Here's something for you Math Wiz's out there.

[CorCentral]

It's not a problem I'm looking solved for school. I graduated 13years ago LOL.
A friend of mine said to use any means possible so here I am


(x+1)*2*343*52352/2-X^33253653=x+32342345235323+X^9999/322

 

[KLin]

the answer is obviously 42.

 

[Vertimus]

A computer can do that anyday.

I'm not seeing anything special in this problem except number crunching.

 

[david46675]

uh
graph them on the TI-83 and see where the lines intersect?

 

[BrokenVisage]

I don't see what the big deal is, just a couple extrapolations here and there to find what x equals.

 

[Vertimus]

You've got to suck majorly at math to find problems like these interesting. It's like back in 2nd grade where people keep asking "what's 9999999 times 999999910101112"?

 

[desteffy]

wow!!! whoever can answer this is the SMARTEST PERSON EVAR!!!!!! OMG!1111

 

[eigen]

Originally posted by: david46675
uh
graph them on the TI-83 and see where the lines intersect?

No look at that first term a TI-83 couldnt handle it.If you havent figured it out in a while I will get the supercomputer to work on it...

 

[brunswickite]

is your friend 10 years old?

 

[david46675]

Originally posted by: eigen

Originally posted by: david46675
uh
graph them on the TI-83 and see where the lines intersect?

No look at that first term a TI-83 couldnt handle it.If you havent figured it out in a while I will get the supercomputer to work on it...

ur right, i didnt look closely...
btw OP, ask ur friend what 99999999999*2394781392847109234 is

 

[IronWing]

Does x = X or are we talking two variables here?

 

[RaWrulez]

Your search - (x+1)*2*343*52352/2-X^33253653=x+32342345235323+X^9999/322 - did not match any documents.

 

[Vertimus]

Point your friend to this thread.

 

[chuckywang]

x and X are the same or different? Do you mean (X^9999)/322 or X^(9999/322)?

 

[Torched]

every time (x+1)*2*343*52352/2-X^33253653=x+32342345235323+X^9999/322 a kitten dies!

Poor, poor kitten

 

[eLiu]

This problem isn't THAT simple. Computers work in finite precision.

The exponent is HUGE. Round off problems in double precision will be significant. Just be glad it's just a monomial. I THINK that w/o "infinite precision" numbers (like in Maple or Mathematica), you won't get a truly accurate number because next to X^33253653, x^1 is gonna have some issues.

I'm not sure if a precanned algorithm in like C could handle this. I mean precanned algorithms have trouble with stuff like (x+19)*(x+18)*...*(x+2)*(x+1)*x. Find all 20 roots of that...it's a more complicated problem than you think.

 

[eigen]

Originally posted by: eLiu
This problem isn't THAT simple. Computers work in finite precision.

The exponent is HUGE. Round off problems in double precision will be significant. Just be glad it's just a monomial. I THINK that w/o "infinite precision" numbers (like in Maple or Mathematica), you won't get a truly accurate number because next to X^33253653, x^1 is gonna have some issues.

I'm not sure if a precanned algorithm in like C could handle this. I mean precanned algorithms have trouble with stuff like (x+19)*(x+18)*...*(x+2)*(x+1)*x. Find all 20 roots of that...it's a more complicated problem than you think.

That why I told him I would send it up to the supercomputer I have acess too.I can allocate up to 6 nodes to work on this...

 

[chuckywang]

Originally posted by: eLiu
This problem isn't THAT simple. Computers work in finite precision.

The exponent is HUGE. Round off problems in double precision will be significant. Just be glad it's just a monomial. I THINK that w/o "infinite precision" numbers (like in Maple or Mathematica), you won't get a truly accurate number because next to X^33253653, x^1 is gonna have some issues.

I'm not sure if a precanned algorithm in like C could handle this. I mean precanned algorithms have trouble with stuff like (x+19)*(x+18)*...*(x+2)*(x+1)*x. Find all 20 roots of that...it's a more complicated problem than you think.

You could always just iterate.