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I like the solution - it’s helpful. Ureal!!!!
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How about (4,6)
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saibharatmba

Bunuel
Official Solution:

If \(m\) and \(n\) are integers, how many pairs of \((m, n)\) are there such that \(3m + 2n = mn\)?

A. 4
B. 5
C. 6
D. 8
E. 10


Arguably the best way to solve this question is to manipulate \(3m + 2n = mn\) such that we can factor it into \((m + a)(n + b) = integer\). Then, benefiting from the fact that \(m\) and \(n\) are integers, we'd be able to find the sets of \((m, n)\) satisfying the equation.

\(3m + 2n = mn\);

\(3m - mn + 2n = 0[\);

\(m(3 - n) + 2n = 0\).

To further factor \((3 - n)\), add -6 to both sides:

\(m(3 - n) - 6 + 2n= -6\);

\(m(3 - n) - 2(3 - n) = -6\);

\((3 - n)(m - 2) = -6\).

Now, the question essentially boils down to finding how many ways -6 can be written as the product of two integers, \((3 - n)\) and \((m - 2)\):

\(-6*1\)

\(-3*2\)

-\(2*3\)

\(-1*6\)

\(1*(-6)\)

\(2*(-3)\)

\(3*(-2)\)

\(6*(-1)\)

Therefore, there are eight sets of integers \((m, n)\) satisfying \((3 - n)(m - 2) = -6\). To find the exact values, though unnecessary, we equate \((3 - n)\) and \((m - 2)\) to each pair of factors above.


Answer: D

How about (4,6)

This is counted when m - 2 = 2 and 3 - n = -3
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how about (5,5)
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Bshubham512

Bunuel
Official Solution:

If \(m\) and \(n\) are integers, how many pairs of \((m, n)\) are there such that \(3m + 2n = mn\)?

A. 4
B. 5
C. 6
D. 8
E. 10


Arguably the best way to solve this question is to manipulate \(3m + 2n = mn\) such that we can factor it into \((m + a)(n + b) = integer\). Then, benefiting from the fact that \(m\) and \(n\) are integers, we'd be able to find the sets of \((m, n)\) satisfying the equation.

\(3m + 2n = mn\);

\(3m - mn + 2n = 0[\);

\(m(3 - n) + 2n = 0\).

To further factor \((3 - n)\), add -6 to both sides:

\(m(3 - n) - 6 + 2n= -6\);

\(m(3 - n) - 2(3 - n) = -6\);

\((3 - n)(m - 2) = -6\).

Now, the question essentially boils down to finding how many ways -6 can be written as the product of two integers, \((3 - n)\) and \((m - 2)\):

\(-6*1\)

\(-3*2\)

\(-2*3\)

\(-1*6\)

\(1*(-6)\)

\(2*(-3)\)

\(3*(-2)\)

\(6*(-1)\)

Therefore, there are eight sets of integers \((m, n)\) satisfying \((3 - n)(m - 2) = -6\). To find the exact values, though unnecessary, we equate \((3 - n)\) and \((m - 2)\) to each pair of factors above.


Answer: D
how about (5,5)

This is counted when m - 2 = 3 and 3 - n = -2.
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Hi Bunuel,
How can one be sure that the 8 sets of integers (m,n) that satisfy (3-n)*(m-2) = -6 will not produce any repeated pairs of (m,n)?

Bunuel
Quote:
Official Solution:

If \(m\) and \(n\) are integers, how many pairs of \((m, n)\) are there such that \(3m + 2n = mn\)?
A. 4
B. 5
C. 6
D. 8
E. 10


Arguably the best way to solve this question is to manipulate \(3m + 2n = mn\) such that we can factor it into \((m + a)(n + b) = integer\). Then, benefiting from the fact that \(m\) and \(n\) are integers, we'd be able to find the sets of \((m, n)\) satisfying the equation.

\(3m + 2n = mn\);

\(3m - mn + 2n = 0[\);

\(m(3 - n) + 2n = 0\).

To further factor \((3 - n)\), add -6 to both sides:

\(m(3 - n) - 6 + 2n= -6\);

\(m(3 - n) - 2(3 - n) = -6\);

\((3 - n)(m - 2) = -6\).

Now, the question essentially boils down to finding how many ways -6 can be written as the product of two integers, \((3 - n)\) and \((m - 2)\):

\(-6*1\)

\(-3*2\)

\(-2*3\)

\(-1*6\)

\(1*(-6)\)

\(2*(-3)\)

\(3*(-2)\)

\(6*(-1)\)

Therefore, there are eight sets of integers \((m, n)\) satisfying \((3 - n)(m - 2) = -6\). To find the exact values, though unnecessary, we equate \((3 - n)\) and \((m - 2)\) to each pair of factors above.


Answer: D
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juliansorre
Hi Bunuel,
How can one be sure that the 8 sets of integers (m,n) that satisfy (3-n)*(m-2) = -6 will not produce any repeated pairs of (m,n)?

Bunuel
Quote:
Official Solution:

If \(m\) and \(n\) are integers, how many pairs of \((m, n)\) are there such that \(3m + 2n = mn\)?
A. 4
B. 5
C. 6
D. 8
E. 10


Arguably the best way to solve this question is to manipulate \(3m + 2n = mn\) such that we can factor it into \((m + a)(n + b) = integer\). Then, benefiting from the fact that \(m\) and \(n\) are integers, we'd be able to find the sets of \((m, n)\) satisfying the equation.

\(3m + 2n = mn\);

\(3m - mn + 2n = 0[\);

\(m(3 - n) + 2n = 0\).

To further factor \((3 - n)\), add -6 to both sides:

\(m(3 - n) - 6 + 2n= -6\);

\(m(3 - n) - 2(3 - n) = -6\);

\((3 - n)(m - 2) = -6\).

Now, the question essentially boils down to finding how many ways -6 can be written as the product of two integers, \((3 - n)\) and \((m - 2)\):

\(-6*1\)

\(-3*2\)

\(-2*3\)

\(-1*6\)

\(1*(-6)\)

\(2*(-3)\)

\(3*(-2)\)

\(6*(-1)\)

Therefore, there are eight sets of integers \((m, n)\) satisfying \((3 - n)(m - 2) = -6\). To find the exact values, though unnecessary, we equate \((3 - n)\) and \((m - 2)\) to each pair of factors above.


Answer: D

Duplicates cannot occur because (3 - n) and (m - 2) are equated to different values in each case. Since each equation directly determines a unique value for n and m, no two cases can produce the same (m, n) pair.
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what about (0,0)?
Bunuel
Official Solution:

If \(m\) and \(n\) are integers, how many pairs of \((m, n)\) are there such that \(3m + 2n = mn\)?

A. 4
B. 5
C. 6
D. 8
E. 10


Arguably the best way to solve this question is to manipulate \(3m + 2n = mn\) such that we can factor it into \((m + a)(n + b) = integer\). Then, benefiting from the fact that \(m\) and \(n\) are integers, we'd be able to find the sets of \((m, n)\) satisfying the equation.

\(3m + 2n = mn\);

\(3m - mn + 2n = 0[\);

\(m(3 - n) + 2n = 0\).

To further factor \((3 - n)\), add -6 to both sides:

\(m(3 - n) - 6 + 2n= -6\);

\(m(3 - n) - 2(3 - n) = -6\);

\((3 - n)(m - 2) = -6\).

Now, the question essentially boils down to finding how many ways -6 can be written as the product of two integers, \((3 - n)\) and \((m - 2)\):

\(-6*1\)

\(-3*2\)

\(-2*3\)

\(-1*6\)

\(1*(-6)\)

\(2*(-3)\)

\(3*(-2)\)

\(6*(-1)\)

Therefore, there are eight sets of integers \((m, n)\) satisfying \((3 - n)(m - 2) = -6\). To find the exact values, though unnecessary, we equate \((3 - n)\) and \((m - 2)\) to each pair of factors above.


Answer: D
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Shivang29
what about (0,0)?
Bunuel
Official Solution:

If \(m\) and \(n\) are integers, how many pairs of \((m, n)\) are there such that \(3m + 2n = mn\)?

A. 4
B. 5
C. 6
D. 8
E. 10


Arguably the best way to solve this question is to manipulate \(3m + 2n = mn\) such that we can factor it into \((m + a)(n + b) = integer\). Then, benefiting from the fact that \(m\) and \(n\) are integers, we'd be able to find the sets of \((m, n)\) satisfying the equation.

\(3m + 2n = mn\);

\(3m - mn + 2n = 0[\);

\(m(3 - n) + 2n = 0\).

To further factor \((3 - n)\), add -6 to both sides:

\(m(3 - n) - 6 + 2n= -6\);

\(m(3 - n) - 2(3 - n) = -6\);

\((3 - n)(m - 2) = -6\).

Now, the question essentially boils down to finding how many ways -6 can be written as the product of two integers, \((3 - n)\) and \((m - 2)\):

\(-6*1\)

\(-3*2\)

\(-2*3\)

\(-1*6\)

\(1*(-6)\)

\(2*(-3)\)

\(3*(-2)\)

\(6*(-1)\)

Therefore, there are eight sets of integers \((m, n)\) satisfying \((3 - n)(m - 2) = -6\). To find the exact values, though unnecessary, we equate \((3 - n)\) and \((m - 2)\) to each pair of factors above.


Answer: D

This is counted when m - 2 = 2 and 3 - n = -3.
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