# Task 8. (16 points)

Let all companies in the country have a certain rank, which is a natural number. When two companies of ranks $m$ and $n$ merge, a new company of rank $(m+n)$ is formed. The profit of the resulting company will be $m \cdot n$ more than the sum of the profits of the companies forming it. The profit of a company of the first rank is 1 d.e. Is there a rank at which the profit of the company will be 2016 d.e.?
# Solution.

Let $p_{n}$ be the profit of the company of rank $n$. Then, according to the problem statement,

$$
p_{m+n}=p_{m}+p_{n}+m n
$$

Notice that for any $n$,

$$
p_{n}=p_{n-1}+p_{1}+(n-1) \cdot 1=p_{n-1}+1+n-1=p_{n-1}+n
$$

We will prove that $p_{n}=1+2+\ldots+n$.

We use mathematical induction.

1) Base case: $p_{1}=1$.
2) Assume $p_{k}=1+2+\ldots+k$. We need to prove that $p_{k+1}=1+2+\ldots+k+(k+1)$.

Indeed, $p_{k+1}=p_{k}+(k+1)=(1+2+\ldots+k)+(k+1)$.

Therefore, $p_{n}=1+2+\ldots+n=\frac{n(n+1)}{2}$.

For a company to have a profit of 2016 units, its rank must satisfy the equation

$$
\frac{n(n+1)}{2}=2016
$$

The last equation has a natural solution $n=63$.

Answer: yes, the rank is 63.