2010 USAJMO Problems/Problem 2
Problem
Let be an integer. Find, with proof, all sequences of positive integers with the following three properties:
- (a). ;
- (b). for all ;
- (c). given any two indices and (not necessarily distinct) for which , there is an index such that .
Solution
The sequence is .
Proof
We will prove that any sequence , that satisfies the given conditions, is an arithmetic progression with as both the first term and the increment. Once this is proved, condition (b) implies that . Therefore , and the sequence is just the even numbers from to . The sequence of successive even numbers clearly satisfies all three conditions, and we are done.
First a degenerate case. If , there is only one element , and condition (b) gives or . Conditions (a) and (c) are vacuously true.
Otherwise, for , we will prove by induction on that the difference for all , which makes all the differences , i.e. the sequence is an arithmetic progression with as the first term and increment as promised.
So first the case. With , exists and is less than by condition (a). Now since by condition (b) , we conclude that , and therefore by condition (c) for some . Now, since , and can only be . So .
Now for the induction step on all values of . Suppose we have shown that for all , . If we are done, otherwise , and by condition (c) for some . This is larger than , but smaller than by the inductive hypothesis. It then follows that , the only element of the sequence between and . This establishes the result for .
So, by induction for all , which completes the proof.
Solution 2
The claim is that in this sequence, if there are elements where , such that , then the sequence contains every number less than .
Proof: Let and be the numbers less than such that . We take this sequence modulo . This means that if is an element in this sequence then is as well. are all elements in the sequence. Clearly, one of and is less than , which means that are in this sequence modulo . Now we want to show every number is achievable. We have already established that and are relatively prime, so by euclidean algorithm, if we take the positive difference of and every time, we will get that is in our sequence. Then, we can simply add or subtract as many times from as desired to get every single number.
We have proved that there are no two numbers that can be relatively prime in our sequence, implying that no two consecutive numbers can be in this sequence. Because our sequence has terms, our sequence must be one of or , the latter obviously fails, so is our only possible sequence.
Solution 3
We will first look at and . Notice that since condition applies to these numbers. Therefore, and must sum to the only index greater than , which is . We also know that . Equating, and gives . We can similarly force and . These both result in . We continue as shown, forcing the first m terms "into" the last m terms. This forms an arithmetic series with common ratio and first term .
Therefore, . Taking the last equality and simplifying, we get . So, the sequence is . Notice that this sequence satisfies all the conditions. ~Leonard_my_dude~
See Also
2010 USAJMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.