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Home » Solution: Let $ x = \fraca + 2ba - 2b $. Then the equation becomes: - Midis

Solution: Let $ x = \fraca + 2ba - 2b $. Then the equation becomes: - Midis

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 Check powers: is it divisible by \( 16 \)? Try 1–5: product \( 120 \), divisible by 8 but not 16. So 16 not guaranteed.

📌 Check 9: need two multiples of 3, but in five consecutive, only one multiple of 3 unless spaced correctly (e.g., 3 and 6: 6 is divisible by 3, 3 by 3 → only one multiple of 3 unless 9 appears). Sequence 7–11: none divisible by 3 → product not divisible by 3? Wait: 9 is divisible by 3, but if no multiple of 3, like 4–8: no multiple of 3 → product \( 4 \times 5 \times 6 \times 7 \times 8 \)? Wait, 6 is included → divisible by 3.

📌 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 \( 360 \):

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