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26 May 2017, 04:42
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What is the greatest common divisor of positive integers \(n\) and \(m\)?
(1) \(n = 1\)
(2) \(n\) and \(m\) are consecutive
(1) \(n = 1\)
(2) \(n\) and \(m\) are consecutive
26 May 2017, 04:42
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Official Solution:
If you look at the original condition, there are 2 variables \((m,n)\). In order to match the number of variables to the number of equations, there must be 2 equations. Therefore, C is most likely to be the answer. By solving con 1) & con 2), you get \((n, m) = (1,2)\), and
Greatest Common Divisor of m and \(n = GCD(m,n) = GCD(1,2) = 1\), hence unique and sufficient. The answer is C. However, this is also an integer question, one of the key questions. Thus, if you apply "CMT 4(A: when you get C too easily, consider A or B)" to con 2), if \(n\) and \(m\) are consecutive, you always get \(GCD(n,m) = 1\). Hence it is unique and sufficient.
In the case of con 1), you must apply "CMT 4(B: if you get A or B too easily, consider D)". If also \(n=1\), no matter which positive integer \(m\) becomes, you always get \(GCD(n,m) = GCD(1,m)=1\). Hence it is unique and sufficient. In this question, con 2) is easy and con 1) is difficult. This is why you must always remember "CMT 4(B: if you get A or B too easily, consider D)".
Answer: D
If you look at the original condition, there are 2 variables \((m,n)\). In order to match the number of variables to the number of equations, there must be 2 equations. Therefore, C is most likely to be the answer. By solving con 1) & con 2), you get \((n, m) = (1,2)\), and
Greatest Common Divisor of m and \(n = GCD(m,n) = GCD(1,2) = 1\), hence unique and sufficient. The answer is C. However, this is also an integer question, one of the key questions. Thus, if you apply "CMT 4(A: when you get C too easily, consider A or B)" to con 2), if \(n\) and \(m\) are consecutive, you always get \(GCD(n,m) = 1\). Hence it is unique and sufficient.
In the case of con 1), you must apply "CMT 4(B: if you get A or B too easily, consider D)". If also \(n=1\), no matter which positive integer \(m\) becomes, you always get \(GCD(n,m) = GCD(1,m)=1\). Hence it is unique and sufficient. In this question, con 2) is easy and con 1) is difficult. This is why you must always remember "CMT 4(B: if you get A or B too easily, consider D)".
Answer: D
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