Competition problem — Number theory

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Number theory Difficulty 5.3 AIME, harder Find the answer

Example 2 Find the solution to the congruence equation $4 x^{2}+27 x-7 \equiv 0(\bmod 15)$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Similarly, direct calculation shows that $x=-7,-2,-1,4$ are solutions. Therefore, the solutions are
$x \equiv-7,-2,-1,4(\bmod 15),$

The number of solutions is 4.

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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.