Question Number 114192 by ZiYangLee last updated on 17/Sep/20

For a positive integer k, we write   (1+x)(1+2x)(1+3x)...(1+kx)=a_0 +a_1 x+a_2 x^2 +...a_k x^k   Let N=a_0 +a_1 +a_2 +...a_k  ,  if N is divisible by 2019, find the   smallest possible value of k.

Answered by Olaf last updated on 17/Sep/20

P_k (x) = Π_(p=1) ^k (1+px) = Σ_(p=0) ^k a_p x^p   Σ_(p=0) ^k a_p  = P_k (1) = Π_(p=1) ^k (1+p) = Π_(p=2) ^(k+1) p = (k+1)!  N = (k+1)!  2019 = 3×673  ⇒ k+1 = 673  k = 672

Commented byZiYangLee last updated on 18/Sep/20

Commented byZiYangLee last updated on 18/Sep/20

nice try!