Please refrain from multiple posts.
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 Z24 that equals 1
Part b. Not sure how to answer that
You need to use the classical theorem:
The equation,
has a solution(s) for in if and only if divides .
The number of solutions for is .
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In your problem ya' got,
in
Notice,
.
Thus, if this equation got a solution then it got 3 of 'em.
Since, the necessary and suffienct conditions state that needs to be divisible by 3 we have,
Yields solutions, otherwise no.
Be easy there and your posts.