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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