# 2. CONDITION

The first and second digits of a two-digit number $N$ are the first and second terms of a certain geometric progression, and the number $N$ itself is three times the third term of this progression. Find all such numbers $N$.
Solution. According to the condition $10 b + b q = 3 b q^{2}$, where $b \neq 0$ is the first term, $q-$ is the common ratio of the progression. From this, $3 q^{2} - q - 10 = 0, q_{1} = 2, q_{2} = -5 / 3$, i.e., $q = 2$. From the inequality $b q \leq 9$ it follows that $b = 1, 2, 3$ or 4.

Answer: $12, 24, 36, 48$.