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 | + | 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>. |
== See also == | == See also == | ||
Revision as of 14:43, 3 August 2020
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 
.
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. 
