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

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jlgdr
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If mn = 3(m + 1) + n and m and n are integers [#permalink] New post   20 Sep 2013, 16:05
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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
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D
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Bunuel
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Re: If mn = 3(m + 1) + n and m and n are integers [#permalink] New post   20 Sep 2013, 16:09
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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, 05:00
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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


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, 08:52
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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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Re: If mn = 3(m + 1) + n and m and n are integers [#permalink] New post   23 Feb 2014, 22:37
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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   16 Feb 2017, 05:23
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\(mn = 3(m + 1) + n; mn=3m+3+n; mn-n=3m+3; n(m-1)=3m+3; n=\frac{3m+3}{m-1}\)
Plug in answer choices into the above equation:
(A) \(2 - n=\frac{(3*2+3)}{2-1}=9\)
(B) \(3 - n=\frac{(3*3+3)}{3-1}=6\)
(C) \(4 - n=\frac{(3*4+3)}{4-1}=5\)
(D) \(5 - n=\frac{(3*5+3)}{5-1}=\) not an integer
(E) \(7 - n=\frac{(3*7+3)}{7-1}=4\)

The correct answer is D.
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Re: If mn = 3(m + 1) + n and m and n are integers [#permalink] New post   21 Feb 2017, 10:22
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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


We are given that mn = 3(m + 1) + n. Let’s simplify that equation:

mn = 3(m + 1) + n

mn - n = 3(m + 1)

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

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

We can see that we can eliminate choices A and C since the denominators will be 1 and 3, respectively. Now let’s check the other answer choices:

B) If m = 3, then n = 3(3+1)/(3-1) = 12/2 = 6, which is an integer.

D) If m = 5, then n = 3(5+1)/(5-1) = 18/4 = 4.5, which is NOT an integer.

Thus choice D is the correct answer.

Answer: D
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Re: If mn = 3(m + 1) + n and m and n are integers [#permalink] New post   15 Oct 2013, 07:24
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   23 Feb 2014, 23:13
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] New post   14 Nov 2023, 00:13
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Re: If mn = 3(m + 1) + n and m and n are integers [#permalink]
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