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Home » 375 as a Fraction? No One Expected This Simple Math Secret! - Dachbleche24

375 as a Fraction? No One Expected This Simple Math Secret! - Dachbleche24

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 But it’s only possible if the 5 positions allow a 3-element non-adjacent subset — which requires that the positions span at least 5 positions (e.g., 1,3,5), and in any 5 consecutive positions, there is exactly **one** such set? Not quite — for positions {1,2,3,4,5}, non-adjacent triples: 1,3,5 only? But 1,4,5 has adjacent, 2,4,5 has adjacent. Yes, only {1,3,5} if starting at 1, but depending on layout.

📌 Actually, the number is not always 1. For example, {1,3,4} has adjacent 3-4, invalid. So only triples with min gap 2.

📌 In a sequence of 5 positions, the number of 3-element subsets with no two adjacent is $ \binom{3}{3} = 1 $? No:

📌 General formula: $ \binom{n - k + 1}{k} = \binom{5 - 3 +

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\(a^2 = 9\), \(b^2 = 16\) gives \((a, b) = (\pm 3, \pm 4)\)
\(a^2 = 16\), \(b^2 = 9\) gives \((a, b) = (\pm 4, \pm 3)\)
\(a^2 = 25\), \(b^2 = 0\) gives \((a, b) = (\pm 5, 0)\)
Counting all combinations for \((a, b)\):
\((0, \pm 5)\) gives 2 points: \((3, 9), (3, -1)\)
\((\pm 4, \pm 3)\) gives 8 points: \((7, 7), (7, 1), (-1, 7), (-1, 1),...
\((\pm 3, \pm 4)\) gives 8 points: \((6, 8), (6, 0), (0, 8), (0, 0), (...
\((\pm 5, 0)\) gives 2 points: \((8, 4), (-2, 4)\)
Total integer points: \(2 + 8 + 8 + 2 = 20\).
Thus, the number of integer coordinate points on the circle is:

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