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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.
my answer is 96, but the real answer is 192.
Please help.
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14 Oct 2015, 11:19
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Sujan Sareen
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 \((3 + 1)(7 + 1)(2 + 1) = 96\).
23 Oct 2015, 13:04
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We have a free video that explains the rule that Bunuel used - https://www.gmatprepnow.com/module/gmat- ... /video/828
Cheers,
Brent
Cheers,
Brent
