Problem - 1120
Let four positive integers $a$, $b$, $c$, and $d$ satisfy $a+b+c+d=2019$. Prove $\left(a^3+b^3+c^3+d^3\right)$ cannot be an even number.
It is easy to show that regardless of integer $n$'s parity, it always hold that $n^3\equiv n\pmod{2}$ because any power of $n$ will not change odd even parity. Therefore, $$a^3+b^3+c^3+d^3\equiv a+b+c+d\equiv 2019\equiv 1\pmod{2}$$
Therefore, it is odd.