Integers

**mathlg** · May 7th 2006, 10:58 AM

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

**Jameson** · May 7th 2006, 12:24 PM

Please refrain from multiple posts.

**ThePerfectHacker** · May 7th 2006, 02:58 PM

You need to use the classical theorem:
The equation,
$ax=b$
has a solution(s) for $x$ in $\mathbb{Z}_n$ if and only if $d=\gcd(a,n)$ divides $b$ .
The number of solutions for $x$ is $d$ .
---------
In your problem ya' got,
$15x=b$ in $\mathbb{Z}_{24}$
Notice,
$d=\gcd(15,24)=3$ .
Thus, if this equation got a solution then it got 3 of 'em.
Since, the necessary and suffienct conditions state that $b$ needs to be divisible by 3 we have,
$b\in \{3,6,9,12,15,18,21\}$
Yields solutions, otherwise no.

Be easy there and your posts.

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