Is the number $6^{1962}+1$ divisible by 31?
I. solution. $1962=6 \cdot 327$, therefore the given numerical expression can be written as:

$$
(6)^{327}-1^{327}+2
$$

The first two terms are divisible by the difference of the bases, which can be transformed as follows:

$$
6^{6}-1=\left(6^{3}\right)^{2}-1=\left(6^{3}-1\right)\left(6^{3}+1\right)=215 \cdot 217=215 \cdot 7 \cdot 31
$$

thus it is divisible by 31. According to this, the given expression is not divisible by 31. Additionally, we found that 2 appears as a remainder in the division.

Krámli György (Szeged, Déri M. mech. eng. tech. inst. I. class)

II. solution. $1962=2 \cdot 981$, therefore

$$
6^{1962}=\left(6^{2}\right)^{981}=36^{981}=(31+5)^{981}
$$

Expanding the binomial in the last form, 31 appears in every term with some positive integer exponent, except in the last term, $5^{981}$. Therefore, when dividing $5^{981}+1$ by 31, the remainder is the same as when dividing the original number.

Proceeding similarly, considering $981=3 \cdot 327$,

$$
5^{981}=\left(5^{3}\right)^{327}=125^{327}=(124+1)^{327}=(4 \cdot 31+1)^{327}
$$

thus, repeating the previous reasoning, we see that the remainder of the division $5^{981}: 31$ is the same as if we were dividing $1^{327}$, which is 1. Therefore, the remainder of the division $\left(6^{1962}+1\right): 31$ is 2.

Hanák Péter (Budapest, Fazekas M. grammar school II. class)