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 GMAT 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.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: If 5400mn = k4, where m, n, and k are positive integers [#permalink]
13 Feb 2011, 13:43

3

This post received KUDOS

Expert's post

banksy wrote:

If 5400mn = k4, 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

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.

Re: If 5400mn = k4, where m, n, and k are positive integers [#permalink]
27 Feb 2011, 02:38

Bunuel wrote:

banksy wrote:

If 5400mn = k4, 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

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.

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. _________________

Re: If 5400mn = k4, where m, n, and k are positive integers [#permalink]
27 Feb 2011, 02:56

Expert's post

Baten80 wrote:

Bunuel wrote:

banksy wrote:

If 5400mn = k4, 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

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.

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, ...).

Re: If 5400mn = k^4, where m, n, and k are positive integers [#permalink]
18 Jun 2014, 14:31

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________