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Given that k is a positive integer less than 6, how many values can k take on such that 4k = k (mod6)has no solutions in x?

 Dec 30, 2020
 #1
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Try some solutions in! After all there are only 5 numbers: 1, 2, 3, 4, 5, 

k=1: 4 = 1 mod 6

k=2: 8 = 2 mod 6

k=3: 12 = 3 mod 6

k=4: 16 = 4 mod 6

k=5: 20 = 5 mod 6

 

Since this question is weirded quite weirdly, i think that it means what values of k do not work instead of x, so therefore 

1, 3, 5 do not work

 

Pls check as I am rusty is modular mathematics

 Dec 30, 2020

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