On Jun 14, 4:12pm, Jimm Wetherbee wrote:
> Subject: Re: Another Challenge
> First: of the original, weigh 6 on each side. Discard the lighter.
All coins on the heavier side could weigh the same if the "different" coin
is lighter than the rest.
>
> Second: of the 6 remaining, weigh 3 on each side. Discard the lighter.
>
> Third: of the 3 remaining, weigh and two. If one is heavier, well there you
> are. If they balance out evenly, the one remaining is the heavy coin.
>
>
>
> At 09:00 AM 6/13/96 +0100, Martin wrote:
>
> >The Problem:
snip...
Jeff Vance
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