Problem 3. Let $n$ and $k$ be natural numbers greater than 2. We will say that $n$ is attracted by $k$ if the following conditions are met:

1) $n$ is divisible by $k$ and has $k-1$ divisors;
2) the sum of the divisors of $n$ is divisible by $k$, but not by $k+1$.

Determine the smallest three-digit number $n$ attracted by 7.
## Solution and Scoring

If $n$ is attracted by 7, then: 1) $n$ is divisible by 7 and has $6=2 \cdot 3$ divisors; 2) the sum of the divisors of $n$ is divisible by 7, but not by 8. We obtain $n=p^{5}$ or $n=p_{1} \cdot p_{2}^{2}$, where $p, p_{1}, p_{2}$ are prime numbers.

2 points

Being a multiple of 7, the number $n$ can be of the form: $n=7^{5}$ or $n=7 \cdot p^{2}$ or $n=7^{2} \cdot p$, where $p$ is a prime factor, $p \neq 7$ 1 point

If $n=7^{5}$, the sum of the divisors of $n$, i.e., $1+7+7^{2}+7^{3}+7^{4}+7^{5}=1+M_{7}$, is not divisible by 7 - condition 2) is not satisfied.

If $n=7 \cdot p^{2}$, the sum of the divisors of $n$, i.e., $1+7+p+7 p+p^{2}+7 p^{2}=8+8 p+8 p^{2}=M_{8}$ - condition 2) is not satisfied.

2 points

If $n=7^{2} \cdot p$, the sum of the divisors of $n$, i.e., $1+7+7^{2}+p+7 p+7^{2} p=57+57 p=57 \cdot(1+p)$ is a multiple of 7, if $1+p$ is a multiple of 7, $p$ being a prime number, minimal. Thus, $p=13$ and $8 \times 57 \cdot 14$.

The number sought is $n=7^{2} \cdot 13=637$.

2 points