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X, Y, and Z are three different Prime numbers, the product XYZ is divisible by how many different positive numbers?

4 6 8 9 12

Please describe method.

MUST KNOW FOR GMAT:

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.

BACK TO THE ORIGINAL QUESTION:

\(n=xyz\) (n=x^1y^1z^1) where \(x\), \(y\), and \(z\) are different prime factors will have \((1+1)(1+1)(1+1)=8\) different positive factors including 1 and xyz itself.

Thanks Bunuel very helpful, I thought I could do a quick calculation here to find out the number of factors but fell for the trap I guess:

2*3*5=30 so 30 is divisible by 1,30,15,6,5,3,2,10 but in my haste forgot to use 10,and 1 in here. One should just stick to the formula for a sure shot.

The approach specified by Bunuel is an efficient, a quick and a standard one. Use the same. Ignore the one specified by me as it is a bit time consuming when compared to that given by Bunuel.

Re: X, Y, and Z are three different Prime numbers, the product [#permalink]
21 Jan 2014, 22:03

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Re: X, Y, and Z are three different Prime numbers, the product [#permalink]
24 Apr 2015, 03:50

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