2024 AMC 10B Problem 7

Pull out the common power: $7^{2024} + 7^{2025} + 7^{2026}$ $= 7^{2024}(1 + 7 + 49)$ $= 7^{2024} \cdot 57.$ And $57 = 3 \cdot 19,$ so the product is a multiple of $19.$ The remainder is $0.$ Thus, A is the correct answer.