1. Find all pairs of natural numbers $a, b$ for which

$$
a+b+D(a, b)+n(a, b)=50
$$

where $D(a, b)$ denotes the greatest common divisor and $n(a, b)$ the least common multiple of natural numbers $a, b$.
1. Let $a=D k, b=D l$, where $D=D(a, b)$ is the greatest common divisor of the numbers $a, b$, so the numbers $k, l$ are coprime. Then $n=n(a, b)=D k l$ and it must hold that $D(k+l+1+k l)=50$, i.e., $(1+k)(1+l) D=50$. Therefore, we find all decompositions of the number 50 into a product of three natural numbers $D, 1+k, 1+l$, where the last two are greater than 1. Without loss of generality, we can assume that $a \leq b$, i.e., $k \leq l$. This gives us the following possibilities:

| $D$ | $1+k$ | $1+l$ | $k$ | $l$ | $a$ | $b$ |
| :---: | :---: | :---: | :---: | ---: | :---: | ---: |
| 1 | 2 | 25 | 1 | 24 | 1 | 24 |
| 1 | 5 | 10 | 4 | 9 | 4 | 9 |
| 5 | 2 | 5 | 1 | 4 | 5 | 20 |

For $D=2$ we get $k=l=4$, but $k, l$ must be coprime.

For $D=10,25$ or 50 we get $k=0$, which does not lead to any solution.

The problem has six solutions: $\{a, b\}=\{1,24\},\{a, b\}=\{4,9\},\{a, b\}=\{5,20\}$.