(C) Given,$150 x \equiv 35 \pmod{31}$.First,simplify the congruence by dividing by the common factor $5$ (since $\gcd(5, 31) = 1$):$30 x \equiv 7 \pmo

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(C) Given,$150 x \equiv 35 \pmod{31}$.First,simplify the congruence by dividing by the common factor $5$ (since $\gcd(5, 31) = 1$):$30 x \equiv 7 \pmod{31}$.We can write $30$ as $-1 \pmod{31}$:$-x \equiv 7 \pmod{31}$.Multiplying both sides by $-1$:$x \equiv -7 \pmod{31}$.To find the positive residue,add $31$:$x \equiv -7 + 31 \pmod{31} \Rightarrow x \equiv 24 \pmod{31}$.Thus,$x = 24$ satisfies the given congruence.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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$A$ value of $x$ satisfying $150 x \equiv 35 \pmod{31}$ is

KCET 2013 All KCET

A
$14$
B
$22$
C
$24$
D
$12$

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4-2.Quadratic Equations and Inequations

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$A$ value of $x$ satisfying $150 x \equiv 35 \pmod{31}$ is

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