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Multiples

by relaxin99 » Wed Feb 04, 2009 10:36 pm
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

Explanation as well, thank you. will post answer soon
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by DanaJ » Thu Feb 05, 2009 12:24 am
Say P = product of numbers 1 to n. First thing you gotta do is break 990 down into factors: 990 = 99 * 10 = 2*(3^2)*5*11. Now since we have that annoying 11 over there (prime number), it can only mean that n is greater than or equal to 11. So the smallest value of P will be 1*2*...*11, with n = 11. Answer B.
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by raunekk » Thu Feb 05, 2009 1:52 am
gr8 explanation Danaj
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by blitz28 » Thu Feb 05, 2009 12:06 pm
I do not know whether I am reading the question right or not.

The Question says Product of all numbers from 1 to N is a mutliple of 990.

However if you take 11 as the answer

1*2*3*4*5*6*&*8*10*11= 39916800..... which is not a multiple of 990, rather the other way is right.

What Am i missing???
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by Hopper39 » Thu Feb 05, 2009 1:34 pm
However if you take 11 as the answer

1*2*3*4*5*6*&*8*10*11= 39916800..... which is not a multiple of 990, rather the other way is right.

What Am i missing???
Sure it is. 990 * 40320 = 11! , i.e. 990n = 11! where n=40320.
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