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

Given that,

• $x \geq y \geq 20 \quad \longrightarrow (1)$
• $2x + 5y = 99 \quad \longrightarrow (2)$

From equation $(2),$ we get.

$x = \frac{99 – 5y}{2}$

Now, we can put the value of $x$ in equation $(1),$ we get.

$\frac{99 – 5y}{2} \geq y \geq – 20 \quad \longrightarrow (3)$

First we take,

$\frac{99 – 5y}{2} \geq y$

$\Rightarrow 2y \leq 99 – 5y$

$\Rightarrow 7y \leq 99$

$\Rightarrow y \leq \frac{99}{7}$

$\Rightarrow y \leq 14.1428$

$\Rightarrow y = \left \lfloor 14.1428 \right \rfloor$

$\Rightarrow \boxed{y = 14}$

So, $\boxed{ – 20 \leq y \leq 14} \quad \longrightarrow (4)$

From equation $(2),$ we get

$\underbrace{2x}_{\text{Always even}} = \underbrace{99 – 5y}_{\text{Even}}$

• $5y \rightarrow \text{odd}$
• $y \rightarrow \text{odd}$

Therefore, we have to find out all the odd integers from the range of $y \in [– 20, 14],$ and for each such value of $y,$ we will find the unique of $x.$

Odd integer of $y : ( \;\underbrace{– 19, – 17, – 15, – 13, – 11, – 9, – 7, – , 5, – 3, – 1}_{\text{Negative integers}}\;, \; \underbrace{1, 3, 5, 7, 9, 11, 13}_{\text{Positive integers}}\;)$

So, the total number of integers that, $y$ can takes is $17.$

$\therefore$ The number of pairs of integers $(x,y)$ is $17.$

Correct Answer $:17$

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