A certain street has 1000 buildings. A sign-maker is contracted to number the houses from 1 to 1000. How many zeroes will he need?

To solve this problem, we need to consider how many times the number '0' appears in each set of numbers:

1. From 1 to 9, there are no zeroes.
2. From 10 to 99, '0' appears 10 times in the units place (10, 20, 30, ..., 90) and 10 times in the tens place (10, 20, ..., 90). So, a total of 20 times.
3. From 100 to 999, '0' appears:
   • 100 times in the units place (100, 110, 120, ..., 990)
   • 100 times in the tens place (100, 101, ..., 199, 200, 201, ..., 299, ..., 900, 901, ..., 999)
   • 10 times in the hundreds place (100, 200, ..., 1000) So, a total of 210 times.
4. In 1000, '0' appears 3 times.

Adding these all up, the sign-maker will need 0 (from step 1) + 20 (from step 2) + 210 (from step 3) + 3 (from step 4) = 233 zeroes.