South African Mathematics Olympiad First Round

Since $13$ is a prime number, the only factors in $169!$ supplying powers of $13$ to the product are $13$, $26$, $39$, $\dots$, $156$, $169$. The first $12$ of these provide one power of $13$ each, but since $169=13^2$, it follows that the total power of $13$ is $12+2=14$.