2024 AMC 10B Problem 7

These powers are far too big to compute, so look for something the three terms have in common.

1. Each term is a power of $7$, and the smallest is $7^{2024}$. The other two are just $7^{2024}$ multiplied by more $7$s:
$$7^{2025} = 7^{2024} \cdot 7, \qquad 7^{2026} = 7^{2024} \cdot 49.$$
2. So pull out $7^{2024}$ from all three:
$$7^{2024} + 7^{2025} + 7^{2026} = 7^{2024} \cdot (1 + 7 + 49) = 7^{2024} \cdot 57.$$
3. Now check $57$: it equals $3 \cdot 19$.
4. So the whole sum equals $19 \cdot (3 \cdot 7^{2024})$. It is $19$ times a whole number, so dividing by $19$ leaves remainder $0$.

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

When several terms are powers of the same number, pull out the smallest power first. What remains is a short, small sum that you can test directly. Here the leftover $1 + 7 + 49 = 57$ was chosen by the problem writers because $19$ divides it.

Look at the remainders of powers of $7$ when divided by $19$:

• $7$ leaves remainder $7$.
• $7^2 = 49 = 2 \cdot 19 + 11$ leaves remainder $11$.
• $7^3$: multiply the last remainder by $7$ to get $77 = 4 \cdot 19 + 1$, remainder $1$.

After that the remainders repeat $7, 11, 1, 7, 11, 1, \ldots$ Three powers in a row, like $7^{2024}, 7^{2025}, 7^{2026}$, always cover one of each, so their remainders add to $7 + 11 + 1 = 19$, which leaves remainder $0$.