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If N = (x^a)*(y^b), where x and y are prime numbers, and a

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pmenon
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If N = (x^a)*(y^b), where x and y are prime numbers, and a [#permalink] New post   11 Jan 2009, 00:11
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If N = (x^a)*(y^b), where x and y are prime numbers, and a and b are positive intergers. Is N^1/2 an integer

1. a+b is even
2. a*b is even

im disputing the OA for this one; got it off another forum.



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Re: If N = (x^a)*(y^b), where x and y are prime numbers, and a [#permalink] New post   11 Jan 2009, 00:57
pmenon wrote:
If N = (x^a)*(y^b), where x and y are prime numbers, and a and b are positive intergers. Is N^1/2 an integer

1. a+b is even
2. a*b is even

im disputing the OA for this one; got it off another forum.


C makes sense.

1: a and b both are either odd or even. insuff...

2: if a is odd, b is even and vice versa.
if a is even, b is also even. insufff...

togather, from 2 at least either a or b is even. from 1 both are either even or odd. suff...
both have to be even..........
if a and b are even and they are +ve, they are at leat 2 or its multiple. so sqrtn is an integer.
C..
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chicagocubsrule
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Re: If N = (x^a)*(y^b), where x and y are prime numbers, and a [#permalink] New post   11 Jan 2009, 01:37
GMAT TIGER wrote:
pmenon wrote:
If N = (x^a)*(y^b), where x and y are prime numbers, and a and b are positive intergers. Is N^1/2 an integer

1. a+b is even
2. a*b is even

im disputing the OA for this one; got it off another forum.


C makes sense.

1: a and b both are either odd or even. insuff...

2: if a is odd, b is even and vice versa.
if a is even, b is also even. insufff...

togather, from 2 at least either a or b is even. from 1 both are either even or odd. suff...
both have to be even..........
if a and b are even and they are +ve, they are at leat 2 or its multiple. so sqrtn is an integer.
C..


Precisely! Thanks, GMAT Tiger.
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Re: If N = (x^a)*(y^b), where x and y are prime numbers, and a [#permalink] New post   11 Jan 2009, 08:44
GMAT TIGER wrote:
1: a and b both are either odd or even. insuff...

2: if a is odd, b is even and vice versa.
if a is even, b is also even. insufff...

togather, from 2 at least either a or b is even. from 1 both are either even or odd. suff...
both have to be even..........


Could you pls elaborate how you came to the conclusion that both a and b are even.
from 2 either a or b is even, in the same way either a or b is odd. From 1 both a and b are either even or odd. How do you come to the conclusion that they both have to be even? It's 50/50... What am I missing?
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Re: If N = (x^a)*(y^b), where x and y are prime numbers, and a [#permalink] New post   11 Jan 2009, 09:04
linau1982 wrote:
GMAT TIGER wrote:
1: a and b both are either odd or even. insuff...

2: if a is odd, b is even and vice versa.
if a is even, b is also even. insufff...

togather, from 2 at least either a or b is even. from 1 both are either even or odd. suff...
both have to be even..........


Could you pls elaborate how you came to the conclusion that both a and b are even.
from 2 either a or b is even, in the same way either a or b is odd. From 1 both a and b are either even or odd. How do you come to the conclusion that they both have to be even? It's 50/50... What am I missing?


From 2:
Suppose: even = 2 (or 4, 6, or so on............) and b = 1 (or 3, 5, or so on)

i. a = 2 and b = 2
ii. a = 2 and b = 1
iii. a = 1 and b = 2
but a = 1 and b = 1 are not possible.

From 1:
i. a = 2 and b = 2
ii: a = 1 and b = 1

Togather: Which options are in both 1 and 2: i. a = 2 and b = 2

Therefore it is C.
Did I clearify your doubts? :wink:
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Re: If N = (x^a)*(y^b), where x and y are prime numbers, and a [#permalink] New post   11 Jan 2009, 09:49
so here is my question: from statements 1 and 2, i think the only option left is for both a and b to be even, right ?

if so, then whats to stop x=y=2 and a=b=2 , in which case you have a YES answer. Or, you could have x=y=2 and a=2, b=3, in which case you have a NO answer.

Thats why I answered E. Where am i mistaken ?
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Re: If N = (x^a)*(y^b), where x and y are prime numbers, and a [#permalink] New post   11 Jan 2009, 11:00
pmenon wrote:
so here is my question: from statements 1 and 2, i think the only option left is for both a and b to be even, right ?

if so, then whats to stop x=y=2 and a=b=2 , in which case you have a YES answer. Or, you could have x=y=2 and a=2, b=3, in which case you have a NO answer.

Thats why I answered E. Where am i mistaken ?


The highlighted part is incorrect. if so, then it violets statement 1 that (a+b) = even because 2+3 = 5.

What you are missing is either one of a or b must be even, in 2, to have a*b even.
If one of a or b is even in 2, then both (a and b) must be even in 1 because in 1 both should be either even or odd. since from 2 either one must be even, ab must be even in 1.

Therefore it is C not E.
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Re: If N = (x^a)*(y^b), where x and y are prime numbers, and a [#permalink] New post   11 Jan 2009, 12:17
pmenon wrote:
If N = (x^a)*(y^b), where x and y are prime numbers, and a and b are positive intergers. Is N^1/2 an integer

1. a+b is even
2. a*b is even

im disputing the OA for this one; got it off another forum.


Statement 1 tells us either:
1) A and B are both odd
2) A and B are bot even
In order for a+b to be even they must both be either even or odd

Statement 2 tells us either:
1) A and B are both even
2) A OR B is even
In order for a*b to be even at least one of these variables needs to be even

Combining the 2 statements leads to answer C



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Re: If N = (x^a)*(y^b), where x and y are prime numbers, and a [#permalink]
11 Jan 2009, 12:17
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