All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
Handshakes (Posted on 2004-06-14) Difficulty: 2 of 5
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?

See The Solution Submitted by Brian Smith    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): 1/2 Surface of a Rectangular Solid... | Comment 9 of 14 |
(In reply to re: 1/2 Surface of a Rectangular Solid... by Larry)

Larry wrote: So if shaking hands is equivalent to finding surface area, a triple handshake (ie one member of each family) is equivalent to finding volume. 

Not quite.  The handshaking problem is as area problem with the number of families, N, denoting the number of dimensions of the rectangular object.

Between 1 family, the result is always 0 (we're talking area, remember?)

Between 2 families it's a simple 2D surface: number in Family A x number in Family B.

Between 3 families it's 1/2 the total surface of a 3D rectangular object: AxB + BxC + CxA.

Between 4 families it becomes 1/4 the total surface of a hyper-rectangle: AB+AC+AD+BC+BD+CD


  Posted by Erik O. on 2004-06-14 13:26:09
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (11)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information