Solution: The product of $ k $ consecutive integers is divisible by $ k! $. For five consecutive integers, this is $ 5! = 120 $. For example, $ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120 $, and $ 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 =

["The Mathematical Wonder: Why the Product of Five Consecutive Integers Is Always Divisible by 120", "The property that the product of any $ k $ consecutive integers is divisible by $ k! $ is one of those elegant number theory facts that reveals deep structure in the integers—quietly surprising, yet logically inevitable. When $ k = 5 $, this means the product of any five consecutive integers is always divisible by $ 5! = 120 $. But more than just a divisibility rule, this principle is a gateway to understanding factorials, combinatorics, and the inherent symmetry in sequences of integers.", "### What Does It Mean for Five Consecutive Integers?", "Let’s define the problem clearly: For any integer $ n $, the product\n$$\nP = n(n+1)(n+2)(n+3)(n+4)\n$$\nis guaranteed to be divisible by $ 5! = 120 $. This holds true regardless of which five consecutive integers you choose.", "For example:\n- $ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120 $, which equals $ 5! $.\n- $ 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 720 $, and $ 720 \div 120 = 6 $, so divisible.\n- $ 7 \cdot 8 \cdot 9 \cdot 10 \cdot 11 = 55440 $, and $ 55440 \div 120 = 462 $, again divisible.", "This pattern continues indefinitely — why? Let’s explore.", "### The Combinatorial Insight: Permutations and Factorials", "The key lies not just in arithmetic but in combinatorics. The product of $ k $ consecutive integers can be expressed as a quotient of factorials:\n$$\nP = \frac{(n+4)!}{(n-1)!} = (n)(n+1)(n+2)(n+3)(n+4)\n$$\nThis simplifies to $ \frac{(n+4)!}{(n-1)!} $, which represents the number of ways to choose 5 items from $ n+4 $, multiplied by the missing $ (n-1)! $. But more importantly, this form exposes divisibility.", "Any integer divisible by $ k! $ must contain all the prime factors needed across $ k $ consecutive terms. In five consecutive integers, the sequence always includes:", "- At least one multiple of 5\n- At least one multiple of 4\n- At least one multiple of 3\n- At least two even numbers (so divisible by $ 2^2 $)\n- Enough flexibility in factorization to include all primes ≤ 5", "Because $ 5! = 120 = 2^3 \cdot 3 \cdot 5 $, and five consecutive integers will invariably contain the necessary combinations of multiplicities for these primes, their product must include all prime powers required for divisibility by 120.", "### Practical Examples Verified", "Let’s confirm with a few more cases:", "- $ 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 = 2520 $, and $ 2520 \div 120 = 21 $ → divisible\n- $ 10 \cdot 11 \cdot 12 \cdot 13 \cdot 14 = 240240 $, $ 240240 \div 120 = 2002 $ → divisible\n- $ (-2)(-1)(0)(1)(2) = 0 $, which is divisible by any integer — including 120", "Even when zero is included (as in negative or zero starting points), the product is technically divisible by 120, since 0 mod 120 is 0.", "### Why This Division Always Happens", "The logic breaks down to this: Among five consecutive integers:", "- There is at least one multiple of 5 → ensures factor of 5\n- At least one of the even numbers is divisible by 4, and another by 2 → ensures at least $ 2^3 $\n- At least one multiple of 3 → ensures factor of 3\n- At least one multiple of 2 (often more) → ensures $ 2^1 $ needed for $ 2^3 $ combined with others", "Thus, regardless of the starting number $ n $, the combination guarantees all prime power factors of $ 5! $.", "### Broader Implications and Applications", "This principle is not just a curiosity:", "- It underpins combinatorial proofs and binomial coefficients, where $ \binom{n+4}{5} $ depends on this divisibility.\n- It inspires easier verification of large factorial-based products in discrete math.\n- It demonstrates how structured sequences (like arithmetic progressions) encode deeper number-theoretic truths.", "### Conclusion", "The fact that the product of any five consecutive integers is divisible by $ 120 $ is a beautiful illustration of hidden order in numbers. It blends arithmetic, combinatorics, and number theory into a single, powerful idea: structure breeds predictability.", "So the next time you multiply five numbers side by side, remember — you’re not just calculating a number, you’re engaging with a fundamental rule that governs the very fabric of integers.", "---", "Key takeaway:\nThe product of any $ k $ consecutive integers is divisible by $ k! $. For $ k = 5 $, this means divisibility by $ 120 $, a principle rooted in factorial structure and combinatorial necessity. This rule holds universally across all five consecutive integer sequences — a quiet yet profound truth waiting in every sequence.", "---", "Keywords: product of consecutive integers, divisibility by k!, k! and factorials, 5! = 120, combinatorial proof, primes in sequences, number theory, divisibility rules.\nMeta description: Understand why the product of five consecutive integers is always divisible by 120, explained using factorials, combinatorics, and number theory."]