Question

What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{1320}$ is defined?

Answer 1

Answer:

$7^{-1} \equiv 943 \,\,\, \text{mod 1320}$

Step-by-step explanation:

According to the information you are looking for the smallest integer n greater than 1 such that

$n^{-1} (\text{mod} \,\,\,\, 1320)$

is defined.

To begin with remember what a coprime means, two numbers are coprimes when their greatest common divisor is 1.

For example, 9,12 are NOT coprime because their greater common divisor is 3. But 5,6 are coprime because their greatest common divisor is 1.

Also what Euler's theorem.  If "a,m" are coprimes then

${\displaystyle a^{\phi(m)} \equiv 1$

Where  ${\displaystyle \phi$ is the Euler's totient function. Remember that the totient function computes the number of coprimes less than  $n$.  Then

The Euler's theorem can also be applied for multiplicative inverses.

The smallest integer coprime to 1320, also

$\phi(1320) = 320$

Then

$7^{-1} \equiv 7 ^{\phi(1320)-1} \,\,\, \text{mod 1320} \equiv 7^{320-1} \,\,\, \text{mod 1320} \\\equiv 7^{319} \,\,\, \text{mod 1320} \equiv 943 \,\,\, \text{mod 1320}$