- Thread starter Denis
- Start date

- Joined
- Dec 30, 2014

- Messages
- 5,086

In the corner for 0minutes, again!\(\displaystyle 0! + 1 + 2 \cdot 3 + 4^5 + 67 - 89 = 1010\)

\(\displaystyle 9 + 87 + 6 + 5 + 43 \cdot 21 = 1010\)

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

987 + 6 - 5 + 43 + 2 + 1*0 = 1033

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:In the corner for 0minutes, again!

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

\(\displaystyle 98 \cdot 7 + 6 \cdot 54 + 3 + 21 + 0 = 1034\)

- Joined
- Apr 22, 2015

- Messages
- 2,060

My thoughts exactly.Why not take the easy way out?

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

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

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

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

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

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

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

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

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

And you, Jomo: you needed to post a solution to 1033....bad boy....

- Joined
- Apr 22, 2015

- Messages
- 2,060

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.)… To the corner Otis: should be t(45) …

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

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

\[\;\]

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