A two-digit number exceeds the sum of the digits of that number by 18. If the digit at the unit's place is double the digit in the ten's place, what is the number?

### Concept & Formula Algebraic representation of two-digit numbers. A two-digit number is written as $10t + u$, where $t$ is the ten's digit and $u$ is the unit's digit. ### Step-by-Step Solution Let the ten's digit be $x$ and the unit's digit be $y$. The number is $10x + y$. Condition 1: The number exceeds the sum of its digits by 18. $10x + y - (x + y) = 18$ $9x = 18$ $x = 2$ So, the ten's digit is 2. Condition 2: The digit at the unit's place is double the digit in the ten's place. $y = 2x$ $y = 2(2) = 4$ The number is $10(2) + 4 = 24$. ### Exam Strategy & Shortcut Option Verification is the fastest route. Check (a) 24: Unit digit (4) is double ten's digit (2). Number (24) minus sum of digits (6) = 18. Matches perfectly. ### Common Pitfall Students might overcomplicate the algebra instead of recognizing that $10x + y - (x + y)$ simply simplifies to $9x$, immediately giving the ten's digit without needing the second condition. ### Final Answer Therefore, the correct answer is **24**.