2007 USAMO Problems/Problem 1
Problem
Let be a positive integer. Define a sequence by setting and, for each , letting be the unique integer in the range for which is divisible by . For instance, when the obtained sequence is . Prove that for any the sequence eventually becomes constant.
Solution
Solution 1
Let and . Thus, because ,
, and by definition, . Thus, . Also, both are integers, so . As the s form a non-increasing sequence of positive integers, they must eventually become constant.
Therefore, for some sufficiently large value of . Then , so eventually the sequence becomes constant.
Solution 2
Let . Since , we have that .
Thus, .
Since , for some integer , we can keep adding to satisfy the conditions, provided that because .
Because , the sequence must eventually become constant.
Solution 3
Define , and . By the problem hypothesis, is an integer valued sequence.
Lemma: The exists a such that .
Proof: Choose any such that . Then: as desired.
Let k be the smallest k such that . Then , and . To make an integer, must be divisible by . Thus, because is divisible by , , and, because , . Then as well. Repeating the same process using instead of gives , and an easy induction can prove that for all , . Thus, becomes a constant function for arbitrarily large values of k.
Note: This solution is a formalization of the second solution. Also, the lemma could have been simplified if I chose k = n, which is exactly the second solution's thought process.
See also
2007 USAMO (Problems • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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