Wait: 2 and 4 → \( 2 \times 4 = 8 \), divisible by 8. Another example: 3–7 → 4 → divisible by 4, 6 → divisible by 2, so again 8.

February 22, 2026

$Wait: 2 and 4 → \( 2 \times 4 = 8 \), divisible by 8. Another example: 3–7 → 4 → divisible by 4, 6 → divisible by 2, so again 8.$

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But more carefully: among five consecutive numbers, the minimal guaranteed prime power factorizations are:
One multiple of 5 → factor of 5
At least two multiples of 2 (possibly one of which is a multiple of 4), but minimal guaranteed is one factor of 4 and one of 2 → ensures \( 2^3 \) is not guaranteed (e.g., sequence 1–5: 2 and 4 → \( 2 \times 4 = 8 \), divisible by 8? 2 and 4 give \( 2^3 \)).
But sequence 4–8: 4 (div by 4), 6 (div by 2), 8 (div by 8) → divisible by 8.
Even sequence 5–9: 8 → div by 8.
Actually, in any five consecutive integers, there is always a multiple of 4 and another even number, so the total power of 2 is at least \( 2 + 1 = 3 \), so divisible by 8.

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