1993 USAMO Problems/Problem 4
Problem 4
Let , be odd positive integers. Define the sequence by putting , , and by letting for be the greatest odd divisor of . Show that is constant for sufficiently large and determine the eventual value as a function of and .
Solution
Part 1) Prove that is constant for sufficiently large .
Note that if there is some for any , then , which is odd. Thus, and by induction, all is constant for .
Also note that since average of positive number is always positive.
Thus, assume for contradiction, , .
Then, ,
Thus, and that means that is a strictly decreasing function and it must reach as , which contradict with the fact that .
Part 1 proven.
Part 2) Show that the constant is .
For any where . for with the same property except with and .
Therefore, if I prove that the constant for any with relatively prime , is , then I have shown that part 2 is true.
Lemma) If , then .
Assume for contradiction that , since both and are odd, is not divisible by .
for some such that is odd.
, where and is another integer.
Thus, is divisible by which contradicts with the assumption that .
Lemma proven
By induction, since .
Since there must exist some where (part 1), .
Resources
1993 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
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