Problem 506

## Subject 1.

| The numbers $a, b, c$ cannot all be odd because 6622 is even. Therefore, either they are all even | $\mathbf{2 p}$ |
| or only one of them is even. | $\mathbf{1 p}$ |
| $a=b=c=2$ does not satisfy the relation | $\mathbf{1 p}$ |
| since $a<b<c \Rightarrow a=2$ | |
| we get $65 b+289 c=6596$. The numbers 289 and 6596 have 17 as a common divisor $\Rightarrow 17$ is | |
| a divisor of $65 b$ and thus of $b$. But $b$ is prime $\Rightarrow b=17$ | $\mathbf{2 p}$ |
| Finally, $c=19$. | $\mathbf{1 p}$ |