1. ## The \$10,000 Puzzle

http://www.joshjames.com/NewGig

*The \$10,000 prize for solving the math equation and puzzle will only be awarded to one person, as determined at our sole discretion. The recipient will be the first person who we determine has found the answer key, signed a non-disclosure agreement, and then finished the equation and puzzle and demonstrated proof of his or her solution after obtaining the key.

I've done the simple but tedious first step of solving for the variables, which I assume is part of the puzzle.

1. 81β - 41Δ + 34μ + 64φ = 801
2. 41β - 41Δ + 56μ + 44φ = 3
3. 25β - 33Δ + 35μ + 54φ = 68
4. 63β - 24Δ + 30μ + 51φ = 6000

Ω cancels out of every equation.

β = 68
Δ = 553
μ = 231
φ = 158

I've thought a bit about the puzzle, but haven't really come up with anything at this point. Since Ω cancels out of each equation, I imagine it has some sort of other significance. I've thought about using congruence, but don't know where the Ω would fit in. Also, the fact that the variables are split up and parentheses are used might have some significance.

Any ideas?

2. Interesting.

Although obviously I'm not going to solve it...but I'll probably have fun thinking about it

3. Ω can be anything. But how can that anything be expressed mathematically since 0 * ∞ is undefined from what I remember and you can't just say ∞ - 1 because that's still ∞ .

It can have any value since it does not exist in the equations, with parentheses or not...the number isn't there, but it can't be ∞ . Strange.

4. You don't have \$10,000...

hand over my \$10,000

6. Can Ω be undefined?

Can that even be an answer?

7. Originally Posted by Jonnyboy

1. 81β - 41Δ + 34μ + 64φ = 801
2. 41β - 41Δ + 56μ + 44φ = 3
3. 25β - 33Δ + 35μ + 54φ = 68
4. 63β - 24Δ + 30μ + 51φ = 6000

Ω cancels out of every equation.

β = 68
Δ = 553
μ = 231
φ = 158
Those equations are correct what is there still to solve ?

I dont see the challenge here yet ?

8. The name of the company (as yet undisclosed, since it's new) is hidden in the answer to the equations.

So you're not even looking for a number, you're looking for a name, as far as I can tell.

9. hm hm

Maybe you can try and rearrange the three equations for one of the greek symbols and then substitute in the last equation and the solution is something like 1 equals 1 what could be the company name. Tho that would be prolly to easy

10. 7/11

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