Difference between revisions of "2005 AMC 12A Problems/Problem 19"
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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. | ||
− | So, when the number <math>2005</math> is expressed in base <math>9</math>, the result should be very similar to the answer to the problem. By basic conversion, <math>2005_9=9^3(2)+9^0(5)=729(2)+1(5)=1458+5=1463</math>. One can see that the result, 1463, has a 4 itself; therefore, since the true answer to the problem does not count <math>4</math>'s, we subtract <math>1</math> from the result to get the true answer to the problem. | + | So, when the number <math>2005</math> is expressed in base <math>9</math>, the result should be very similar to the answer to the problem. By basic conversion, <math>2005_9=9^3(2)+9^0(5)=729(2)+1(5)=1458+5=1463</math>. One can see that the result, <math>1463</math>, has a <math>4</math> itself; therefore, since the true answer to the problem does not count <math>4</math>'s, we subtract <math>1</math> from the result to get the true answer to the problem. |
<math>1463-1=\boxed{1462}</math> | <math>1463-1=\boxed{1462}</math> |
Revision as of 17:38, 20 May 2008
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. So, when the number is expressed in base , the result should be very similar to the answer to the problem. By basic conversion, . One can see that the result, , has a itself; therefore, since the true answer to the problem does not count 's, we subtract from the result to get the true answer to the problem.
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 |