A Yearly Remainder

As $n$ has the same remainder of 2019 when divided by both 2020 and 2021, $n$ must be 2019 more than an integer multiple of the LCM of 2020 and 2021. Since any two consecutive integers greater than 2 are relatively prime, their LCM is simply their product. This means that $n=2020 \cdot 2021 \cdot k+2019$ where $k$ is a positive integer. We then divide this equation by 235. As the product of 2020 and 2021 is divisible by 235, we need only find the remainder of $\frac{2019}{235}$ which we find to be 139.

$n = 2019$ has a remainder of $2019$ when divided by both $2020$ and $2021$ , and when $2019$ is divided by $235$ there is a remainder of $\boxed{139}$ .

N mod 2021=2019

N mod 2020=2019

N=4084439

N mod 235 =139

Notice that $2020=2\cdot 2\cdot 5\cdot 101$ and $2021=43\cdot 47$ . We are interested in the remainder of $n$ modulo $235=5\cdot 47$ .

$2020$ has a factor of $5$ and $2021$ has a factor of $47$ , so we can consider $n$ modulo $5$ and $47$ . This gives: $n\equiv 2019 \equiv 45 (\textrm{mod}\enspace 47),$ $n\equiv 2019 \equiv 4 (\textrm{mod}\enspace 5)$ From this, the Chinese Remainder Theorem gives us the solution $n\equiv 139 (\textrm{mod}\enspace 235)$ .