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If a,b,c are integers from the set of positive integers less than 7 such that

\(\begin{align*} abc&\equiv 1\pmod 7,\\ 5c&\equiv 2\pmod 7,\\ 6b&\equiv 3+b\pmod 7, \end{align*}\)

then what is the remainder when a+b+c is divided by 7?

RektTheNoob  Aug 8, 2017
 #1
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By simple iteration:

a = 3, b = 2, c=6

{3+2+6} mod 7 = 4

Guest Aug 8, 2017
 #2
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If a,b,c are integers from the set of positive integers less than 7 such that

\(\begin{align*} abc&\equiv 1\pmod 7,\\ 5c&\equiv 2\pmod 7,\\ 6b&\equiv 3+b\pmod 7, \end{align*}\)

then what is the remainder when a+b+c is divided by 7 ?

 

\(\begin{array}{rcll} \mathbf{Formula}: \\\\ 6b & \equiv & 3+b \pmod 7 \quad & | \quad -b \\ 6b-b & \equiv & 3+b-b \pmod 7 \\ 5b & \equiv & 3 \pmod 7 \\ \begin{array}{|rcll|} \hline \mathbf{5b} &\mathbf{\equiv}& \mathbf{3 \pmod 7} \\ \hline \end{array} & (1) \\\\ \begin{array}{|rcll|} \hline \mathbf{5c} &\mathbf{\equiv}& \mathbf{2 \pmod 7} \\ \hline \end{array} & (2) \\\\ \begin{array}{|rcll|} \hline \mathbf{abc} &\mathbf{\equiv}& \mathbf{1 \pmod 7} \\ \hline \end{array} & (3) \\\\ \begin{array}{|rcll|} \hline \mathbf{a+b+c} &\mathbf{\equiv}& \mathbf{x \pmod 7} \\ \hline \end{array} & (4) \\\\ (1)+(2): \qquad 5b+5c & \equiv & 3+2 \pmod 7\\ 5(b+c) & \equiv & 5 \pmod 7 \quad & | \quad :5 \\ \frac{5(b+c)}{5} & \equiv & \frac{5}{5} \pmod 7 \\ b+c & \equiv & 1 \pmod 7 \\ \begin{array}{|rcll|} \hline \mathbf{b+c} &\mathbf{\equiv}& \mathbf{1 \pmod 7} \\ \hline \end{array} & (5) \\\\ (5) \text{ in }(4): \qquad a+b+c & \equiv & x \pmod 7 \\ a+ (b+c) & \equiv & x \pmod 7 \\ a+ (1) & \equiv & x \pmod 7 \\ a+ 1 & \equiv & x \pmod 7 \\ \begin{array}{|rcll|} \hline \mathbf{a+ 1 } &\mathbf{\equiv}& \mathbf{x \pmod 7} \\ \hline \end{array} & (6) \\\\ (1) \times (2): \qquad 5b\times 5c & \equiv & 3\times 2 \pmod 7 \\ 25bc & \equiv & 6 \pmod 7 \quad & | \quad 6 \equiv -1 \pmod 7 \\ 25bc & \equiv & -1 \pmod 7 \\ \begin{array}{|rcll|} \hline \mathbf{25bc} &\mathbf{\equiv}& \mathbf{-1 \pmod 7} \\ \hline \end{array} & (7) \\\\ \mathbf{a=\ ? } \\ (3) \div (7): \qquad \frac{abc}{25bc} & \equiv & \frac{1}{-1} \pmod 7 \\ \frac{a}{25} & \equiv & -1 \pmod 7 \quad & | \quad \times 25 \\ \frac{25a}{25} & \equiv & (-1)\cdot 25 \pmod 7 \\ a & \equiv & -25 \pmod 7 \\ a & \equiv & -25 + 4\times 7 \pmod 7 \\ a & \equiv & -25 + 28 \pmod 7 \\ a & \equiv & 3 \pmod 7 \\ a-3 &=& n\cdot 7 \\ a &=& 3+ n\cdot 7 \qquad n \in \mathbb{Z} \\ \begin{array}{|rcll|} \hline \mathbf{a} &\mathbf{=}& \mathbf{ 3+ n\cdot 7 \qquad n \in \mathbb{Z} } \\ \hline \end{array} & (8) \\\\ \mathbf{x=\ ? } \\ (6): \qquad a+1 & \equiv & x \pmod 7 \quad & | \quad a = 3+ n\cdot 7 \\ 3+ n\cdot 7 + 1 & \equiv & x \pmod 7 \\ 4+ n\cdot 7 & \equiv & x \pmod 7 \quad & | \quad n\cdot 7 \pmod 7 = 0 \\ 4 & \equiv & x \pmod 7 \\ 4 \pmod 7 &=& x \\ 4 &=& x \\ \begin{array}{|rcll|} \hline \mathbf{x} &\mathbf{=}& \mathbf{ 4 } \\ \hline \end{array} \\ \end{array}\)

 

The remainder when \(a+b+c\) is divided by \(7\) is \(\mathbf{4}\)

 

laugh

heureka  Aug 10, 2017
edited by heureka  Aug 10, 2017
edited by heureka  Aug 10, 2017
edited by heureka  Aug 10, 2017

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