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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:
25%
(medium)
Question Stats:
75% (01:25) correct
25%
(01:20)
wrong
based on 1250
sessions
History
Date
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Not Attempted Yet
What is the greatest common factor of positive integers a and b?
(1) a = b + 4
(2) b/4 is an integer
(1) a = b + 4
(2) b/4 is an integer
As OA is not given, I got C as an answer. Please let me know if its correct.
Statement 1 --> Clearly insufficient as we a & b can have several values.
Statement 2 --> b is a multiple of 4. i.e. 4, 8 , 12..... but as we don't know what a is, this statement is insufficient.
Combing the 2 -> Statement 1 will give values of a as 8, 12, 16 and from statement 2 we will have several values of b.
So considering a as 12 and b as 16
12 => 2^2 * 3
16 ==> 2^4
GCF = 2^2 = 4
Another value of a as 8 and b as 12
a = 2^3
b = 2^2 * 3
GCF = 2^2 = 4.
Therefore, both statement together (c) are sufficient to answer this question.
Statement 1 --> Clearly insufficient as we a & b can have several values.
Statement 2 --> b is a multiple of 4. i.e. 4, 8 , 12..... but as we don't know what a is, this statement is insufficient.
Combing the 2 -> Statement 1 will give values of a as 8, 12, 16 and from statement 2 we will have several values of b.
So considering a as 12 and b as 16
12 => 2^2 * 3
16 ==> 2^4
GCF = 2^2 = 4
Another value of a as 8 and b as 12
a = 2^3
b = 2^2 * 3
GCF = 2^2 = 4.
Therefore, both statement together (c) are sufficient to answer this question.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
27 Jan 2012, 16:50
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What is the greatest common factor of positive integers a and b?
(1) a = b + 4. Clearly insufficient.
(2) b/4 is an integer --> \(b=4k\) --> \(b\) is a multiple of 4, though still insufficient as no info about a.
(1)+(2) \(a=b+4=4k+4=4(k+1)\) --> useful property:
if \(a\) and \(b\) are multiples of \(k\) and are \(k\) units apart from each other then \(k\) is greatest common divisor of \(a\) and \(b\). For example if \(a\) and \(b\) are multiples of 7 and \(a=b+7\) then 7 is GCD of \(a\) and \(b\).
So, as we have that both \(a\) and \(b\) are multiples of 4 and are 4 units apart each other (\(a=b+4\) ), then 4 is GCD of \(a\) and \(b\).
Or another way if you are not familiar with above property: we have \(a=4(k+1)\) and \(b=4k\), now \(k\) and \(k+1\) are consecutive integers thus they do not share any common factor but 1, which means that GCD of \(a\) and \(b\) is 4.
Answer: C.
Similar questions (with explanation of this property):
gcd-of-a-b-126427.html
what-is-the-greatest-common-factor-of-x-and-y-1-x-and-y-are-109273.html
if-x-and-y-are-positive-integers-such-that-x-8y-12-what-101196.html
Hope it helps.
(1) a = b + 4. Clearly insufficient.
(2) b/4 is an integer --> \(b=4k\) --> \(b\) is a multiple of 4, though still insufficient as no info about a.
(1)+(2) \(a=b+4=4k+4=4(k+1)\) --> useful property:
if \(a\) and \(b\) are multiples of \(k\) and are \(k\) units apart from each other then \(k\) is greatest common divisor of \(a\) and \(b\). For example if \(a\) and \(b\) are multiples of 7 and \(a=b+7\) then 7 is GCD of \(a\) and \(b\).
So, as we have that both \(a\) and \(b\) are multiples of 4 and are 4 units apart each other (\(a=b+4\) ), then 4 is GCD of \(a\) and \(b\).
Or another way if you are not familiar with above property: we have \(a=4(k+1)\) and \(b=4k\), now \(k\) and \(k+1\) are consecutive integers thus they do not share any common factor but 1, which means that GCD of \(a\) and \(b\) is 4.
Answer: C.
Similar questions (with explanation of this property):
gcd-of-a-b-126427.html
what-is-the-greatest-common-factor-of-x-and-y-1-x-and-y-are-109273.html
if-x-and-y-are-positive-integers-such-that-x-8y-12-what-101196.html
Hope it helps.
27 Jan 2012, 16:53
Kudos
Bookmarks
Thanks for the property. You told me about this before too, and I will make sure I remember it. Thanks again B 
