Science:Math Exam Resources/Courses/MATH312/December 2005/Question 05/Solution 1

The correct answer is c = 37.

Following the hints, we seek to count the number of factors of 10 in the number $153!$. To do this we count the number of 5s occuring in its expansion. The number of 5s occurring in the expansion of $n!$ is given by

$\lfloor \frac{n}{5}\rfloor +\lfloor \frac{n}{5^2}\rfloor +...+\lfloor \frac{n}{5^{\lfloor {\log }_5(n)\rfloor }}\rfloor$

This is true since you get a factor of 5 every 5 integers. You get another factor of 5 every 25 integers and so on. Counting this, we see that

$\lfloor \frac{153}{5}\rfloor +\lfloor \frac{153}{5^2}\rfloor +\lfloor \frac{153}{5^3}\rfloor =30+6+1=37$.