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

Home > Just Math
e pluribus unum (Posted on 2003-12-08) Difficulty: 3 of 5
Below are three groups of three numbers each. Combine the numbers in each group using the standard binary operations (addition, subtraction, multiplication, division, and exponentiation) so that each group yields the same number (there is one unique solution).
  1. 1, 6, 11
  2. 13, 20, 33
  3. 20, 33, 40
For example, given:
15, 19, 24          11, 30, 36          20, 22, 36
you could make:
24(19-15)=6       (30+36)11=6         20+22-36=6

See The Solution Submitted by DJ    
Rating: 3.8462 (13 votes)

Previous (in Sequences) : Sequence Sum
    Next (in Just Math) : The Ladder
Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionMore than one solution...Dej Mar2012-09-23 06:42:17
SolutionPraneeth Yalavarthi2007-07-17 12:59:13
SolutionSolutionK Sengupta2007-03-16 10:56:49
agreeing about solutionEmily2005-08-26 21:39:19
Don't start that againMinion1232003-12-09 01:29:00
Please don't:Gamer2003-12-08 22:04:09
a wild stabken2003-12-08 21:56:15
re: This one was really hard....Victor Zapana2003-12-08 20:58:51
Some ThoughtsThis one was really hard....Penny2003-12-08 18:54:31
Solutionethan2003-12-08 16:38:47
SolutionSolution (No computer program used)Penny2003-12-08 16:22:14
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 (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2014 by Animus Pactum Consulting. All rights reserved. Privacy Information