Last visit was: 25 Apr 2024, 05:23 It is currently 25 Apr 2024, 05:23

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Manager
Manager
Joined: 10 Dec 2020
Posts: 50
Own Kudos [?]: 82 [6]
Given Kudos: 955
GRE 1: Q166 V162
Send PM
Most Helpful Community Reply
Intern
Intern
Joined: 09 Jul 2018
Posts: 49
Own Kudos [?]: 77 [2]
Given Kudos: 0
Send PM
General Discussion
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Own Kudos [?]: 354 [0]
Given Kudos: 64
GRE 1: Q160 V152
Send PM
Moderator
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1085
Own Kudos [?]: 883 [1]
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
Send PM
m and n are positive integers such that mn is a multiple of 16 [#permalink]
1
Given that \(m\) and \(n\) are positive integers such that \(m*n\) is a multiple of \(16\) and \(\frac{m}{n}\) is a multiple of \(12\) and we need to find Which of the following integers could be the value of \(m\)

Given that \(\frac{m}{n}\) is a multiple of 12
=> \(\frac{m}{n}\) = 12*k (where k is an integer)
=> m = 12k * n
=> n = \(\frac{m}{12k}\)

m*n is a multiple of 16
=> m * \(\frac{m}{12k}\) is a multiple of 16
=> \(\frac{m^2}{12k}\) is a multiple of 16

Even if we take k =1 we get
\(\frac{m^2}{12}\) is a multiple of 16

\(m^2\) minimum is a multiple of = 12 * 16 = \(2^2 * 3 * 2^4\) = \(2^6 * 3^1\)

Since we have even power on left hand side (2) so on right hand side also we should have even power
=> \(m^2\) minimum is a multiple of = \(2^6 * 3^2\)
=> m minimum is a multiple of = \(2^3 * 3^1\) = 24

=> m is a multiple of 24
=> 24 and 48 are possible

So, Answer will be A and D
Hope it helps!
Intern
Intern
Joined: 22 Aug 2023
Posts: 8
Own Kudos [?]: 7 [1]
Given Kudos: 11
Send PM
Re: m and n are positive integers such that mn is a multiple of 16 [#permalink]
1
Given, m * n = 16a , where a is any integer
m/n = 12b , where b is any integer
Multiply both

m^2 = 16 x 12 ab
ab = m^2/192
Now, a and b both are integers so there product should also be an integer
Only for values of m in option a and d , m would be an integer and hence are the answers
Prep Club for GRE Bot
[#permalink]
Moderators:
Moderator
1085 posts
GRE Instructor
218 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne