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. $\boxed{6}$

["Understanding Why Any Three Consecutive Integers Yield a Product Divisible by 6", "Mathematics is filled with elegant patterns and unavoidable truths, and one of the simplest yet most powerful insights lies in the behavior of three consecutive integers. Their product is always divisible by 6—a fact grounded in the fundamental properties of numbers, specifically divisibility by 2 and 3. This property not only demonstrates a key number theory principle but also reveals why 6 emerges as the greatest common divisor of all such products.", "### The Structure of Three Consecutive Integers", "Let the three consecutive integers be ( n ), ( n+1 ), and ( n+2 ), where ( n ) is any whole number. These three numbers cover every residue class modulo 3 and span two or more even numbers, ensuring rich divisibility by 2. Simultaneously, among any three consecutive numbers, at least one must be divisible by 3, due to the repeating pattern of remainders when divided by 3.", "### Divisibility by 2: At Least One Even Number", "Among any three consecutive integers, at least one is even. This follows simply from modular arithmetic: any integer ( n ) falls into one of the residue classes mod 2—either 0 or 1. In a group of three consecutive numbers, at least one will be congruent to 0 mod 2. Therefore, each such product contains a factor of 2.", "### Divisibility by 3: One Number Must Be Multipled by 3", "Dividing integers by 3 yields three possible remainders: 0, 1, or 2. In any set of three consecutive numbers, one of them must leave a remainder of 0 when divided by 3. This is because each full set of three consecutive integers cycles through all residue classes mod 3. So one is divisible by 3, guaranteeing that their product is divisible by 3.", "### Conclusion: Product Divisible by ( 2 \ imes 3 = 6 )", "Since the product of any three consecutive integers contains at least one multiple of 2 and one multiple of 3, and since 2 and 3 are prime and coprime, their product ( 2 \ imes 3 = 6 ) must divide every such product. This makes 6 the greatest common divisor (GCD) of all such products—no larger integer can divide every example, since adjusting ( n ) produces different products with no common factor beyond 6.", "### A Quick Example", "Take ( 1, 2, 3 ):\n( 1 \ imes 2 \ imes 3 = 6 ), clearly divisible by 6.", "Now ( 2, 3, 4 ):\n( 2 \ imes 3 \ imes 4 = 24 ), and ( 24 \div 6 = 4 ), confirming divisibility.", "( 3, 4, 5 ):\n( 3 \ imes 4 \ imes 5 = 60 ), and ( 60 \div 6 = 10 ), again divisible.", "No matter the trio, 6 divides the product—this invariant is both simple and profound.", "### Real-World Relevance", "This property isn’t just theoretical. It appears in algorithms involving number theory, cryptography, and combinatorics. Understanding why consecutive products are divisible by 6 strengthens foundational mathematical reasoning and underscores why patterns involving multiples of small primes are essential in diverse applications.", "---", "In summary: Any three consecutive integers yield a product divisible by 6 because among them, one is divisible by 2 and another by 3—ensuring divisibility by ( 2 \ imes 3 = 6 ). This proven principle makes 6 the greatest common divisor of all such products, a beautiful and practical truth in number theory.", "Keywords: three consecutive integers, divisibility by 6, product of integers, GCD of consecutive products, divisibility by 2 and 3, number theory insight."]