How many three digit numbers contain the digit 3?

Answer 1

I think this will work:

Since the numbers have to at least have a digit in the hundreds position, if you find how many of them there are in all the one-hundreds, you can assume the same amount in the other hundreds, except for the three-hundreds, in which every number has a 3.

For the one-hundreds:

3 in the last position: 103, 113, 123, 133, 143, 153, 163, 173, 183, 193 (10 numbers)

3 in the second position: 130, 131, 132, 134, ...139 (9 numbers, because we already counted the 133 above, where it was in the last position, so we don't want to count that number twice)

For the two-hundreds (notice that it's the same as for the one-hundreds):

3 in the last position: 203, 213, 223..293 (10 numbers)

3 in the second position: 230, 231, 232, ...239 (9 numbers, because we already counted the 233 above, where it was in the last position, so we don't want to count that number twice)

So, for the one-hundreds, there are 19, and for the two-hundreds, there are 19. This is the same for the other hundreds, except for the three-hundreds, which will have 100 numbers, since all of the three-hundreds have at least one 3 in them (in the first position).

Here is a list of how many each hundred has:

1-hundreds: 19

2-hundreds: 19

3-hundreds: 100

4-hundreds: 19

5-hundreds: 19

6-hundreds: 19

7-hundreds: 19

8-hundreds: 19

9-hundreds: 19

I'm assuming you are not including numbers that begin with a zero in the hundreds position. If you are including them, then add another 19 to the total.

The total, without zeroes in the first position, is 8 x 19 + 100 = 252.

The total, with zeroes, is 9 x 19 + 100 = 271.