perplexus.info :: Just Math : new operations

new operations (Posted on 2004-03-31)

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))=

Submitted by Victor Zapana

Rating: 3.3000 (10 votes)

Solution: (Hide)

The answer is 30.
Just to let you know...
x @ y = (x + y) / y^2
x $ y = (√(xy)) / (5xy)
x _ y = [x(x + 1)y] / 2
= y[x + (x-1) + (x-2) + ... + 2 + 1]

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Answer	K Sengupta	2008-03-03 10:53:02
re: Very poor	Jack McBarn	2004-04-05 14:35:31
re: Very poor	Federico Kereki	2004-04-01 12:29:02
re(3): And $ Makes 30 ... ?	Tristan	2004-04-01 04:05:23
Thank you	Robert Mendez	2004-04-01 03:29:50
re(2): And $ Makes 30 ... ?	Victor Zapana	2004-03-31 22:06:58
re(2): And $ Makes 30 ...er...6 I mean	Richard	2004-03-31 20:09:52
re: And $ Makes 30 ... ?	Tristan	2004-03-31 18:33:18
And $ Makes 30	Richard	2004-03-31 16:40:37
re: ERRing Thoughts	Ady TZIDON	2004-03-31 15:57:04
The _ operator	Brian Smith	2004-03-31 14:52:43
Solution for the @ operator	Larry	2004-03-31 14:39:51
Beginning Thoughts	TomM	2004-03-31 14:36:59
Very poor	e.g.	2004-03-31 13:57:37

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