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📅 May 3, 2026 👤 scraface

May 03, 2026

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R(5) &= 3(25) + 10 + 1 = 75 + 10 + 1 = 86, \\
R(6) &= 3(36) + 12 + 1 = 108 + 12 + 1 = 121, \\
R(7) &= 3(49) + 14 + 1 = 147 + 14 + 1 = 162, \\
R(8) &= 3(64) + 16 + 1 = 192 + 16 + 1 = 209, \\
R(9) &= 3(81) + 18 + 1 = 243 + 18 + 1 = 262, \\
R(10) &= 3(100) + 20 + 1 = 300 + 20 + 1 = 321.
We are to find \( \gcd(R(1), R(2), \ldots, R(10)) = \gcd(6, 17, 34, 57...
Notice:
\( R(2) = 17 \), a prime number.
Check whether 17 divides any of the other terms modulo 17:

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