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Updated on: 26 Feb 2019, 06:07
Originally posted by GMATinsight on 26 Feb 2019, 03:54.
Last edited by GMATinsight on 26 Feb 2019, 06:07, edited 1 time in total.
Last edited by GMATinsight on 26 Feb 2019, 06:07, edited 1 time in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
14% (02:15) correct
86%
(01:42)
wrong
based on 66
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Not Attempted Yet
Is integer x square of an integer?
1) Sum of all the factors of x is odd
2) \(x = a^p*b^q\) where a and b are prime numbers and, p and q are odd integers
Source: https://www.GMATinsight.com
1) Sum of all the factors of x is odd
2) \(x = a^p*b^q\) where a and b are prime numbers and, p and q are odd integers
Source: https://www.GMATinsight.com
26 Feb 2019, 05:34
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Is integer x square of an integer?
1) Sum of all the factors of x is odd
Just this knowledge is sufficient to eliminate this choice
Logically if only prime factor is 2, the factors will always be odd..
If x is 2, factors are 1 and 2, and sum = 1+2=3. 2 is not a square
If x is 4, factors are 1, 2 and 4, and sum = 1+2+4=7. 4 is a square.
Insuff.
2) \(x = a^p*b^q\) where a and b are prime numbers and, p and q are odd integers
a and b are prime numbers, and their power p and q are odd,
If a=b, answer is yes, otherwise No
Not sufficient
Combined..
a=b=2, and p=q=3......YES
I can't get any other value which can fit in both so C
1) Sum of all the factors of x is odd
Just this knowledge is sufficient to eliminate this choice
Logically if only prime factor is 2, the factors will always be odd..
If x is 2, factors are 1 and 2, and sum = 1+2=3. 2 is not a square
If x is 4, factors are 1, 2 and 4, and sum = 1+2+4=7. 4 is a square.
Insuff.
2) \(x = a^p*b^q\) where a and b are prime numbers and, p and q are odd integers
a and b are prime numbers, and their power p and q are odd,
If a=b, answer is yes, otherwise No
Not sufficient
Combined..
a=b=2, and p=q=3......YES
I can't get any other value which can fit in both so C
26 Feb 2019, 05:45
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chetan2u
Second statement doesn't mention that the Prime numbers are distinct... so \(x =2^3*2^5\) is also acceptable making the number as perfect square.
. and I am sure, you too got tricked by your question by realizing there was more to what you thought the question actually meant. 
