Math: So you like M&M's...

[Monte314]

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[enWTFp]

To be honest, I don't like M&Ms, hope I'm still allowed to present my solution.

Answer: 0 distinct solutions, which is finite. All variables here are positive integers.

Suppose there exists distinct positive integer solution (N,M). We'll show contradiction.

Note: if a^m = b^n, m>n (wlog), then b = a^(m/n). Hence we can assume M = N^x for some x>1 (wlog).

N^(N^x)^N = N^(xN^(N^x)) , N^xN = xN^(N^x) , x=N^y for some y , N^xN = (N^y)N^(N^x) ,

xN = y + N^x , y=Nt for some t , x = t + N^(x-1) , hence x > N^(x-1) .

Since N>1 and x>1, N^(x-1) >= 2^(x-1) >= x , contradiction.

For some reason this problem reminds me of an old joke. A professor meets his colleague on the street.
- Aren't you having an exam right now? And you left the students all alone?! They're gonna cheat!
- Cheat? Are you crazy, even I can't solve the problems I've given them, you say they're gonna cheat...

[Monte314]

Wow. enWTFp has produced an elegant solution. It is correct, of course.

There are many ways to attack this problem. Anyone else?

[Wufnu]

At first I said, "Man, if M=N then there is an infinite set." I realized that he must have meant M is not equal to N so I took a second look at the problem set and said "lol, there is no solution." That word finite was there, though, so I had to make sure.

I was racking my brain because whar ahm frum (engineering/physics) 0 is definitely not finite. On the other hand, from my abstract algebra course, I knew that 0 was a real number and in a set of real numbers that is not infinite it's... finite. In fact, I was so frustrated with the issue that I looked up the definition of finite which solidified my understanding that 0 was, in fact, not finite (multiple definitions, with those saying 0 wasn't finite outnumbering those that said it was). I came up with multiple solutions to this problem and all of them had as a solution the zero set.

One method was to compare which side would "grow" faster. If M>N, then the M^N^M side would grow faster and vice versa if M<N. What this meant was that unless M=N they would never be the same. Ever.

Finding no solution, and putting my blind faith in Monte to not base the answer on such a gray area, I sallied forth.

Using algebruh, I realized that any N^M^N must be some multiple of M (durr) so set N=T*M where T was some positive integer variable. I ended up with:
1=(N/M)*M^([M^M]*(M^[N/M]-[N/M]^M).

I even tried working out the M^T-T^M and ended up with M^T-T^M = mth root of M^(N/M) - N^M/M^N. I thought I was on to something there and when I worked it out I ended up with M^T-T^M.

Alot of help that did me.

I tried it again, using logs. It was long and complicated, so I'll just show how I worked one side then show you my end result. Replacing M^N^M with M*logbaseM(N), and realizing that M*logbaseM(N) = M*LN(N)/LN(M) blah blah blah so that (this will be ugly):
LN([LN(N^M^N)/LN(N)]/LN(N) = M

At this point I says to myself, "screw logs."

I differentiated both sides with respect to M and N. Here, both sides are represented by x^y^x:
w/ respect to x: x^(y^x)*(y^x*log(y)*log(x)+y^x/x)
w/ respect to y: x^(y^x)*y^x*x/y*log(x)

Gez wut? It told me nothing.

I made a Matlab code of the following format:

q=1;
for p=1:100
m(q,1)=p;
q=q+1;
end

i=1;
for i= 1:100
T(1,i)=m(i,1);
T(2,i) = solve('n^(m^n)=m^(n^m)','n');
end
T


The end result? Matlab couldn't solve it without a gajillion dollar symbolic solver toolbox addon. Screw you, Mathworks!

I came here to say "You know what? I give up. There's no solution and anyone that says there is is a liar" but enWTFp beat me to it.

[TheLastMohican]

*Applause for enWTFp and Wufnu*
I puzzled over this for a while, thinking, "that doesn't make any sense"...but it never occurred to me that it was a trick question. In retrospect, the simplest things are the hardest.

[Monte314]

wufnu, your approaches make perfect sense, and can be turned into solutions. Your observation that the growth rate of the exponential is greater than the growth rate of polynomials is the key point, so you've touched the heart of the problem.

And, zero is finite.

[muguly]

Why are you all so smart?!?!?!?!?!? I suck at math.

I like peanut M&M's if that counts.

[Wufnu]

Enjoying math doesn't make you smart, at least no more than knowing how to paint or rebuild a carburetor. It's a skill, like anything else :)

[Monte314]

Why are you all so smart?!?!?!?!?!? I suck at math.

I like peanut M&M's if that counts.

Uh... you get 2 points (three if you like the blue ones.)

[Mozzes]

So this is basically asking for a solution to N^(M*N) = M^(M*N). Taking the Nth root of each side of the equation would give N^M = M^M and taking the Mth root would only leave N = M so I'd guess the only possible solutions are where N and M are equal. I guess this isn't as fancy as some of the other responses but the reasoning seems fairly simply for why the answer is what it is.

I guess you could call this the 7th grade algebra solution.

[Monte314]

So this is basically asking for a solution to N^(M*N) = M^(M*N). Taking the Nth root of each side of the equation would give N^M = M^M and taking the Mth root would only leave N = M so I'd guess the only possible solutions are where N and M are equal. I guess this isn't as fancy as some of the other responses but the reasoning seems fairly simply for why the answer is what it is.

I guess you could call this the 7th grade algebra solution.


Mozzes, I'm afraid I don't follow your solution.

[Autoptic]

N^(MN)=M^(M^(NM) != N^M^N=N^M^N

A base to a power isn't equivalent to a base times it's exponent.

[Mozzes]

Mozzes, I'm afraid I don't follow your solution.

Oops, I didn't read the instructions closely (I was considering N^M^N = N^MN which appears to have same solution when N!=M). I'll have to revise.

edit: Here's another shot:

If the only condition is N != M then let's just set N = M + C where C != 0 (otherwise N = M which we can't have). In that case we have

(M + C) ^ [M ^ (M + C)] = M ^ [(M + C) ^ M]

For this to be true M + C = M and since C != 0 we have a contradiction.

[curiousjane]

Uh... you get 2 points (three if you like the blue ones.)

I like M&M's, but not math. Perhaps THIS book will help me?
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Or this one?
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I'm amazed by mathematicians. To the nth degree. ;)
Carry on. And delete this post if it is too off topic.

[TheLastMohican]

And, zero is finite.

Don't rub it in...;D

I like M&M's, but not math. Perhaps THIS book will help me?


:laugh:

[Monte314]

I like M&M's, but not math. Perhaps THIS book will help me?
To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

Or this one?
To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.-link

I'm amazed by mathematicians. To the nth degree. ;)
Carry on. And delete this post if it is too off topic.

Ten points! No, Twelve!