Difference between revisions of "2007 AMC 12A Problems/Problem 16"
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This gives us a total of <math>2 + 4 + 6 + 8 = 20</math> sequences. There are <math>3! = 6</math> to permute these, for a total of <math>120</math>. | This gives us a total of <math>2 + 4 + 6 + 8 = 20</math> sequences. There are <math>3! = 6</math> to permute these, for a total of <math>120</math>. | ||
| − | However, we note that the conditions of the problem require | + | However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are <math>2! \cdot 4 = 8</math> numbers which start with zero, so our answer is <math>120 - 8 = 112 \Longrightarrow \mathrm{(C)}</math>. |
== See also == | == See also == | ||
Revision as of 08:18, 5 January 2009
Problems
How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?
Solution
We can find the number of increasing arithmetic sequences of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences.
| Common difference | Sequences possible | Number of sequences |
| 1 |  |
8 |
| 2 |  |
6 |
| 3 |  |
4 |
| 4 |  |
2 |
This gives us a total of 
sequences. There are 
to permute these, for a total of 
.
However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are 
numbers which start with zero, so our answer is 
.
See also
| 2007 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 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 | |
