Just another trailing zeroes problem

The number of trailing zeros $n$ is equal to the number of factors of $5$ (seeing as there are at least as many factors of $2$ ), therefore:

$n =\left\lfloor\frac{5201}{5}\right\rfloor+\left\lfloor\frac{5201}{25}\right\rfloor+\left\lfloor\frac{5201}{125}\right\rfloor+\left\lfloor\frac{5201}{625}\right\rfloor+\left\lfloor\frac{5201}{3125}\right\rfloor =1040+208+41+8+1=\boxed{1298}.$

Instead of finding the different powers of 5 we can divide the remainder left again and again by 5. 5201/5 = 1040 , 1040/5 = 208 , 208/5 = 41 , 41/5 = 8 , 8/5 = 1 . Answer = 1040 + 208 + 41 + 8 + 1 = 1298 . This is much easier.