Now mod this!

We are interested in the number $X \; = \; \sum_{n=1}^{2019} \frac{10^n-1}{3} \; = \; \tfrac{10}{27}\big(10^{2019} - 1\big) - 673$ Now $10^{672} \equiv 1 \pmod{673}$ , so $10^{2019} \equiv 10^{3 \times 673} \equiv 10^3 \pmod{673}$ , so $10^{2019} - 1 \equiv 999 = 27 \times 37 \pmod{673}$ , and hence $X \equiv \tfrac{10}{27}\big(10^{2019} - 1\big) \equiv 10 \times 37 \equiv \boxed{370} \pmod{673}$

We split this problem into two parts: (I) find a closed form expression of the given sum; (II) mod it by $673$ .

Part I: Find a closed form expression of the sum

For all positive integer $n$ , define $a_n = \underbrace{333 \dots 3}_{n \text{ digits}},$ and the partial sum of sequence $(a_n)$ $\begin{aligned} S_n &:= a_1 + a_2 + a_3 + \cdots + a_n \\[0.5em] &\phantom{:}= 3 + 33 + 333 + \cdots + 333 \! \dots \! 3 . \end{aligned}$ The key is to observe that $3a_k + 1 = 10^k, \quad \text{ for all } k \ge 1,$ and sum this identity over $k = 1, 2, 3, \dots, n$ to get $3S_n + n = \underbrace{\sum_{k=1}^{n} 10^k = \frac{10}{9}(10^n - 1)}_{\text{geometric series}},$ which means, ${\color{darkred} S_n = \frac{10}{27}(10^n - 1) - \frac{n}{3}}.$

Part II: Mod the desired sum by $673$

Now we proceed to find the remainder of $S_{2019}$ divided by $p = 673$ .

Firstly, by the Euclidean algorithm together with Bézout’s lemma , we can find the modular inverse of $27$ modulo $p$ to be $\color{#0C6AC7} 349$ .

Specifically, $1 = \text{gcd}(27, 673) = {\color{#0C6AC7} 349} \cdot 27 - 14 \cdot \underbrace{673}_{p} \equiv {\color{#0C6AC7} 349} \cdot 27 \pmod{p},$ or put simply, $27^{-1} \equiv {\color{#0C6AC7}349} \pmod{p}.$ Hence, $\begin{aligned} S_{2019} &= \frac{10}{27} (10^{2019} - 1) - \frac{2019}{3} \\[0.7em] &= 10 \cdot 27^{-1} \cdot (10^{2019} - 1) - 673 \\[0.7em] &\equiv 10 \cdot {\color{#0C6AC7}349} \cdot (10^{2019} - 1) \\[0.7em] &\equiv 125 \cdot (10^{2019} - 1) \pmod{p}. \end{aligned}$ Finally, we tackle the term $10^{2019} - 1$ modulo $p$ with the aide of Fermat’s little theorem : $10^{p} \equiv 10 \pmod{p},$ or, as $2019 = 3p$ , $10^{2019} - 1 = (10^p)^3 - 1 \equiv 10^3 - 1 = 999 \equiv 326 \pmod{p}.$ Therefore, the desired value is $S_{2019} \equiv 125 \cdot 326 = 40750 \equiv \boxed{\mathbf{370}} \pmod{p}.$ And this conclude the solution.