Difference between revisions of "2005 AMC 12A Problems/Problem 19"
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<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> | ||
− | 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. | + | 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>2005_9=9^3(2)+9^0(5)=729(2)+1(5)=1458+5=1463</math>. |
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<math>1463-1=\boxed{1462}</math> | <math>1463-1=\boxed{1462}</math> |
Revision as of 20:30, 24 September 2010
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:
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 |