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Thread: Inverse of 12x\equiv 1 (mod 15)

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

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

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

    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!

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

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

    Suppose a solution exists . . .


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

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

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


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

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

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

    And we see that $\displaystyle a$ 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
    $\displaystyle 12x\equiv 1 \ \mbox{(mod 15)}$

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