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If 5400mn = k^4, where m, n, and k are positive integers
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Updated on: 14 Jun 2013, 04:24
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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
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Originally posted by banksy on 13 Feb 2011, 14:20.
Last edited by Bunuel on 14 Jun 2013, 04:24, edited 1 time in total.
Edited the question.




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Re: If 5400mn = k4, where m, n, and k are positive integers
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13 Feb 2011, 14:43
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.
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Re: If 5400mn = k4, where m, n, and k are positive integers
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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 \(10+15=25\).
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Re: If 5400mn = k4, where m, n, and k are positive integers
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27 Feb 2011, 03:56
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, ...). Hope it's clear.
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Re: If 5400mn = k4, where m, n, and k are positive integers
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27 Feb 2011, 07:10



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Re: If 5400mn = k^4, where m, n, and k are positive integers
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14 Jun 2013, 05:28
Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE
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Re: If 5400mn = k^4, where m, n, and k are positive integers
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19 Jun 2014, 02:19
\(k^4 = 5400* mn\) \(k^4 = 3^3 . 2^3 . 5^2 . mn\) 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
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Re: If 5400mn = k^4, where m, n, and k are positive integers
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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
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Re: If 5400mn = k^4, where m, n, and k are positive integers
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10 Mar 2019, 00:58
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
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