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

For smaller numbers like this you can just test solutions, but I'd like to know how to do these problems when the numbers are scaled up.

 

Thanks in advance!

:P

 Jan 6, 2021
 #1
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Note that the left-hand side is divisible by 3. Therefore, since 3 divides 6, the left-hand side will be equivalent to a multiple of 3. Therefore, if x were to be 1, 2, 4, or 5, then the given congruence would have no solutions. 4 values.

:D

 Jan 6, 2021
 #2
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Thank you so much Aven!

CoolStuffYT  Jan 6, 2021

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