13. Leshа wrote on the board in ascending order all natural divisors of a natural number $n$, and Dima erased several first and several last numbers of the resulting sequence, leaving 151 numbers. What is the maximum number of these 151 divisors that could be fifth powers of natural numbers?

(M. Achtipov)
13. Answer: 31.

Lemma. If $n$ is divisible by $a^{5}$ and $b^{5}$, then $n$ is also divisible by $a^{4} b, a^{3} b^{2}, a^{2} b^{3}, a b^{4}$.

Proof of the lemma. Note that $n^{5}=n^{4} \cdot n$ is divisible by $a^{20} \cdot b^{5}$; taking the fifth root, we get that $n$ is divisible by $a^{4} b$. Similarly, $n^{5}=n^{3} \cdot n^{2}$ is divisible by $a^{15} \cdot b^{10}$, from which it follows that $n$ is divisible by $a^{3} b^{2}$. The proof for the other two divisors is completely analogous.

Let's proceed to solving the problem. We will prove the estimate, i.e., that the number of fifth powers is always no more than 31. Arrange the listed divisors in a row in ascending order. We will prove that there are at least four numbers between any two fifth powers in this row. Indeed, if $a^{5}, b^{5}$ are divisors of the number $n$, then the four divisors mentioned in the lemma are located in the row between $a^{5}$ and $b^{5}$: if, for example, $a > b$, then $a^{5} > a^{4} b > a^{3} b^{2} > a^{2} b^{3} > a b^{4} > b^{5}$.

If there were at least 32 fifth powers in the row, then there would be at least $4 \cdot 31 = 124$ numbers in the intervals between them, i.e., there would be at least 156 numbers in total in the row—contradicting the condition! Therefore, there cannot be more than 31 fifth powers.

Let's provide an example. Suppose $n=2^{150}$. This number has exactly 151 divisors (including 1 and $n$), and the fifth powers among them are the numbers $2^{5 k}, 0 \leqslant k \leqslant 30$—exactly 31 numbers.