Difference between revisions of "2005 AMC 12A Problems/Problem 19"
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===Solution 3=== | ===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>. | 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=== | ||
+ | |||
+ | We seek to find the amount of numbers that contain at least one <math>4,</math> and subtract this number from <math>2005.</math> | ||
+ | |||
+ | We can simply apply casework to this problem. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | The amount of three digit numbers with at least one <math>4</math> is <math>8*19+100=252.</math> | ||
+ | |||
+ | The amount of four digit numbers with at least one <math>4</math> is <math>252+1+19=272</math> | ||
+ | |||
+ | This, our answer is <math>2005-19-252-272=162,</math> or <math>B.</math> | ||
== See also == | == See also == |
Revision as of 22:06, 1 September 2020
Contents
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,
Solution 3
Since any numbers containing one or more s were skipped, we need only to find the numbers that don't contain a at all. First we consider - . 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 - , we have possibilities for the thousands place, and possibilities for the hundreds, tens, and ones places. This is possibilities (because doesn't count) or numbers. From - there are numbers, of which don't contain a . Therefore the total is , or .
Solution 4
We seek to find the amount of numbers that contain at least one and subtract this number from
We can simply apply casework to this problem.
The amount of numbers with at least one that are one or two digit numbers are which gives numbers.
The amount of three digit numbers with at least one is
The amount of four digit numbers with at least one is
This, our answer is or
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.