+7 If \(a, b, c\) are positive integers less than \(13\) such that
\(\begin{align*} 2ab+bc+ca&\equiv 0\pmod{13}\\ ab+2bc+ca&\equiv 2abc\pmod{13}\\ ab+bc+2ca&\equiv 8abc\pmod {13} \end{align*}\)
then determine the remainder when \(a+b+c\) is divided by \(13\).
I've tried to solve this but I just don't know how
If are positive integers less than such thatthen determine the remainder when is divided by .
We know that the integers modulo 13 form a field, so we can operate on systems almost like usual. Equalities will just mean congruence modulo 13 .
From the first equation we obtain ca=−2ab−bc and therefore the second equation gives
ab+2bc−2ab−bc=2b(−2ab−bc)
that is
b(a−c)=2b2(2a+c)
The third equation gives
ab+bc−4ab−2bc=8b(−2ab−bc)
that is:
b(3a+c)=8b^2(2a+c)
We know that b≠0 (because b is a positive integer less than 13 ), so we can simplify it off. If 2a+c=0 , then also a−c=0 , but this implies a=c=0 , which is excluded. Thus we obtain
2b=(a−c) / (2a+c)
8b=(3a+c) / (2a+c)
Thus we need
4(a−c) / (2a+c)=(c+3a) / (2a+c)
and therefore (4a−4c)=(c+3a) , that is, a=5c .
Now we can substitute in 2ab+bc+ca=0 to get
10bc+bc+5c^2=0
Since c≠0 we obtain 11b+5c=0 , so 2b=5c Multiplying by 7 yields 14b=35c and so b=9c
Now we can substitute in
ab+2bc+ca=2abc
to get:
45c^2+18c^2+5c^2=90c^3
Thus:
3c^2=12c^3
and so −c=3 , that is, c=10 . Hence a=5c=11 and b=9c=12
Solution:
a=11,b=12,c=10
and so a+b+c≡7(mod13)
+2440 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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We know that the integers modulo 13 form a field, so we can operate on systems almost like usual. Equalities will just mean congruence modulo 13 .
From the first equation we obtain ca=−2ab−bc and therefore the second equation gives
ab+2bc−2ab−bc=2b(−2ab−bc)
that is
b(a−c)=2b2(2a+c)
The third equation gives
ab+bc−4ab−2bc=8b(−2ab−bc)
that is:
b(3a+c)=8b^2(2a+c)
We know that b≠0 (because b is a positive integer less than 13 ), so we can simplify it off. If 2a+c=0 , then also a−c=0 , but this implies a=c=0 , which is excluded. Thus we obtain
2b=(a−c) / (2a+c)
8b=(3a+c) / (2a+c)
Thus we need
4(a−c) / (2a+c)=(c+3a) / (2a+c)
and therefore (4a−4c)=(c+3a) , that is, a=5c .
Now we can substitute in 2ab+bc+ca=0 to get
10bc+bc+5c^2=0
Since c≠0 we obtain 11b+5c=0 , so 2b=5c Multiplying by 7 yields 14b=35c and so b=9c
Now we can substitute in
ab+2bc+ca=2abc
to get:
45c^2+18c^2+5c^2=90c^3
Thus:
3c^2=12c^3
and so −c=3 , that is, c=10 . Hence a=5c=11 and b=9c=12
Solution:
a=11,b=12,c=10
and so a+b+c≡7(mod13)
+2440 Guest’s solution is correct. Mine is NOT.
My partial solution (copied from my archive of solved equations) is based on similar equations with a different set of congruencies (mod 13).
\(\begin{align*} 2ab+bc+ca&\equiv abc \pmod{13}\\ ab+2bc+ca&\equiv 3abc\pmod{13}\\ ab+bc+2ca&\equiv 5abc\pmod {13} \end{align*} \)
I failed to modify the congruencies to match the question. 
However, this solution method will work for the equations in the OP’s question.
GA
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