<p>The number of times the digit 3 will be written when listing the integers from 1 to 1000, is</p>
Step-by-Step Solution
Key Concept: Count the digit 3 by considering cases based on how many positions contain the digit 3, then multiply by the number of occurrences in each case.
<p><strong>Solution:</strong> Since 3 does not occur in 1000, we count from 1 to 999.</p><p>Any number between 1 and 999 is of the form $xyz$, where $0 \leq x, y, z \leq 9$.</p><p><strong>Case 1:</strong> 3 occurs exactly once: $^3C_1 \times 9 \times 9 = 243$ numbers</p><p><strong>Case 2:</strong> 3 occurs exactly twice: $^3C_2 \times 9 = 27$ numbers</p><p><strong>Case 3:</strong> 3 occurs in all three digits: 1 number</p><p>Total occurrences of digit 3: $1 \times 243 + 2 \times 27 + 3 \times 1 = 243 + 54 + 3 = 300$</p>
Correct Answer: C