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X, Y, and Z are three different Prime numbers, the product

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X, Y, and Z are three different Prime numbers, the product [#permalink] New post 09 Dec 2010, 21:03
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X, Y, and Z are three different Prime numbers, the product XYZ is divisible by how many different positive numbers?

A. 4
B. 6
C. 8
D. 9
E. 12

Please describe method.
[Reveal] Spoiler: OA
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Re: xyz prime [#permalink] New post 09 Dec 2010, 22:08
gettinit wrote:
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.


PRIME # is the # that has only 2 factors: One is 1 and another is the # itself.

infer that

FOR any given # 1 and the # iteself are the definite factors.

Knowing above conepts:

product of X,Y, and Z = XYZ

divisible by 1, xyz, x, y, z, xy,yz,xz ==> total 8

ANSWER "C"

ADDITIONAL INFO.

if question has 5 constants a,b,c,d,e, we do not have to count in the above way

Basically we are selecting, from the product, one constant, set of two constants, set of 3 constants ....and so on set of all the # of constants.

so if 5 varibales are given, total # ways to select is 5C1+5C2+5C3+5C4+5C5 = 5+10+10+5+1 = 31

And answer will be 31+1 =32 (as "1" is a factor for every #)

Regards,
Murali.

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Re: xyz prime [#permalink] New post 09 Dec 2010, 23:46
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gettinit wrote:
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.

Answer: C.

For more on number properties check: math-number-theory-88376.html

Hope it helps.
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Re: xyz prime [#permalink] New post 10 Dec 2010, 06:52
Thanks for the equation
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Re: xyz prime [#permalink] New post 10 Dec 2010, 11:55
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.
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Re: xyz prime [#permalink] New post 10 Dec 2010, 23:14
Friends,

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.

Regards,
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Re: xyz prime [#permalink] New post 20 Oct 2011, 06:16
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made a stupid mistake!

Yes the answer is C

1- x- y- z- xy- xz- yz- xyz
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Re: xyz prime [#permalink] New post 20 Oct 2011, 10:07
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Re: X, Y, and Z are three different Prime numbers, the product [#permalink] New post 21 Jan 2014, 22:03
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Re: X, Y, and Z are three different Prime numbers, the product [#permalink] New post 21 Jan 2014, 22:33
gettinit wrote:
X, Y, and Z are three different Prime numbers, the product XYZ is divisible by how many different positive numbers?

A. 4
B. 6
C. 8
D. 9
E. 12

Please describe method.


Let us say X = 2, Y = 3 and Z = 5. Then XYZ = 30 - it is divisible by

1, 2, 3, 5, 6, 10, 15, 30 - 8 different numbers
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Re: X, Y, and Z are three different Prime numbers, the product   [#permalink] 21 Jan 2014, 22:33
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