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Source: — Problem Solving |

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by Brent@GMATPrepNow » Sat Dec 13, 2008 9:32 pm
Given: xy + z = x(y+z)
Expand --> xy + z = xy + xz
Subract xy from both sides --> z = xz
Set equal to zero --> xz - z = 0
Factor --> z(x-1) = 0
z=0 or z=1
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by krazzy4 » Sun Dec 14, 2008 12:02 am
If n is a positive integer and the product of all the integers from 1to 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 mrsmarthi » Sun Dec 14, 2008 8:32 pm
I would go with B - ie 11.
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by Brent@GMATPrepNow » Sun Dec 14, 2008 9:00 pm
It helps to write out the prime factorization of 990 --> 990 = 2x3x3x5x11

So, if a number is to be a multiple of 990, that number must have in its prime factorization a 2, two 3's, a 5 and an 11

To get an 11 in the prime factorization n must go as high as 11

So, the answe is B
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