Find the remainder of
\(3 + 33 + 333 + 3333 + \cdots + \underbrace{333 \dots 3}_{2019 \text{ digits}}\) divided by 673.
a=2019;b=(10^a - 1) / 9 * 3;c=c+b; a=a - 1; if(a>=1, goto1, 0);printc
1.8513703703 E+2019 mod 673 = 500
Sorry! I did it two different ways and got 2 different answers!! However, I believe this one is the accurate one.
3.7037037037 E+2018 mod 673 =370