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Need a Math Solution!

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posted on Sep, 30 2004 @ 06:32 PM
My son brought home an interesting math problem today (he is in the third grade). I'm clueless.

Here it is:

A(n)+B(n)+C(n)=?

This should read a to the nth power plus b to the nth power etc.

Supposedly the hardest problem ever?

I know the solution to Asquared plus bsquared and a to the nth plus b to the nth, but I am a little befuddled on this one. Help? Is this the right forum?

Admin, mods, smods, feel free to move to the appropriate topic thread.

posted on Sep, 30 2004 @ 06:58 PM
I'm not math-magician, but I believe you have to have an equation to solve for unknowns. This is not an equation. All quantities are unknown. There is no "solution."

posted on Sep, 30 2004 @ 07:02 PM
Is the problem written like this:

A(n)+B(n)+C(n)=

OR

is it written with superscripts (i.e. the n is above the A, B and C)?

posted on Sep, 30 2004 @ 07:04 PM
Here's a reference to it... google to the rescue!

www.math.niu.edu...

Don't know if that will help... he might understand more of it than I do.

Also see www.pims.math.ca...

Good luck....

posted on Sep, 30 2004 @ 07:21 PM
I guess you mean another way of writing this equation? Don't know the answer, but I remember what you are talking about.

Here's another math problem for everyone.

What is 1+1/2+1/4+1/8+1/16..... ? and write out how you find it. The second part is hardest.

[edit on 30-9-2004 by Jamuhn]

posted on Sep, 30 2004 @ 07:29 PM
it looks like your kid's teacher mentioned fermat's last theorem: mathworld.wolfram.com...,

which is the claim that x^n + y^n = z^n (x to the n plus y to the n equals z to the n) has no positive-integer solutions for integer n > 2. Ie, for n =2 this is equation just the pythagorean theorem:

that in a right triangle with sides a, b, and hypotenuse c, then a^2 + b^2 = c^2. Moreover, there are examples of integer a,b,c that make this work: 3^2 + 4^2 = 9 + 16 = 25 = 5^2, for example.

Fermat's last theorem says that for n > 2, you can no longer find any three integers x,y,z such that x^n + y^n = z^n, and the proof of that claim has taken almost three hundred years to find, and was definitely among the most difficult math problems of all time (thus far).

posted on Sep, 30 2004 @ 08:15 PM

Originally posted by everlastingnoitall
A(n)+B(n)+C(n)=?

Maybe: n(A+B+C)
Since it can't be solved the only logical thing to do is simplify it. Being a fourth grade level problem, it must be simple.

He's in third grade, I doubt it's Fermat's Last Theorum.

posted on Sep, 30 2004 @ 08:20 PM
yeah, i don't think the teacher assigned it; i'm thinking the teacher mentioned it in class.

posted on Sep, 30 2004 @ 08:23 PM
Well wouldn't the math problem have to be set up like this

A(n)+B(n)+C(n) so that would be saying n*n*n
Which would be assuming that n=1 because you have to do all () first. then it would be
A times 1 plus b plus c right?
Could be wrong but thats what I recall.

posted on Sep, 30 2004 @ 08:27 PM
Heck I don't know
you would have to group them, o well wouldn't it just end up being a+b+c because you are multiplying them all by one?

posted on Sep, 30 2004 @ 08:56 PM
hmm i erno im in the 8th grade and i havent gotten a problem like that ever!

posted on Sep, 30 2004 @ 08:59 PM
Actually, it looks as though it's just common a common distribution problem meant to be simplified. A times n=AN + B times N= BN... and so on. It could be a more complicated theorem, but if she really assigned it to a third grade class, I doubt they were supposed to look into it any further than that

posted on Sep, 30 2004 @ 09:09 PM
OMG, i am soooo tired that I read it wrong. Disregard what I just said about distributing. The n's should not be in parantheses, but as exponents. Very sorry about the error.

posted on Sep, 30 2004 @ 09:38 PM

Originally posted by everlastingnoitall

I know the solution to Asquared plus bsquared and a to the nth plus b to the nth, but I am a little befuddled on this one.

well what is the solution to Asquared plus bsquared and a to the nth plus b to the nth

I'm curious to know where this is going, because as stand alone variables the answer could be anywhere between [-infinity, infinity].

[edit on 30-9-2004 by websurfer]

posted on Sep, 30 2004 @ 10:48 PM
Jamun, your problem looks like an infinite series that converges to 2. you can just keep adding the stuff, or you can use calculus. (summation, i=0) 1/(2^i)

that definitely isnt fourth grade math. fourth grade i hadnt even learned long division, much less powers.

posted on Sep, 30 2004 @ 10:50 PM
Ah, but you cannot divide 1/0.

Showing the solution is much harder than the answer.

posted on Oct, 1 2004 @ 10:37 AM
1/(2^0) is not 1/0. x^0 = 1, so 1/(2^0)=1

I agree with everyone who says A^n + B^n + C^n isn't a grade 3 problem. It'd probably take me a day of research relearning things like Riemann sums and convergence of discrete systems to get an answer for that. I'll stick with the crazy triple integrations by parts I have to do for class, thank you very much.

posted on Oct, 1 2004 @ 03:28 PM
What makes you guys assume it's a power function? Parenthesis have always meant multiplication in my math books, without exception. Multiplication definately is a fourth grade thing and I still think my solution is probably what they're looking for.

posted on Oct, 1 2004 @ 03:41 PM
Okay, first take the....umm. Well, then...er. Uhhh.

AARRRGHHH! DON'T ASK ME I WENT TO ART SCHOOL!!!

good luck with that.

posted on Oct, 1 2004 @ 03:51 PM

Originally posted by Ut
1/(2^0) is not 1/0. x^0 = 1, so 1/(2^0)=1

I agree with everyone who says A^n + B^n + C^n isn't a grade 3 problem. It'd probably take me a day of research relearning things like Riemann sums and convergence of discrete systems to get an answer for that. I'll stick with the crazy triple integrations by parts I have to do for class, thank you very much.

LOL, oh yea, forgot that n^0 = 1. I had read about the problem in a math book, and the guy who discovered it and solved it used algebra. The way I posted the problem may have been an intermediate step of the way and not the original problem, oh well.

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