B1. Andrew and Beatrice practice their free throws in basketball. One day, they attempted a total of 105 free throws between them, with each person taking at least one free throw. If Andrew made exactly $1 / 3$ of his free throw attempts and Beatrice made exactly $3 / 5$ of her free throw attempts, what is the highest number of successful free throws they could have made between them?
Correct answer: 59 free throws.
Solution 1: From their success rates we conclude that each of them must have made a multiple of 15 throws. Specifically, from Andrew's success rate, his number of throws must be a multiple of 3 . Since the total number of throws (105) is also a multiple of 3 , Beatrice's number of throws must be a multiple of 3 too. From Beatrice's success rate, her number of throws must be a multiple of 5 , and thus must in fact be a multiple of 15 . Similarly, since 105 is a multiple of 5 , Andrew's number of throws must be a multiple of 5 and thus a multiple of 15 too.
Since $1 / 30$, $b>0$. Let $M$ be the number of successful free throws. We have
$$
M=\frac{a}{3}+\frac{3 b}{5}=\frac{a}{3}+\frac{3(105-a)}{5}=\frac{945-4 a}{15}=63-\frac{4 a}{15} .
$$
$M$ is maximal when $\frac{4 a}{15}$ is minimal. That is, $a=15$ and so $M=59$.
The maximum possible number of successful free throws they could have made between them is 59 .