The smallest number of $5$ digits beginning with $3$ and ending with $5$ will be

### Concept & Logic To construct the absolute minimum value for a number with fixed boundaries (starting and ending digits), all free internal positions must be filled with the lowest possible digit, which is $0$. ### Step-by-Step Solution **Deduction:** 1. The number has 5 places: [Ten-Thousands] [Thousands] [Hundreds] [Tens] [Units]. 2. Condition 1: It begins with $3$. So the Ten-Thousands place is $3$. (Format: $3 \_ \_ \_ \_$) 3. Condition 2: It ends with $5$. So the Units place is $5$. (Format: $3 \_ \_ \_ 5$) 4. Condition 3: It must be the smallest possible number. Therefore, fill the Thousands, Hundreds, and Tens places with the minimum digit $0$. 5. The final number is $30005$. ### Exam Strategy & Shortcut You do not need to construct the number manually. Simply scan the multiple-choice options. Since all options start with $3$ and end with $5$, compare their magnitudes directly: $30005 < 30015 < 30025 < 31005$. Option C is strictly the smallest. ### Common Pitfall A common mistake is repeating the starting digit in the middle (e.g., $33335$) or using $1$s ($31115$) instead of $0$s, failing to realize that $0$ is permissible and necessary for a minimum value. ### Final Answer Therefore, the correct answer is **30005**.