Problem Solving

hey fellaz,

This is a question that i find very tedious to work out and definitely will eat up a good chunk of my time on the exam if it were to come. Is there perhaps an easier way to do it? Thank you very much in advance.

If n is a positive integer and the product of all integers from 1 to n inclusive, is a multiple of 990. What is the least possible value of n?

A. 10
B. 11
C. 12
D. 13
E. 14

Thanks again.

With multiple and divisibility problems, the trick is usually factoring the number to it's prime factorization

990 = 99*10 = 9*11*2*5 (= 2*3^2*5*11)

So n! (product of 1 to n inclusive) must contain all of these factors. 11! is the smallest which contains all of these factors and the answer is B.


Manhattan GMAT Number Properties book is a good resource for learning how to approach these types of problems.

Thank you very much