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Difference between revisions of "2001 IMO Shortlist Problems/N6"
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== Solution ==
 
== Solution ==
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Solution: ''note, please mind me for my notation, I am not familiar to LATEX''
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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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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.
 
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== Resources ==
 
  
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== See also ==
 
* [[2001 IMO Shortlist Problems]]
 
* [[2001 IMO Shortlist Problems]]
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17475 Discussion on AOPS/MathLinks]
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17475 Discussion on AOPS/MathLinks]
  
 
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]

Revision as of 19:34, 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

For example, let's say that the integers are, $k, k+a_1, k+a_2, ..., k+a_{99}$. Now this turns into a problem of solving for the $99$ integers $a_i$. This then each ai takes on the form, $j+b_1, j+b_2,..., j+b_{98}$. Then we must find the $98$ $b$ integers. By doing this process over and over again, we obtain the last $3$ numbers, $y, y+u_1, y+u_2$. 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 $a_i, b_i, c_i$, etc.) is $99+98+\ldots+2$, not exceeding $25000$.

See also

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