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 base 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 $2005-19-252-272=1462,$ or $\boxed{B}.$

~coolmath2017

Solution 5(Super fast)

This is very analogous to base $9$ . But, in base $9$ , we don't have a $9$ . So, this means that these are equal except for that base 9 will be one more than the operation here. $2005_9 = 5+0+0+1458 = 1463$ . $1463 - 1 = 1462$

Therefore, our answer is $\boxed {1462}$

~Arcticturn

2005 AMC 12A (Problems • Answer Key • Resources)
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

Page

Toolbox

Search