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Home » C Sharp String Secrets: Unlock Faster Performance in Seconds! - SUpost

C Sharp String Secrets: Unlock Faster Performance in Seconds! - SUpost

📅 July 5, 2026 👤 scraface

Jul 05, 2026

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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:
R(1) & = 6 \mod 17 = 6, \\
R(3) & = 34 \equiv 0 \mod 17, \\
R(4) & = 57 \equiv 57 - 3\times17 = 57 - 51 = 6 \mod 17, \\
R(5) & = 86 \equiv 86 - 5\times17 = 86 - 85 = 1 \mod 17, \\
R(6) & = 121 \equiv 121 - 7\times17 = 121 - 119 = 2 \mod 17, \\

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