Difference between revisions of "2005 AMC 12A Problems/Problem 19"
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===Solution 2=== | ===Solution 2=== | ||
| − | Alternatively, consider that counting without the number <math>4</math> is almost equivalent to counting in base <math>9</math>; only, in base <math>9</math>, the number <math>9</math> is not counted. Since <math>4</math> is skipped, the symbol <math>5</math> represents <math>4</math> miles of travel, and we have traveled <math>2004_9</math> miles. By basic conversion, <math> | + | Alternatively, consider that counting without the number <math>4</math> is almost equivalent to counting in base <math>9</math>; only, in base <math>9</math>, the number <math>9</math> is not counted. Since <math>4</math> is skipped, the symbol <math>5</math> represents <math>4</math> miles of travel, and we have traveled <math>2004_9</math> miles. By basic conversion, <math>2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}</math> |
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== See also == | == See also == | ||
Revision as of 19:59, 16 April 2017
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?
Solutions
Solution 1
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:
Solution 2
Alternatively, consider that counting without the number is almost equivalent to counting in base 
; only, in base , the number is not counted. Since is skipped, the symbol 
represents miles of travel, and we have traveled 
miles. By basic conversion, 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
