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For Plan A (compounded annually): - Upplift

📅 February 28, 2026 👤 scraface

Feb 28, 2026

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📌 Let \( k = 5a + 3 $. Substitute into the second:

📌 5a + 3 \equiv 4 \pmod{7} \implies 5a \equiv 1 \pmod{7}

📌 Multiply both sides by the modular inverse of 5 modulo 7. Since \( 5 \cdot 3 = 15 \equiv 1 \mod 7 \), the inverse is 3:

📌 a \equiv 3 \cdot 1 = 3 \pmod{7} \implies a = 7b + 3

📌 k = 5(7b + 3) + 3 = 35b + 15 + 3 = 35b + 18

📌 So \( k \equiv 18 \pmod{35} \)

📌 Now include the third congruence: \( k \equiv 2 \pmod{9} \)

📌 Let \( k = 35b + 18 \). Then:

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