There are 11 plates of which each weighs a natural number of grams and no two plates have the same weight. We know that no matter how we put the plates (it doesn't have to be all of them) on a scale with two sides, that the side with more plates is always heavier (we don't know which side is heavier if they have an equal number of plates). Prove that one plate has to weigh more than 35 grams.

I think looking at the case where the plates weigh $35,34,33,32,31,30,29,28,27,26,25$ is worth it. We can notice that the sum of the last $6$ numbers is $165$ while the sum of our first $5$ numbers is also $165$, a contradiction. We can also notice if we were to add any smaller plate (as we don't want to use any plate that weighs more than $35$) that we can yet again sort the plates in descending order, take the first $5$ and they will never weigh less than the last $6$. My question is, this seems like a far reach for a proof. Does anyone else have any decent ideas that might help me out? Thanks in advance!

  • $\begingroup$ I think that you nailed it. I also think that your analysis matches that intended by the problem composer. $\endgroup$ May 7, 2022 at 7:49
  • $\begingroup$ I applaud you for correctly evaluating that your idea (while a good first attempt) is far from a proof. Of course, user2661923's comment is completely wrong. The answer Robin Balean gave you is indeed a correct one. $\endgroup$
    – user21820
    May 7, 2022 at 16:43

1 Answer 1


Here's one way to see it. Label the weights of the plates $x_1,\dots,x_{11}$, such that we have $x_{i+1}\geq x_i + 1$. We know that $x_1+\dots+x_6 > x_7+\dots +x_{11}$ and so $x_1 > (x_{11}-x_6) + (x_{10}-x_5) + (x_9-x_4) + (x_8-x_3) + (x_7-x_2)\geq 5+5+5+5+5=25$. So $x_1>25$ and hence $x_{11}> 25+10=35$.

  • $\begingroup$ Thanks - just fixed the typo $\endgroup$ May 7, 2022 at 9:22

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