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anonymous
 4 years ago
QUESTION FOR FUN
SOLVE FOR n:
n^n +n^(n1) + n^(n2)+...+n^2+n^1+n^0 = n^n^n
anonymous
 4 years ago
QUESTION FOR FUN SOLVE FOR n: n^n +n^(n1) + n^(n2)+...+n^2+n^1+n^0 = n^n^n

This Question is Closed

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.01 + n + n^2 .. n^n = n^n^n n=1 is a clear solution for n not equal to 1, we have => (n^(n+1) 1)/(n1) = n^n^n unsure what will come next.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01^1 +1^0 = 1^1^1 2=1 which is FALSE

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, n=1 is not a solution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Should I post my solution?

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0Oh,,Sorry about that.

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0Well I failed to do it doesn't mean no one will do it. Many geniuses are there on OS. I'd recommend you to wait. :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02 minutes, I'm having a crack.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Before I spend the next hour looking at it to find I cant find a solution, is there definately solutions?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Apologise if it is obvious to have solutions, its been a long day of uni

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Actually, NO SOLUTION

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But u need to PROVE it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We sure? Do you have a explanation/proof to that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I guess I have got one proof...

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0BY no solution, you mean no integral solution right ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The proof that I have got is very simple

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I will message u if u want

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'd like to see it. I feel there is a solution as n^n^n grows faster than a quadratic and the quadratic is smaller than the exponent around n=0

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0WOlfram tells me a solution, You must be meaning no integral solutions are there hmm ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How did you enter that into wolfram? I couldnt get it to work.

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=%28n%5E%28n%2B1%29+1%29%2F%28n1%29+%3D+n%5En%5En

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, I mean integer solution of n

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0hmm, that is what i assumed.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well from the question n^n +n^(n1) + n^(n2)+...+n^2+n^1+n^0 = n^n^n it is clear n>=0 and n is integer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So yea we agreed there is no solution? as the only intersection is irrational.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now, u just need to prove it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Surely this would suffice: n^n +n^(n1) + n^(n2)+...+n^2+n^1+n^0>n^n^n \[\forall n \le 1\] and n^n +n^(n1) + n^(n2)+...+n^2+n^1+n^0<n^n^n \[\forall n ge 2\] Q.E.D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ge is supposed to read as >=

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But u directly said it without any proof

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let A=n^n +n^(n1) + n^(n2)+...+n^2+n^1+n^0, and let B=n^n^n. Note: n is belonging arbitrarily to the set of integers. Then since A<B for all n>1, i.e. all n>=2, and A>B for all n<=1 then since there exists no integer between 1 and 2, there is no possible integer solutions to A=B

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since R.H.S is divisible by n, L.H.S must also be divisible by n. Now, all the terms except n^0 of L.H.S is divisible by n for all values of n. So,L.H.S is divisible by n only if n^0 is also divisible by n. n^0 is divisible by n only when n=1 but n=1 is not the solution. THUS, NO SOLUTION

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Interesting how you look at it mroe from the abstract whilst i can see it only from the geometric side of things.

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0neatly done @sauravshakya
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