SOLUTION: Find number of positive integers n such that n+2n^2+3n^3+...+2019n^2019 is divisible by natural number (n-1).

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Question 1172343: Find number of positive integers n such that n+2n^2+3n^3+...+2019n^2019
is divisible by natural number (n-1).
Answer by ikleyn(52898) About Me  (Show Source):
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It is well known, obvious and self-evident fact that every degree  n%5Ek  gives the remainder 1, when is divided by (n-1).


Therefore, the sum  n + 2n^2 + 3n^3 + . . . + 2019n^2019  gives the remainder

    1 + 2 + 3 + . . . + 2019

when is divided by (n-1).



In turn, the sum  1 + 2 + 3 + . . . + 2019  is the sum of the first 2019 natural numbers, and, therefore, is equal to  

    %28%281%2B2019%29%2A2019%29%2F2 = %282020%2A2019%29%2F2 = 1010*2019.



The number  1010*2019  has THIS DECOMPOSITION into the product of prime numbers

            1010*2019 = (2*5*101)*(3*673).


So, its decomposition is the product of 5 prime numbers with multiplicities 1 for each participating prime.


Therefore, the number of divisors of the number  1 + 2 + 3 + . . . 2019 = 1010*2019  is  

    (1+1)*(1+1)*(1+1)*(1+1)*1+1) = 2*2*2*2*2 = 2%5E5 = 32.



Each such divisor is the potential number (n-1).


THEREFORE, the answer to the problem's question is  32.


Solved.