Question 20
For a positive integer $$n$$, let $$n\bmod13$$ denote its remainder on division by $$13$$. Integers $$a,b,c$$ satisfy $$4a+5b+6c\equiv1\pmod{13}$$, $$a-b-7c\equiv3\pmod{13}$$ and $$3a-4b+5c\equiv9\pmod{13}$$. Then $$(a+b+c)\bmod13$$ is
Correct Answer: 5
Solution
From the second congruence, $$a\equiv3+b+7c\pmod{13}$$. Substitution into the other two gives $$9b+8c\equiv2\pmod{13}$$ and $$b\equiv0\pmod{13}$$. Hence $$c\equiv10$$ and $$a\equiv8$$, so $$a+b+c\equiv18\equiv5\pmod{13}$$.
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