1. Find all natural numbers $m$ and $n$ for which the sum of the greatest common divisor and the least common multiple is equal to 101.
II/1. Let $d$ be the greatest common divisor of the numbers $m$ and $n$. Then $m = d m_{1}$ and $n = d n_{1}$, and the numbers $m_{1}$ and $n_{1}$ are coprime. The least common multiple of $m$ and $n$ is $d m_{1} n_{1}$. We have

$$
101 = d + d m_{1} n_{1} = d(1 + m_{1} n_{1})
$$

Since $1 + m_{1} n_{1} \geq 2$ and 101 is a prime number, the only possibility is $d = 1$ and $m_{1} n_{1} = 100$. The numbers $m = m_{1}$ and $n = n_{1}$ are coprime. The one that is divisible by 2 is therefore divisible by 4. The one that is divisible by 5 is therefore divisible by 25. The possible pairs of solutions are $(1,100), (4,25), (25,4)$, and $(100,1)$.

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-08.jpg?height=51&width=1636&top_left_y=954&top_left_x=216)

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-08.jpg?height=48&width=1631&top_left_y=1004&top_left_x=218)

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-08.jpg?height=57&width=1633&top_left_y=1051&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-08.jpg?height=52&width=1633&top_left_y=1102&top_left_x=220)

(If not all solutions are provided, one point is awarded for one of the solutions $(1,100)$ or $(100,1)$ and one point for one of the solutions $(4,25)$ or $(25,4)$.) Justification that there are no other possibilities (that is, considering the coprimality of $m_{1}$ and $n_{1}$ when solving the equation $m_{1} n_{1} = 100$)

1 point

If the contestant does not introduce the numbers $m_{1}$ and $n_{1}$, they can earn a maximum of 1 point for recognizing that $d$ divides the least common multiple of $m$ and $n$ or for proving that $m$ and $n$ have different parities.