Problem:
For each positive integer k, let Sk denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is k. For example, S3 is the sequence 1,4,7,… For how many values of k does Sk contain the term 2005?
Solution:
Suppose that the nth term of the sequence Sk is 2005 . Then 1+(n−1)k=2005 so k(n−1)=2004=22⋅3⋅167. The ordered pairs (k,n−1) of positive integers that satisfy the last equation are (1,2004),(2,1002),(3,668),(4,501),(6,334), (12,167),(167,12),(334,6),(501,4),(668,3),(1002,2), and (2004,1). Thus the requested number of values is 12 . Note that the number of divisors of 22⋅3⋅167 can also be found to be (2+1)(1+1)(1+1)=12 by using the formula for the number of divisors.
The problems on this page are the property of the MAA's American Mathematics Competitions