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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]

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27 Feb 2011, 03:38

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 \(10+15=25\).
_________________

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]

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18 Jun 2014, 15:31

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).

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

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17 Sep 2015, 01:01

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).

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

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24 Oct 2016, 06:56

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.
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Re: If 5400mn = k^4, where m, n, and k are positive integers [#permalink]

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28 Dec 2016, 10:33

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 _________________

Give me a hell yeah ...!!!!!

gmatclubot

Re: If 5400mn = k^4, where m, n, and k are positive integers
[#permalink]
28 Dec 2016, 10:33

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