2016 AIME II Problems/Problem 9
The sequences of positive integers and are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let . There is an integer such that and . Find .
Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for . When we get to and , we have and , which works, therefore, the answer is .
Solution by Shaddoll
Using the same reasoning ( isn't very big), we can guess which terms will work. The first case is , so we assume the second and fourth terms of are and . We let be the common ratio of the geometric sequence and write the arithmetic relationships in terms of .
The common difference is , and so we can equate: . Moving all the terms to one side and the constants to the other yields
, or . Simply listing out the factors of shows that the only factor less than a square that works is . Thus and we solve from there to get .
Solution by rocketscience
|2016 AIME II (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|