Maths Olympiad Prep

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Number theory Difficulty 5.3 AIME, harder Find the answer

7. Use the previous problem to solve (i) 3x≡1( mod 125)3 x \equiv 1(\bmod 125); (ii) 5x≡1( mod 243)5 x \equiv 1(\bmod 243).

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

7. (i) 3x≡1( mod 53),3⋅2≡1( mod 5),1−(1−3⋅2)3=1263 x \equiv 1\left(\bmod 5^{3}\right), 3 \cdot 2 \equiv 1(\bmod 5), 1-(1-3 \cdot 2)^{3}=126, so x=42x=42 is a solution to the original congruence equation. (ii) 5x≡1( mod 35).5⋅(−1)≡1( mod 3).1−(1−5⋅(−1))5=5 x \equiv 1\left(\bmod 3^{5}\right) .5 \cdot(-1) \equiv 1(\bmod 3) .1-(1-5 \cdot(-1))^{5}= −7775.x=−1555-7775 . x=-1555 is a solution.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.