$\displaystyle 12x\equiv 1 \ \mbox{(mod 15)}$
I am having trouble find the inverse.
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.
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.