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If \(a, b, c\) are positive integers less than \(13\) such that

 \(\begin{align*} 2ab+bc+ca&\equiv 0\pmod{13}\\ ab+2bc+ca&\equiv 2abc\pmod{13}\\ ab+bc+2ca&\equiv 8abc\pmod {13} \end{align*}\)

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 .

 Nov 13, 2021

Best Answer 

 #5
avatar
+4

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)       
 

 Nov 17, 2021
 #2
avatar+115358 
0

I'd like to see an answer for this too.    indecision

 Nov 13, 2021
 #3
avatar+2228 
+3

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

--. .-

 Nov 15, 2021
edited by Guest  Nov 15, 2021
 #4
avatar+115358 
0

Thanks Ginger.

I get the drift.  I will try it later. 

Melody  Nov 15, 2021
 #5
avatar
+4
Best Answer

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)       
 

Guest Nov 17, 2021
 #6
avatar+2228 
+3

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.  Someone should troll me for it.

However, this solution method will work for the equations in the OP’s question.

 

 

GA

--. .-

 Nov 17, 2021
 #7
avatar+115358 
0

Thanks Guest and Ginger   laugh

 Nov 17, 2021
 #8
avatar+7 
0

Thanks, these were really helpful

 Nov 21, 2021

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