
Integers
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

Please refrain from multiple posts.

You need to use the classical theorem:
The equation,
$\displaystyle ax=b$
has a solution(s) for $\displaystyle x$ in $\displaystyle \mathbb{Z}_n$ if and only if $\displaystyle d=\gcd(a,n)$ divides $\displaystyle b$.
The number of solutions for $\displaystyle x$ is $\displaystyle d$.

In your problem ya' got,
$\displaystyle 15x=b$ in $\displaystyle \mathbb{Z}_{24}$
Notice,
$\displaystyle d=\gcd(15,24)=3$.
Thus, if this equation got a solution then it got 3 of 'em.
Since, the necessary and suffienct conditions state that $\displaystyle b$ needs to be divisible by 3 we have,
$\displaystyle b\in \{3,6,9,12,15,18,21\}$
Yields solutions, otherwise no.
Be easy there and your posts.