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? $(\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804$

Solutions

Solution 1

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:

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

Solution 2

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. Since $4$ is skipped, the symbol $5$ represents $4$ miles of travel, and we have traveled $2004_9$ miles. By basic conversion, $2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}$

Solution 3

Since any numbers containing one or more $4$s were skipped, we need only to find the numbers that don't contain a $4$ at all. First we consider $1$ - $1999$. Single digits are simply four digit numbers with a zero in all but the ones place (this concept applies to double and triple digits numbers as well). From $1$ - $1999$, we have $2$ possibilities for the thousands place, and $9$ possibilities for the hundreds, tens, and ones places. This is $2 \cdot 9 \cdot 9 \cdot 9-1$ possibilities (because $0000$ doesn't count) or $1457$ numbers. From $2000$ - $2005$ there are $6$ numbers, $5$ of which don't contain a $4$. Therefore the total is $1457 + 5$, or $1462$ $\Rightarrow$ $\boxed{\text{B}}$.

Solution 4

We seek to find the amount of numbers that contain at least one $4,$ and subtract this number from $2005.$

We can simply apply casework to this problem.

The amount of numbers with at least one $4$ that are one or two digit numbers are $4,14,24,34,40-49,54,\cdots,94$ which gives $19$ numbers.

The amount of three digit numbers with at least one $4$ is $8*19+100=252.$

The amount of four digit numbers with at least one $4$ is $252+1+19=272$

This, our answer is $2005-19-252-272=162,$ or $\boxed{B}.$

~coolmath2017

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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