If you still want to try, then don't be discouraged by my scepticism. After all, Fermat's Last Theorem doesn't work with n=2 either!

]]>(abcd..........)^n = a^n+b^n+c^n+d^n..................

Here, both a,b,c,d...etc and n go on and on.

It is easy to say there would always exist a n digit number whose sum of digits raised to the power n is equal to the number. To prove that may be quite difficult. As n becomes higher, the combinations available increase, thereby increasing the possibility of such a number existing. ]]>

So, are we back to a^3 + b^3 + c^3 = a×10^2 + b×10^1 + c×10^0 ?

This can be rearranged to: a^3 + b^3 + c^3 - a×10^2 - b×10^1 - c×10^0 = 0

Then: a^3 - a×10^2 + b^3 - b×10^1 + c^3 - c×10^0 = 0

Then: a(a^2 - 10^2) + b(b^2 - 10^1) + c(c^2 - 10^0) = 0

Is this helping or making it worse I wonder.

]]>i was stuck at n=8, where there were 100 million possibilities.. i was using Javascript, and my computer sort of died..

now for the ultimate question:

Prove, using induction or otherwise, that there are solutions for all n element of Z+, n>2.

(i don't know the answer myself )

]]>It comes under the heading "havin' a little fun with us" I think.

Good find, by the way. Interesting website, too.

]]>How dare you!!!

Even though we all do it.

]]>How did you do it?]]>

I was thinking along the lines:

Start with: 1^3+5^3+3^3=153

Expand decimal: 1^3+5^3+3^3=1×10^2 + 5×10^1 + 3×10^0

Algebraically: a^3+b^3+b^3=a×10^2 + b×10^1 + c×10^0

But that was as far as I got ...

]]>There are a few more where n=7: 1,741,725; 4,210,818; 9,800,817

There's an n=8: 24,678,050

n=9: 146,511,208; 472,335,975; 534,494,836; 912,985,153

n=10: 4,679,307,774

A bit different: 3435= 3^3+4^4+3^3+5^5 and 40585= 4!+0!+5!+8!+5!

And for the grand finale, wait, this is too amazing to just put here. I'll put it in this

]]>