2^m+2019 = |n-2020|+n-2020, where m and n are non-negative integers an

A question like this is tricky because it doesn’t test you on rules, it tests you on exceptions to that rule.
\(2^m\) is perceived by a lot of test takers to be always even. The exception is \(2^0\) which is 1 and is odd. A clue to this effect is given by telling you that m and n are non-negative integers –> note that 0 is non-negative.

The RHS can be simplified in two ways by using the definition of modulus.

|n-2020| = n-2020 if n≥ 2020
|n-2020| = 2020 – n if n<2020.

When we substitute the second case, we have \(2^m\) + 2019 = 2020 – n + n -2020. Simplifying, we have \(2^m\) + 2019 = 0 or \(2^m\) = -2019.
No integral power of 2 can yield -2019. As such, we can rule out the possibility that n<2020.

If n≥ 2020, \(2^m\) + 2019 = n-2020 + n – 2020. Simplifying, we have,

\(2^m\) + 2019 = 2n – 4040 OR

\(2^m\) = 2n – 6059.

2n is even and 6059 is odd. Even – Odd = Odd.

This means \(2^m\) is odd. This is possible only in one case –> when m = 0.

Substituting this value of m = 0 in the equation and simplifying, we have 2n = 6059 + 1 which yields us n = 3030.
The value of m+n = 3030 + 0 = 3030.

The correct answer option is D.

Hope that helps!