A person writes down every 4-digit number from 1000 to 9999. How many times in…

Concept: To count how many times a specific digit appears across a full range of numbers, count its occurrences position by position (units, tens, hundreds, thousands, ...). For each position, the count equals the number of ways every other position can independently take any valid value, remembering that a leading digit can never be 0.

Applying this to all 4-digit numbers (1000 to 9999 — 9000 numbers in all, with the thousands digit running from 1 to 9):

1. Thousands place: the digit 2 occupies the thousands place only for 2000-2999, i.e. 1000 numbers, giving 1000 occurrences.

2. Hundreds place: the digit 2 occupies the hundreds place whenever the thousands digit is any of the 9 valid nonzero digits (1-9) and the tens and units digits are each any of the 10 digits (0-9): 9 × 10 × 10 = 900 occurrences.

3. Tens place: by the same reasoning (thousands digit 9 choices, hundreds and units digits 10 choices each): 9 × 10 × 10 = 900 occurrences.

4. Units place: by the same reasoning: 9 × 10 × 10 = 900 occurrences.

5. Total: adding all four positions gives 1000 + 900 + 900 + 900 = 3700.

Cross-check: the 9000 four-digit numbers together contain 9000 × 4 = 36000 digit-slots in total. By the same position-wise reasoning, every nonzero digit (1 through 9) totals 3700 occurrences, while digit 0 (which can never lead) totals only 900 × 3 = 2700 occurrences. Summing: 9 × 3700 + 2700 = 33300 + 2700 = 36000, which matches the total digit-slots exactly — confirming 3700 is correct.