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Saurabh Malpani
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Post subject: Manhattan Challenge Problem Posted: Sat Apr 28, 2007 8:15 pm |
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12/01/03
Question
How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
I understand the the solution except the last part where the answer is arrived at by Subtracting the 81 from 144 (I mean I understand what are the steps followed but I don't understand why My answer is wrong)
My Apporach was:
As known that to be multiple of 6 we need to have atleast one 2 and one 3. In my apporach I try to calculate, just as we do in Combinations,the number of factors when I have ONLY 2s and Only 3s in additon to 5 and 7
2^4 * 5^2 *7^2 --- i.e multipy 3*3*3 =27
2^4 * 5^2 *7^2------- i.e multipy 3*3*3 =27
Sum them up I get 54.
So what's wrong with my apporach..???
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Saurabh Malpani
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Post subject: Posted: Sat Apr 28, 2007 8:18 pm |
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Well as I was typing the problem the solution clicked to me the problem in my apporach is that i am not adding 7^2 * 5^2---3*3=9 . 9+ 54 =63 which is the correct answer.
Thanks
Saurabh Malpani
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StaceyKoprince
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Post subject: ManhattanGMAT Challenge Problem Dec 1 2003 Posted: Sun Apr 29, 2007 11:35 pm |
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| ManhattanGMAT Staff |
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Posts: 6077 Location: San Francisco
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Great! It's always the best when you figure it out for yourself! (And do remember - the challenge problems are likely to be harder than anything, or almost anything, you'll see on the test. Just make sure you've mastered absolutely everything else - otherwise, it's not a wise use of your time to use these extensively.)
_________________ Stacey Koprince Instructor Director of Online Community ManhattanGMAT
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atayamir
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Post subject: Re: Manhattan Challenge Problem Posted: Mon Feb 28, 2011 9:18 pm |
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Posts: 1
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jnelson0612
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Post subject: Re: Manhattan Challenge Problem Posted: Thu Mar 03, 2011 10:11 am |
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| ManhattanGMAT Staff |
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Posts: 1857
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_________________ Jamie Nelson ManhattanGMAT Instructor
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