The total number of integers between 200 and 400, each of which either begins with 3 or ends with 3 or both is

### Concept & Strategy This problem relies on set theory, specifically the Principle of Inclusion-Exclusion, or simple logical division of a range to avoid double counting. We must count numbers that start with $3$, numbers that end with $3$, and ensure the overlap (numbers that start AND end with $3$) is only counted once. ### Step-by-Step Solution * **Given:** The target range is between $200$ and $400$. We are looking for integers that either begin with $3$, end with $3$, or both. * **Step 1: Count numbers that begin with 3.** Within the $200$ to $400$ range, all numbers in the $300$s begin with $3$. These span from $300$ to $399$. Total numbers = $100$. *(Note: This block already includes all numbers that begin AND end with $3$, like $303, 313$, etc.)* * **Step 2: Count numbers that end with 3, but do NOT begin with 3.** Since we already completely covered the $300$s, we only need to look at the remaining part of the range: the $200$s (from $200$ to $299$). Numbers in this block ending in $3$ are: $203, 213, 223, 233, 243, 253, 263, 273, 283, 293$. Total numbers = $10$. * **Step 3: Combine the counts.** Total valid integers = (Numbers starting with $3$) + (Numbers ending with $3$ in the remaining range) Total = $100 + 10 = 110$. ### Exam Strategy & Shortcut Break the problem into mutually exclusive sets to bypass inclusion-exclusion formulas entirely. Set A: The entire $300$ series ($100$ numbers). Set B: The $200$ series ending in $3$. Since it's exactly one $100$-number block, any specific unit digit appears exactly $10$ times. $100 + 10 = 110$. This mental math takes less than 10 seconds. ### Common Pitfall Using the union formula $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ but miscalculating the sets. Students might incorrectly say there are $20$ numbers ending in $3$ between $200$ and $400$, add that to $100$, and then subtract the $10$ overlaps. While mathematically sound ($100 + 20 - 10 = 110$), it introduces unnecessary steps and room for arithmetic error. ### Final Answer **Therefore, the correct answer is 110.**