Problem Solving

[Math Revolution GMAT math practice question]

x and y are non-negative integers. If xy+2x+3y=0, then y=?

A. 0
B. 1
C. 2
D. 3
E. 4

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

x and y are non-negative integers. If xy+2x+3y=0, then y=?

A. 0
B. 1
C. 2
D. 3
E. 4

$$\left\{ \matrix{
x,y \ge 0\,\,{\rm{ints}}\,\,\,\left( * \right) \hfill \cr
xy + 2x + 3y = 0\,\,\left( {**} \right) \hfill \cr} \right.\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,? = y$$

You should realize the pair (x,y)=(0,0) satisfies all above, hence y=0 must be the right answer.

On the other hand, let´s solve the problem throughly, for didactic reasons:
$$\left( {**} \right)\,\,\, \Rightarrow \,\,\,x\left( {y + 2} \right) + 3y + \underline 6 = \underline 6 \,\,\,\,\, \Rightarrow \,\,\,\,x\left( {y + 2} \right) + 3\left( {y + 2} \right) = 6$$
$$\left( {y + 2} \right)\left( {x + 3} \right) = 6\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,y + 2\,\,{\rm{and}}\,\,x + 3\,\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{pair}}\,\,{\rm{of}}\,\,{\rm{positive}}\,\,{\rm{divisors}}\,\,{\rm{of}}\,\,6$$
$$\left\{ \matrix{
y + 2 = 1 \hfill \cr
x + 3 = 6 \hfill \cr} \right.\,\,\,\, \Rightarrow \,\,\,\,y < 0$$
$$\left\{ \matrix{
y + 2 = 2 \hfill \cr
x + 3 = 3 \hfill \cr} \right.\,\,\,\, \Rightarrow \,\,\,\,\left( {x,y} \right) = \left( {0,0} \right)\,\,\,{\rm{viable}}\,\,\,\, \Rightarrow \,\,\,\,\,? = y = 0$$
$$\left\{ \matrix{
y + 2 = 3 \hfill \cr
x + 3 = 2 \hfill \cr} \right.\,\,\,\, \Rightarrow \,\,\,\,x < 0$$
$$\left\{ \matrix{
y + 2 = 6 \hfill \cr
x + 3 = 1 \hfill \cr} \right.\,\,\,\, \Rightarrow \,\,\,\,x < 0$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

x and y are non-negative integers. If xy+2x+3y=0, then y=?

A. 0
B. 1
C. 2
D. 3
E. 4

NONNEGATIVE implies that we should test ZERO.
Let x=0.
Substituting x=0 into xy + 2x + 3y = 0, we get:
0*y + 2*0 + 3y = 0
3y = 0
y = 0

The correct answer is A.

=>
xy+2x+3y=0
=> xy+2x+3y+6=6
=> (x+3)(y+2)=6
Since x and y are integers, x+3 and y+2 are integers.
The possible pairs (x+3,y+2) are as follows:
( x+3, y+2 ) = (1,6), (2,3), (3,2), (6,1), (-1,-6), (-2,-3), (-3,-2) and (-6,-1).
The corresponding pairs (x,y) are:
(x,y) = (-2,4), (-1,2), (0,0), (3,-1), (-4,-8), (-5,-5), (-6,-4) and (-9,-3).
The unique pair with non-negative x and y is (x,y)=(0,0).
Thus, y = 0.

Therefore, the answer is A.
Answer: A