In the numbers from 1 to 500, both included, how many contain the digit 7 at least once?

Explanation

1. 01

   Count the numbers with NO 7. Write 0 to 499 as three digits (000 to 499): hundreds digit 0 to 4 (5 ways), tens and units each 9 ways (any digit except 7).

2. 02

   That gives $5 \times 9 \times 9 = 405$ strings. Drop 000 and add 500 (which has no 7): still 405 numbers from 1 to 500 without a 7.

3. 03

   Numbers with at least one 7 $= 500 - 405 = 95$.