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by Baldini » Fri Mar 06, 2009 8:19 am
If n is a positive integer and the product of all the 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
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by Mr2Bits » Fri Mar 06, 2009 8:24 am
11. Least prime number that is a multiple of 990.

Whats the OA?
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by Baldini » Fri Mar 06, 2009 9:03 am
OA is B. Can you explain how you got the answer?
thanks
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by bmlaud » Fri Mar 06, 2009 9:08 am
990 has 4 prime factors 2,3,5,11
the minimum no. that n can take should be 11.
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by scoobydooby » Fri Mar 06, 2009 10:22 am
n*(n-1)*(n-2)*.......1 =n!
=>n!=990k (k an integer)
=>n!=9*2*5*11 (for the smallest n, lets take k=1)

if n=10, 10!=10*9*8..1 (we cant get 990, which has a 11)
so the least value of n could be 11, so that 11! would give the 11 we need to make a 990 atleast.

n=11
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by Mr2Bits » Fri Mar 06, 2009 10:23 am
Baldini wrote:OA is B. Can you explain how you got the answer?
thanks
When you see the following : product of all the integers from 1 to n


Think PRIME numbers. Of the numbers listed, 11 is the smallest prime number that is a multiple of 990
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