Track / Stage 5 / 87 of 400 #687 of 1964

Problem 687

AIME late

Number theory Difficulty 5.3 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, $ signs and \frac vs / are all fine.

Official 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.

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