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The number of pairs of integers \((x , y)\) satisfying \(x|\geq y\geq-20\) and \(2x+5y=99\) is

Updated On: Jan 15, 2026
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Solution and Explanation

Step 1: Express \(x\) in terms of \(y\)

Given the equation: \[ 2x + 5y = 99 \Rightarrow x = \frac{99 - 5y}{2} \] For \(x\) to be an integer, \(99 - 5y\) must be an even number. Since 99 is odd, \(5y\) must also be odd for their difference to be even. \(5y\) is odd only when \(y\) is odd. Therefore: \[ y \text{ must be odd} \]

Step 2: Determine valid integer values for \(y\)

Test case for \(y = -19\): \[ x = \frac{99 - 5(-19)}{2} = \frac{99 + 95}{2} = \frac{194}{2} = 97 \] Test case for \(y = 13\): \[ x = \frac{99 - 5(13)}{2} = \frac{99 - 65}{2} = \frac{34}{2} = 17 \] The valid values for \(y\) are odd integers ranging from \(-19\) to \(13\).

Step 3: Calculate the quantity of valid integer pairs

The sequence of valid \(y\) values forms an arithmetic progression: \[ a = -19,\quad d = 2,\quad t_n = 13 \] Using the formula for the nth term: \[ t_n = a + (n - 1)d \] Substituting the values: \[ 13 = -19 + (n - 1) \cdot 2 \Rightarrow 32 = 2(n - 1) \Rightarrow n = 17 \] There are 17 distinct integer values for \(y\), each yielding an integer \(x\) such that \(x \geq y\).

✅ Final Answer: 17 integer pairs

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