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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.

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..
_________________

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..

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?

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?
_________________

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.

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.