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

• Jul 2nd 2010, 03:23 PM
dwsmith
Inverse of 12x\equiv 1 (mod 15)
$12x\equiv 1 \ \mbox{(mod 15)}$

I am having trouble find the inverse.
• Jul 2nd 2010, 03:27 PM
undefined
Quote:

Originally Posted by dwsmith
$12x\equiv 1 \ \mbox{(mod 15)}$

I am having trouble find the inverse.

gcd(12,15) does not equal 1, so 12 has no inverse mod 15.
• Jul 2nd 2010, 08:31 PM
Soroban
Hello, dwsmith!

Quote:

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

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.

• Jul 2nd 2010, 08:41 PM
melese
Quote:

Originally Posted by dwsmith
$12x\equiv 1 \ \mbox{(mod 15)}$

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.