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If n=s^a*t^b, where a, b, s and t are integers, is √n an integer? [#permalink]
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Updated on: 29 May 2017, 10:33
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If \(n=s^a t^b\), where \(a\), \(b\), \(s\) and \(t\) are integers, is \(\sqrt{n}\) an integer? 1) \(a+b\) is an even number 2) \(a\) is an even number
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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, 00:22
Hi
I think there is some typo. Do you refer to 'r' or to 't'?



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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, 03:23
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!



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If n=s^a*t^b, where a, b, s and t are integers, is √n an integer? [#permalink]
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31 May 2017, 00:53
==>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. Answer: E
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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, 00:56
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 a=even, b=even, s=2, t=3 then n=s^at^b is a integer so the answer is E Hope it is clear... Kudo if you like the explanation...



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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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15 Jun 2017, 02:30
lovenish wrote: 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 a=even, b=even, s=2, t=3 then n=s^at^b is a integer so the answer is E Hope it is clear... Kudo if you like the explanation... The highlighted is not valid as prompt stats s & t are Integers.



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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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15 Jun 2017, 03:20
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
so the answer is E
Hope it is clear...
Kudo if you like the explanation...



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If n=s^a*t^b, where a, b, s and t are integers, is √n an integer? [#permalink]
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17 Jul 2017, 19:25
Pay attention to SPECIAL CASE in which a or b is NEGATIVE integer!!! Some SPECIAL CASES: x = y x = 0, +1, 1
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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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02 Aug 2017, 05:50
lovenish wrote: 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 a=even, b=even, s=2, t=3 then n=s^at^b is a integer so the answer is E Hope it is clear... Kudo if you like the explanation... The question clearly mentions that a,b,s,t are INTEGERS. The answer should be C. chetan2u please help a bit here.
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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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02 Aug 2017, 09:11
rekhabishop wrote: lovenish wrote: 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 a=even, b=even, s=2, t=3 then n=s^at^b is a integer so the answer is E Hope it is clear... Kudo if you like the explanation... The question clearly mentions that a,b,s,t are INTEGERS. The answer should be C. chetan2u please help a bit here. Hi Rekhabishop 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



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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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12 Aug 2017, 08:15
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 . Hence the answer should be E
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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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12 Aug 2017, 08:32
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?



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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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12 Aug 2017, 09:14
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. For more check here: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative MegathreadHope it helps.
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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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12 Aug 2017, 09:20
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. For more check here: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative MegathreadHope it helps. Yes, thanks , makes sense ! Sent from my iPhone using GMAT Club Forum mobile app



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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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13 Aug 2017, 10:01
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 nonprime!!! Thus E




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