Solution:
\(\text{Modular arithmetic theorem: }\\ \text {If (a, b, c) are positive integers and} \pmod{m} \text { is relatively prime, then there exists }\\ (a^{-1}, b^{-1}, c^{-1}) \text { such that } \left(a*a^{-1} \equiv b*b^{-1} \equiv c*c^{-1} \pmod{m} \right), \\\text{ with a, b, c, relatively prime within the ring of modulo (m); i.e. } 1\leq [a’, b’, c’] < 13. \)
\(\begin{array}{|rccc|} \hline a^{-1}b^{-1}c^{-1} (2ab + bc + ca) \equiv 1 \pmod{13}\\ a^{-1}b^{-1}c^{-1} (ab + 2bc + ca) \equiv 3 \pmod{13} \\ a^{-1}b^{-1}c^{-1} (ab + bc + 2ca) \equiv 5 \pmod{13}\\ \hline\end{array} \Rightarrow \begin{array}{|rccc|} \hline 2c^{-1} + a^{-1} + b^{-1} \equiv 1 \pmod{13}\\ c^{-1} + 2a^{-1} + b^{-1} \equiv 3 \pmod{13}\\ c^{-1} + a^{-1} + 2b^{-1} \equiv 5 \pmod{13}\\ \hline\end{array} \)
\(\text {add equations: }\\ 4c^{-1} + 4a^{-1} + 4b^{-1} \equiv 9 \pmod{13}\\ 4(c^{-1} + a^{-1} + b^{-1}) \equiv 9 \pmod{13}\\ \)
\(\text {Note that} -4\pmod{13}\equiv 9, \text{ so } \left (a^{-1} + b^{-1} + c^{-1}\right) \equiv -1\\ \text {and } -1 \equiv 12 \pmod{13},\text { so search for three numbers where } 1 \leq a,b,c < 12.\\ \text {using the above inverse equations to find which value belongs to each variable. }\\ \text{Then take the Modular Multiplicative Inverse of these numbers, sum them to (S) and find } \mathrm S\pmod{13}. \\ \text {if you do this correctly, you will find the answer (remainder) is } \boxed {2}\\ \)
GA
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