In a number of three digits, the digits in the unit's place and in the hundred's place are equal and the sum of all the digits is 8. The number of such numbers is

### Concept & Logic To find the possible numbers, we can represent the three-digit number algebraically and use the given constraints to find integer solutions. A three-digit number can be written based on its place values, but here we only need to focus on the individual digits. Let the three-digit number be represented by the digits $x y x$, where $x$ is the digit in the hundred's and unit's place, and $y$ is the digit in the ten's place. ### Step-by-Step Solution * **Given:** The sum of the digits is 8. * **Equation:** $x + y + x = 8$ * **Simplification:** $2x + y = 8$ * **Deduction:** Since $x$ is the leading digit of a three-digit number, it cannot be $0$. Thus, $x \ge 1$. Both $x$ and $y$ must be single-digit integers (0 to 9). * Let's test possible values for $x$: * If $x = 1$: $2(1) + y = 8 \implies y = 6$. The number is $161$. * If $x = 2$: $2(2) + y = 8 \implies y = 4$. The number is $242$. * If $x = 3$: $2(3) + y = 8 \implies y = 2$. The number is $323$. * If $x = 4$: $2(4) + y = 8 \implies y = 0$. The number is $404$. * If $x = 5$: $2(5) + y = 8 \implies 10 + y = 8 \implies y = -2$. (Not possible as digits cannot be negative). There are exactly 4 valid combinations. ### Exam Strategy & Shortcut For digit-sum problems with symmetrical conditions, systematically plug in values for the repeated digit starting from $1$. You can quickly write out $2x + y = 8$ and list the pairs $(1,6), (2,4), (3,2), (4,0)$ in your head within seconds. Stop as soon as $2x \ge 8$. ### Common Pitfall Students often forget that the hundred's digit cannot be zero. If $x=0$, $y=8$, giving the number $080$, which is technically a two-digit number ($80$), not a three-digit number. Always remember the leading digit constraint $x \neq 0$. ### Final Answer **Therefore, the correct answer is 4.**