jmc

The easiest approach is to consider how many times 8 can appear in the units place, how many times in the tens place, and how many times in the hundreds place. If we put a 8 in the units place, there are 10 choices for the tens place and 10 choices for the hundreds place (including having no hundreds digit) for a total of $10\times 10=100$ options, which means that 8 will appear in the ones place 100 times. (If we choose the hundreds place to be equal to 0, we can just think of that as either a two digit or a one digit number.) Likewise, if we put an 8 in the tens place, there are 10 choices for the units place and 10 choices for the hundreds place for a total of 100 options and 100 appearances of 8 in the tens place. Finally, if we put a 8 in the hundreds place, we have 10 options for the units place and 10 options for the tens place for another 100 options and 100 appearances of 8. Since $100+100+100=300$, there will be a total of $\boxed{300}$ appearances of 8.