Competition problem — Number theory

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Number theory Difficulty 5.5 AIME, harder Find the answer

1. Find all solutions of each of the following linear congruences.
a) $3 x \equiv 2(\bmod 7)$
d) $15 x \equiv 9(\bmod 25)$
b) $6 x \equiv 3(\bmod 9)$
e) $128 x \equiv 833(\bmod 1001)$
c) $17 x \equiv 14(\bmod 21)$
f) $\quad 987 x \equiv 610(\bmod 1597)$

Your answer

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

Solution

1. a) $x \equiv 3(\bmod 7)$
e) $x \equiv 812(\bmod 1001)$
b) $x \equiv 2,5,8(\bmod 9)$
f) $x \equiv 1596(\bmod 1597)$
c) $x \equiv 7(\bmod 21)$
d) no 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.