Divisor Function of Sums in Fractions

Question

I have a question that I've been working on for a while now. It says, "Let $A=\{0,1,2,\dots,2018\}$. Prove that $\forall n\in\mathbb{N},\exists\{a_0,a_1,a_2,\dots,a_{2018}\}\subseteq\mathbb{N}$, $$(a_0<a_1<\dots<a_{2018})\wedge\frac{(\sum^{2018}_{k=0}a_k)div2019}{n+\sum^{2018}_{k=0}(a_kdiv2019)}=1."$$ I have no clue which $a_k$ to construct, and how they'll operate in that nasty fraction. I have been trying to find an upper bound for $a_{2018}$ and working from there, but I seem to be getting nowhere. I also tried setting the numerator equal to the denominator (since the fraction is equal to one) and working from there, but again, I didn't get very far. Help would be appreciated.

Answer 1

I figured out my own question: Choose $a_k=2019k+n$. Then, in the nemerator, we have that $$\left(\sum^{2018}_{k=0}a_k\right)div2019=\left(\sum^{2018}_{k=0}(2019k+n)\right)div2019 =\sum^{2018}_{k=0}k+n.$$ In the denominator, we get $$n+\sum^{2018}_{k=0}(a_kdiv2019)=n+\sum^{2018}_{k=0}\big((2019k+n)div2019\big)=n+\sum^{2018}_{k=0}k.$$ It's now easy to see that the ratio between the numerator and denominator is $1$.