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

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== Solution ==
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== Solution 1==
 
We find the number of numbers with a <math>4</math> and subtract from <math>2005</math>. Quick counting tells us that there are <math>200</math> numbers with a 4 in the hundreds place, <math>200</math> numbers with a 4 in the tens place, and <math>201</math> numbers with a 4 in the units place (counting <math>2004</math>). Now we apply the [[Principle of Inclusion-Exclusion]]. There are <math>20</math> numbers with a 4 in the hundreds and in the tens, and <math>20</math> for both the other two [[intersection]]s. The intersection of all three sets is just <math>2</math>. So we get:
 
We find the number of numbers with a <math>4</math> and subtract from <math>2005</math>. Quick counting tells us that there are <math>200</math> numbers with a 4 in the hundreds place, <math>200</math> numbers with a 4 in the tens place, and <math>201</math> numbers with a 4 in the units place (counting <math>2004</math>). Now we apply the [[Principle of Inclusion-Exclusion]]. There are <math>20</math> numbers with a 4 in the hundreds and in the tens, and <math>20</math> for both the other two [[intersection]]s. The intersection of all three sets is just <math>2</math>. So we get:
  

Revision as of 22:41, 3 November 2013

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 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, $2005_9=9^3(2)+9^0(5)=729(2)+1(5)=1458+5=1463$.

$1463-1=\boxed{1462}$

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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