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Divisibility

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by Anurag@Gurome » Mon May 21, 2012 3:21 am
GmatKiss 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
The number as the product of primes as a^x * b^y * c^z, where a, b, and c are prime factors and x, y, and z are their exponents.

The number of factors the number contains is expressed by the formula (x + 1)(y + 1)(z + 1).

Now 264,600 = 2^3 * 3^3 * 5² * 7²

Number of factors = (3 + 1)(3 + 1)(2 + 1)(2 + 1) = 144. So our number contains 144 distinct factors.

Number of factors, which contain 2 and 3 is 3 * 3 = 9 (2 * 3, 2² * 3, 2^3 * 3, 2 * 3^2, 2^2 * 3², 2^3 * 3², 2 * 3^3, 2^2 * 3^3, 2^3 * 3^3, which are 9)
Powers of 5 and 7 are 2 each, so number of factors that contain 5 and 7 = (2 + 1) * (2 + 1) = 9
So, 9 * 9 = 81
Required number of numbers = 144 - 81 = 63.

The correct answer is D.
Last edited by Anurag@Gurome on Mon May 21, 2012 4:18 am, edited 1 time in total.
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by GmatKiss » Mon May 21, 2012 3:45 am
Anurag@Gurome wrote:
GmatKiss 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
264,600 = 2^3 * 3^3 * 5² * 7^2

Number of factors = (3 + 1)(3 + 1)(2 + 1)(2 + 1) = 144. So our number contains 144 distinct factors. Number of factors, which contain 2 and 3 is 3 * 3 = 9 (2 * 3, 2² * 3, 2^3 * 3, 2 * 3^2, 2^2 * 3², 2^3 * 3², 2 * 3^3, 2^2 * 3^3, 2^3 * 3^3, which are 9) multiplied by (2 + 1) * (2 + 1) = 9 (powers of 5 and 7 plus 1) implies 9 * 9 = 81
Required number of numbers = 144 - 81 = 63.

The correct answer is D.
Could you pls explain the blue part.
I am unable to follow how you got to second 9.

Thanks,
GK
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by kullayappayenugula » Mon May 21, 2012 3:48 am
hi Anurag,,

My approach is as follows:

First prime factorization of 264600 gives 2^3*3^3*5^2*7^2.

Now as we have to see that the numbers dividing(i.e. factors) 264600 should not contain 6

we ensure that the 2 and 3 don't appear in the factors.

Now the factors that don't have 6 can we as below

2^3*5^2*7^2 => no.of factors that can be formed are (3+1)(2+1)(2+1) = 36

similary 3^3*5^2*7^2 => no.of factors that can be formed are (3+1)(2+1)(2+1) = 36

=> total facotrs is 72.

Can you please let me know what other multiples I am missing to exclude?
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by Anurag@Gurome » Mon May 21, 2012 4:09 am
kullayappayenugula wrote:hi Anurag,,

My approach is as follows:

First prime factorization of 264600 gives 2^3*3^3*5^2*7^2.

Now as we have to see that the numbers dividing(i.e. factors) 264600 should not contain 6

we ensure that the 2 and 3 don't appear in the factors.

Now the factors that don't have 6 can we as below

2^3*5^2*7^2 => no.of factors that can be formed are (3+1)(2+1)(2+1) = 36

similary 3^3*5^2*7^2 => no.of factors that can be formed are (3+1)(2+1)(2+1) = 36

=> total facotrs is 72.

Can you please let me know what other multiples I am missing to exclude?
Till here it is correct. But there are some repetitions here. Like factors which are not divisible by 6 are counted twice. Both 36's count the factors which have only 5's and/or 7's.
5 * 7 = 35, 5 * 7² = 245, 5² * 7 = 175, 5 * 7^0 = 5, ..., which means factors of 5² * 7² are counted twice.
Now 5² * 7² have (2 + 1) * (2 + 1) = 9 factors.
So, this 9 should be subtracted from 72, which means 72 - 9 = 63.

Hope it helps.
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by Anurag@Gurome » Mon May 21, 2012 4:19 am
GmatKiss wrote: Could you pls explain the blue part.
I am unable to follow how you got to second 9.

Thanks,
GK
I have edited and explained better in my previous reply. Hope that helps.
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by bubbliiiiiiii » Tue May 22, 2012 12:16 am
Number of factors, which contain 2 and 3 is 3 * 3 = 9 (2 * 3, 2² * 3, 2^3 * 3, 2 * 3^2, 2^2 * 3², 2^3 * 3², 2 * 3^3, 2^2 * 3^3, 2^3 * 3^3, which are 9)
Powers of 5 and 7 are 2 each, so number of factors that contain 5 and 7 = (2 + 1) * (2 + 1) = 9
So, 9 * 9 = 81
Hi Sir,

Considering the part in Blue, shouldn't the part in red be (3+1)*(3+1)?

Please let me know what am I missing?
Regards,

Pranay
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