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Home » $ k = 2 $: $ 405^\circ $ — too large - Anadea

$ k = 2 $: $ 405^\circ $ — too large - Anadea

📅 May 14, 2026 👤 scraface

May 14, 2026

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📌 What is the largest integer that must divide the product of any five consecutive integers?

📌 To solve this problem, consider five consecutive integers \( n, n+1, n+2, n+3, n+4 \). Among any set of five consecutive integers, at least one is divisible by 5, at least one is divisible by 2, and at least one is divisible by 3. This guarantees that the product is divisible by \( 5 \times 2 \times 3 = 30 \).

📌 To see if 30 is the largest such integer, observe:

📌 One of these integers is divisible by 5.

📌 At least one is divisible by 2, and at least one by 4 (since there are five numbers), ensuring divisibility by \( 2^3 = 8 \). However, the product is already guaranteed by \( 2 \times 3 = 6 \), and typically one is divisible by 4 but not consistently ensuring a higher power.

📌 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 \( 2^3 \)).

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