+0

# Hey guys!

0
164
3
+1222

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
Sort:

#1
0

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
+18767
+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:

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

heureka  Apr 5, 2017
#3
+1222
0

Thanks guys!

tertre  Apr 5, 2017

### 15 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details