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Joined: 10 Dec 2020
Posts: 50
Given Kudos: 955
GRE 1: Q166 V162
m and n are positive integers such that mn is a multiple of 16
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22 Aug 2021, 06:46
22 Aug 2021, 06:46
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\(m\) and \(n\) are positive integers such that \(m*n\) is a multiple of \(16\) and \(\frac{m}{n}\) is a multiple of \(12\). Which of the following integers could be the value of \(m\)?
Indicate all such integers.
A)24
B)32
C)36
D)48
Indicate all such integers.
A)24
B)32
C)36
D)48
m and n are positive integers such that mn is a multiple of 16
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07 Jul 2023, 11:07
07 Jul 2023, 11:07
2
XavierAlexander wrote:
\(m\) and \(n\) are positive integers such that \(m*n\) is a multiple of \(16\) and \(\frac{m}{n}\) is a multiple of \(12\). Which of the following integers could be the value of \(m\)?
Indicate all such integers.
A)24
B)32
C)36
D)48
Indicate all such integers.
A)24
B)32
C)36
D)48
As a GRE tutor, I strive to find the least math-y way to solve hard problems.
I suspect that many test-takers will find it daunting to solve this problem abstractly.
For such students, a more practical approach might be to PLUG IN THE ANSWERS.
When a correct answer plugged in, m/n will be a multiple of 12, and mn will be divisible by 16.
A: m=24
Since m/n must be a multiple of 12, let 24/n = 12, with the result that n=2 and mn = 24*2 = 48.
Use the calculator or old-fashioned arithmetic to determine that mn/16 = 48/16 = 3.
Since mn is divisible by 16, A is a viable answer choice.
B: m=32
Since m/n must be a multiple of 12, let 32/n = 12, with the result that n is not an integer.
Eliminate B.
C: m=36
Since m/n must be a multiple of 12, let 36/n = 12, with the result that n=3 and mn = 36*3 = 108.
Use the calculator or old-fashioned arithmetic to determine that mn/16 = 108/16 = 6.75.
Since mn is NOT divisible by 16, C is not a viable answer choice.
D: m=48
Since m/n must be a multiple of 12, let 48/n = 12, with the result that n=4 and mn = 48*4 = 192.
Use the calculator or old-fashioned arithmetic to determine that mn/16 = 192/16 = 12.
Since mn is divisible by 16, D is a viable answer choice.
Show: ::
A and D
General Discussion
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
m and n are positive integers such that mn is a multiple of 16
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22 Aug 2021, 09:11
22 Aug 2021, 09:11
this question is of would/could be type
given \(\frac{m}{12n}\) is integer \(m\) must be divisible by 12
possible answer choices for \(m={24, 36, 48}\) must be divisible by 16 as the product of \(m*n\)
if \(m=24\), then \(n={2,4,8,10, ...N\), where N is positive integer and multiple of 2
this logic violates the choice for \(m=36\)
if \(m=36\), then \(n={4,8,12,...K}\), where K is positive integer and multiple of 4
\(\frac{36}{4}\) cannot be divided by 12
neither \(\frac{36}{12}\) can be divided by 12
hence. answer is 24 and 48, (A, D)
given \(\frac{m}{12n}\) is integer \(m\) must be divisible by 12
possible answer choices for \(m={24, 36, 48}\) must be divisible by 16 as the product of \(m*n\)
if \(m=24\), then \(n={2,4,8,10, ...N\), where N is positive integer and multiple of 2
this logic violates the choice for \(m=36\)
if \(m=36\), then \(n={4,8,12,...K}\), where K is positive integer and multiple of 4
\(\frac{36}{4}\) cannot be divided by 12
neither \(\frac{36}{12}\) can be divided by 12
hence. answer is 24 and 48, (A, D)
