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# hard number theory

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If $a,b,c$ are positive integers less than $13$ such that\begin{align*} 2ab+bc+ca&\equiv 0\pmod{13}\\ ab+2bc+ca&\equiv 3abc\pmod{13}\\ ab+bc+2ca&\equiv 8abc\pmod {13} \end{align*}then determine the remainder when $a+b+c$ is divided by $13$

Sep 9, 2023

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Adding the three congruences, we get [5ab + 5bc + 5ca \equiv 11abc \equiv 0 \pmod{13},]so ab+bc+ca≡0(mod13). Subtracting the second congruence from the first, we get [ab + 3bc + 3ca \equiv 0 \pmod{13},]so ab+bc+ca≡−3abc≡0(mod13). Subtracting the third congruence from the first, we get [4ab + 4bc + 4ca \equiv -8abc \pmod{13},]so ab+bc+ca≡−4abc≡0(mod13).

Since ab+bc+ca≡0(mod13), we can write ab+bc+ca=13m for some integer m. Then −4abc=13m, so abc=−3m. Since m is an integer, abc must be divisible by 3. However, a,b,c are positive integers less than 13, and the only multiples of 3 less than 13 are 3, 6, and 9. But abc cannot be 3, because −3m is negative. Also, if abc=6, then m=−2, which is not an integer. Therefore, abc=9, and a+b+c=12​.

Sep 9, 2023
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See the answer here. Modify it if necessary: https://web2.0calc.com/questions/modular-number-theory#r6

Sep 9, 2023