Yes, this is the solution for the original puzzle.

]]>You lost me in all this analysis … I couldn’t even find what’s the final solution!

I, personally, have done this: After exhausting the 2-3 possibilities of having a single coin in one cup, I could easily find a solution in which the coin with the recognizable weight is the only one left out: You put coins 1,2,3,4 and 5 in one cup and 7,8 in the other. The cups are balanced, so the coin is #6.

And this is the only combination with which we can get such a result weiging 7 of the 8 coins.

(Of course, this doesn’t provide a formula for n coins, but it does what is expected to do: it solves the puzzle! :))

Alkis

]]>1) the sums of two triangular numbers

2) the sums of two triangular numbers plus 1

3) the sums of two triangular numbers plus 2

4) the triangular numbers minus one

is not more than 2. ]]>