Begunni A.

What can be the product of several different prime numbers if it is divisible by each of them, decreased by 1?

Find all possible values of this product.

#
Let $N=p_{1} p_{2} \ldots p_{k}(k \geq 2)$ be the product of several distinct prime numbers $p_{1}<p_{2}<\ldots<p_{k}$, satisfying the condition of the problem. Since by condition $N$ is divisible by the even number $p_{2}-1$, it is itself even and $p_{1}=2$. The number $N=p_{1} p_{2} \ldots p_{k}$ has a unique divisor $p_{1}$ from the interval $\left(1, p_{2}\right)$, but $p_{2}-1$ also belongs to this interval, so $p_{2}-1=p_{1}=2$. Thus, $p_{2}=3$, and $N$ can take the value $2 \cdot 3=6$.

If $k \geq 3$, then by condition the number $p_{3}-1$, belonging to the interval $\left(p_{2}, p_{3}\right)$, is a divisor of the number $N$. The only divisor of $N$ that can belong to this interval is the number $p_{1} p_{2}=6$. Therefore, $p_{3}=p_{1} p_{2}+1=7$. The number $2 \cdot 3 \cdot 7=42$ satisfies the condition of the problem.

If $k \geq 4$, then by condition the even number $p_{4}-1$, belonging to the interval $(p_{3}, p_{4})$, is also a divisor of $N$. From the even divisors of the number $N$, only the numbers $p_{1} p_{3}=14$ and $p_{1} p_{2} p_{3}=42$ can belong to this interval. The number $15=p_{1} p_{3}+1$ is composite, so $p_{4}=p_{1} p_{2} p_{3}+1=43$. The number $2 \cdot 3 \cdot 7 \cdot 43=1806$ satisfies the condition of the problem.

If $k \geq 5$, then by condition the even number $p_{5}-1$, belonging to the interval $\left(p_{4}, p_{5}\right)$, must also be a divisor of $N$. From the even divisors of the number $N$, only the numbers $p_{1} p_{4}=86, p_{1} p_{2} p_{4}=258, p_{1} p_{3} p_{4}=602$, and $p_{1} p_{3} p_{3} p_{4}=1806$ can belong to this interval. Each of the numbers $87, 259, 603, 1807$ is composite. Therefore, the number $N$ cannot have more than four distinct prime divisors.

## Answer

$6, 42, 1806$.