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.

Answer with solution will be appreciated

Imcool Dec 1, 2022

#1**0 **

I have seen this solved. Search for it by putting it in the "Search Box" at the top right-hand corner.

Guest Dec 1, 2022

#2**0 **

Notice that if b = 2a and c = 3a the left hand side of the first equation becomes 4a2 + 6a2 + 3a2 = 13a2

In other words it is a multiple of 13 and hence satisfies the first equation.

For this situation, if a, b and c are all positive integers less than 13, then a can only be one of 1, 2, 3 or 4.

Trying these in turn we find that only a = 3 (hence b = 6 and c = 9) satisfy the second and third equations.

Hence a + b + c = 5 mod(13)

Guest Dec 1, 2022