[ Prime numbers and their properties $]$ [ Examples and counterexamples. Constructions ]

Find the smallest natural number $n$, for which the following condition is satisfied: if the number $p-$ is prime and $n$ is divisible by $p-1$, then $n$ is divisible by $p$.
Let $n$ satisfy this condition. Since $n$ is divisible by $1=2-1$, it must be divisible by 2, but then it is divisible by $3=2+1$, by

$7=2 \cdot 3+1$ and by $43=2 \cdot 3 \cdot 7+1$. Therefore, $n$ must be divisible by $1806=2 \cdot 3 \cdot 7 \cdot 43$. Hence, the minimum $n$ (if it exists) is no less than 1806.

On the other hand, for 1806 the condition of the problem is satisfied. Here are all the divisors of the number 1806: 1, 2, 3, 6, 7, 14, 21, 42, 43, 86, 129, 301, 1806. Now increase each of them by one, to get: 2, 3, 4, 7, 8, 15, 22, 43, 44, 87, 130, 302, 1807.

It remains to select all the prime numbers from the second set and verify that they are included in the first set (1807 is not prime, it is divisible by 13).

## Answer

$n=1806$.