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

I am having trouble find the inverse.

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- Jul 2nd 2010, 02:23 PMdwsmithInverse of 12x\equiv 1 (mod 15)
$\displaystyle 12x\equiv 1 \ \mbox{(mod 15)}$

I am having trouble find the inverse. - Jul 2nd 2010, 02:27 PMundefined
- Jul 2nd 2010, 07:31 PMSoroban
Hello, dwsmith!

Quote:

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

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.

- Jul 2nd 2010, 07:41 PMmelese
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.