2007 AMC 12A Problems/Problem 25
Problem
Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of including the empty set, are spacy?
Solution
Solution 1
Let denote the number of spacy subsets of . We have .
The spacy subsets of can be divided into two groups:
- those not containing . Clearly .
- those containing . We have , since removing from any set in produces a spacy set with all elements at most equal to and each such spacy set can be constructed from exactly one spacy set in .
Hence,
From this recursion, we find that
1 | 2 | 3 | 4 | 6 | 9 | 13 | 19 | 28 | 41 | 60 | 88 | 129 |
Solution 2
Since each of the elements of the subsets must be spaced at least two apart, a divider counting argument can be used.
From the set we choose at most four numbers. Let those numbers be represented by balls. Between each of the balls there are at least two dividers. So for example, o | | o | | o | | o | | represents .
For subsets of size there must be dividers between the balls, leaving dividers to be be placed in spots between the balls. The number of way this can be done is .
Therefore, the number of spacy subsets is .
Solution 3
As a last resort, we can brute force the result by repeated casework. Luckily, 12 is not a very large number, so solving it this way is still possible.
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
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