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Math Help - Inverse of 12x\equiv 1 (mod 15)

  1. #1
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    Inverse of 12x\equiv 1 (mod 15)

    12x\equiv 1 \ \mbox{(mod 15)}

    I am having trouble find the inverse.
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    12x\equiv 1 \ \mbox{(mod 15)}

    I am having trouble find the inverse.
    gcd(12,15) does not equal 1, so 12 has no inverse mod 15.
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    Hello, dwsmith!

    12x\equiv 1 \text{ (mod 15)}

    I am having trouble find the inverse.
    No wonder . . . it doesn't exist.

    Suppose a solution exists . . .


    We have: . 12x \:\equiv\:1\text{ (mod 15)}

    This means: . 12x \:=\:15a + 1\:\text{ for some integer }a.

    Solve for x\!:\;\;x \:=\:\dfrac{15a+1}{12} \:=\:a + \dfrac{3a+1}{12}


    Since x is an integer, 3a+1 must be a multiple of 12.

    This means: . 3a+1 \:=\:12b\:\text{ for some integer }b.

    Solve for a\!:\;\;a \:=\:\dfrac{12b-1}{3} \:=\:4b - \frac{1}{3}

    And we see that a is not an integer.

    . . Therefore, there is no solution.

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  4. #4
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    Quote Originally Posted by dwsmith View Post
    12x\equiv 1 \ \mbox{(mod 15)}

    I am having trouble find the inverse.
    By definition an integer b is called an inverse of a modulo n if b is a solution to the congruence ax\equiv 1 (mod\ n). And a solution exists only if gcd(a,n)|1, which is not the case here.
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