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 can be divided into the subsets containing
and the ones not containing
. The latter is just
, whereas the former is
(since removing
from any of these sets produces a spacy set with maximal element
). Hence,
![$S_{n + 1} = S_{n} + S_{n - 2}$](http://latex.artofproblemsolving.com/6/7/c/67cef40285a255e1c2cdf97100c25e31affcf407.png)
From this recursion, we find that
S(1) | S(2) | S(3) | S(4) | S(5) | S(6) | S(7) | S(8) | S(9) | S(10) | S(11) | S(12) | |
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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