## Task 5 - 170835

Determine all ordered triples $\left[P_{1} ; P_{2} ; P_{3}\right]$ of prime numbers $P_{1}, P_{2}, P_{3}$ with $P_{2}>P_{3}$, that satisfy the equation:

$$
P_{1}\left(P_{2}+P_{3}\right)=165
$$
Assume there are three such prime numbers. Then, because $165=3 \cdot 5 \cdot 11$, the prime number $P_{1}$ can only be one of the numbers $3,5,11$.

1. Case: Let $P_{1}=3$, then $P_{2}+P_{3}=55$.

The sum of two natural numbers is only odd if one summand is even and the other is odd. Therefore, because $P_{2}>P_{3}$ and 2 is the only even prime number, the only solution to (2) can be $P_{2}=53$; $P_{3}=2$.

Since these two numbers are prime, the triplet $[3 ; 53 ; 2]$ is a solution.

2. Case: Let $P_{1}=5$, then $P_{2}+P_{3}=33$. (3)

Similarly to Case 1, the only solution to (3) can be $P_{2}=31 ; P_{3}=2$. Since 31 and 2 are prime numbers, the triplet $[5 ; 31 ; 2]$ is also a solution.

3. Case: Let $P_{1}=11$, then $P_{2}+P_{3}=15$ : (4)

Similarly to Case 1, the only solution to (4) can be $P_{2}=13 ; P_{3}=2$. Since 13 and 2 are prime numbers, the triplet $[11 ; 13 ; 2]$ also satisfies equation (1).

Thus, the three triplets $[3 ; 53 ; 2],[5 ; 31 ; 2],[11 ; 13 ; 2]$ are the exact solutions to (1).