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

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 \(10+15=25\).

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.

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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

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

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

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