Question

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

Answer 1

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.

Answer 2

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

Answer 3

So, n=1 is not a solution

Answer 4

Should I post my solution?

Answer 5

Oh,,Sorry about that.

Answer 6

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. :)

Answer 7

As u say

Answer 8

2 minutes, I'm having a crack.

Answer 9

U can take your time

Answer 10

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

Answer 11

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

Answer 12

Actually, NO SOLUTION

Answer 13

But u need to PROVE it

Answer 14

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

Answer 15

Ahh okay

Answer 16

I guess I have got one proof...

Answer 17

BY no solution, you mean no integral solution right ?

Answer 18

The proof that I have got is very simple

Answer 19

Shoot.

Answer 20

I will message u if u want

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

Sorry, I mean integer solution of n

Answer 26

hmm, that is what i assumed.

Answer 27

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

Answer 28

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

Answer 29

Now, u just need to prove it

Answer 30

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<n^n^n \[\forall n ge 2\] Q.E.D

Answer 31

ge is supposed to read as >=

Answer 32

But u directly said it without any proof

Answer 33

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<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

Answer 34

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

Answer 35

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

Answer 36

neatly done @sauravshakya