Problem 8.8. Masha wrote on the board in ascending order all natural divisors of some number $N$ (the very first divisor written is 1, the largest divisor written is the number $N$ itself). It turned out that the third from the end divisor is 21 times greater than the second from the beginning. What is the largest value that $N$ can take?
Answer: 441.

Solution. The second divisor from the beginning is the smallest prime divisor of the number $N$, let's denote it as $p$. The third divisor from the beginning is either $p^{2}$ or the second largest prime divisor of the number $N$, let's denote it as $q$.

Case 1. The third divisor from the beginning is $p^{2}$. Then the third divisor from the end is $\frac{N}{p^{2}}$. According to the problem,

$$
\begin{aligned}
& \frac{N}{p^{2}}=21 p \\
& N=21 p^{3}
\end{aligned}
$$

It is clear that 3 and 7 are divisors of the number $N$, so $p \leqslant 3$. If $p=2$, then the third largest divisor of the number $N$ is 3; if $p=3$, then the third largest divisor of the number $N$ is no more than 7, i.e., not $3^{2}$. Contradiction.

Case 2. The third divisor from the beginning is $q$. Then the third divisor from the end is $\frac{N}{q}$. According to the problem,

$$
\begin{aligned}
\frac{N}{q} & =21 p \\
N & =21 p q
\end{aligned}
$$

It is clear that 3 and 7 are divisors of the number $N$, so $p \leqslant 3, q \leqslant 7$. From this, we get that $N \leqslant 441$. It is not difficult to check that this number satisfies the condition.

## 9th grade