Counting zeros in product of numbers

Question

This is surprising a simple asked question...

How many zeros does the product $25^5$,$150^4$ and $2008^3$ end with?

(A)5 (B)9 (C)10 (D)12 (E)13

The problem is,I am not allowed to use calculator and using pure handwritten long multiplication working is going to be tedious.This question is a question trained for mathematics Olympiad so I have to solve this fast.

The one and only number I can evaluate is $150^4$.The end is 0,multiplied it self 4 times so therefore $150^4$ have 4 "0".

Answer 1

The idea is to find the greatest power $10^n$ dividing the product of the given integers, first you have to find the greatest powers $2^n$ and $5^n$ dividing each number :

• $25^5$ gives you $2^{0}$ and $5^{10}$ because $25=5^2$
• $150^4$ gives you $2^4$ and $5^{8}$ because $150=3\cdot 5^2\cdot 2$
• $2008^3$ gives you $2^{9}$ and $5^{0}$ because $2008=2^3\cdot 251$

To sum up the greatest power of $10$ dividing the product is $10^{13}$