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12 Jun 2014, 07:37
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
56% (02:09) correct
44%
(01:58)
wrong
based on 1066
sessions
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If \(a\) and \(b\) are positive integers, what is the greatest common divisor of \(a\) and \(b\)?
(1) \(a + 3b = 61\)
(2) \(5a - b = 1\)
Kudos for a correct solution.
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12 Jun 2014, 07:37
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SOLUTION
If \(a\) and \(b\) are positive integers, what is the greatest common divisor of \(a\) and \(b\)?
Notice that two statements together are obviously sufficient to answer the question. When you see such question you should be extremely cautious when choosing C for an answer. Chances are that the question is a "C trap" question ("C trap" is a problem which is VERY OBVIOUSLY sufficient if both statements are taken together).
(1) \(a + 3b = 61\).
Let the greatest common divisor of \(a\) and \(b\) be \(d\), then \(a=md\) and \(b=nd\), for some positive integers \(m\) and \(n\). So, we'll have \((md)+3(nd)=d(m+3n)=61\). Now, since 61 is a prime number (61=1*61) then \(d=1\) and \(m+3n=61\) (vice versa is not possible because \(m\) and \(n\) are positive integers and therefore \(m+3n\) cannot equal to 1). Hence we have that the GCD(x, y)=d=1. Sufficient.
(2) \(5a - b = 1\) --> \(5a=b+1\).
\(5a\) and \(b\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1 (for example 20 and 21 are consecutive integers, thus only common factor they share is 1). So, \(5a\) and \(b\) don't share any common factor but 1, thus \(a\) and \(b\) also don't share any common factor but 1. Hence, the GCD(x, y) is 1. Sufficient.
Answer: D.
Try NEW divisibility PS question.
If \(a\) and \(b\) are positive integers, what is the greatest common divisor of \(a\) and \(b\)?
Notice that two statements together are obviously sufficient to answer the question. When you see such question you should be extremely cautious when choosing C for an answer. Chances are that the question is a "C trap" question ("C trap" is a problem which is VERY OBVIOUSLY sufficient if both statements are taken together).
(1) \(a + 3b = 61\).
Let the greatest common divisor of \(a\) and \(b\) be \(d\), then \(a=md\) and \(b=nd\), for some positive integers \(m\) and \(n\). So, we'll have \((md)+3(nd)=d(m+3n)=61\). Now, since 61 is a prime number (61=1*61) then \(d=1\) and \(m+3n=61\) (vice versa is not possible because \(m\) and \(n\) are positive integers and therefore \(m+3n\) cannot equal to 1). Hence we have that the GCD(x, y)=d=1. Sufficient.
(2) \(5a - b = 1\) --> \(5a=b+1\).
\(5a\) and \(b\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1 (for example 20 and 21 are consecutive integers, thus only common factor they share is 1). So, \(5a\) and \(b\) don't share any common factor but 1, thus \(a\) and \(b\) also don't share any common factor but 1. Hence, the GCD(x, y) is 1. Sufficient.
Answer: D.
Try NEW divisibility PS question.
12 Jun 2014, 10:58
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Bunuel
Let k be the greatest common divisor of a and b such that a=kx and b=ky, where x and y are co-prime positive integers
St. 1; a+3b=61;
kx+3ky=61;
x+3y=(61/k)
now since 61 is a prime number therefore maximum value of k is 1
hence sufficient
st.2
5a-b=1
5kx-ky=1
5x-y=(1/k) since x and y are integers, therefore 5x-y will only be integer if k=1
hence sufficient
therefore answer should be D
