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 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))=

 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

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