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QUESTION FOR FUN
SOLVE FOR n:
n^n +n^(n1) + n^(n2)+...+n^2+n^1+n^0 = n^n^n
 one year ago
 one year ago
QUESTION FOR FUN SOLVE FOR n: n^n +n^(n1) + n^(n2)+...+n^2+n^1+n^0 = n^n^n
 one year ago
 one year ago

This Question is Closed

shubhamsrgBest ResponseYou've already chosen the best response.0
1 + 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.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
1^1 +1^0 = 1^1^1 2=1 which is FALSE
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
So, n=1 is not a solution
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Should I post my solution?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
Oh,,Sorry about that.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
Well 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. :)
 one year ago

ceb105Best ResponseYou've already chosen the best response.0
2 minutes, I'm having a crack.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
U can take your time
 one year ago

ceb105Best ResponseYou've already chosen the best response.0
Before I spend the next hour looking at it to find I cant find a solution, is there definately solutions?
 one year ago

ceb105Best ResponseYou've already chosen the best response.0
Apologise if it is obvious to have solutions, its been a long day of uni
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Actually, NO SOLUTION
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
But u need to PROVE it
 one year ago

ceb105Best ResponseYou've already chosen the best response.0
We sure? Do you have a explanation/proof to that?
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
I guess I have got one proof...
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
BY no solution, you mean no integral solution right ?
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
The proof that I have got is very simple
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
I will message u if u want
 one year ago

ceb105Best ResponseYou've already chosen the best response.0
I'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
 one year ago

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

ceb105Best ResponseYou've already chosen the best response.0
How did you enter that into wolfram? I couldnt get it to work.
 one year ago

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

sauravshakyaBest ResponseYou've already chosen the best response.1
Sorry, I mean integer solution of n
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
hmm, that is what i assumed.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Well 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
 one year ago

ceb105Best ResponseYou've already chosen the best response.0
So yea we agreed there is no solution? as the only intersection is irrational.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Now, u just need to prove it
 one year ago

ceb105Best ResponseYou've already chosen the best response.0
Surely 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
 one year ago

ceb105Best ResponseYou've already chosen the best response.0
ge is supposed to read as >=
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
But u directly said it without any proof
 one year ago

ceb105Best ResponseYou've already chosen the best response.0
Let 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
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Since 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
 one year ago

ceb105Best ResponseYou've already chosen the best response.0
Interesting how you look at it mroe from the abstract whilst i can see it only from the geometric side of things.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
neatly done @sauravshakya
 one year ago
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