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Number of factors

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Number of factors  [#permalink]

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New post 14 Oct 2015, 06:29
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Number of factors of A = 2^3 * 5^7 * 7^2

my answer is 96, but the real answer is 192.
Please help.
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Re: Number of factors  [#permalink]

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New post 14 Oct 2015, 11:19
Sujan Sareen wrote:
Number of factors of A = 2^3 * 5^7 * 7^2

my answer is 96, but the real answer is 192.
Please help.


Finding the Number of Factors of an Integer:

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

Check for more here: math-number-theory-88376.html

So, the number of factors of \(2^3 * 5^7 * 7^2\) is \((3 + 1)(7 + 1)(2 + 1) = 96\).
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Re: Number of factors  [#permalink]

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New post 23 Oct 2015, 13:04
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We have a free video that explains the rule that Bunuel used - http://www.gmatprepnow.com/module/gmat- ... /video/828

Cheers,
Brent
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Re: Number of factors  [#permalink]

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New post 29 Oct 2015, 17:52
I'm curious why you say that 192 is the real answer. Do you have a copy of the full question? (If that's all there is, then it's simply an error. Bunuel, of course, is correct.)
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Re: Number of factors  [#permalink]

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New post 29 Jan 2018, 00:51
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Re: Number of factors   [#permalink] 29 Jan 2018, 00:51
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