World's Hardest Math Problem: 1

[wrightfan123]

Okay, I was bored in Algebra today, so I devised an insanly hard math problem. Whoever wins gets a prize:

(abc)+(d-e)(fg-hij)+(klmnop-qrs)
(tu)-(v)+(wxyz)

Here's the key:
a=1
b=2
c=pi
d=bCUBED
e=7
f=41
g-bf
h=6
i=100
j=k
k=hSQUARED=6
l=12
m=-b*6
n=-12
o=47*pi
p=piCUBED
q=1/a=6
r=iSQUARED
s=106
t=w
u=z
v=(e-f)SQUARED
w=v
x=60
y=xSQUARED
z=y

...Good luck math geeks!

[Splash]

what do you expect us to do?

[EvilPurpleMonkey]

Considering you gave us the whole bloody key, it's not very hard, It would just take a long time to finish. In fact I'll do it in a couple hours.

[pancakemix]

Considering you gave us the whole bloody key, it's not very hard, It would just take a long time to finish. In fact I'll do it in a couple hours.

I'd do it right now if I had my calculator.

[Iz Man]

Considering you gave us the whole bloody key, it's not very hard, It would just take a long time to finish. In fact I'll do it in a couple hours.

stole my thunder.
plus, I assume the "=(wxyz)" would be on the right side of the equation and not in the denominator?

[Koesen]

That's not really that hard. It's a lot of work to do all the multiplications and so on, but other than that, it's very basic. And I studied French, not math

Edit: I see a bunch of other people arrived at the same conclusion while I was typing

[Jafnhár]

Look, anyone can create problems that take long time to do because they have many numbers and letters. There is nothing original about this one, it just takes time to do.

[GreecePwns]

π^4=6494112581993339....

Obviously the answer to this is NO!

[Koesen]

If you want a hard math problem, try proving that if an integer n is greater than 2, then the equation an + bn = cn has no solutions in non-zero integers a, b, and c.

[edit: sorry, that should read a to the power of n, b to the power of n and c to the power of n)

[The1exile]

Instead of typing SQUARED or CUBED use ^2 or ^3 respectively.

[ranck3]

Its not hard, just takes time to do, i am busy right now so im not doing it. but i may later.

[ParadiceCity9]

I liked algebra last year...it's the only math i'm good at..im really bad at geometry this year

[mandalorian2298]

If you want a hard math problem, try proving that if an integer n is greater than 2, then the equation an + bn = cn has no solutions in non-zero integers a, b, and c.

[edit: sorry, that should read a to the power of n, b to the power of n and c to the power of n)

You didn't state that the following is false:
a=b
b=c
c=a

Thus, here is my counter example: a=b=c=1

[mibi]

figured it out.

(wtf)^2/st+fu=gfy

[heavycola]

figured it out.

(wtf)^2/st+fu=gfy

my god i think he's cracked it

[Splash]

The problem isn't hard, reciting pie is.

3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679
8214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196
4428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273
724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609

[wrightfan123]

oops, there was a typo: it's supposed2b +(wxyz), not +(wxyz)

[GreecePwns]

oops, there was a typo: it's supposed2b +(wxyz), not +(wxyz)

Those are the same thing...

[Puff]

hey right, since you made the problem maybe you can provide the answer.

and whats the prize anyway.

[Syzygy]

There is no prize.

[Puff]

There is no prize.

that sucks

and here I was hoping it was a year of premium

[Syzygy]

Anyone feel like offering a year's free premium for answering some hard puzzle?

[Puff]

Anyone feel like offering a year's free premium for answering some hard puzzle?

meh only a mathematician could figure it out anyway.

[Syzygy]

Anyone feel like offering a year's free premium for answering some hard puzzle?

meh only a mathematician could figure it out anyway.

It's not really hard, just tedious though.

[Dancing Mustard]

figured it out.

(wtf)^2/st+fu=gfy

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