Problem:
Consider sequences of positive real numbers of the form x,2000,y,…, in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of x does the term 2001 appear somewhere in the sequence?
Answer Choices:
A. 1
B. 2
C. 3
D. 4
E. more than 4
Solution:
If a,b, and c are three consecutive terms of such a sequence, then ac−1=b, which can be rewritten as c=(1+b)/a. Applying this rule recursively and simplifying yields
…,a,b,a1+b,ab1+a+b,b1+a,a,b,…
This shows that at most five different terms can appear in such a sequence. Moreover, the value of a is determined once the value 2000 is assigned to b and the value 2001 is assigned to another of the first five terms. Thus, there are four such sequences that contain 2001 s a term, namely
2001,2000,1,10001,10001001,2001,…1,2000,2001,10001001,10001,1,…,40019992001,2000,4001999,2001,40019992002,40019992001,…, and 4001999,2000,40019992001,40019992002,2001,4001999,…,
respectively. The 4 values of x are 2001, 1,40019992001, and 4001999 .
The problems on this page are the property of the MAA's American Mathematics Competitions