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📅 March 11, 2026 👤 scraface

Mar 11, 2026

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Start with the first two:
Step 1: Solve \( N \equiv 3 \pmod{4} \) and \( N \equiv 2 \pmod{5} \)
Let \( N = 4a + 3 \). Substitute into second:
4a + 3 \equiv 2 \pmod{5} \Rightarrow 4a \equiv -1 \equiv 4 \pmod{5} \R...
So \( a = 5b + 1 \), then \( N = 4(5b + 1) + 3 = 20b + 7 \)
Thus, \( N \equiv 7 \pmod{20} \)
Step 2: Incorporate \( N \equiv 4 \pmod{7} \)
Let \( N = 20b + 7 \). Then:
20b + 7 \equiv 4 \pmod{7} \Rightarrow 20b \equiv -3 \equiv 4 \pmod{7}
Since \( 20 \equiv 6 \equiv -1 \pmod{7} \), we have:

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