Talk:2007 USAMO Problems/Problem 1

I am proposing to delete Solution 2. It is essentially the same as Solution 1. If there were two significantly different solutions, then it would be appropriate to list them both, but the two solutions are nearly identical. I'll leave the matter open for discussion for a few days.

I also think the argument can be simplified considerably. We observe that

$b_{k+1} = \left(\frac{k}{k+1}\right)b_k + \frac{a_{k+1}}{k+1}$ .

The first term is strictly less than $b_k$ , and the second term is strictly less than 1, and their sum is an integer. Therefore, $b_{k+1} \le b_k$ . Therefore, the $b_k$ 's must eventually become constant (since they form a non-increasing sequence of positive integers). From there, we can finish as in the existing solutions.

Do you think this is a significant improvement, or it is simply a trivial restatement of the existing solutions? --DPatrick 20:53, 30 April 2007 (EDT)

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Talk:2007 USAMO Problems/Problem 1