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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.
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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.
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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>.  
  
 
<math>1463-1=\boxed{1462}</math>
 
<math>1463-1=\boxed{1462}</math>

Revision as of 19:27, 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? $(\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804$

Solution

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:

$2005-(200+200+201-20-20-20+2) = 1462 \Longrightarrow \mathrm{(B)}$

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. So, when the number $2005$ is expressed in base $9$, the result should be very similar to the answer to the problem. 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 (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
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