Difference between revisions of "2001 IMO Shortlist Problems/N6"
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=== First Solution === | === First Solution === | ||
− | + | For example, let's say that the integers are, <math>k, k+a_1, k+a_2, ..., k+a_{99}</math>. Now this turns into a problem of solving for the <math>99</math> integers <math>a_i</math>. This then each ai takes on the form, <math>j+b_1, j+b_2,..., j+b_{98}</math>. Then we must find the <math>98</math> <math>b</math> integers. By doing this process over and over again, we obtain the last <math>3</math> numbers, <math>y, y+u_1, y+u_2</math>. 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 <math>a_i, b_i, c_i</math>, etc.) is <math>99+98+\ldots+2</math>, not exceeding <math>25000</math>. | |
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=== Second Solution === | === Second Solution === |
Latest revision as of 12:08, 17 August 2011
Problem
Is it possible to find 100 positive integers not exceeding 25,000, such that all pairwise sums of them are different?
Solution
The answer is yes.
First Solution
For example, let's say that the integers are, . Now this turns into a problem of solving for the integers . This then each ai takes on the form, . Then we must find the integers. By doing this process over and over again, we obtain the last numbers, . 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 , etc.) is , not exceeding .
Second Solution
Lemma: We try to show that for a prime , there are such integers less than or equal to . Then it suffices because and is enough.
Proof: Define to be and . Then, we show that the integers does the trick for to . If , then . Thus, we may say that and, since . From , we have . From , we infer that which gives us . Factorizing,
This yields or . But , and this implies that . The other case is similar. So, we are done.
Therefore, every sum must be distinct, as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.