Problem:
Arithmetic sequences (an) and (bn) have integer terms with a1=b1=1<a2≤ b2 and anbn=2010 for some n. What is the largest possible value of n?
Answer Choices:
A. 2
B. 3
C. 8
D. 288
E. 2009
Solution:
Because an=1+(n−1)d1 and bn=1+(n−1)d2 for some integers d1 and d2, it follows that n−1 is a factor of gcd(an−1,bn−1). The ordered pair (an,bn) must be one of (2,1005),(3,670),(5,402),(6,335),(10,201),(15,134), or (30,67). For every pair except the sixth pair, the numbers an−1 and bn−1 are relatively prime, so n=2. In the exceptional case, gcd(15−1,134−1)=7. The sequences defined by an=2n−1 and bn=19n−18 satisfy the conditions, so n=8.
The problems on this page are the property of the MAA's American Mathematics Competitions