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

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

Answered by Olaf last updated on 17/Sep/20

$${\mathrm{P}}_k\left(x\right)=\underset{p=1}{\prod ^k}(1+px)=\underset{p=0}{\sum ^k}{a}_p{x}^p$$$$\underset{p=0}{\sum ^k}{a}_p={\mathrm{P}}_k\left(1\right)=\underset{p=1}{\prod ^k}(1+p)=\underset{p=2}{\prod ^{k+1}}p=(k+1)!$$$$\mathrm{N}=(k+1)!$$$$2019=3\times 673$$$$\Rightarrow k+1=673$$$$k=672$$

Commented by ZiYangLee last updated on 18/Sep/20

Commented by ZiYangLee last updated on 18/Sep/20

$$\mathrm{nice}\mathrm{try}!$$

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