Abstract Alg

[Vlg]

Question:

Find b such that the equation 15x = b in Z₂₄ has

a. More than one solution for x

b. No solution for x


Part a. Since im working in Z₂₄ would I have more than one answer for part A.

I would guess the answer would be 8, 16, 24, 32
15(8) = 120 and in ₂₄ that equals 1


Part b. Not sure how to answer that

 

[pka]

It seems to me your definitions are not standard.
Usually in Z<SUB>24</SUB> the equation 15x=3 has solutions {5,15,23}.
That is, Mod(75,24)=3: moreover, 15*8=0 not 1.

So for part b, look for a b such Mod(15x,24)≠b.

 

[Vlg]

OK thanks. I understand now how you got the solution for part a but for part b I think the only way it could be no solution is with a negative number. Not sure if that would be the correct way

 

[pka]

See if you can find a solution to 15x=4 in Z<SUB>24</SUB>.

 

[Vlg]

oh ok I see. I could not find a solution to 15x=4 in Z24.
So 4 would be my answer

Thanks for helping me

 

[pka]

That would be one value for b for which there are no solutions.
There are others!