Difference between revisions of "2005 AMC 12A Problems/Problem 19"

Problem

A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005, how many miles has the car actually traveled? $(\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804$

Solution

We find the number of numbers with a $4$ and subtract from $2005$. Quick counting tells us that there are $200$ numbers with a 4 in the hundreds place, $200$ numbers with a 4 in the tens place, and $201$ numbers with a 4 in the units place (counting $2004$). Now we apply the Principle of Inclusion-Exclusion. There are $20$ numbers with a 4 in the hundreds and in the tens, and $20$ for both the other two intersections. The intersection of all three sets is just $2$. So we get:

$2005-(200+200+201-20-20-20+2) = 1462 \Longrightarrow \mathrm{(B)}$