A person writes down all the 4-digit numbers, one after another. How many…

Concept: To count how many times a particular digit appears across a full range of numbers, count its occurrences place by place (units, tens, hundreds, thousands, ...) and add them up. In a complete range where every value of a place cycles through 0-9 equally often, a fixed digit occupies that place in exactly one-tenth of the numbers - except the leading (highest) place of the range, which never takes the value 0, so it must be counted directly rather than as one-tenth of the block.

1. The 4-digit numbers run from 1000 to 9999 - a block of 9000 numbers with four place values: thousands, hundreds, tens, and units.

2. Thousands place: this digit ranges only over 1-9 (never 0). Fixing it at 2, the other three places (hundreds, tens, units) vary freely over 0-9 each, giving 10 x 10 x 10 = 1000 numbers - so digit 2 occupies the thousands place 1000 times.

3. Hundreds place: the thousands digit ranges over its 9 possible values (1-9), the hundreds digit is fixed at 2, and the tens and units digits vary freely over 0-9 each, giving 9 x 10 x 10 = 900 numbers - so digit 2 occupies the hundreds place 900 times.

4. Tens place: by the same reasoning (thousands free over 1-9, hundreds free over 0-9, tens fixed at 2, units free over 0-9), digit 2 occupies the tens place 9 x 10 x 10 = 900 times.

5. Units place: symmetric to the tens place, digit 2 occupies the units place 9 x 10 x 10 = 900 times.

6. Adding the four place-wise counts: 1000 + 900 + 900 + 900 = 3700 total occurrences of the digit 2.

Cross-check: every digit 1-9 appears exactly 1000 times in the thousands place (since that place cycles through 1-9 equally across the 9000 numbers), so every digit from 1 to 9 has the same total of 1000 + 900 + 900 + 900 = 3700; only digit 0 is different, with total 900 + 900 + 900 = 2700 because it never occupies the thousands place. Summing across all ten digits - 2700 (for 0) + 3700 x 9 (for 1 through 9) - gives 36000, which equals the total digit count 9000 numbers x 4 digits each = 36000, confirming the breakdown is consistent.

So the digit 2 is written 3700 times in total.