5.1. Among all the simple fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{3}{4}$. In your answer, specify its numerator.
Answer: 73. Solution. It is required to find a fraction $\frac{a}{b}$ such that $\frac{a}{b}-\frac{3}{4}=\frac{4 a-3 b}{4 b}$ reaches a minimum. Therefore, the maximum two-digit $b$ is sought, for which $4 a-3 b=1$. If in this case $b \geq 50$, then the fraction $\frac{a}{b}-\frac{3}{4}=\frac{1}{4 b}$ will always be less than any other fraction with a larger integer numerator and a different two-digit $b$.

Solve the equation $4 a-3 b=1$. Since $b=\frac{4 a-1}{3}=a+\frac{a-1}{3}-$ is an integer, then $a=1+3 k$, where $k-$ is any integer. Therefore, $b=1+4 k$. The maximum $k$, for which $a$ and $b$ are two-digit, will be $k=24$. Therefore, $b=97$ and $a=73$, that is, the required fraction: $\frac{a}{b}=\frac{73}{97}$.