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by Ian Stewart » Sun Jun 22, 2008 6:19 am
The question really asks:

If n! is divisible by 990, what is the smallest possible value of n?

Prime factorize 990:
990 = 11*90 = 2*3^2*5*11

We know n! is divisible by 990, and therefore is divisible by the divisors of 990 -- including the primes 2, 3, 5 and 11. If n! is divisible by 11, n must be 11 or larger. Notice also that 11! is divisible by 2, 3^2 and 5, so n=11 is the correct answer.
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by rosh26 » Sun Jun 22, 2008 8:47 am
Hi,

you said that "We know n! is divisible by 990, and therefore is divisible by the divisors of 990"

But 990 is also divisible by 10 - so why is n! not divisible by 10??
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by Stuart@KaplanGMAT » Sun Jun 22, 2008 9:00 am
rosh26 wrote:Hi,

you said that "We know n! is divisible by 990, and therefore is divisible by the divisors of 990"

But 990 is also divisible by 10 - so why is n! not divisible by 10??
n! is divisible by 10.

Ian broke 990 down into it's prime factors; since both 2 and 5 appear on the list of prime factors, 990 is also divisible by 2*5 = 10. However, since we already have 2 and 5 on our list, including 10 would be redundant and lead to problems.
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by rosh26 » Sun Jun 22, 2008 10:16 am
Right, so why isnt the solution 10 is my question?
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by Ian Stewart » Sun Jun 22, 2008 2:21 pm
rosh26 wrote:Right, so why isnt the solution 10 is my question?
Because 10! isn't divisible by 11.
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by rosh26 » Mon Jun 23, 2008 2:26 pm
ok...but since 10 is divisible by 990, why did we choose to prime factorize 990 (which gave that it was divisible by 11), instead of just factoring normally?

Thanks so much for your help!
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by Ian Stewart » Mon Jun 23, 2008 6:49 pm
rosh26 wrote:ok...but since 10 is divisible by 990, why did we choose to prime factorize 990 (which gave that it was divisible by 11), instead of just factoring normally?

10 is not divisible by 990; the opposite is true: 990 is divisible by 10. And I'm not sure what you mean by 'factoring normally'; prime factorization is 'normal'; in fact, it's one of the most common techniques in all of mathematics.
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by rosh26 » Mon Jun 23, 2008 6:56 pm
Sorry, I guess I wasn't clear...

I meant to factor 990 with not just prime numbers:

990 = 10 * 99
2*5 11*9
3*3

When I did the problem, this was how I factored, which is why I chose 10...I see now that I read the solution incorrectly...thanks for your help
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