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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: 167
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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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

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09 Dec 2010, 23:08
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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

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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10 Dec 2010, 00:46
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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

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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10 Dec 2010, 07:52
Thanks for the equation
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Manager
Joined: 13 Jul 2010
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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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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.
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Joined: 25 May 2011
Posts: 152

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20 Oct 2011, 07:16
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1- x- y- z- xy- xz- yz- xyz
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20 Oct 2011, 11:07
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1
x
y
z
xy
yz
zx
xyz
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Re: X, Y, and Z are three different Prime numbers, the product [#permalink]

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

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15 May 2016, 09:45
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Re: X, Y, and Z are three different Prime numbers, the product [#permalink]

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16 Jan 2017, 00:03
Since X,Y,Z are different primes=>
Number of factors of X*Y*Z => 2*2*2=> 8
Hence C.

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

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20 Jan 2017, 01:51
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

Suppose
X = 2,
Y = 3 and
Z = 5.

Then XYZ = 30 - it is divisible by

1, 2, 3, 5, 6, 10, 15, 30 -
so ther are 8 different numbers

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

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03 Apr 2017, 11:01
Given:x,y,z are prime factors that means they have no factors other than 1 and the no. itself.

Therefore the power of these prime nos. is 1.

The total no. of factors (thus) = 2*2*2=8
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Re: X, Y, and Z are three different Prime numbers, the product [#permalink]

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06 Apr 2017, 10:03
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

To determine the number of factors of XYZ, or (X^1)(Y^1)(Z^1), we add 1 to each exponent of each unique prime factor and multiply those values together. The result will equal the number of factors of the given number.

Thus, XYZ has (1 + 1)(1 + 1)(1 + 1) = 2 x 2 x 2 = 8 factors.

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

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07 Apr 2017, 13:53
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 p=number of prime numbers=3
total positive factors=2(p+1)=8
C
Re: X, Y, and Z are three different Prime numbers, the product   [#permalink] 07 Apr 2017, 13:53
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