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If m and n are positive integers, each of the following coul

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If m and n are positive integers, each of the following coul  [#permalink]

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New post Updated on: 12 Jan 2014, 17:13
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If m and n are positive integers, each of the following could be equivalent to \(\sqrt{800mn}\) EXCEPT

A. \(80\sqrt{m}\)
B. \(40\sqrt{10m}\)
C. \(40m\sqrt{5}\)
D. \(20\sqrt{5m}\)
E. \(20\sqrt{2m}\)

OE
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File comment: OE
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Hi, I want to know if this OE is the best way to solve this question, please.

Originally posted by goodyear2013 on 12 Jan 2014, 13:13.
Last edited by Bunuel on 12 Jan 2014, 17:13, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If m and n are positive integers, each of the following coul  [#permalink]

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New post 01 Jun 2014, 23:11
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Consider the options one by one:

A:

\(80\sqrt{m} = \sqrt{6400m}\)

\(\sqrt{6400m} = \sqrt{800mn}\)

If n = 8, then the condition satisfies

B:

\(40\sqrt{10m} = \sqrt{16000m}\)

\(\sqrt{16000m} = \sqrt{800mn}\)

If n = 20, then the condition satisfies

C:

\(40m\sqrt{5} = \sqrt{1600 m^2 * 5} = \sqrt{8000 m^2}\)

\(\sqrt{8000 m^2} = \sqrt{800mn}\)

If m = 1 & n = 10, then the condition satisfies

D:

\(20\sqrt{5m} = \sqrt{2000m}\)

\(\sqrt{2000m} = \sqrt{800mn}\)

2000m = 800mn

2000 = 800n

We will require to make n a fraction to satisfy the condition; m & n are +ve integers, so the condition fails

E:

\(20\sqrt{2m} = \sqrt{800m}\)

\(\sqrt{800m} = \sqrt{800mn}\)

If n = 1, then the condition satisfies

Answer = D
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Re: each of following equivalent to sqrt(800mn) EXCEPT  [#permalink]

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New post 12 Jan 2014, 16:02
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I think OE solution is quickest method given the constraint that m and n need to be positive integers. Basically, whatever number is outside the square root in the answer choices is a perfect square and needs to be squared and multiplied by whatever is inside the square root....for n to be a positive integer, this number has to be divisible by 800.

Sqrt800mn = Sqrt20^2*2*m*n

solving for n

C
40m*Sqrt5 = Sqrt40^2*m^2*5 = Sqrt20^2*2^2*m^2*5
Sqrt20^2*5*2^2*m^2 = Sqrt20^2*2*m*n
n=5m (m=1, n=5...ok)

D
20sqrt5m = Sqrt5m*20^2
5m = 2*m*n
5=2n (n=(5/2)....not pos int)
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Re: If m and n are positive integers, each of the following coul  [#permalink]

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New post 01 Jun 2014, 23:36
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goodyear2013 wrote:
If m and n are positive integers, each of the following could be equivalent to \(\sqrt{800mn}\) EXCEPT

A. \(80\sqrt{m}\)
B. \(40\sqrt{10m}\)
C. \(40m\sqrt{5}\)
D. \(20\sqrt{5m}\)
E. \(20\sqrt{2m}\)

OE
Attachment:
OE.png


Hi, I want to know if this OE is the best way to solve this question, please.


Take all the numbers inside the root

(A) 6400m which can be written as 800m(8)
(B) 16000m can be written as 800m(20)
(C) 8000m(m) can be written as 800m(10m)
(D) 2000m cannot be written as 800m x integer as 2000 is not divisible by 800
(E) 800m can be written as 800m(1)

Hence the answer is D.
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Re: If m and n are positive integers, each of the following coul  [#permalink]

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New post 12 Sep 2018, 10:02
m,n are positive integers

sqrt(800mn)= 20*sqrt(2*m*n)

Equate this to 20*sqrt(5*m) which gives n=5/2 which is incorrect as m,n are positive integers.
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Re: If m and n are positive integers, each of the following coul &nbs [#permalink] 12 Sep 2018, 10:02
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If m and n are positive integers, each of the following coul

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