QUESTION FOR FUN
SOLVE FOR n:
n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0 = n^n^n

- anonymous

QUESTION FOR FUN
SOLVE FOR n:
n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0 = n^n^n

- Stacey Warren - Expert brainly.com

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- schrodinger

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- shubhamsrg

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)/(n-1) = n^n^n
unsure what will come next.

- anonymous

1^1 +1^0 = 1^1^1
2=1 which is FALSE

- anonymous

So, n=1 is not a solution

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## More answers

- anonymous

Should I post my solution?

- shubhamsrg

Oh,,Sorry about that.

- shubhamsrg

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. :)

- anonymous

As u say

- anonymous

2 minutes, I'm having a crack.

- anonymous

U can take your time

- anonymous

Before I spend the next hour looking at it to find I cant find a solution, is there definately solutions?

- anonymous

Apologise if it is obvious to have solutions, its been a long day of uni

- anonymous

Actually, NO SOLUTION

- anonymous

But u need to PROVE it

- anonymous

We sure? Do you have a explanation/proof to that?

- anonymous

Ahh okay

- anonymous

I guess I have got one proof...

- shubhamsrg

BY no solution, you mean no integral solution right ?

- anonymous

The proof that I have got is very simple

- anonymous

Shoot.

- anonymous

I will message u if u want

- anonymous

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

- shubhamsrg

WOlfram tells me a solution, You must be meaning no integral solutions are there hmm ?

- anonymous

How did you enter that into wolfram? I couldnt get it to work.

- shubhamsrg

http://www.wolframalpha.com/input/?i=%28n%5E%28n%2B1%29+-1%29%2F%28n-1%29+%3D+n%5En%5En

- anonymous

Sorry, I mean integer solution of n

- shubhamsrg

hmm, that is what i assumed.

- anonymous

Well from the question n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0 = n^n^n
it is clear n>=0 and n is integer

- anonymous

So yea we agreed there is no solution? as the only intersection is irrational.

- anonymous

Now, u just need to prove it

- anonymous

Surely this would suffice: n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0>n^n^n \[\forall n \le 1\] and n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0

- anonymous

ge is supposed to read as >=

- anonymous

But u directly said it without any proof

- anonymous

Let A=n^n +n^(n-1) + n^(n-2)+...+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

**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

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

- anonymous

Interesting how you look at it mroe from the abstract whilst i can see it only from the geometric side of things.

- shubhamsrg

neatly done @sauravshakya

**
**

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