2009 AIME I Problems/Problem 13
Problem
The terms of the sequence defined by for are positive integers. Find the minimum possible value of .
Solution
This question is guessable but let's prove our answer
let put into now
and set them equal now
let's rewrite it
Let make it looks nice and let
Since and are integer, we can see is divisible by
But we can't have an infinite sequence of proper factors, unless
Thus,
So now, we know
To minimize , we need
Thus, answer
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |