1. ## 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

2. Please refrain from multiple posts.

3. 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 $\displaystyle d$.
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$\displaystyle 15x=b$ in $\displaystyle \mathbb{Z}_{24}$
$\displaystyle d=\gcd(15,24)=3$.
Since, the necessary and suffienct conditions state that $\displaystyle b$ needs to be divisible by 3 we have,
$\displaystyle b\in \{3,6,9,12,15,18,21\}$