Solution: Any three consecutive integers include at least one multiple of 2 and one multiple of 3. Thus, their product is divisible by $2 \times 3 = 6$. For example, $1 \times 2 \times 3 = 6$, $2 \times 3 \times 4 = 24$, and $3 \times 4 \times 5 = 60$. The greatest common divisor of all such products is 6. $\boxed{6}$

Mon, 04 May 2026 13:17:33 +0000

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