m and n are positive integers such that m*n is a multiple of 16 and

Question

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?

I. 24
II. 36
III. 48

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

chetan2u · Answer

\frac{m}{n}=12a………..m*n=16b……..n=\frac{16b}{m}

\frac{m}{\frac{16b}{m}}=12a 
 m^2=12*16ab=8^2*3*ab 
This means m= 8*3*z=24z

m is a multiple of 24.

Only I and III possible

D

Alphabetrium · Answer

Hello, i wanted to understand if i am going wrong somewhere - if n = 1, then m*n being a multiple of 16 is not satisfied in 24, right? So the answer should only be III?

chetan2u · Answer

We are looking at COULD be true. 
This means we have to look for values of m, say here 24, and some corresponding value of n that can fit into both the equations.

When n =2, 24*2=48=16*3 and 24/2=12. 
Thus both equations are possible, and hence 24 is a possible value of m.

Crytiocanalyst · Answer

I. 24
for n=4 the condition is satisfied

II. 36
no value of n is satisfied since whatever is taken as n cannot be divided to yield a multiple of 12 therefore out

III. 48
n=1 it's satisfied

Therefore IMO D

sallyvn · Answer

the answer is D 
I & III 
m= 24 =>n= 2, so m*n is a multiple of 16 
m= 48 => n = 4, so m*n is a multiple of 16

BrushMyQuant · Answer

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 D 
Hope it helps!

m and n are positive integers such that m*n is a multiple of 16 and

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m and n are positive integers such that m*n is a multiple of 16 and

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