2024 CSAT — Q35

Write $X = 10a + b$ (tens digit $a$, units $b$); the interchange gives $Y = 10b + a$. Then

The greatest two-digit number is $99$, so $11(a+b) = 99 \Rightarrow a + b = 9$.

Now count valid digit pairs. $X$ is two-digit so $a \ge 1$. $Y$ must also be two-digit, so its tens digit $b \ge 1$. With $a + b = 9$ and both $\ge 1$:

That is $8$ values of $X$. (The pair $a=9, b=0$ is excluded because $Y = 09$ is not a two-digit number.)

Answer: (d) 8.

The identity $X + Y = 11(a+b)$ is the whole solve: a two-digit number plus its reverse is always a multiple of $11$, so “greatest two-digit sum” forces $a+b=9$ (since $11\times9=99$). The only subtlety is the two-digit constraint on $Y$: $b\ge1$ kills the $(9,0)$ pair, trimming the count from $9$ to $8$. Each ordered $X$ counts separately ($18\ne81$), so do not halve. Recognising the $11(a+b)$ structure converts a brute-force digit hunt into a one-line divisibility read.