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
re: Solution by Charlie)
I used my disney guide to programming and came up with a similar set of code and unsurprisingly the same solution.
I'm curious to know if somebody can come up with a non-brute-force method of solving this.
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Posted by fwaff
on 2004-06-14 08:57:15 |
re(2): Solution - no brute force?
