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Very Tedious Problem :(

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by zico20 » Wed Jun 10, 2009 6:33 am
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.
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by raleigh » Wed Jun 10, 2009 7:27 am
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.
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by zico20 » Wed Jun 10, 2009 7:50 am
Thank you very much
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