1. The number n is the product of three (not necessarily distinct) prime numbers. If we increase each of them by 1, the product of the increased numbers will be 963 more than the original product. Determine the original number $n$.
SOLUTION. We are looking for $n=p \cdot q \cdot r$ with prime numbers $p \leqq q \leqq r$, which satisfy the equation

$$
(p+1)(q+1)(r+1)=p q r+963
$$

The right-hand side of this equation is an odd number in the case of the smallest prime $p=2$, so then the factors $q+1$ and $r+1$ from the product on the left side must also be odd numbers. For prime numbers $q, r$ this means that $q=r=2$, but the triplet $p=q=r=2$ does not satisfy equation (1). Therefore, $p \geqq 3$.

We will now show that it is necessary that $p=3$. In the opposite case, all three prime numbers $p$, $q, r$ are greater than 3, and thus neither the product $p q r$, nor the right side of (1) are numbers divisible by three (the number 963 is, after all, a multiple of three). From the condition that the product $(p+1)(q+1)(r+1)$ on the left side of (1) is not divisible by three, it follows that none of the prime numbers $p, q, r$ can give a remainder of 2 when divided by three. And since a remainder of 0 is excluded in the considered case, the prime numbers $p, q, r$ can only give a remainder of 1 when divided by three. From this, we further obtain that the difference $(p+1)(q+1)(r+1)-p q r$ gives the same remainder as the number $2 \cdot 2 \cdot 2-1 \cdot 1 \cdot 1=7$ when divided by three, i.e., a remainder of 2. This is a contradiction, however, because according to (1) we have $(p+1)(q+1)(r+1)-p q r=963$. The equality $p=3$ is thus proven.

After substituting $p=3$ into (1), we get the equation $4(q+1)(r+1)=3 q r+963$, which we will gradually transform into a "product" form:

$$
\begin{aligned}
4 q r+4 q+4 r+4 & =3 q r+963 \\
q r+4 q+4 r & =959, \\
(q+4)(r+4) & =975 .
\end{aligned}
$$

The number 975 has the prime factorization $975=3 \cdot 5^{2} \cdot 13$, so with regard to the inequalities $3 \leqq q \leqq r$ or $7 \leqq q+4 \leqq r+4$ for the smaller factor of the derived factorization, we must have $q+4 \leqq \sqrt{975}<32$, so $q+4 \in\{13,15,25\}$. This is satisfied by the only prime number $q=11$, for which $q+4=15$. For the second factor, we then have $r+4=5 \cdot 13=65$, from which $r=61$, which is indeed a prime number. The sought number $n=p \cdot q \cdot r$ is thus unique and has the value

$$
n=3 \cdot 11 \cdot 61=2013 .
$$

GUIDING AND ADDITIONAL PROBLEMS: