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There are 64 stones of different weights and a 2-sided scale. Explain how one can find two heaviest stones in 68 trials on this scale.

Do you start by taking any two and then comparing it to all the others? It seems to take up too much trials.

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    This is problem 42 in Chapter 8 of Genkin and Fomin, Mathematical Circles (Russian Experience). – 2017-02-04

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Think of an elimination tournament. To get the heaviest, weigh the stones in $32$ pairs, keeping the heaviest. Keep going. That gets you the heaviest in how many weighings? Now which ones could be in second place?

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    Better than 68? Second heaviest can be any of the six beaten by the heaviest, and it takes five weighings to find the heaviest of six, no? – 2017-02-04
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    The second heaviest will be eliminated on the first round if it is matched against the heaviest. – 2017-02-04
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    @JohnWaylandBales: I didn't say throw all the losers away. You need to keep track of them for the search for the second. – 2017-02-04
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    @GerryMyerson: you are right. I glitched thinking of the number of rounds, not the number of weighings in the search for second. – 2017-02-04
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    63 trials to find heaviest? – 2017-02-09
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    @GerardL.: Yes, there are $32$ in the first round, then $16$, and so on. Each weighing eliminates one coin as a potential heaviest, so we need $63$. Then there are only six coins which could be second. – 2017-02-09
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    @RossMillikan How are the six determined? – 2017-02-10
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    They are the six that lost to the winner. Every other stone has already lost to one of these, but it is possible that the second heaviest is the first stone that lost to the winner, or the second, or ... – 2017-02-10