2012/13 AHSMC Part 1 Problem 16

Since $13$ divides $6(m + 11n) = (6m + n) + 13(5n)$, $13$ divides $6m + n$. Since $11$ divides $6(m + 13n) = (6m + n) + 11(7n)$, $11$ also divides $6m + n$. Hence $11 \times 13 = 143$ divides $6m + n$, so that $6m + n = 143k$ for some integer $k$. Since $6(m + n) = 143k + 5n = 6(24k + n) - (k + n)$, $6$ divides $k + n$ so that $k + n \ge 6$. Now $6(m + n) = 143k + 5n = 138k + 5(k + n) \ge 138 + 30 = 168$. Consequently $m + n \ge 28$, and this is attained if $m = 23$ and $n = 5$. The answer is (C).