Problem 1103

I. Solution. Since 71 is a prime number, according to Wilson's theorem (see, for example, the university textbook "Number Theory" by Freud R. and Gyarmati E.), $70!+1$ is divisible by 71. It is therefore sufficient to show that 61! gives the same remainder when divided by 71 as 70!. Let's examine the product $A=62 \cdot 63 \cdot \ldots \cdot 70$ accordingly.

$$\begin{aligned} A & \equiv(71-9) \cdot(71-8) \cdot(71-7) \cdot \ldots \cdot(71-1) \equiv(-9) \cdot(-8) \cdot(-7) \cdot \ldots \cdot(-1)= \\ & =72 \cdot(-7!) \equiv-7!=-70 \cdot 72 \equiv 1 \quad(\bmod 71) \end{aligned}$$

Thus, $70!=A \cdot 61!\equiv 61!(\bmod 71)$, which is what we wanted to prove.

Remark. The received solutions can be categorized into two groups: those using Wilson's theorem and those in which the remainder of 61! was calculated in multiple steps. The advantage of the latter solution is that it can provide all numbers $n$ for which 71 is a divisor of $n!+1-n$. These $n$ values are 7, 9, 19, 51, 61, 63. This was pointed out by Ágnes Bartha, Tamás Birkner, Tímea Haszpra, and Ferenc Visnovitz in their papers.

II. Solution. Let's calculate for which $n$ numbers $n!+1$ is divisible by 71 ($n$ is a positive integer). The remainder of $n!$ when divided by 71, denoted as $r_{n}$, can be obtained from the remainder of $(n-1)!$ when divided by 71, denoted as $r_{n-1}$, as follows: $r_{n}=r_{n-1} \cdot n-71\left[\frac{r_{n-1} \cdot n}{71}\right]$, where $[x]$ denotes the integer part of $x$.

| $n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | $\mathbf{7}$ | 8 | $\mathbf{9}$ | $\ldots$ | $\mathbf{1 9}$ | $\ldots$ | $\mathbf{5 1}$ | $\ldots$ | $\mathbf{6 1}$ | 62 | $\mathbf{6 3}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $r_{n}$ | 1 | 1 | 2 | 6 | 24 | 49 | 10 | $\mathbf{7 0}$ | 63 | $\mathbf{7 0}$ | $\ldots$ | $\mathbf{7 0}$ | $\ldots$ | $\mathbf{7 0}$ | $\ldots$ | $\mathbf{7 0}$ | 9 | $\mathbf{7 0}$ |

Numbers that give a remainder of 70 when divided by 71, when 1 is added to them, result in numbers divisible by 71. These are the numbers mentioned in the remark, among which the last one before 61 is 61. This also proves the statement of our problem.