# 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)

I agree that both solutions essentially say the same thing (mine just happens to be solution 1 since I typed it up first ;) ), so I went ahead and merged them. Official solutions has the aforementioned argument as the first solution, so I assume it is an improvement over the other one; while typing it up, it felt like a significant improvement too. Azjps (talk) 22:12, 30 April 2007 (EDT)