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

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Manager
Joined: 10 Feb 2011
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If 5400mn = k^4, where m, n, and k are positive integers [#permalink]

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13 Feb 2011, 13:20
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Question Stats:

70% (01:54) correct 30% (02:15) wrong based on 619 sessions

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Jun 2013, 03:24, edited 1 time in total.
Edited the question.
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Joined: 02 Sep 2009
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Re: If 5400mn = k4, where m, n, and k are positive integers [#permalink]

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13 Feb 2011, 13:43
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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!

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 $$10+15=25$$.

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

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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!

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 $$10+15=25$$.

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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27 Feb 2011, 02:56
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!

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 $$10+15=25$$.

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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27 Feb 2011, 06:10
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Re: If 5400mn = k^4, where m, n, and k are positive integers [#permalink]

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14 Jun 2013, 04:28
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Re: If 5400mn = k^4, where m, n, and k are positive integers [#permalink]

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19 Jun 2014, 01:19
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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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28 Dec 2016, 09:33
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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14 Jan 2018, 12:22
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Re: If 5400mn = k^4, where m, n, and k are positive integers   [#permalink] 14 Jan 2018, 12:22
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