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Given that k  is a positive integer less than 6, how many values can k take on such that \(3x \equiv k \pmod{6}\)  has no solutions in x ?

tertre  Apr 5, 2017
 #1
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By simple inspection, k can take only the following values:

k =1, 2, 4 and 5, so that it has no solution in x.

Guest Apr 5, 2017
 #2
avatar+20016 
+1

Given that k  is a positive integer less than 6,

how many values can k take on such that

\(3x \equiv k \pmod{6}\) has no solutions in x ?

 

Rewrite:

\(\begin{array}{|lrcll|} \hline 3x \equiv k \pmod{6} \\ \text{rewrite...} \\ \begin{array}{|rcll|} \hline 3x-k &=& n\cdot 6 \quad &| \quad n \in \mathbf{Z} \qquad (n \text{ is a integer}) \\ 3x &=& n\cdot 6 +k \\ x &=& \frac{n\cdot 6 +k}{3} \\ x &=& 2n+\frac{k}{3} \quad &| \quad \frac{k}{3} \text{ is a integer, if } k = 0 \text{ or } k = 3 \\ & & \quad &| \quad \frac{k}{3} \text{ is not a integer, if } k = 1,\ k = 2,\ k=4,\ k= 5 \qquad 0 < k < 6 \\ \hline \end{array} \\ \hline \end{array} \)

 

laugh

heureka  Apr 5, 2017
 #3
avatar+3048 
0

Thanks guys!

tertre  Apr 5, 2017

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