Solution 9fe6547b-4d4a-4836-b92c-3d4f958887dc

Points: 1 pt

Factoring out $7^{2024}$, we obtain
\[
7^{2024}+7^{2025}+7^{2026}
=7^{2024}(1+7+49)
=57\cdot 7^{2024}.
\]
Since $57=19\cdot 3$, we have $19\mid 57$, and hence
\[
19\mid 57\cdot 7^{2024}.
\]
Therefore, the remainder is $0$.

Final answer:
0