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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  [#permalink]

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New post 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

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  [#permalink]

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New post 13 Feb 2011, 14:43
11
26
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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 27 Feb 2011, 07:10
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Re: If 5400mn = k^4, where m, n, and k are positive integers  [#permalink]

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New post 14 Jun 2013, 05:28
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Re: If 5400mn = k^4, where m, n, and k are positive integers  [#permalink]

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New post 19 Jun 2014, 02:19
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\(k^4 = 5400* mn\)

\(k^4 = 3^3 . 2^3 . 5^2 . mn\)

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  [#permalink]

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New post 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  [#permalink]

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New post 14 Jan 2018, 13:22
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Re: If 5400mn = k^4, where m, n, and k are positive integers &nbs [#permalink] 14 Jan 2018, 13:22
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