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I am having difficulty with questions concerning multiples

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Manager
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I am having difficulty with questions concerning multiples [#permalink] New post 17 Jun 2007, 17:40
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I am having difficulty with questions concerning multiples and factors. Here is a question I got today:

If n is a positive integer and the product of all 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

I set up the equation n!/990, but I don't know how to solve. Can anyone explain? Thanks!
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 [#permalink] New post 17 Jun 2007, 17:58
Answer 11

Break down 990 in to prime factors 3x3x2x5x11 => 11 is the min number required to form a multiple of 990
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 [#permalink] New post 17 Jun 2007, 18:23
n! = 990x (where x is an integer)
= 11*90 = 30*3*11 = 3*2*15*11*x
= 1*2*3*3*5*11*x
= 1*3*5*6*11*x

The rest of the factors, 2,4,7,8,9,10 should be in x. Minimum value of n = 11
  [#permalink] 17 Jun 2007, 18:23
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I am having difficulty with questions concerning multiples

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