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krazzy4: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Sat Apr 26, 2008 7:42 am

by krazzy4 » Sat Dec 13, 2008 9:22 pm

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Source: Beat The GMAT — Problem Solving | Sat Dec 13, 2008 9:22 pm

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

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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krazzy4: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Sat Apr 26, 2008 7:42 am

Re:

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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mrsmarthi: Senior | Next Rank: 100 Posts; Posts: 45; Joined: Sun Nov 30, 2008 8:41 pm; Thanked: 4 times

by mrsmarthi » Sun Dec 14, 2008 8:32 pm

I would go with B - ie 11.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

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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