Conquer Club

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!

what do you expect us to do?

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.

EvilPurpleMonkey wrote: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.

EvilPurpleMonkey wrote: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?

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

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.

π^4=6494112581993339....

Obviously the answer to this is NO!

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)

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

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

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

Koesen wrote: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

figured it out.

(wtf)^2/st+fu=gfy

mibi wrote:figured it out.

(wtf)^2/st+fu=gfy

my god i think he's cracked it

The problem isn't hard, reciting pie is.

3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679
8214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196
4428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273
724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609

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

wrightfan123 wrote:oops, there was a typo: it's supposed2b +(wxyz), not +(wxyz)

Those are the same thing...

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

and whats the prize anyway.

There is no prize.

Syzygy wrote:There is no prize.

that sucks

and here I was hoping it was a year of premium

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

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

meh only a mathematician could figure it out anyway.

Puff wrote:

Syzygy wrote: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.

mibi wrote:figured it out.

(wtf)^2/st+fu=gfy

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