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\[ \fracd60 + \fracd40 = 5 \] - Ready Digital AB

📅 March 7, 2026 👤 scraface

Mar 07, 2026

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So, \( n^2 \equiv 1 \pmod{5} \) when \( n \equiv 1 \) or \( 4 \pmod{5}...
Step 3: Combine using Chinese Remainder Theorem
We seek the smallest positive \( n \) such that:
\( n \equiv 1 \text{ or } 2 \pmod{3} \)
\( n \equiv 1 \text{ or } 4 \pmod{5} \)
\( n \equiv 1 \pmod{3}, n \equiv 1 \pmod{5} \Rightarrow n \equiv 1 \pm...
\( n \equiv 1 \pmod{3}, n \equiv 4 \pmod{5} \): Solve \( n = 5k + 4 \)...
\( n \equiv 2 \pmod{3}, n \equiv 1 \pmod{5} \): Let \( n = 5k + 1 \)....
\( n \equiv 2 \pmod{3}, n \equiv 4 \pmod{5} \): \( n = 5k + 4 \). Then...
So possible solutions: 1, 4, 11, 14. The smallest positive \( n > 1 \)...

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