## Task 15/89

Determine all natural numbers $n$ that satisfy the equation $n!+1=(10 n+1)^{2}$!
The following considerations lead quickly to the goal:

1. For $n=1$, $1!+1101$ and $7 \cdot 6 \cdot 5>101$ certainly hold $10!+1>(10 \cdot 10+1)^{2}$. From this it follows: If the equation has a solution $n \in N$, then $1<n<10$.
4. We bracket $n$ by halving the intervals in question (approximately). For $n=5$, $5!+1<(10 \cdot 5+1)^{2}$, so $5<n<10$. For $n=7$, $7!+1=5041=71^{2}=(10 \cdot 7+1)^{2}$. Therefore, $n=7$ is the (only) solution to the given equation.