There is at least one multiple of 3 → $ 3 \mid $ product
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- Among any four consecutive integers:
- There are at least two even numbers, and one of them is divisible by 4 → total factor of $ 2^3 = 8 $
- There is exactly one multiple of 4 and another even, so $ 8 \mid $ product
- There is no guaranteed factor of 5 or higher
- So the product is always divisible by $ 8 \cdot 3 = 24 $. But can we do better?
- Check more precisely:
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