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Home » = 5m \Rightarrow m = 1 - Esdistancia

= 5m \Rightarrow m = 1 - Esdistancia

📅 March 17, 2026 👤 scraface

Mar 17, 2026

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📌 Any five consecutive integers must include at least one multiple of 3? No: e.g., 1–5: no multiple of 3? 1,2,3,4,5 → 3 is there. 4–8: 6 → yes. 7–11: 9 → yes. 8–12: 9 and 12 → yes. Actually: in any 5 consecutive integers, the distance is less than 6, so by pigeonhole, since residues mod 3 are 0,1,2, and 5 > 3, so at least one residue class must repeat or cover all? Actually, in any 3 consecutive, one divisible by 3; the span 5 covers at least one full residue cycle. Minimal case: the set must cover at least one multiple of 3. In fact, the maximum gap between multiples of 3 is 3, so 5 numbers span more than one cycle, so at least one number is divisible by 3. Similarly for 5: the gap is 5, so one number is divisible by 5.

📌 Thus, product always divisible by \( 2^3 = 8 \), \( 3 \), \( 5 \), so by \( \text{lcm}(8,3,5) = 120 \).

📌 And since for the sequence 1,2,3,4,5: product is \( 120 \), which is exactly 120. So no larger fixed integer divides all such products.

📌 Therefore, the largest integer that must divide the product of any five consecutive integers is \( \boxed{120} \).

📌 Question: What is the sum of all positive divisors of \( 360 \) that are perfect squares?

📌 First, factor \( 360 \):

📌 = 36 \times 10 = (2^2 \cdot 3^2) \cdot (2 \cdot 5) = 2^3 \cdot 3^2 \cdot 5^1

📌 We seek divisors of \( 360 \) that are perfect squares. A divisor is a perfect square if all exponents in its prime factorization are even.

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