A sequence a1, a2, . . . of non-negative integers is
defined by the rule an+2 = |an+1 − an| for n ≥ 1.
If a1 = 999, a2 < 999, and a2006 = 1, how many
different values of a2 are possible?
(A) 165
(B) 324
(C) 495
(D) 499
(E) 660
2006 AMC 12 B, Problem #25— “The condition an+2 =
|an+1 − an | implies that an and an+3 have the same parity for
all n ≥ 1 .”
Solution (B) The condition an+2 = |an+1 − an | implies that an and an+3
have the same parity for all n ≥ 1. Because a2006 is odd, a2 is also odd.
Because a2006 = 1 and an is a multiple of gcd(a1 , a2 ) for all n, it follows
that 1 = gcd(a1 , a2 ) = gcd(33 · 37, a2 ). There are 499 odd integers in the
interval [1, 998], of which 166 are multiples of 3, 13 are multiples of 37, and
4 are multiples of 3 · 37 = 111. By the Inclusion-Exclusion Principle, the
number of possible values of a2 cannot exceed 499 − 166 − 13 + 4 = 324.
To see that there are actually 324 possibilities, note that for n ≥ 3, an <
max(an−2 , an−1 ) whenever an−2 and an−1 are both positive. Thus aN = 0
for some N ≤ 1999. If gcd(a1 , a2 ) = 1, then aN −2 = aN −1 = 1, and for
n > N the sequence cycles through the values 1, 1, 0. If in addition a2 is
odd, then a3k+2 is odd for k ≥ 1, so a2006 = 1.
Difficulty: Hard
NCTM Standard: Number and Operations Standard: understand meanings of operations and
how they relate to one another
Mathworld.com Classification: Number Theory > Sequences > Sequence