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Re: If n=s^a*t^b, where a, b, s and t are integers, is √n an integer? [#permalink]

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29 May 2017, 02:23

Can someone please explain why the answer isn't C?

My logic is as follows:

We know that all variables are integers. If n = (s^a)(t^b), then for sqrt(n) to be an integer (s^(a/2)) and (t^(b/2)) must both be integers, which means that "a" and "b" must be divisible by 2.

Statement 1: a+b is an even number. Two scenarios are possible. Either "a" and "b" are both even or both odd. If both even then N must be an integer, if both odd then N will not be an integer. Insufficient

Statement 2: "a" is an even number. This tells us nothing about "b". Insufficient.

Combining the two statements: If "a" is even, then for a+b to be even, "b" must also be even. If "a" and "b" are both even, it means that both are divisible by 2, which means that "s^a" and "t^b" both have integer roots. Sufficient.

Can someone please check my logic and let me know if I'm not seeing something? Thanks!

==>In the original condition, there are 5 variables (n,s,t,a,b) and 1 equation (n=s^at^b). In order to match the number of variables to the number of equations, there must be 5 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), s=t=2 and a=b=2 yes, but s=t=2 and a=b=-2 no, hence it is not sufficient. Therefore, the answer is E.

Re: If n=s^a*t^b, where a, b, s and t are integers, is √n an integer? [#permalink]

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15 Jun 2017, 02:20

2

This post received KUDOS

1

This post was BOOKMARKED

n=s^at^b

Statement 1:

a+b=even it can happen only with below case

case 1: odd+odd = even case 2: even+even= even

but since we are not sure what is a and b i.e even or odd so statement 1 is insufficient

Statement 2:

a is even

we don't know about b if it is even or odd.

so statement 2 is also insufficient

combining statement+ statement 2

now we know that since a is even then b is also even for a+b to be even.

BUT even if we know the values of a and b we are not aware about the values of S and T.

a = even, b= even, s=1/4 , t=1/2 then n=s^at^b is not an integer ----- yes this case is not possible as it is stated that s and t are integers {Thanks M02men for correcting me}. a=even, b=even, s=2, t=3 then n=s^at^b is a integer

case : since a and b are even integer after combining both statement 1 and statement 2. We need to consider the negative even integers also i.e -2, -4, -6 And something raise to negative power will turn into fraction example : 2^(-1)=1/2 So in equation n=(s^a)(t^b) If a and b are negative even integers then equation will be n=s^(-a)t^(-b)------- n=1/(s^a)(t^b) so now if consider some random even values for a and b like a= -2, b=-2 , s=3, t=4 then n=3^(-2)4^(-2)-------n=1/(3^2)(4^2) now square root of n is not an integer a=2 , b=2, s=3, t=4 then n=(3^2)(4^2)- now square root of n is a integer so both combined together is also insufficient

I've done some correction in my answer to my other post in the same thread

have a look at the below cases.

case : since a and b are even integer after combining both statement 1 and statement 2. We need to consider the negative even integers also i.e -2, -4, -6 And something raise to negative power will turn into fraction example : 2^(-1)=1/2 So in equation n=(s^a)(t^b) If a and b are negative even integers then equation will be n=s^(-a)t^(-b)------- n=1/(s^a)(t^b) so now if consider some random even values for a and b like a= -2, b=-2 , s=3, t=4 then n=3^(-2)4^(-2)-------n=1/(3^2)(4^2) now square root of n is not an integer a=2 , b=2, s=3, t=4 then n=(3^2)(4^2)- now square root of n is a integer so both combined together is also insufficient

so the answer is E

Hope it is clear...

Kudo if you like the explanation...

chetan2u Please put your comments on this and correct me if i am missing something

Re: If n=s^a*t^b, where a, b, s and t are integers, is √n an integer? [#permalink]

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12 Aug 2017, 07:15

Consider the both case when a,b is positive and a,b is negative then apply the condition given in the equation . Note : a, b can be even and positive OR a,b can be even and negative . when negative , then n will not be integer .

Re: If n=s^a*t^b, where a, b, s and t are integers, is √n an integer? [#permalink]

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12 Aug 2017, 07:32

TheKingInTheNorth wrote:

Consider the both case when a,b is positive and a,b is negative then apply the condition given in the equation . Note : a, b can be even and positive OR a,b can be even and negative . when negative , then n will not be integer .

Hence the answer should be E

So in GMAT odd and even integers are not neccesarily always positive? Only prime numbers have to be positive. Correct?

Consider the both case when a,b is positive and a,b is negative then apply the condition given in the equation . Note : a, b can be even and positive OR a,b can be even and negative . when negative , then n will not be integer .

Hence the answer should be E

So in GMAT odd and even integers are not neccesarily always positive? Only prime numbers have to be positive. Correct?

GMAT does not have its own math, the above is generally true.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. So, ..., -4, -2, 0, 2, 4, ... are all even integers.

An odd number is an integer that is not evenly divisible by 2. So, ..., -3, -1, 1, 3, 5, ... are all odd integers.

A Prime number is a positive integer with exactly two distinct divisors: 1 and itself.

Re: If n=s^a*t^b, where a, b, s and t are integers, is √n an integer? [#permalink]

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12 Aug 2017, 08:20

Bunuel wrote:

SOUMYAJIT_ wrote:

TheKingInTheNorth wrote:

Consider the both case when a,b is positive and a,b is negative then apply the condition given in the equation . Note : a, b can be even and positive OR a,b can be even and negative . when negative , then n will not be integer .

Hence the answer should be E

So in GMAT odd and even integers are not neccesarily always positive? Only prime numbers have to be positive. Correct?

GMAT does not have its own math, the above is generally true.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. So, ..., -4, -2, 0, 2, 4, ... are all even integers.

An odd number is an integer that is not evenly divisible by 2. So, ..., -3, -1, 1, 3, 5, ... are all odd integers.

A Prime number is a positive integer with exactly two distinct divisors: 1 and itself.

Re: If n=s^a*t^b, where a, b, s and t are integers, is √n an integer? [#permalink]

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13 Aug 2017, 09:01

NOTE: we can also see this problem as IS n = perfect square => Thus n = (s^a)*(t^b) MUST have powers (a + b) whose sum is even and ALSO factors s and t be +ve prime numbers. Now we know that (a + b) = even such that both a and b are even, but we don’t know anything about s and t except that s and t are integers. They can be +ve or -ve or non-prime!!! Thus E