Modular

Number Theory Level 2

Find x x 150 x ≡ 35 ( m o d 31 ) . 150 x \equiv 35 \pmod{31}.

14 22 12 24

Ravneet Singh by Ravneet Singh , 30, India

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1 solution

Dividing through by 5 5 gives us that 30 x ≡ 7 ( m o d 31 ) 30x \equiv 7 \pmod{31} . Next, we can write this as

31 x − x ≡ 7 ( m o d 31 ) ⟹ − x ≡ 7 ( m o d 31 ) ⟹ x ≡ − 7 ( m o d 31 ) = 24 31x - x \equiv 7 \pmod{31} \Longrightarrow -x \equiv 7 \pmod{31} \Longrightarrow x \equiv -7 \pmod{31} = \boxed{24} .

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