Problem A
Let $a$ and $b$ be positive whole numbers such that $\frac{4.5}{11}<\frac{a}{b}<\frac{5}{11}$. Find the fraction $\frac{a}{b}$ for which the sum $a+b$ is as small as possible. Justify your answer.
Solution
By multiplying numerators and denominators by 7 , we can rewrite the inequalities as follows:
$$
\frac{7 \cdot 4.5}{7 \cdot 11}6$, i.e., $b=7$ is the smallest possible denominator for $a=3$, yielding the sum $a+b=3+7=10$.

Finally, if $a \geq 4, a+b>a+2 a=3 a \geq 3 \cdot 4=12>10$. Thus, $a$ cannot be 4 or more and yield the smallest possible sum.

We conclude that the fraction $\frac{3}{7}$ fits the two inequalities and has the smallest sum of numerator and denominator.