## Task 1 - 110731

Determine all prime numbers $p$ that simultaneously satisfy the following conditions:

(1) $\mathrm{p}<100$.

(2) $p$ leaves a remainder of 2 when divided by both 3 and 5.

(3) $p$ leaves a remainder of 1 when divided by 4.
Assume that $p$ is a prime number satisfying conditions (1), (2), and (3).

Due to (2), $p-2$ is divisible by both 3 and 5. Since 3 and 5 are coprime, $p-2$ is therefore divisible by $3 \cdot 5=15$, meaning that $p$ is of the form $n \cdot 15+2$ (where $n$ is a natural number). Due to (3), $p$ and thus also $n$ must be odd.

Therefore, due to (1), only the numbers 17, 47, and 77 can satisfy the conditions of the problem. Among them, 77 is not a prime number, and 47 does not satisfy condition (3). Thus, only 17 can be the solution to the problem.

Indeed, 17 satisfies conditions (1), (2), and (3), and is thus the only such prime number.