My attempt was as follows, I took all values of y starting from 12and -12 (largest absolute given), and placed those values for which x was an integer.

x y
0 12
12 -12
1 10
11 -10
2 8
10 -8
3 6
9 -6
4 4
8 -4
5 2
7 -2
6 0

I realized there was a pattern which was following, so I din make the whole table, just realized it would be 13.....
but here , I have made the whole table...

Lets hope the answer is 13 !

I would approach like this..

-12 <= y <= 12

For x to be an integer, Y must be even.

So out of 25 values of y, only 13 are even ( -12,-10 ......., 0 , 2, 4 , .... 10,12)

So there are 13 possible pairs that both x and y are integers.

Mani_mba wrote:I would approach like this..

-12 <= y <= 12

For x to be an integer, Y must be even.

So out of 25 values of y, only 13 are even ( -12,-10 ......., 0 , 2, 4 , .... 10,12)

So there are 13 possible pairs that both x and y are integers.

Nice solution.

Very real to GMAT type question.

Really good approach

Mani_mba wrote:I would approach like this..

-12 <= y <= 12

For x to be an integer, Y must be even.

So out of 25 values of y, only 13 are even ( -12,-10 ......., 0 , 2, 4 , .... 10,12)

So there are 13 possible pairs that both x and y are integers.

2x+y = 12, or x = 6- y/2, that implies, even y will give integer x
mody<=12, or 12<=y<=12, or y ranges from -12,-10,-8,.....0,............12
its an AP, -12+(n-1)2 = 12, that implies, n = 13 and hence 13 pairs.

Given 2x+y=12 and |y|<=12, it means -12 <= y <= 12. For this range, x is between 12 and 0, inclusive. There are only 13 values for x. Hence I can have 13 ordered pairs.