Problem 6.4. All divisors of a natural number $n$ were written in a row in ascending order from 1 to $n$ itself. It turned out that the second-to-last number in the row is 101 times greater than the second. For what largest $n$ is this possible?
Answer: $101^{3}$.

Solution. Let $p$ be the smallest divisor of the number $n$ different from 1, then $p$ is a prime number, otherwise there would be smaller divisors. Thus, the first two numbers in the row will be equal to 1 and $p$, and the last two - equal to $\frac{n}{p}$ and $n$. According to the problem, we have $\frac{n}{p}=101 p$, or, in other words, $n=101 p^{2}$.

Since $p$ is the smallest prime divisor of $n$, the inequality $p \leqslant 101$ holds. Therefore, $n \leqslant 101^{3}$. The number $101^{3}$ fits.

## Criteria

One largest suitable criterion is used:

## 76 . Any correct solution to the problem.

4 6. The equation of the form $\frac{n}{p}=101 p$ is obtained.

## 16 . The correct answer is obtained.