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sauravshakya

  • 2 years ago

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

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  1. shubhamsrg
    • 2 years ago
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    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.

  2. sauravshakya
    • 2 years ago
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    1^1 +1^0 = 1^1^1 2=1 which is FALSE

  3. sauravshakya
    • 2 years ago
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    So, n=1 is not a solution

  4. sauravshakya
    • 2 years ago
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    Should I post my solution?

  5. shubhamsrg
    • 2 years ago
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    Oh,,Sorry about that.

  6. shubhamsrg
    • 2 years ago
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    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. :)

  7. sauravshakya
    • 2 years ago
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    As u say

  8. ceb105
    • 2 years ago
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    2 minutes, I'm having a crack.

  9. sauravshakya
    • 2 years ago
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    U can take your time

  10. ceb105
    • 2 years ago
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    Before I spend the next hour looking at it to find I cant find a solution, is there definately solutions?

  11. ceb105
    • 2 years ago
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    Apologise if it is obvious to have solutions, its been a long day of uni

  12. sauravshakya
    • 2 years ago
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    Actually, NO SOLUTION

  13. sauravshakya
    • 2 years ago
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    But u need to PROVE it

  14. ceb105
    • 2 years ago
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    We sure? Do you have a explanation/proof to that?

  15. ceb105
    • 2 years ago
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    Ahh okay

  16. sauravshakya
    • 2 years ago
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    I guess I have got one proof...

  17. shubhamsrg
    • 2 years ago
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    BY no solution, you mean no integral solution right ?

  18. sauravshakya
    • 2 years ago
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    The proof that I have got is very simple

  19. ceb105
    • 2 years ago
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    Shoot.

  20. sauravshakya
    • 2 years ago
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    I will message u if u want

  21. ceb105
    • 2 years ago
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    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

  22. shubhamsrg
    • 2 years ago
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    WOlfram tells me a solution, You must be meaning no integral solutions are there hmm ?

  23. ceb105
    • 2 years ago
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    How did you enter that into wolfram? I couldnt get it to work.

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

  25. sauravshakya
    • 2 years ago
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    Sorry, I mean integer solution of n

  26. shubhamsrg
    • 2 years ago
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    hmm, that is what i assumed.

  27. sauravshakya
    • 2 years ago
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    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

  28. ceb105
    • 2 years ago
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    So yea we agreed there is no solution? as the only intersection is irrational.

  29. sauravshakya
    • 2 years ago
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    Now, u just need to prove it

  30. ceb105
    • 2 years ago
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    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

  31. ceb105
    • 2 years ago
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    ge is supposed to read as >=

  32. sauravshakya
    • 2 years ago
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    But u directly said it without any proof

  33. ceb105
    • 2 years ago
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    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

  34. sauravshakya
    • 2 years ago
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    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

  35. ceb105
    • 2 years ago
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    Interesting how you look at it mroe from the abstract whilst i can see it only from the geometric side of things.

  36. shubhamsrg
    • 2 years ago
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    neatly done @sauravshakya

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