sum of its digits must also be divisible by 9.

The sum of the digits of the given number is

$$
\begin{aligned}
4+0+3+2+9+1+x+6+1+ & 1+2+6+6+0+5+6+3+5+5+8+x \\
& =73+2 x .
\end{aligned} \quad \begin{aligned}
4 \text { points }
\end{aligned}
$$

For the number $73+2 x=8 \cdot 9+1+2 x$ to be divisible by 9, $1+2 x$ must be divisible by 9, which is only possible if $2 x=8$, or $x=4$.

The sought digits are $x=4$ and $y=0$.