The Smiths, the Andrings and the Cliffords all hold a big party. Everyone shakes hands with every member of the other two families (no one shakes hands with members of their own family), 142 handshakes in all.

Assuming that there at least as many Andrings as Smiths, and at least as many Cliffords as Andrings, how many of each family are present?

(In reply to 1/2 Surface of a Rectangular Solid... by Erik O.)

Eric O. wrote: "The problem boils down to finding an integer value for 1/2 the surface area of a rectangular solid. "

So if shaking hands is equivalent to finding surface area, a triple handshake (ie one member of each family) is equivalent to finding volume.  Which brings us back to Ady's post entitled "ONE FOR ALL", and all for one.  Athos, Porthos, and Aramis doing the 3 Musketeer handshake.   Sorry, no time for d'Artagnan. 

Posted by Larry on 2004-06-14 12:29:12