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Difference between revisions of "2005 AMC 12A Problems/Problem 19"

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== Solution ==
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==Solutions==
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=== Solution 1===
 
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:
 
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>
  
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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. 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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===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>2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}</math>
 +
 
 +
===Solution 3===
 +
Since any numbers containing one or more <math>4</math>s were skipped, we need only to find the numbers that don't contain a <math>4</math> at all. First we consider <math>1</math> - <math>1999</math>. Single digits are simply four digit numbers with a zero in all but the ones place (this concept applies to double and triple digits numbers as well). From <math>1</math> - <math>1999</math>, we have <math>2</math> possibilities for the thousands place, and <math>9</math> possibilities for the hundreds, tens, and ones places. This is <math>2 \cdot 9 \cdot 9 \cdot 9-1</math> possibilities (because <math>0000</math> doesn't count) or <math>1457</math> numbers. From <math>2000</math> - <math>2005</math> there are <math>6</math> numbers, <math>5</math> of which don't contain a <math>4</math>. Therefore the total is <math>1457 + 5</math>, or <math>1462</math> <math>\Rightarrow</math> <math>\boxed{\text{B}}</math>.
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===Solution 4===
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We seek to find    the amount of numbers that contain at least one <math>4,</math> and subtract this number from <math>2005.</math>
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 +
We can simply apply casework to this problem.
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The amount of numbers with at least one <math>4</math> that are one or two digit numbers are <math>4,14,24,34,40-49,54,\cdots,94</math> which gives <math>19</math> numbers.
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The amount of three digit numbers with at least one <math>4</math> is <math>8*19+100=252.</math>
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The amount of four digit numbers with at least one <math>4</math> is <math>252+1+19=272</math>
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This, our answer is <math>2005-19-252-272=1462,</math> or <math>\boxed{B}.</math>
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~coolmath2017
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===Solution 5(Super fast)===
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This is very analogous to base <math>9</math>. But, in base <math>9</math>, we don't have a <math>9</math>. So, this means that these are equal except for that base 9 will be one more than the operation here. <math>2005_9 = 5+0+0+1458 = 1463</math>. <math>1463 - 1 = 1462</math>
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Therefore, our answer is <math>\boxed {1462}</math>
  
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<math>1463-1=\boxed{1462}</math>
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~Arcticturn
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 15:39, 21 October 2024

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? $(\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804$

Solutions

Solution 1

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:

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

Solution 2

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. Since $4$ is skipped, the symbol $5$ represents $4$ miles of travel, and we have traveled $2004_9$ miles. By basic conversion, $2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}$

Solution 3

Since any numbers containing one or more $4$s were skipped, we need only to find the numbers that don't contain a $4$ at all. First we consider $1$ - $1999$. Single digits are simply four digit numbers with a zero in all but the ones place (this concept applies to double and triple digits numbers as well). From $1$ - $1999$, we have $2$ possibilities for the thousands place, and $9$ possibilities for the hundreds, tens, and ones places. This is $2 \cdot 9 \cdot 9 \cdot 9-1$ possibilities (because $0000$ doesn't count) or $1457$ numbers. From $2000$ - $2005$ there are $6$ numbers, $5$ of which don't contain a $4$. Therefore the total is $1457 + 5$, or $1462$ $\Rightarrow$ $\boxed{\text{B}}$.

Solution 4

We seek to find the amount of numbers that contain at least one $4,$ and subtract this number from $2005.$

We can simply apply casework to this problem.

The amount of numbers with at least one $4$ that are one or two digit numbers are $4,14,24,34,40-49,54,\cdots,94$ which gives $19$ numbers.

The amount of three digit numbers with at least one $4$ is $8*19+100=252.$

The amount of four digit numbers with at least one $4$ is $252+1+19=272$

This, our answer is $2005-19-252-272=1462,$ or $\boxed{B}.$

~coolmath2017

Solution 5(Super fast)

This is very analogous to base $9$. But, in base $9$, we don't have a $9$. So, this means that these are equal except for that base 9 will be one more than the operation here. $2005_9 = 5+0+0+1458 = 1463$. $1463 - 1 = 1462$

Therefore, our answer is $\boxed {1462}$

~Arcticturn

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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