2005 AMC 12A Problems/Problem 19

Revision as of 17:37, 20 May 2008 by MathAndKnowledge (talk | contribs) (Solution)

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.

$1463-1=\boxed{1462}$

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 12 Problems and Solutions
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