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Senior Manager
Joined: 03 May 2007
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2x + y = 12 |y| <= 12 For how many ordered pairs (x , y) [#permalink]
 30 Aug 2007, 20:56
2x + y = 12
|y| <= 12
For how many ordered pairs (x , y) that are solutions of the system above are x and y both integers?
A. 7
B. 10
C. 12
D. 13
E. 14
P.S. please no answers like "i think it's B or C" 
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Manager
Joined: 17 Aug 2007
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Re: I have no clue [#permalink]
30 Aug 2007, 21:23
Sergey_is_cool wrote: 2x + y = 12 |y| <= 12 For how many ordered pairs (x , y) that are solutions of the system above are x and y both integers? A. 7 B. 10 C. 12 D. 13 E. 14 P.S. please no answers like "i think it's B or C" since |y| is 12 so y lies between -12 to 12. now if we substitute -12 in the equation we get x = 12 , if substitute -11 we get a fraction. again if we substitute y = -10 we get x = 11 if we substitute y = -9 we get x as a fraction. so since we are interested only in integers, there would half as many pairs as the range of Y which 24/2 = 12. so answer is C
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Director
Joined: 03 May 2007
Posts: 903
Schools: University of Chicago, Wharton School
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Re: I have no clue [#permalink]
30 Aug 2007, 21:25
Sergey_is_cool wrote: 2x + y = 12 |y| <= 12 For how many ordered pairs (x , y) that are solutions of the system above are x and y both integers? A. 7 B. 10 C. 12 D. 13 E. 14 P.S. please no answers like "i think it's B or C" D. 13 
cus y can olny be even integer between -12 to 12 inncluding.
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Senior Manager
Joined: 19 Feb 2007
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when y= 12, x = 0
when y = -12, x = 12
so x can have values from 0 to 12 inclusive..
so (D)
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Manager
Joined: 17 Aug 2007
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sidbidus wrote: when y= 12, x = 0
when y = -12, x = 12
so x can have values from 0 to 12 inclusive..
so (D)
yep .. my mistake.. missed the godamm zero in between...
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VP
Joined: 08 Jun 2005
Posts: 1172
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hello sergey ! nice to see you back in the forum.
see here:
http://www.gmatclub.com/forum/7/46071
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GMAT Club Legend
Joined: 07 Jul 2004
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x = (12-y)/2 = 6 - y/2
So we are going for even values of y. That's 12,10,8,6,4,2,0,-2,-4,-6,-8,-10,-12. Total 13 solutions.
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Intern
Joined: 31 Aug 2007
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Re: I have no clue [#permalink]
01 Sep 2007, 00:27
Keeping in mind the second rule of the problem, start by trying 1, 0, and -1 for x (simple, quick numbers) and you notice that x>=0.
Now try to find the top end for x, I started with 6, too low. 12, that's as high as you can go for x, maxing out y.
12-(-1) = 13.
And just because I am not actually taking the test, and am kind OCD:
0,12
1,10
2,8
3,6
4,2
5,2
6,0
7,-2
8,-4
9,-6
10,-8
11,-10
12,-12
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Re: I have no clue
[#permalink]
01 Sep 2007, 00:27
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