Equation Advanced (II)

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Equation Advanced (II) [#permalink]

04 May 2009, 19:41

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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?

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Re: Equation Advanced (II) [#permalink]

04 May 2009, 20:27

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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Joined: 17 Nov 2007

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Concentration: Entrepreneurship, Other

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Kudos [?]: 1300 [0], given: 346

Re: Equation Advanced (II) [#permalink]

04 May 2009, 21:01

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]

05 May 2009, 19:48

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, 19:48

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Equation Advanced (II)

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Equation Advanced (II)

Equation Advanced (II)

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