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by cans » Thu Jun 09, 2011 7:41 am
n! is multiple of 990
990=11*10*9
11 is a prime number, and thus n! must contain 11 so that it is multiple of 990
thus least value of n=11
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by manpsingh87 » Thu Jun 09, 2011 7:43 am
divya23 wrote:if n is a +ve int and product of all d int from 1 to n inclusive is a multiple of 990 find least value of n

[spoiler]ans = 11[/spoiler]
product of all integers from 1 to n=n!;
990=11*3^2*2*5;
now largest prime no. in 990 is 11, also since 990, is a factor of n! therefore minimum value of n is 11
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by [email protected] » Fri Jun 10, 2011 10:18 pm
Guyzzz... please tell me why can't 2 or 3 be the answer for this question.
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by Anurag@Gurome » Fri Jun 10, 2011 10:37 pm
[email protected] wrote:Guyzzz... please tell me why can't 2 or 3 be the answer for this question.
Let's say n = 3
Then, the product of all the integers from 1 to n inclusive = 1*2*3 = 6
Now 6 is not a multiple of 990.

The product of all the integers from 1 to n inclusive, i.e. n! will be a multiple of 990 if and only if all the factors of 990 is contained within n!

Now, 990 = 9*10*11 = 2*(3^2)*5*11
Hence, largest prime factor of 990 is 11.
Therefore, n! must contain 11 within itself.
Now, if n = 11, n! = 11*10*9*8*...*2*1 = 990*(Something)

Hence, least possible value of n is 11.
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