11.2. Is the number $4^{2019}+6^{2020}+3^{4040}$ prime?
Answer: No.

Solution. Transform the expression: $\quad 4^{2019}+6^{2020}+3^{4040}=\left(2^{2}\right)^{2019}+2^{2020} \cdot 3^{2020}+3^{2 \cdot 2020}=$ $\left(2^{2019}\right)^{2}+2 \cdot 2^{2019} \cdot 3^{2020}+\left(3^{2020}\right)^{2}=\left(2^{2019}+3^{2020}\right)^{2}$. Therefore, the original number is the square of the number $2^{2019}+3^{2020}$, i.e., it is not a prime number.

Remark: Note that the number $2^{2019}+3^{2020}$ is not prime. Its smallest prime divisor is 2699, and thus it is also a divisor of the original number.

Criteria: Only the answer - 0 points. The idea to transform the expression by substituting 4 with $2^{2}$, 6 with $2 \cdot 3$, and 4040 with $2 \cdot 2020$, but the solution is not completed or has an error - 3 points. 7 points for a complete and correct solution (including, possibly, obtained by examining remainders from division).