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Don't overthink it, set up a table and keep increasing x from 0 upwards until the second equations perimeters are no longer met. Also go below 0 just to be certain.
Correct pairs
0,12
1,10
2,8
3,6
4,4
5,2
6,0
7,-2
8,-4
9,-6
10,-8
11,-10
12,-12
Correct pairs
0,12
1,10
2,8
3,6
4,4
5,2
6,0
7,-2
8,-4
9,-6
10,-8
11,-10
12,-12
Bunuel
12 Sep 2025, 07:36
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Let me help you tackle this integer solution counting problem systematically.
Understanding the Constraints:
We have \(2x + y = 12\) and \(|y| \leq 12\), where we need both x and y to be integers.
The absolute value constraint \(|y| \leq 12\) means \(-12 \leq y \leq 12\), giving y a range from -12 to +12 inclusive.
Key Insight - Finding Integer Solutions:
From the equation \(2x + y = 12\), we can express:
\(x = \frac{12-y}{2}\)
For x to be an integer, it must be divisible by 2. This happens when \(x\) is even.
Since 12 is even, \(x\) is even when y is even!
Applying the Pattern:
Let me demonstrate with a few values:
When \(y = -12\): \(x = \frac{12-(-12)}{2} = \frac{24}{2} = 12\) ✓ (integer)
When \(y = -11\): \(x = \frac{12-(-11)}{2} = \frac{23}{2} = 11.5\) ✗ (not integer)
When \(y = -10\): \(x = \frac{12-(-10)}{2} = \frac{22}{2} = 11\) ✓ (integer)
When \(y = 0\): \(x = \frac{12-0}{2} = 6\) ✓ (integer)
When \(y = 12\): \(x = \frac{12-12}{2} = 0\) ✓ (integer)
Counting Valid Pairs:
The even values of y in the range \([-12, 12]\) are:
-12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12
That's 13 even values, and each gives us an integer value for x.
Answer: (D) 13
Master this problem type systematically! Check out the complete solution on Neuron by e-GMAT to see:
Access detailed explanations for all official GMAT questions and build pattern recognition skills here on Neuron.
Understanding the Constraints:
We have \(2x + y = 12\) and \(|y| \leq 12\), where we need both x and y to be integers.
The absolute value constraint \(|y| \leq 12\) means \(-12 \leq y \leq 12\), giving y a range from -12 to +12 inclusive.
Key Insight - Finding Integer Solutions:
From the equation \(2x + y = 12\), we can express:
\(x = \frac{12-y}{2}\)
For x to be an integer, it must be divisible by 2. This happens when \(x\) is even.
Since 12 is even, \(x\) is even when y is even!
Applying the Pattern:
Let me demonstrate with a few values:
When \(y = -12\): \(x = \frac{12-(-12)}{2} = \frac{24}{2} = 12\) ✓ (integer)
When \(y = -11\): \(x = \frac{12-(-11)}{2} = \frac{23}{2} = 11.5\) ✗ (not integer)
When \(y = -10\): \(x = \frac{12-(-10)}{2} = \frac{22}{2} = 11\) ✓ (integer)
When \(y = 0\): \(x = \frac{12-0}{2} = 6\) ✓ (integer)
When \(y = 12\): \(x = \frac{12-12}{2} = 0\) ✓ (integer)
Counting Valid Pairs:
The even values of y in the range \([-12, 12]\) are:
-12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12
That's 13 even values, and each gives us an integer value for x.
Answer: (D) 13
Master this problem type systematically! Check out the complete solution on Neuron by e-GMAT to see:
- Why this even-odd pattern emerges algebraically
- Alternative approaches that save time on similar problems
- Common variations where the pattern changes
- A framework for handling any constraint-based integer counting problem
Access detailed explanations for all official GMAT questions and build pattern recognition skills here on Neuron.
