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

Home > Just Math
new operations (Posted on 2004-03-31) Difficulty: 3 of 5
Mathematicians have just created three new operation symbols, @, $,and _, each doing a different thing to a pair of numbers (All of these symbols can be derived from already-known operations and notations). The following equations using the new symbols are correct:

5@5= 2/5
2@2= 1
2@3= 5/9
3@4= 7/16
2@4= 3/8

1$0= undefined
1$4= 1/10
4$1= 1/10
10$10= 1/50
(1@1) $ 1= (√2)/10

2_2= 6
2_3= 9
6_2= 42
3_5= 30
5_6= 90

If the previous equations are correct, compute the following equation.

1 @ (1 $ (1_1))=

See The Solution Submitted by Victor Zapana    
Rating: 3.3000 (10 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): And $ Makes 30 ...er...6 I mean | Comment 8 of 14 |
(In reply to re: And $ Makes 30 ... ? by Tristan)

I wrote "a $ b = 1/(5*sqrt(a*b))  if  a @ b = (a+b)/b*b.  Thus if also a _ b = b*a*(a+1)/2 we have 1 @ (1 $ (1 _ 1)) = 30."

Let's take it slowly and see what we get.

1 _ 1 = 1*1*2/2 = 1

1 $ 1 = 1/(5*sqrt(1)) = 1/5

1 @ (1/5) = (6/5)/(1/5)^2 = (6/5)/(1/25) = 6*25/5 = 30.

I think I did it right, but there is probably some other set of formulas that work and make the answer come out to be 6. What is more important here than the value of the answer is whether the three formulas give the values specified by the problem.  I hope mine for $ and the ones I got from earlier comments for _ and @ do give the specified values.

Edited on March 31, 2004, 9:04 pm
  Posted by Richard on 2004-03-31 20:09:52

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 (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information