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Home » 2v_3 - 3(6 + 2v_1) = 4 \implies 2v_3 - 18 - 6v_1 = 4 \implies 2v_3 = 22 + 6v_1 \implies v_3 = 11 + 3v_1. - AMAZONAWS

2v_3 - 3(6 + 2v_1) = 4 \implies 2v_3 - 18 - 6v_1 = 4 \implies 2v_3 = 22 + 6v_1 \implies v_3 = 11 + 3v_1. - AMAZONAWS

📅 March 11, 2026 👤 scraface
Mar 11, 2026
2v_3 - 3(6 + 2v_1) = 4 \implies 2v_3 - 18 - 6v_1 = 4 \implies 2v_3 = 22 + 6v_1 \implies v_3 = 11 + 3v_1.

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📌 Solution: Solve $x \equiv 3 \pmod{7}$ and $x \equiv 5 \pmod{9}$. Let $x = 7k + 3$. Substitute into the second congruence: $7k + 3 \equiv 5 \pmod{9} \Rightarrow 7k \equiv 2 \pmod{9}$. The modular inverse of 7 modulo 9 is 4 (since $7 imes 4 = 28 \equiv 1 \pmod{9}$), so $k \equiv 2 imes 4 = 8 \pmod{9}$. Thus, $k = 9m + 8$, and $x = 7(9m + 8) + 3 = 63m + 59$. The solutions are $59, 122, 185, \dots$. Within 1–150, only 59 and 122 satisfy. The count is $oxed{2}$.

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