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:

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

Alternatively, consider that counting without the number $4$ is almost equivalent to counting in base $9$ ; only, in base $9$ , the number $9$ is not counted. So, when the number $2005$ is expressed in base $9$ , the result should be very similar to the answer to the problem. By basic conversion, $2005_9=9^3(2)+9^0(5)=729(2)+1(5)=1458+5=1463$ . One can see that the result, $1463$ , has a $4$ itself; therefore, since the true answer to the problem does not count $4$ 's, we subtract $1$ from the result to get the true answer to the problem.

$1463-1=\boxed{1462}$

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2005 AMC 12A Problems/Problem 19

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