metman82 wrote:990 = 9 * 110 = 9 * 2 * 55 = 9 * 2 * 5 * 11

No other primefactors.

Yes, to be clear the prime factorization is 3 * 3 * 2 * 5 * 11.

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- jnelson0612
- ManhattanGMAT Staff
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metman82 wrote:990 = 9 * 110 = 9 * 2 * 55 = 9 * 2 * 5 * 11

No other primefactors.

Yes, to be clear the prime factorization is 3 * 3 * 2 * 5 * 11.

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- JJ32
- Course Students
**Posts:**4**Joined:**Tue Jul 14, 2009 2:45 am

Sorry I understand how to get the prime factors of 990, but I am just confused of what the question is actually asking.

Is it asking for the largest prime factor or if there was another prime factor of 990 in the answer choices could that also be the answer? Just trying to figure out what the question asking?

Thanks and Regards

Is it asking for the largest prime factor or if there was another prime factor of 990 in the answer choices could that also be the answer? Just trying to figure out what the question asking?

Thanks and Regards

JJ32

- RonPurewal
- ManhattanGMAT Staff
**Posts:**19654**Joined:**Tue Aug 14, 2007 8:23 am

JJ32 wrote:Sorry I understand how to get the prime factors of 990, but I am just confused of what the question is actually asking.

Is it asking for the largest prime factor or if there was another prime factor of 990 in the answer choices could that also be the answer? Just trying to figure out what the question asking?

Thanks and Regards

the question is asking for the smallest n such that 990 is contained in the product 1 x 2 x 3 x 4 x ... x n.

if n were a smaller factor of 990 then it wouldn't work, because in that case the product 1 x 2 x 3 x 4 x ... x n wouldn't be divisible by 11, and so, a fortiori, it wouldn't be divisible by 990.

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- JJ32
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RonPurewal wrote:JJ32 wrote:Sorry I understand how to get the prime factors of 990, but I am just confused of what the question is actually asking.

Is it asking for the largest prime factor or if there was another prime factor of 990 in the answer choices could that also be the answer? Just trying to figure out what the question asking?

Thanks and Regards

the question is asking for the smallest n such that 990 is contained in the product 1 x 2 x 3 x 4 x ... x n.

if n were a smaller factor of 990 then it wouldn't work, because in that case the product 1 x 2 x 3 x 4 x ... x n wouldn't be divisible by 11, and so, a fortiori, it wouldn't be divisible by 990.

Thanks Ron. It took a while for me, but now I understand.

JJ32

- RonPurewal
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**Posts:**19654**Joined:**Tue Aug 14, 2007 8:23 am

good!

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- vijaykumar.kondepudi
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**Posts:**24**Joined:**Sat Dec 12, 2009 9:23 pm

jnelson0612 wrote:madhan, that's certainly one way to look at it. Thanks!

I fail to understand the LCM approach. LCM (Least Common Multiple) is for atleast (minimum) 2 numbers. We have only one number here: 990.

How can we conclude the answer to the question is 11 by this approach ?

- tim
- ManhattanGMAT Staff
**Posts:**5665**Joined:**Tue Sep 11, 2007 9:08 am**Location:**Southwest Airlines, seat 21C

the poster who mentioned LCM used the term incorrectly. you haven't missed anything.. :)

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- sahilk934
- Students
**Posts:**1**Joined:**Sat Mar 14, 2015 8:08 am

I found this one useful

We are told that n!=990∗k=2∗5∗32∗11∗k --> n!=2∗5∗32∗11∗k which means that n! must have all factors of 990 to be the multiple of 990, hence must have 11 too, so the least value of n is 11 (notice that 11! will have all other factors of 990 as well, otherwise the least value of n would have been larger)..

We are told that n!=990∗k=2∗5∗32∗11∗k --> n!=2∗5∗32∗11∗k which means that n! must have all factors of 990 to be the multiple of 990, hence must have 11 too, so the least value of n is 11 (notice that 11! will have all other factors of 990 as well, otherwise the least value of n would have been larger)..

- tim
- ManhattanGMAT Staff
**Posts:**5665**Joined:**Tue Sep 11, 2007 9:08 am**Location:**Southwest Airlines, seat 21C

Thanks for sharing. Let us know if there are any further questions on this one!

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