Problem 1. Let $n$ be a non-zero natural number. For each natural number $k$, we denote by $a(k, n)$ the number of natural divisors $d$ of $k$ such that $k \leq d^{2} \leq n^{2}$. Calculate $\sum_{k=1}^{n^{2}} a(k, n)$.
Solution. ${ }^{1}$ For $1 \leq k \leq n^{2}$, let $A(k, n)$ be the set of natural divisors $d$ of $k$ such that $k \leq d^{2} \leq n^{2}$. A natural number $d \in\{1,2, \ldots, n\}$ belongs to the sets

$$
A(d, n), A(2 d, n), \ldots, A\left(d^{2}, n\right)
$$

and only these.

.3 points

Therefore, the contribution of each $d$ in the sum $\sum_{k=1}^{n^{2}} a(k, n)=\sum_{k=1}^{n^{2}}|A(k, n)|$ is $\underbrace{1+1+\cdots+1}_{d \text { times }}=d$.

2 points

Thus $\sum_{k=1}^{n^{2}} a(k, n)=\sum_{d=1}^{n} d=n(n+1) / 2$.