Problem 606

Solution:

The answer is (B). Let $a < b < c < d$ be the four prime numbers corresponding to the ages. Then
$$60 = \frac{a + b + c + d}{4} < \max \{a, b, c, d\} = d < 70.$$
So we have only two possibilities: $d = 61$ or $d = 67$. If $d$ were $61$, then $a < b < c < 60$ so that
$$240 = a + b + c + d < 3 \cdot 59 + 61 < 240$$
and this is impossible. Otherwise $d = 67$ and since $a < b < c < d$ we have $a \leq 53$, $b \leq 59$, $c \leq 61$. On the other hand
$$240 = a + b + c + d \leq 53 + 59 + 61 + 67 = 240,$$
so that $(a, b, c, d) = (53, 59, 61, 67)$ is the only solution.