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Thus, the probability that at least two channels allow sufficient flow is: - Upplift

📅 February 28, 2026 👤 scraface

Feb 28, 2026

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We solve \( n^3 \equiv 1 \pmod{8} \) and \( n^3 \equiv 1 \pmod{125} \)...
Step 1: Solve \( n^3 \equiv 1 \pmod{8} \)
Try \( n = 1 \): \( 1^3 = 1 \equiv 1 \mod 8 \)
\( n = 3 \): \( 27 \equiv 3 \mod 8 \)
\( n = 5 \): \( 125 \equiv 5 \mod 8 \)
\( n = 7 \): \( 343 \equiv 7 \mod 8 \)
\( n = 1 \) works.
Actually, solutions are \( n \equiv 1 \pmod{8} \), since the multiplic...
Actually, check:
\( 1^3 = 1 \)

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