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 #4
avatar+28125 
+1
Jul 21, 2019
 #2
avatar+22892 
+4

If a, b, and c are positive integers less than 13 such that
\(\large{\begin{align*} 2ab+bc+ca&\equiv 0\pmod{13}\\ ab+2bc+ca&\equiv 6abc\pmod{13}\\ ab+bc+2ca&\equiv 8abc\pmod {13} \end{align*}}\)
then determine the remainder when a+b+c is divided by 13.

 

\(\begin{array}{|lrcll|} \hline & 2ab+bc+ca &\equiv& 0\pmod{13} \quad |\quad : (abc)\\ (1) & \dfrac{2}{c} + \dfrac{1}{a} + \dfrac{1}{b} &\equiv& 0 \pmod{13} \\ \hline & ab +2bc+ ca &\equiv& 6abc\pmod{13} \quad |\quad : (abc)\\ (2) & \dfrac{1}{c} + \dfrac{2}{a} + \dfrac{1}{b} &\equiv& 6 \pmod{13} \\ \hline & ab + bc+2ca &\equiv& 8abc\pmod{13} \quad |\quad : (abc)\\ (3) & \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{2}{b} &\equiv& 8 \pmod{13} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline 3*(1)-(2)-(3): & 3*\left( \dfrac{2}{c} + \dfrac{1}{a} + \dfrac{1}{b}\right) \\ & -\left( \dfrac{1}{c} + \dfrac{2}{a} + \dfrac{1}{b}\right) \\ & -\left( \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{2}{b}\right) \\ & &\equiv& 3*0 - 6 - 8 \pmod{13} \\ & \dfrac{6}{c} + \dfrac{3}{a} + \dfrac{3}{b} \\ & -\dfrac{1}{c} - \dfrac{2}{a} - \dfrac{1}{b} \\ & -\dfrac{1}{c} - \dfrac{1}{a} - \dfrac{2}{b} \\ & &\equiv& -14 \pmod{13} \\ & \dfrac{4}{c} &\equiv& -14 \pmod{13} \\ & \dfrac{4}{c} &\equiv& 13-14 \pmod{13} \\ & \dfrac{4}{c} &\equiv& -1 \pmod{13} \quad | \quad +1 \\ & \dfrac{4}{c}+1 &\equiv& 0 \pmod{13} \quad | \quad *c \\ & 4+c &\equiv& 0 \pmod{13} \quad | \quad -4 \\ & c &\equiv& -4 \pmod{13} \\ & c &\equiv& 13-4 \pmod{13} \\ & \mathbf{ c } &\equiv& \mathbf{ 9 \pmod{13} } \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline 3*(3)-(2)-(1): & 3*\left( \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{2}{b}\right) \\ & -\left( \dfrac{1}{c} + \dfrac{2}{a} + \dfrac{1}{b}\right) \\ & -\left( \dfrac{2}{c} + \dfrac{1}{a} + \dfrac{1}{b}\right) \\ & &\equiv& 3*8 - 6 - 0 \pmod{13} \\ & \dfrac{3}{c} + \dfrac{3}{a} + \dfrac{6}{b} \\ & -\dfrac{1}{c} - \dfrac{2}{a} - \dfrac{1}{b} \\ & -\dfrac{2}{c} - \dfrac{1}{a} - \dfrac{1}{b} \\ & &\equiv& 18 \pmod{13} \\ & \dfrac{4}{b} &\equiv& 18-13 \pmod{13} \\ & \dfrac{4}{b} &\equiv& 5 \pmod{13} \quad | \quad -5 \\ & \dfrac{4}{b}-5 &\equiv& 0 \pmod{13} \quad | \quad *b \\ & 4-5b &\equiv& 0 \pmod{13} \quad | \quad *(-1) \\ & 5b &\equiv& 0 \pmod{13} \quad | \quad +4 \\ & 5b &\equiv& 4 \pmod{13} \quad | \quad : 5 \\ & b &\equiv& 4*\dfrac{1}{5} \pmod{13} \\ & && \boxed{\dfrac{1}{5} \pmod{13} \\ \equiv 5^{\varphi(13)-1}\pmod{13} \\ \equiv 5^{12-1}\pmod{13} \\ \equiv 5^{11}\pmod{13} \\ \equiv 48828125\pmod{13} \\ \equiv 8\pmod{13} } \\ & b &\equiv& 4*8 \pmod{13} \\ & b &\equiv& 32 \pmod{13} \\ & b &\equiv& 32-2*13 \pmod{13} \\ & b &\equiv& 6 \pmod{13} \\ & \mathbf{ b } &\equiv& \mathbf{ 6 \pmod{13} } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline 2ab+bc+ca &\equiv& 0\pmod{13} \quad & | \quad b=6, \ c=9 \\ 2a*6+6*9+9*a &\equiv& 0\pmod{13} \\ 21a+54 &\equiv& 0\pmod{13} \quad & | \quad 21\equiv 8\pmod{13},\ \quad 54\equiv 2\pmod{13} \\ 8a +2 &\equiv& 0\pmod{13} \quad & | \quad :2 \\ 4a+1 &\equiv& 0 \pmod{13} \quad & | \quad -1 \\ 4a &\equiv& -1 \pmod{13} \quad | \quad : 4 \\ a &\equiv& (-1)*\dfrac{1}{4} \pmod{13} \\ && \boxed{\dfrac{1}{4} \pmod{13} \\ \equiv 4^{\varphi(13)-1}\pmod{13} \\ \equiv 4^{12-1}\pmod{13} \\ \equiv 4^{11}\pmod{13} \\ \equiv 4194304\pmod{13} \\ \equiv 10\pmod{13} } \\ a &\equiv& (-1)*10 \pmod{13} \\ a &\equiv& -10 \pmod{13} \\ a &\equiv& 13-10 \pmod{13} \\ \mathbf{ a } &\equiv& \mathbf{ 3 \pmod{13} } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{a+b+c \pmod{13}} \\ &\equiv& 3+6+9 \pmod{13} \\ &\equiv& 18 \pmod{13} \\ &\equiv& 18-13 \pmod{13} \\ &\equiv& \mathbf{ 5 \pmod{13} } \\ \hline \end{array}\)

 

laugh

Jul 21, 2019
Jul 20, 2019
 #2
avatar
+1
Jul 20, 2019
 #1
avatar+8579 
+3

 

By the SAS congruence theorem,  △PQT ≅ △SRT  So...

 

m∠QPT  =  m∠RST  =  37°

 

m∠QTP  =  180° - 50°   =   130°

 

m∠PQT  =  180° - 37° - 130°  =  13°

 

Since  QT = RT  we can substitute  QT  in for  RT in the next equation.


\(RT+TP\ =\ 11\\~\\ QT+TP\ =\ 11\\~\\ QT\ =\ 11 - TP\)

 

Now we can substitute  11 - TP  in for  QT in the next equation.

 

By the Law of Sines,

 

\( \frac{TP}{\sin13^\circ} \ =\ \frac{QT}{\sin37^\circ}\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}\ =\ QT\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}\ =\ 11-TP\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}+TP\ =\ 11\\~\\ TP\cdot(\frac{\sin37^\circ}{\sin13^\circ}+1)\ =\ 11\\~\\ TP\ =\ 11\div(\frac{\sin37^\circ}{\sin13^\circ}+1)\\~\\ TP\ =\ \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ} \)

 

Now that we know the length of  TP ,  we can find the length of  QT.

 

\(QT\ =\ 11-TP\\~\\ QT\ =\ 11-\frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}\)

 

Finally, we can use the Law of Sines again to find  QP.

 

\(\frac{QP}{\sin130^\circ}\ =\ \frac{TP}{\sin13^\circ}\\~\\ QP\ =\ TP\cdot\frac{\sin130^\circ}{\sin13^\circ}\\~\\ QP\ =\ \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}\cdot\frac{\sin130^\circ}{\sin13^\circ}\\~\\ QP\ =\ \frac{11\sin130^\circ}{\sin37^\circ+\sin13^\circ}\)

 

So we have found:

 

\(\begin{array}{c} TP&=& \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 2.993\\~\\ QT& =& 11-\frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 8.007\\~\\ QP& =&\frac{11\sin130^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 10.192 \end{array}\)

 

And all lengths are in meters.

Jul 20, 2019
 #2
avatar+682 
+4

or go click here to see a diffrent method

Jul 20, 2019
 #1

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