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What i get is that $a_1$ divides $a_j-a_i$ , but this doesnt tell me anything more . How to show there are two numbers that one evenly divides the other wih this approach .

  • Note : i know the other proof of n boxes i would like to see how the accepted answer show us the proof of the claim . ( Also it should be $a_1 \leq 100$ not $99$)
RobPratt
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ProblemDestroyer
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    Not only are you correct that $a_1\leq 100$ but I believe that there are only 100 remainders $r_2,\dots r_{101}$ (notice it starts from $r_2$) not 101! – PCeltide Sep 05 '22 at 13:01
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    Also as $a_1\leq 100$ the possible NONZERO remainders that may be obtained by division of $a_1$ are $1,\dots, 99$, that is in total, $99$. – PCeltide Sep 05 '22 at 13:03
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    The poster of that question did just ask for hints, but I'm with you - I don't see how that method leads to a final answer. – JonathanZ Sep 05 '22 at 13:04
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    I see that was just hint , hmm understood . But yeah that hint doesnt gonna work much here – ProblemDestroyer Sep 05 '22 at 13:10
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    Yes I agree... Though I wonder how it got accepted, the hint does not lead anywhere it seems, the other answer should get accepted instead – PCeltide Sep 05 '22 at 13:17
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    I agree, the second answer makes a lot more sense. Here is a more elegant way of writing it down: https://math.stackexchange.com/questions/2271176/choosing-101-numbers-from-1-2-dots-200?rq=1 – Andreas Tsevas Sep 05 '22 at 13:25
  • That's beautiful! Thank you! – PCeltide Sep 05 '22 at 13:30
  • Yeah indeed , thank you @Andreas – ProblemDestroyer Sep 05 '22 at 13:40

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