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16 Sep 2014, 01:45
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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) \(5a - b = 1\)
(1) \(a + 3b = 61\)
(2) \(5a - b = 1\)
16 Sep 2014, 01:45
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Official Solution:
If \(a\) and \(b\) are positive integers, what is the greatest common divisor of \(a\) and \(b\)?
It's important to note that two statements together are obviously sufficient to answer the question. When encountering such a question, be extremely cautious when choosing C as an answer. There's a high likelihood that the question is a "C trap" question ("C trap" refers to a problem that 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 1). Hence we have that the \(GCD(x, y)=d=1\). Sufficient.
(2) \(5a - b = 1\). Rearrange: \(5a=b+1\).
\(5a\) 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 the 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
If \(a\) and \(b\) are positive integers, what is the greatest common divisor of \(a\) and \(b\)?
It's important to note that two statements together are obviously sufficient to answer the question. When encountering such a question, be extremely cautious when choosing C as an answer. There's a high likelihood that the question is a "C trap" question ("C trap" refers to a problem that 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 1). Hence we have that the \(GCD(x, y)=d=1\). Sufficient.
(2) \(5a - b = 1\). Rearrange: \(5a=b+1\).
\(5a\) 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 the 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
11 Feb 2017, 00:58
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My way to solve this question:
If \(a\) and \(b\) are positive integers, what is the greatest common divisor of \(a\) and \(b\)?
(1) \(a + 3b = 61\)
a =61 -3b ..........Plug in numbers
b=1 ......a=58.......GCD= 1
b=2.......a=55.......GCD= 1
b=3.......a=52.......GCD= 1
b=4.......a=49.......GCD= 1
b=5.......a=46.......GCD= 1
It is safe to conclude that GCD=1
Sufficient
(2) \(5a - b = 1\)
b =5a -1
a=1 ........b=4.......GCD= 1
a=2.........b=9.......GCD= 1
a=3.........b=14.....GCD= 1
a=4.........b=19.....GCD= 1
It is safe to conclude that GCD=1
Sufficient
Answer: D
If \(a\) and \(b\) are positive integers, what is the greatest common divisor of \(a\) and \(b\)?
(1) \(a + 3b = 61\)
a =61 -3b ..........Plug in numbers
b=1 ......a=58.......GCD= 1
b=2.......a=55.......GCD= 1
b=3.......a=52.......GCD= 1
b=4.......a=49.......GCD= 1
b=5.......a=46.......GCD= 1
It is safe to conclude that GCD=1
Sufficient
(2) \(5a - b = 1\)
b =5a -1
a=1 ........b=4.......GCD= 1
a=2.........b=9.......GCD= 1
a=3.........b=14.....GCD= 1
a=4.........b=19.....GCD= 1
It is safe to conclude that GCD=1
Sufficient
Answer: D
