JEE Mathematics 2023 Question with Solution

Step-by-step solution

Common mistakes

• Mistake: Treating the selection as unordered and dividing the count by $2$. Why it is wrong: $a$ and $b$ come from two different sets, so each valid pair is naturally an ordered pair $(a,b)$. What to do instead: Count all valid choices of $a$ with corresponding valid $b$ directly.

• Mistake: Using only $a+b=25$ and forgetting other values congruent to $2 \pmod{23}$. Why it is wrong: The condition is on the remainder modulo $23$, so all feasible sums of the form $23k+2$ within the allowed range must be considered. What to do instead: List every admissible sum value before counting pairs.

• Mistake: Ignoring the parity restriction from the sets. Why it is wrong: $a$ is always even and $b$ is always odd, so $a+b$ must be odd; any even candidate sum is impossible. What to do instead: Check compatibility of the required sum with the parity of the chosen sets first.