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
May 7th 2006, 12:24 PM
Please refrain from multiple posts.
May 7th 2006, 02:58 PM
You need to use the classical theorem:
has a solution(s) for in if and only if divides .
The number of solutions for is .
In your problem ya' got, in
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,