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If mn = 3(m + 1) + n and m and n are integers

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If mn = 3(m + 1) + n and m and n are integers [#permalink] New post 20 Sep 2013, 15:05
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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
[Reveal] Spoiler: OA
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Re: If mn = 3(m + 1) + n and m and n are integers [#permalink] New post 20 Sep 2013, 15:09
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jlgdr 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


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

n won't be an integer only if m = 5 (from the options).

Answer: D.

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Re: If mn = 3(m + 1) + n and m and n are integers [#permalink] New post 22 Sep 2013, 04:00
jlgdr 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


Alternative way of looking at it

mn = 3(m+1) + n
n = 3(m+1)/(m-1)
n = 3[(m-1)+2)]/(m-1)
n = 3(1 + 2/(m-1))
n = 3 + 6/(m-1)

=> given answer choice - 1 should divide 6.

Only 5 does not satisfy the above condition.
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Re: If mn = 3(m + 1) + n and m and n are integers [#permalink] New post 15 Oct 2013, 06:24
This is how I solved the question mn=3(m+1)+n => mn= 3m+3 +n => mn-3m=n+3 => m(n-3)=n+3 => m=n+3/n-3 In this cast 7 fits the bill. Where am I going wrong Bunuel?
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Re: If mn = 3(m + 1) + n and m and n are integers [#permalink] New post 15 Oct 2013, 07:52
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mohnish104 wrote:
This is how I solved the question mn=3(m+1)+n => mn= 3m+3 +n => mn-3m=n+3 => m(n-3)=n+3 => m=n+3/n-3 In this cast 7 fits the bill. Where am I going wrong Bunuel?


If m=7, then 7=(n+3)/(n-3) --> n=4=integer. Thus m could be 7.

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COLLECTION OF QUESTIONS:
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Re: If mn = 3(m + 1) + n and m and n are integers [#permalink] New post 23 Feb 2014, 21:37
mn = 3(m + 1) + n and m and n are integers-

Solved by placing the values directly from the options:

For m=2, 2n = 9 + n .......... OK
For m=3, 3n = 12 +n..........OK
For m=4, 4n = 15 + n..........OK
For m=5, 5n = 18 + n.......... 4n = 18... then n becomes a non-integer
So Answer = D = 5

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Re: If mn = 3(m + 1) + n and m and n are integers [#permalink] New post 23 Feb 2014, 22:13
jlgdr 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


Let us plug in:

m = 2 , 2(n) = 3(2 + 1) + n then n will be 9
m = 3, 3(n) = 3(4) + n, then n will be an integer
m = 4, 4(n) = 3(5) + n, then n will be an integer
m = 5, 5(n) = 3(6) + n, then n will be 18/4 (not an integer) (BINGO!)
m = 7, 7(n) = 3(8) + n, still n will be an integer

Answer is D.

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Re: If mn = 3(m + 1) + n and m and n are integers   [#permalink] 23 Feb 2014, 22:13
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