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

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Manager
Joined: 13 Jul 2010
Posts: 169
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Kudos [?]: 48 [2] , given: 7

X, Y, and Z are three different Prime numbers, the product [#permalink]

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09 Dec 2010, 22:03
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Difficulty:

25% (medium)

Question Stats:

66% (01:46) correct 34% (00:38) wrong based on 186 sessions

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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

[Reveal] Spoiler: OA
Manager
Joined: 30 Aug 2010
Posts: 91
Location: Bangalore, India
Followers: 5

Kudos [?]: 134 [0], given: 27

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09 Dec 2010, 23:08
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

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

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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Joined: 02 Sep 2009
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Kudos [?]: 68351 [3] , given: 9797

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10 Dec 2010, 00:46
3
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Expert's post
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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

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.

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

Hope it helps.
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Senior Manager
Status: Bring the Rain
Joined: 17 Aug 2010
Posts: 406
Location: United States (MD)
Concentration: Strategy, Marketing
Schools: Michigan (Ross) - Class of 2014
GMAT 1: 730 Q49 V39
GPA: 3.13
WE: Corporate Finance (Aerospace and Defense)
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Kudos [?]: 43 [0], given: 46

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10 Dec 2010, 07:52
Thanks for the equation
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Manager
Joined: 13 Jul 2010
Posts: 169
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Kudos [?]: 48 [0], given: 7

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10 Dec 2010, 12:55
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.
Manager
Joined: 30 Aug 2010
Posts: 91
Location: Bangalore, India
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Kudos [?]: 134 [0], given: 27

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11 Dec 2010, 00:14
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,
Murali.
Manager
Joined: 25 May 2011
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Kudos [?]: 53 [1] , given: 71

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20 Oct 2011, 07:16
1
KUDOS

1- x- y- z- xy- xz- yz- xyz
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GMAT Date: 01-06-2012
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Kudos [?]: 6 [1] , given: 8

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20 Oct 2011, 11:07
1
KUDOS
Ans 8

1
x
y
z
xy
yz
zx
xyz
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Kudos [?]: 115 [0], given: 0

Re: X, Y, and Z are three different Prime numbers, the product [#permalink]

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21 Jan 2014, 23:03
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Re: X, Y, and Z are three different Prime numbers, the product [#permalink]

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21 Jan 2014, 23:33
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

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]

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24 Apr 2015, 04:50
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: X, Y, and Z are three different Prime numbers, the product   [#permalink] 24 Apr 2015, 04:50
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