Maths Olympiad Prep

Track / Stage 5 / 127 of 400 #727 of 1964

Problem 727

AIME late
Number theory Difficulty 5.3 Find the answer

Find the solution to the congruence equation 4x2+27x−7≡0( mod 15)4 x^{2}+27 x-7 \equiv 0(\bmod 15).

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

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Official solution

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

The number of solutions is 4.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.