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by thephoenix » Sat Feb 13, 2010 2:20 am
(prime factors) of 264,600 = 2^3 * 3^3 * 5^2 *7^2
Numbers that will divide 264600 and that are not divisible by 6. will be the factors which contain only 2,5,7 and 3,5,7

number of factors= (3+1)(3+1)(2+1)(2+1)=144

now 2*3=6----> all factors where 2*3 is a part of factor will be divided by 6
there are 3 two's and 3 three's----->number of factors which contain 2 and 3 is 3*3=9
now when these no's r mutliplied to 5 and 7 we have to multiply 9 by (2+1)*(2+1)=9 (powers of 5 and 7 plus 1) --> 9*9=81 no's which are divisible by 6 so ---> 144-81=63. is the reqd no.

oooffff!!!!!!!

it took me 15 mins....IMO D
and on the exam day i wud have guessed after a max of 2 min

eager to c any shorter approach
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by komal » Tue Feb 16, 2010 8:29 am
gmatnmein2010 wrote:How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72

how we can do this one quickly
We start of by factorizing 264,600

=2^3 * 3^3 * 5^2 * 7^2

To create numbers from these factors we basically separate multiples of 2 & 3, since any combination of these will be divisible by 6.

Hence we find the number of factors for
2^3 * 5^2 * 7^2

and add it to the factors of

3^3 * 5^2 * 7^2

In case someone doesn't know how to calculate the number of factors of a given number - add the powers of it's prime factors by 1 and multiply them.

In our case it is (3+1)*(2+1)*(2+1) = 36
similarly for 3^3 * 5^2 * 7^2 it is (3+1)*(2+1)*(2+1) = 36

Now if we add the two numbers above we end up double counting the factors of 5^2*7^2 = (2+1)*(2+1) = 9

Hence the answer is 36+36-9 = 63.
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