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29 Sep 2017, 03:56
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Difficulty:
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Question Stats:
58% (02:01) correct
42%
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based on 97
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If x > 1, is positive integer b a factor of positive integer a?
(1) \(x^{(a + b)} = x^{(ab)}\)
(2) \(x^{(\frac{a}{b})} = x^{(\frac{a}{2})}\)
(1) \(x^{(a + b)} = x^{(ab)}\)
(2) \(x^{(\frac{a}{b})} = x^{(\frac{a}{2})}\)
29 Sep 2017, 04:42
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Bunuel
The question stem implies that whether \(a=b*Integer\)
Statement 1: this implies \(ab=a+b\) or \(ab-b=a\)
Hence \(a=b*(a-1)= b*Integer\) as \(a\) & \(b\) are integers, hence \(a\) is a multiple of \(b\) or \(b\) is a factor of \(a\). Sufficient
Statement 2: implies \(\frac{a}{b}=\frac{a}{2}\) or \(b =2\). But we cannot find the value of \(a\). Hence Insufficient
Option A
23 Jul 2019, 02:54
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Hi,
Lets breakdown the question stem here :
We have x > 1 and a and b to be positive integers. The question asks us whether b is a factor of a, in other words, Is a/b an integer
When given questions where we need to prove some term to be an integer, it always makes sense to manipulate the statements algebraically rather than plug in values.
Statement 1 : x^(a+b) = x^(ab)
a + b = ab
Dividing by b we get
(a/b) + 1 = a -----> (a/b) = a - 1
Since a is an integer, even a - 1 is also an integer, so a/b is also an integer. Sufficient.
Statement 2 : x^(a/b) = x^(a/2)
(a/b) = a/2
Cancelling a on both sides we get (if the sign of a was not given, then we cannot cancel a on both sides since a could also take a value of 0. This is a common trap that the GMAT will lay for you, for example if ab = bc, we should never cancel b on both sides, but instead take things to the LHS and factorize i.e. ab - bc = 0 ----> b(a - c) = 0 ----> b = 0 or a = c)
1/b = 1/2 ----> b = 2. We have no information about a, insufficient.
Answer : A
Lets breakdown the question stem here :
We have x > 1 and a and b to be positive integers. The question asks us whether b is a factor of a, in other words, Is a/b an integer
When given questions where we need to prove some term to be an integer, it always makes sense to manipulate the statements algebraically rather than plug in values.
Statement 1 : x^(a+b) = x^(ab)
a + b = ab
Dividing by b we get
(a/b) + 1 = a -----> (a/b) = a - 1
Since a is an integer, even a - 1 is also an integer, so a/b is also an integer. Sufficient.
Statement 2 : x^(a/b) = x^(a/2)
(a/b) = a/2
Cancelling a on both sides we get (if the sign of a was not given, then we cannot cancel a on both sides since a could also take a value of 0. This is a common trap that the GMAT will lay for you, for example if ab = bc, we should never cancel b on both sides, but instead take things to the LHS and factorize i.e. ab - bc = 0 ----> b(a - c) = 0 ----> b = 0 or a = c)
1/b = 1/2 ----> b = 2. We have no information about a, insufficient.
Answer : A
