Call a set of integers spacy if it contains no more than
one out of any three consecutive integers. How many
subsets of {1, 2, 3, . . . , 12}, including the empty set,
are spacy?
(A) 121
(B) 123
(C) 125
(D) 127
(E) 129
2007 AMC 12 A, Problem #25—
“A set Sn contains n do not contain either n − 1 or
n − 2.”
Solution
Answer (E): For each positive integer n, let Sn = {k : 1 ≤ k ≤ n},
and let cn be the number of spacy subsets of Sn . Then c1 = 2, c2 = 3,
and c3 = 4. For n ≥ 4, the spacy subsets of Sn can be partitioned
into two types: those that contain n and those that do not. Those that
do not contain n are precisely the spacy subsets of Sn−1 . Those that
contain n do not contain either n − 1 or n − 2 and are therefore in oneto-one correspondence with the spacy subsets of Sn−3 . It follows that
cn = cn−3 + cn−1 . Thus the first twelve terms in the sequence (cn ) are
2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, and there are c12 = 129 spacy
subsets of S12 .
OR
Note that each spacy subset of S12 contains at most 4 elements. For each
such subset a1 , a2 , . . . , ak , let b1 = a1 −1, bj = aj −aj−1 −3 for 2 ≤ j ≤ k,
and bk+1 = 12 − ak . Then bj ≥ 0 for 1 ≤ j ≤ k + 1, and
b1 + b2 + · · · + bk+1 = 12 − 1 − 3(k − 1) = 14 − 3k.
¡
¢
The number of solutions for (b1 , b2 , . . . , bk+1 ) is 14−2k
for 0 ≤ k ≤ 4, so
k
the number of spacy subsets of S12 is
µ ¶ µ ¶ µ ¶ µ ¶ µ ¶
14
12
10
8
6
+
+
+
+
= 1 + 12 + 45 + 56 + 15 = 129.
0
1
2
3
4
Difficulty: Hard
NCTM Standard: Algebra Standard: Understand patterns, relations, and functions.
Mathworld.com Classification: Foundations of Mathematics > Set Theory > Sets