2024 AMC 10B Problem 7

All three terms share the factor $7^{2024}$:
$$7^{2024} + 7^{2025} + 7^{2026} = 7^{2024}\left(1 + 7 + 7^2\right) = 7^{2024} \cdot 57.$$
Since $57 = 3 \cdot 19$, the entire expression is $19 \cdot \left(3 \cdot 7^{2024}\right)$, a multiple of $19$. Its remainder on division by $19$ is $0$.

The answer is $\boxed{\textbf{(A)}\ 0}$.

A sum of consecutive powers is a geometric block; factor out the smallest power and the remaining short sum $1 + r + r^2$ is a small number whose divisibility is easy to read. Here the problem setters chose $1 + 7 + 49 = 57$ precisely because $19$ divides it. Look for common factors before reaching for cycles of powers.

Powers of $7$ mod $19$ cycle with period $3$: $7, 11, 1$. The exponents $2024, 2025, 2026$ are three consecutive integers, so they hit each residue once and the sum is $7 + 11 + 1 = 19 \equiv 0$.

• Trying to compute each of the three powers mod 19 separately without noticing the common factor, and making a cycle-length error along the way.
• Concluding the remainder is $57 \bmod 19$ "plus something" from $7^{2024}$, forgetting that a zero factor kills the whole product mod $19$.