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Math Help - Integers

  1. #1
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    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
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  2. #2
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    Please refrain from multiple posts.
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  3. #3
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    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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