Aspire Faculty ID #13265 · Topic: JEE Main 2023 (30 January Evening Shift) · Just now
JEE Main 2023 (30 January Evening Shift)

The number of ways of selecting two numbers $a$ and $b$, $a\in\{2,4,6,\ldots,100\}$ and $b\in\{1,3,5,\ldots,99\}$ such that $2$ is the remainder when $a+b$ is divided by $23$ is:

Solution

$a = 2x,; x=1,2,\dots,50$
 
$b = 2y-1,; y=1,2,\dots,50$

$a+b = 2x + (2y-1) = 2(x+y) - 1$

$2(x+y) - 1 = 23k + 2$

$2(x+y) = 23k + 3$

$23k + 3$ must be even ⟹ $k$ is odd

Let $k = 2t + 1$

$2(x+y) = 23(2t+1) + 3 = 46t + 26$

$x+y = 23t + 13$

Now $1 \le x,y \le 50$ 
⟹ $2 \le x+y \le 100$
So possible values:
$x+y = 13,;36,;59,;82$
For $x+y=13 \Rightarrow 12$
For $x+y=36 \Rightarrow 35$
For $x+y=59 \Rightarrow 42$
For $x+y=82 \Rightarrow 19$
$12+35+42+19 = 108$
$\boxed{108}$
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