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Prime number factors

Expert replies
by vinayreguri » Sun Jun 12, 2011 5:39 am
If x,y,z are prime numbers, which if the following integers have the same number of factors
I. xy
II. xyz
III. (x^2)y or x square * y

A. I & II
B. II & III
C. I & III
D. I,II & III
E. Cannot determine

Last edited by vinayreguri on Sun Jun 12, 2011 6:26 am, edited 1 time in total.
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Source: — Problem Solving |

by Frankenstein » Sun Jun 12, 2011 5:45 am
Hi,
Number of factors of xy is(1+1)*(1+1) = 4
Number of factors of xyz is(1+1)*(1+1)*(1+1) = 8
Number of factors of x^2y is(2+1)*(1+1) = 6

Can you recheck the question. I don't find answer in the options. Is the third one x^(2y) or (x^2).y?
Last edited by Frankenstein on Sun Jun 12, 2011 5:50 am, edited 2 times in total.
Cheers!

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by Anurag@Gurome » Sun Jun 12, 2011 5:48 am
vinayreguri wrote:If x,y,z are prime numbers, which if the following integers have the same number of factors
I. xy
II. xyz
III. x^2y
Note that we cannot definitely determine the number of factors of x^2y, as we don't know the value of y. Factors of x^2y are : 1, x, x^2, ..., x^(y - 1), x^y, x^(y + 1), ...., x^2y

If, y = 2 --> Number of factors of x^2y is 3
If, y = 3 --> Number of factors of x^2y is 4 etc

Hence, we cannot definitely say which of them has same number of factors.

The correct answer is E.
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by Brent@GMATPrepNow » Sun Jun 12, 2011 7:02 am
vinayreguri wrote:If x,y,z are prime numbers, which if the following integers have the same number of factors
I. xy
II. xyz
III. (x^2)y or x square * y

A. I & II
B. II & III
C. I & III
D. I,II & III
E. Cannot determine

We can apply a nice rule regarding the number of divisors/factors number has:

If N = (p^a)(q^b)(r^c)..., where p, q, r (etc.) are prime numbers, then the total number of positive divisors of N is equal to (a+1)(b+1)(c+1)

Example: Since 400 = (2^4)(5^2), the total number of positive divisors of 400 equals (4+1)(2+1)=15

Aside: this rule works well for this question, since we are told that x, y and z are prime numbers

Now let's examine our answer choices:

Note: We are not explicitly told that x, y and z are different prime numbers, but we'll that the answer is E either way. Let's begin by saying the numbers are all different.

I. Since xy = (x^1)(y^1), the number of factors is (1+1)(1+1)=4
II. Since xyz = (x^1)(y^1)(z^1), the number of factors is (1+1)(1+1)(1+1)=8
III. Since (x^2)y = (x^2)(y^1), the number of factors is (2+1)(1+1)=6

So it looks like no values share the same number of factors, which means the answer is E by process of elimination.

Now what if x, y and z are NOT different?

For example, if x=y=z=2, then xyz=(x^2)y=8, in which case they would have the same number of factors.

So if looks like II and III could have the same number of factors, but the question does not say "could," which means I'll still with my answer of E
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by cans » Tue Jun 14, 2011 4:00 am
x,y,z prime numbers.
xy has 4 factors: 1,x,y,xy
xyz has 8 factors: 1,x,y,z,xy,yz,xz,xyz
(x^2)y 6 has 1,x,x^2,y,xy,(x^2)y
IMO E
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