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Home » もはや古典！ Nintendo Switchを完全コントロールする驚愕スイッチ改造! - Dyverse

もはや古典！ Nintendo Switchを完全コントロールする驚愕スイッチ改造! - Dyverse

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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$ n = 1 $: $ 1^2 = 1 \equiv 1 \pmod{12} $ ✅
Thus, the smallest such $ n $ is $ \boxed{1} $.
Solution: Consider four consecutive integers: $ n, n+1, n+2, n+3 $.
The product is $ P = n(n+1)(n+2)(n+3) $.
Among any four consecutive integers, one is divisible by 4.
At least two are even, so the product is divisible by $ 2^3 = 8 $.
Thus, $ P $ is divisible by $ \text{lcm}(8, 3) = 24 $.
We check if a larger fixed divisor is guaranteed:
Not always divisible by 5 (e.g., $ n = 1 $: product = 24, not divisibl...
Not always divisible by higher powers (e.g., $ n = 1 $: product = 24,...

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