Match the post#

-0! + 1234 - 5*6*7 - 8 + (√9)! = 1021

987 + 6*5 + 4 - 3 + 2 + 1.0 = 1021
 
0! + 12 + (3! + 4)/(-.5 + .6)/(-.7 + .8) + 9 = 1022

(9 + 8 - 7 - 6)^5 - 4 + 3 - 2 + 1 - 0 = 1022
 
0! + 1234 - 5*6*7 - 8 + (√9)! = 1023

987 + 6*5 + 4 + 3 - 2 + 1.0 = 1023
 
0! + 1 + 2.3 + 4^5 + 6.7 - 8 - √9 = 1024

9 + 8 - 7 - 6 - 5 + 4 - 3 + 2^10 = 1024
 
\(\displaystyle 0 + 1 \cdot 2^3 \cdot 4 \cdot 5 \cdot 6 + 7 \cdot 8 + 9 = 1025\)

\(\displaystyle 9 \cdot 87 + 6 + 5 \cdot 43 + 21 + 0 = 1025\)
 
\(\displaystyle 0! + 1 \cdot 2^3 \cdot 4 \cdot 5 \cdot 6 + 7 \cdot 8 + 9 = 1026\)

\(\displaystyle 9 \cdot 87 + 6 + 5 \cdot 43 + 21 + 0! = 1026\)

:)
 
0! + 1 + 2 - 3 + 4^5 - 6 + 7 - 8 + 9 = 1027

-9 + 8.7 - 6 + 5 + 4.3 + 2^10 = 1027
 
\(\displaystyle 0 + 1 - 2 + 3 + 4^5 - 6 + 7 - 8 + 9 = 1028\)

\(\displaystyle (9 + 8 - 7 - 6)^5 - 4 + 3! + 2 + 1 - 0! = 1028\)
 
\(\displaystyle 0! + 1 - 2 + 3 + 4^5 - 6 + 7 - 8 + 9 = 1029\)

\(\displaystyle (9 + 8 - 7 - 6)^5 - 4 + 3! + 2 + 1 - 0 = 1029\)

:)
 
\(\displaystyle 0! - 1 - 2 + 3! + 4^5 - 6 + 7 - 8 + 9 = 1030\)

\(\displaystyle (9 + 8 - 7 - 6)^5 - 4 + 3! + 2 + 1 + 0! = 1030\)
 
0 - 1 - 2 - 3 + T(45) - 6 + 7 - 8 + 9 = 1031

F(9 + 8 - 7 + 6) + 54 - 3^2 - 1 + 0 = 1031

T(45) = 45th Triangular number = 1035
F(16) = 16th Fibonacci number = 987
 
Why not take the easy way out?

\(\displaystyle 0! - 1 - 2 - 3 + T(45) - 6 + 7 - 8 + 9 = 1032\)

\(\displaystyle F(9 + 8 - 7 + 6) + 54 - 3^2 - 1 + 0! = 1032\)
 
ksdhart2 said:
\(\displaystyle 0! + 1 + 2 \cdot 3 + 4^5 + 67 - 89 = 1010\)

\(\displaystyle 9 + 87 + 6 + 5 + 43 \cdot 21 = 1010\)
Click to expand...
In the corner for 0minutes, again!

Moderator Edit: Denis lends Jomo a hand:

0 - 1 + 234 + 5 + 6 + 789 = 1033

987 + 6 - 5 + 43 + 2 + 1*0 = 1033
 
Jomo said:
In the corner for 0minutes, again!
Click to expand...

Sure, sure, but only if you go to the corner for 1033 minutes - you forgot to post a solution for 1033! Here's mine for 1034. I tried something a bit unorthodox, and I hope it's allowed:

\(\displaystyle \int\limits_{0}^{1} 234 \: \text{dx} + 5 + 6 + 789 = 1034\)

\(\displaystyle 98 \cdot 7 + 6 \cdot 54 + 3 + 21 + 0 = 1034\)
 
Otis said:
My thoughts exactly.

\(\displaystyle f(0+1+2+3+4+5+6+7+8+9) = 1035\)

\(\displaystyle f(9+8+7+6+5+4+3+2+1+0) = 1035\)
Click to expand...

If \(f(x)\) is the same as \(T(x)\) and denotes the \(n^{th}\) triangular number, then sure.

\(\displaystyle 0 + 1 \cdot 23 + 4 \cdot 56 + 789 = 1036\)

\(\displaystyle 987 + (6 + 5) \cdot 4 + 3 + 2 \cdot 1 + 0\)
 
\(\displaystyle 0! + 1 \cdot 23 + 4 \cdot 56 + 789 = 1037\)

\(\displaystyle 987 + (6 + 5) \cdot 4 + 3 + 2 \cdot 1 + 0! = 1037\)

To the corner Otis: should be t(45) , not f(45)

And you, Jomo: you needed to post a solution to 1033....bad boy....
 
Denis said:
… To the corner Otis: should be t(45) …
Click to expand...
In post #35, function f is any function that works. (That's one of the many tricks of the trade, and you said we could use any of them.)

\(\displaystyle g(0+1+2+3+4+5+6+7+8+9) = 1038\)

\(\displaystyle g(9+8+7+6+5+4+3+2+1+0) = 1038\)
\[\;\]
 
Ahhh...I see...all my fault:
I meant T = Triangular number, F = Fibonacci number;
not functions as such.
HOKAY: now changing the rules:
no functions or stuff like T and F allowed!

-(0! + 1 + 2 + 3) + 4^5 - 67 + 89 = 1039

98 - 76 + 5 - 4*3 + 2^10 = 1039

Edit: Fixed Typo
 
Excuse me Sir Denise, but -(0! + 2 + 2 + 3) + 4^5 - 67 + 89 = 1038 NOT 1039. Go to the senior corner!

Let f(x) =1040,
Then f(0*1 +23-4*5+6789=1040
Hmm, maybe f(9-8+765-432-10) = 1040
 
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