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Given that k is a positive integer less than 6, how many values can k take on such that 3k\(\equiv\)k (mod6)has no solutions in x?

 Sep 14, 2018
 #1
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where is x in the equation?

 Sep 14, 2018
 #2
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Given that k is a positive integer less than 6, how many values can k take on such that 3k=k (mod6)has no solutions in x?

 

\(\frac{3k}{6}=a+\frac{n}{6}\quad and \quad \frac{k}{6}=b+\frac{n}{6}\quad \\\text{where a, b and n are pos integers and }k<6\)

 

 

 

if k=1   it doesn't work

if k=2   it doesn't work

if k=3    it works

if k= 4  it doesn't work

if k= 5  it doesn't work

so k=3 

 

k can only take on 1 value and that value is 3

 Sep 15, 2018

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