# Question #2c645

$n = 637$
We know that ${5}^{5} = 3125 > 2557$ so the number of trailing zeroes is given by
$n = {\sum}_{k = 1}^{4} \left\lfloor \frac{2557}{5} ^ k \right\rfloor = 511 + 102 + 20 + 4 = 637$
which is the number of $5$ factors appearing in $2557$ Those $5$ multiplied by the $2$ factors are responsible by the trailing zeroes.