If 5400mn = k^4, where m, n, and k are positive integers

Question

If 5400mn = k^4, where m, n, and k are positive integers, what is the least possible value of m + n?

A. 11
B. 18
C. 20
D. 25
E. 33

Bunuel · Accepted Answer

banksy please format the questions properly!

Question should read:
 If 5400mn = k^4, where m, n, and k are positive integers, what is the least possible value of m + n? 
A. 11
B. 18
C. 20
D. 25
E. 33

Note that m, n, and k are positive integers.

First of all:  5,400=2^3*3^3*5^2 . Now, in order  5,400mn=2^3*3^3*5^2*m*n  to be equal to the integer in fourth power then  mn  must complete the powers of 2, 3 and 5 to the fourth power, hence the least value of  mn  for which  2^3*3^3*5^2*m*n=k^4  is for  mn=2*3*5^2=150 . In this case  5,400mn=2^3*3^3*5^2*(2*3*5^2)=(2*3*5)^4=k^4 .

So we have that the least value of  mn  is  2*3*5^2 . Next: in order to minimize  m+n  we should break  2*3*5^2  into two multiples which are closest to each other:  2*5=10  and  3*5=15 , their sum is  10+15=25 .

Answer: D.

PareshGmat · Answer

k^4 = 5400* mn

k^4 = 3^3 . 2^3 . 5^2 . mn

In order to make RHS a perfect power of 4, we require it to be multiplied by  3, 2 & 5^2

mn = 3 . 2 . 5^2

mn = 150 = 10 * 15 (Least possible)

Answer = 10 + 15 = 25 = D

MHIKER · Answer

I did not understand the following parts. 
m*n=2*3*5^2=150, and 
 So we have that the least value of  mn  is  2*3*5^2 .  Next: in order to minimize  m+n  we should break  2*3*5^2  into two multiples which are closest to each other:  2*5=10  and  3*5=15 , their sum is  10+15=25 .

Bunuel · Answer

As explained before in order  5,400mn=2^3*3^3*5^2*m*n  to be equal to the integer in fourth power then  mn  must complete the powers of 2, 3 and 5 to the fourth power er (well generally to the multiple of 4, though as we need the least value of mn then to 4), hence the least value of  mn  for which  2^3*3^3*5^2*m*n=k^4  is for  mn=2*3*5^2=150 . In this case  5,400mn=2^3*3^3*5^2*(2*3*5^2)=(2*3*5)^4=k^4 :  mn must have one 2 to complete 2^3 to 2^4, one 3 to complete 3^3 to 3^4 and two 5's to complete 5^2 to 5^4 .

So we have that the least value of  mn  is  2*3*5^2 . Next: in order to minimize  m+n  (taking into account that  mn=2*3*5^2 ) we should break  2*3*5^2  into two multiples which are closest to each other:  2*5=10  and  3*5=15 , their sum is  10+15=25  (all other break downs of  mn=2*3*5^2=150  will have the greater sum: 1+150=151, 2+75=77, 3+50=52, ...).

Hope it's clear.

Bunuel · Answer

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Bunuel · Answer

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stonecold · Answer

Excellent Question from KAPLAN.
Here is my response to this one =>
As k is a postive integer => K^4 will be a perfect fourth power
5400=2^3*5^2*3^3
Least value of m*n=> 2*3*5^2

Hence Least value of m+n => 10+15 = 25

Hence D

Mattia73 · Answer

We know that K^4= 5400*m*n

5400= 2^3*3^3*5
Least value of m*n= 2*3*5^3

m*n= 6*25 6+25= 31
m*n=10*15 10+15= 25
m*n=30*5 30+5= 35
m*n=50*3 50+3= 53
m*n=75*2 75+2= 77

25 is the least value of m+n
  Correct D

Sonia2023 · Answer

chetan2u   Bunuel  - for minimum values of two numbers whose product is fixed , we generally need the numbers to be as close as possible.

So, if the number is a perfect square, we can easily take out minimum values. But for the numbers which are not perfect squares, can we not take square root of the product and then solve it as follows:

\sqrt{150}  = 5  \sqrt{6}  
5 \sqrt{ 6} + 5 \sqrt{ 6}  = minimum value of the sum
2*5*  \sqrt{6} 
10* \sqrt{6} 
25

Did this method worked only for this question or can we use this one for other questions as well?

Thank you

chetan2u · Answer

The method would require knowing square roots and can become calculation intensive. 
 If the options had 24 too, what would you have answered.

A quicker way using your approach would be. 
The number is  5\sqrt{6} 
Now, we know  \sqrt{6}  should be between  \sqrt{4}  or 2 and  \sqrt{9}  or 3.
Thus, look for factors to the in the range or closest to range 5*2 to 5*3. 
150 has factors 10 and 15 close to that range.

However best is to look for prime factors as shown above 
OR 
Or 150 lies close to 144, which is square of 12. So, the closest factors should lie on either side of the 12.
Which factor is closest to 12 => you can make out that it is 10, and from 10, you can find the other factor 10*15

Spectre26 · Answer

5400 = 2^3*3^3*5^2*mn

Since, m, n, k are positive integers
mn must be at least  2*3*5^2 , so that (5400mn)^(1/4) is also an integer, as k is an integer

AM \geq{GM} 
 \frac{(m + n)}{2} \geq mn

m+n must be atleast  2*\sqrt{mn} = 2*\sqrt{150} 
2*150 lies between 24 and 26, hence  Option D

(144<150<169 => 12^2 < 150 < 13^2)

siddhantvarma · Answer

Hi  Spectre26

Interesting solution. Can you apply AM>=GM in all such questions? Is there a relevant theory around this that can help understand the concept better?

Spectre26 · Answer

May be applied in following scenarios:
 
 If AB is given or can be deduced and minimum value of A+B is required. 
 If A+B is given or can be deduced and maximum value of AB is required.

Also depends on the values and option choices, here in this question option choices were convenient.

For theory, you may read AP and GP from any standard source.

However a basic proof involving only two real nos. can be as follows:

(a-b)^2 \geq 0 
 a^2 - 2ab + b^2 \geq 0 
adding  4ab  on both sides
 a^2 + 2ab + b^2 \geq 4ab 
 (a+b)^2 \geq 4ab 
 ((a+b)/2)^2 \geq ab 
 (a+b)/2 \geq \sqrt{ab} 
 AM \geq GM

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If 5400mn = k^4, where m, n, and k are positive integers

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If 5400mn = k^4, where m, n, and k are positive integers

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