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lahoosaher
Joined: 16 Jan 2009
Last visit: 20 Apr 2018
Posts: 236
Given Kudos: 16
Concentration: Technology, Marketing
Schools: Ross '16 (D) Tepper '16 (D) McCombs '16 (D) ISB '15 (D) Marshall '16 (WL) McDonough '16 (WL) Goizueta '16 (WL) Kelley '16 (D) Kenan-Flagler '16 (D) Rotman '16 (S)
GMAT 1: 700 Q50 V34
GPA: 3
WE:Sales (Telecommunications)
Schools: Ross '16 (D) Tepper '16 (D) McCombs '16 (D) ISB '15 (D) Marshall '16 (WL) McDonough '16 (WL) Goizueta '16 (WL) Kelley '16 (D) Kenan-Flagler '16 (D) Rotman '16 (S)
GMAT 1: 700 Q50 V34
Posts: 236
Updated on: 06 Jun 2019, 08:23
Originally posted by lahoosaher on 07 Jun 2009, 11:12.
Last edited by Bunuel on 06 Jun 2019, 08:23, edited 3 times in total.
Last edited by Bunuel on 06 Jun 2019, 08:23, edited 3 times in total.
Edited the question and added the OA
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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:
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Question Stats:
67% (01:38) correct
33%
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based on 2975
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What is the greatest common divisor of positive integers m and n ?
(1) m is a prime number
(2) 2n = 7m
(1) m is a prime number
(2) 2n = 7m
02 Mar 2012, 12:20
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What is the greatest common divisor of positive integers m and n?
(1) m is a prime number --> if \(m=2=prime\) and \(n=1\) then \(GCD(m,n)=1\) but if \(m=2=prime\) and \(n=4\) then \(GCD(m,n)=2\). Two different answers, hence not sufficient.
(2) 2n=7m --> \(\frac{m}{n}=\frac{2}{7}\) --> \(m\) is a multiple of 2 and \(n\) is a multiple of 7, but this is still not sufficient: if \(m=2\) and \(n=7\) then \(GCD(m,n)=1\) (as both are primes) but if \(m=4\) and \(n=14\) then \(GCD(m,n)=2\) (basically as \(\frac{m}{n}=\frac{2x}{7x}\) then as 2 and 7 are primes then \(GCD(m, n)=x\)). Two different answers, hence not sufficient.
(1)+(2) Since from (1) \(m=prime\) and from (2) \(\frac{m}{n}=\frac{2}{7}\) then \(m=2=prime\) and \(n=7\), hence \(GCD(m,n)=1\). Sufficient.
Answer: C.
(1) m is a prime number --> if \(m=2=prime\) and \(n=1\) then \(GCD(m,n)=1\) but if \(m=2=prime\) and \(n=4\) then \(GCD(m,n)=2\). Two different answers, hence not sufficient.
(2) 2n=7m --> \(\frac{m}{n}=\frac{2}{7}\) --> \(m\) is a multiple of 2 and \(n\) is a multiple of 7, but this is still not sufficient: if \(m=2\) and \(n=7\) then \(GCD(m,n)=1\) (as both are primes) but if \(m=4\) and \(n=14\) then \(GCD(m,n)=2\) (basically as \(\frac{m}{n}=\frac{2x}{7x}\) then as 2 and 7 are primes then \(GCD(m, n)=x\)). Two different answers, hence not sufficient.
(1)+(2) Since from (1) \(m=prime\) and from (2) \(\frac{m}{n}=\frac{2}{7}\) then \(m=2=prime\) and \(n=7\), hence \(GCD(m,n)=1\). Sufficient.
Answer: C.
22 Dec 2017, 11:54
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lahoosaher
Target question: What is the GCD of m and n?
Statement 1: m is a prime number
If m is a prime number, it has exactly 2 divisors (1 and m), so this tells us that the GCD of m and n must be either 1 or m.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT.
Statement 2: 2n = 7m
If 2n = 7m then we can rearrange the equation to get n = (7/2)m
IMPORTANT: Notice that if m were to equal an ODD number, then n would not be an integer. For example, if m = 3, then n = 21/2 (n is not an integer). Similarly, if m = 11, then n = 77/2 (n is not an integer). So, in order for n to be an INTEGER, m must be EVEN.
If m must be EVEN, there are several possible values for m and n. Consider these two cases:
case a: m = 2 and n = 7, in which case the GCD = 1
case b: m = 4 and n = 14, in which case the GCD=2
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT.
Statements 1 & 2 combined
From statement 1, we know that m is prime, and from statement 2, we know that m is even.
Since 2 is the only even prime number, we can conclude that m must equal 2.
If m = 2, then n must equal 7, which means that the GCD must be 1.
Since we are able to answer the target question with certainty, statements 1 & 2 combined are sufficient, and the answer is C
Cheers,
Brent
