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Equation Advanced (II)

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Equation Advanced (II) [#permalink] New post 04 May 2009, 18:41
00:00
A
B
C
D
E

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(N/A)

Question Stats:

100% (02:08) correct 0% (00:00) wrong based on 0 sessions
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If mn = 3(m+1) + n and m and n are integers, m could be any of the following values EXCEPT:

(A) 2
(B) 3
(C) 4
(D) 5
(E) 7

Is there any shortcut to this?
Expert Post
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Posts: 3574
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
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Kudos [?]: 2346 [0], given: 359

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Re: Equation Advanced (II) [#permalink] New post 04 May 2009, 19:27
Expert's post
mn = 3(m+1) + n --> mn+n = 3(m+1) + 2n --> (m+1)(n-3) = 2n --> m = n+3 / n - 3

or m=1+6/x, where x = n-3

we have maximum of m at x=1: m = 7
at x=2: m=4
When x increases, m decreases. So, m=5 is impossible value for the equation.
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Expert Post
CEO
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Joined: 17 Nov 2007
Posts: 3574
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 432

Kudos [?]: 2346 [0], given: 359

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Re: Equation Advanced (II) [#permalink] New post 04 May 2009, 20:01
Expert's post
typhoidX wrote:
Can you explain how you got "m=1+6/x, where x = n-3"? I don't quite get the process.


m = n+3 / n - 3 ---(x = n -3 )----> m= (x+6) / x --- > m = 1 +6/x
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Re: Equation Advanced (II) [#permalink] New post 05 May 2009, 18:48
joyseychow wrote:
If mn = 3(m+1) + n and m and n are integers, m could be any of the following values EXCEPT:

(A) 2
(B) 3
(C) 4
(D) 5
(E) 7

Is there any shortcut to this?


Quick way is plugging-in in this case.

Note that m and n are integers.

mn = 3(m+1) + n
mn - n = 3(m+1)
n (m -1) = 3(m+1)
n = 3(m+1)/(m -1)

While pluging-in, 5 doesnot fit to the equation in producing an integer valuefor n.

n = 3(m+1)/(m -1)
n = 3(5+1)/(5 -1)
n = 3x6/4
n = 9/2

So it is D.
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Re: Equation Advanced (II)   [#permalink] 05 May 2009, 18:48
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