Question: What is the remainder when $ 2023^2 + 2024^2 + 2025^2 $ is divided by 6?
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- Is 48 always a divisor? $ 24 $ is not divisible by 48 → no
- So maximum guaranteed divisor is 24.
- Thus, the largest integer that must divide the product is $ \boxed{24} $.
- Solution: We compute modulo 6. Note that $ 2023 \equiv 1 \pmod{6} $, $ 2024 \equiv 2 \pmod{6} $, $ 2025 \equiv 3 \pmod{6} $.
- 2023^2 \equiv 1^2 =
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