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990

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by Frankenstein » Wed Jun 29, 2011 10:53 am
Hi,
990 = 9*10*11
11 is a prime number so we need n! to be at least 11! to be divisible by 990.

Hence, B
Cheers!

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by winniethepooh » Wed Jun 29, 2011 12:12 pm
Hey Frankenstine, where do 7 and 8 divide 990 with no remainder?
And if this is not the question wants then why haven't u selected 10 as the least possible value?
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by Frankenstein » Thu Jun 30, 2011 2:28 am
winniethepooh wrote:Hey Frankenstine, where do 7 and 8 divide 990 with no remainder?
And if this is not the question wants then why haven't u selected 10 as the least possible value?
Hi,
I think you have misinterpreted the question. What the question says is:
Find the least value of n such that 990 is a factor of n!.
990 = 2*3^2*5*11
Now for 990 to be a factor of n!, 11 should be a factor of n!. The least value of n that satisfies this condition is 11.
Consider 10! = 1.2.3...10
11 is not a factor of this right?
Cheers!

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by amit2k9 » Thu Jun 30, 2011 2:44 am
990 = 11*10*9
thus 11!.
B it is.
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by winniethepooh » Thu Jun 30, 2011 6:32 am
Thanks Frankenstine!
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by Buix0065 » Fri Jul 01, 2011 2:43 pm
Frankenstein wrote:
winniethepooh wrote:Hey Frankenstine, where do 7 and 8 divide 990 with no remainder?
And if this is not the question wants then why haven't u selected 10 as the least possible value?
Hi,
I think you have misinterpreted the question. What the question says is:
Find the least value of n such that 990 is a factor of n!.
990 = 2*3^2*5*11
Now for 990 to be a factor of n!, 11 should be a factor of n!. The least value of n that satisfies this condition is 11.
Consider 10! = 1.2.3...10
11 is not a factor of this right?
Hi Frankenstein,

Just to make sure I understand your reasoning.

1. a number is a factor if all primes pair
2. 990 has 2, 5, and 11 that need to pair
3. the smallest that n can be to ensure n! will pair 2, 5, 11 is 11?

Thanks for your insights!
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