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

 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.)
 Beginning Thoughts | Comment 2 of 14 |
I found @, but have only random thoughts for the other two:

@:
5 @ 5 = 2/5 = 10/25 = (5+5)/5²
2 @ 2 = 1 = 4/4 = (2+2)/2²
2 @ 3 = 5/9 = (2+3)/3²
3 @ 4 = 7/16 = (3+4)/4²
2 @ 4 = 3/8 = 6/16 = (2+4)/4²

\$:
1\$0 undefined - presumably division by 0 is involved
1\$4 = 1/10 (?) 1+4=5; 1*4=4; 2/(4*5)= 1/10
4\$1 = 1/10 (?) 1+4=5; 1*4=4; 2/(4*5)= 1/10
10\$10 = 1/50 (?) 10+10=20; 10*10=100; 8/(20*100) = 1/50

_:
2_2 = 6 = 2*3
2_3 = 9 = 3*3
6_2 = 42 = 6*7 or 2*3*7
3_5 = 30 = 2*3*5
5_6 = 90 = 6*3*5 or 2*3*3*5

 Posted by TomM on 2004-03-31 14:36:59

 Search: Search body:
Forums (0)