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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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Expert's post
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
Expert's post
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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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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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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