5.1. Among the numbers exceeding 2013, find the smallest even number $N$ for which the fraction $\frac{15 N-7}{22 N-5}$ is reducible.
Answer: 2144. Solution: The presence of a common factor in the numbers $15 N-7$ and $22 N-5$ implies that the same factor is present in the number $(22 N-5)-(15 N-7)=7 N+2$, and subsequently in the numbers $(15 N-7)-2 \cdot(7 N+2)=N-11,(7 N+2)-7 \cdot(N-11)=79$. Since 79 is a prime number, the fraction is reducible by 79, so $N-11=79 m, N=11+79 m$. Given that $N$ is even, we have $N=90+158 p$. The required value is achieved at $p=13$.