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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: This one was really hard.... | Comment 4 of 11 |
(In reply to This one was really hard.... by Penny)

hmm mebe ill do the problem if it gets publicized by the site (not if its posted on a post) tch tch dun criticize. easy and hard questions all are welcomed in this site. Wow, do Sylvester and you come up with problems to give each other every day?
  Posted by Victor Zapana on 2003-12-08 20:58:51

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