Difference between revisions of "2005 AMC 12A Problems/Problem 19"
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== Problem == | == 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? | + | A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads <tt>002005</tt>, how many miles has the car actually traveled? |
<math> | <math> | ||
(\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804 | (\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804 | ||
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== Solution == | == Solution == | ||
| − | 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 | + | 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: |
<div style="text-align:center;"><math>2005-(200+200+201-20-20-20+2) = 1462 \Longrightarrow \mathrm{(B)}</math></div> | <div style="text-align:center;"><math>2005-(200+200+201-20-20-20+2) = 1462 \Longrightarrow \mathrm{(B)}</math></div> | ||
Revision as of 21:30, 21 September 2007
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?
Solution
We find the number of numbers with a 
and subtract from 
. Quick counting tells us that there are 
numbers with a 4 in the hundreds place, numbers with a 4 in the tens place, and 
numbers with a 4 in the units place (counting 
). Now we apply the Principle of Inclusion-Exclusion. There are 
numbers with a 4 in the hundreds and in the tens, and for both the other two intersections. The intersection of all three sets is just 
. So we get:
See also
| 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 | |
