Difference between revisions of "2001 IMO Shortlist Problems/N6"
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== Solution == | == Solution == | ||
| − | + | Solution: ''note, please mind me for my notation, I am not familiar to LATEX'' | |
| − | + | For example, let's say that the integers are, k, k+a1, k+a2, ..., k+a99. Now this turns into a problem of solving for the 99 integers "a". This then each ai takes on the form, j+b1, j+b2,..., j+b98. Then we must find the 98 "b" integers. By doing this process over and over again, we obtain the last 3 numbers, y, y+u1, y+u2. Obviously these 3 integers can have different sums, and the number of different "parts" in every sequence (the number of terms that are different for ai, bi, ci, etc.) is 99+98+...+2, not exceeding 25000. | |
== Resources == | == Resources == | ||
Revision as of 17:56, 19 March 2010
Problem
Is it possible to find 100 positive integers not exceeding 25,000, such that all pairwise sums of them are different?
Solution
Solution: note, please mind me for my notation, I am not familiar to LATEX For example, let's say that the integers are, k, k+a1, k+a2, ..., k+a99. Now this turns into a problem of solving for the 99 integers "a". This then each ai takes on the form, j+b1, j+b2,..., j+b98. Then we must find the 98 "b" integers. By doing this process over and over again, we obtain the last 3 numbers, y, y+u1, y+u2. Obviously these 3 integers can have different sums, and the number of different "parts" in every sequence (the number of terms that are different for ai, bi, ci, etc.) is 99+98+...+2, not exceeding 25000.
