multiples Q.

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godemol: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sat Jun 20, 2009 6:06 pm; Thanked: 1 times; GMAT Score:720

by godemol » Mon Aug 10, 2009 2:07 pm

The product of 3 consecutive positive multiples of 4 must be divisible by each of the following EXCEPT

a.16
b.36
c.48
d.128
e.192

Does anybody know a fast way of solving this question? Thanks.

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Source: Beat The GMAT — Data Sufficiency | Mon Aug 10, 2009 2:07 pm

tohellandback: Legendary Member; Posts: 752; Joined: Sun May 17, 2009 11:04 pm; Location: Tokyo; Thanked: 81 times; GMAT Score:680

by tohellandback » Mon Aug 10, 2009 7:04 pm

IMO B

since there are three consecutive positive multiples of 4

4K*4(k+1)*4(k+2)
A- out
C-48=16*3, we already have 4*4=16. and of any three consecutive integers, one must be a multiple of 3. -out

D-128=4*4*4*2. we already have 4*4*4. and of three consecutive integers at least one must be a multiple of 2. --out

E. 192= 4*4*4*3. same reasoning as C- out

B-36=4*3*3. we may or may not get two 3's
answer B

The powers of two are bloody impolite!!

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