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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] New post 13 Feb 2011, 13:20
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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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Re: If 5400mn = k4, where m, n, and k are positive integers [#permalink] New post 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!

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

Hope it's clear.
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Re: If 5400mn = k4, where m, n, and k are positive integers [#permalink] New post 27 Feb 2011, 06:10
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Re: If 5400mn = k^4, where m, n, and k are positive integers [#permalink] New post 14 Jun 2013, 04:28
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Re: If 5400mn = k^4, where m, n, and k are positive integers [#permalink] New post 18 Jun 2014, 14:31
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Re: If 5400mn = k^4, where m, n, and k are positive integers [#permalink] New post 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] 19 Jun 2014, 01:19
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