5. How many ordered pairs $(m, n)$ of natural numbers satisfy $m n+2 m-2 n=2020$?
First method:

We have $\quad m n+2 m-2 n-4+4=2020$ 1 POINT

Furthermore, $\quad m(n+2)-2(n+2)+4=2020, \quad 1$ POINT

which means $(m-2)(n+2)=2016$. 1 POINT

The number $n+2$ is a divisor of the number 2016. 1 POINT

Let's factorize the number 2016 into prime factors: $2016=2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 7=2^{5} \cdot 3^{2} \cdot 7 . \quad 1$ POINT

Every divisor of the number 2016 is of the form $2^{a} \cdot 3^{b} \cdot 7^{c}$, where

$a \in\{0,1,2,3,4,5\}, b \in\{0,1,2\}, c \in\{0,1\} . \quad 1$ POINT

The number of natural divisors of the number 2016 is equal to $6 \cdot 3 \cdot 2=36$. 2 POINTS

The divisors of the number 2016 that do not satisfy the conditions of the problem are the numbers 1 and 2, because if $n+2=1$ and $n+2=2$, then $n$ is not a natural number, and in all other cases the number $n$ is natural, so there are 34 natural numbers $n$ that satisfy the conditions of the problem.

1 POINT

For each such $n$, the natural number $m$ that satisfies the original equation is uniquely determined because the equation is

equivalent to $m=2+\frac{2016}{n+2}$.

Thus, there are 34 ordered pairs $(m, n)$ of natural numbers that satisfy the equation

$m n+2 m-2 n=2020$.

1 POINT

.TOTAL 10 POINTS

Note: If the student knows that the number of divisors of the number $2016=2^{5} \cdot 3^{2} \cdot 7^{1}$ is equal to $(5+1)(2+1)(1+1)=6 \cdot 3 \cdot 2=36$, grade with 3 POINTS.