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**How 2026 Federal

**How 2026 Federal
At least one of the four is divisible by $ 3 $
Now consider $ 15 $. Since $ 3 $ and $ 5 $ always divide the product, the product is divisible by $ 15 $.
But we can go further: among four consecutive odd numbers, at least one is divisible by $ 3 $, and at least one divisible by $ 5 $. Also, at least two of them are congruent modulo $ 4 $, so their product includes multiple factors of $ 2 $, but since all are odd, no power of $ 2 $ is introduced.
So the largest integer that must divide the product of any four consecutive odd integers is $ 105 $.
Solution: Let the number be $ x $. We are given:
Let $ x = 7k + 1 $. Substitute into second congruence:
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