Consider the provided qustion,

If a \(\displaystyle\in{Z}_{{n}}\) then |a|=n/(gcd(a,n))ZSK

here given \(\displaystyle{Z}_{{12}}\). So, n=12

Given a = 6, b = 2

So, a + b = 6 + 2 = 8

\(\displaystyle{\left|{a}\right|}={\left|{3}\right|}=\frac{{12}}{{{\gcd{{\left({3},{12}\right)}}}}}=\frac{{12}}{{3}}={4}\)

\(\displaystyle{\left|{b}\right|}={\left|{8}\right|}=\frac{{12}}{{{\gcd{{\left({8},{12}\right)}}}}}=\frac{{12}}{{4}}={3}\)

\(\displaystyle{\left|{a}+{b}\right|}={\left|{11}\right|}=\frac{{12}}{{{\gcd{{\left({11},{12}\right)}}}}}=\frac{{12}}{{1}}={12}\)

If a \(\displaystyle\in{Z}_{{n}}\) then |a|=n/(gcd(a,n))ZSK

here given \(\displaystyle{Z}_{{12}}\). So, n=12

Given a = 6, b = 2

So, a + b = 6 + 2 = 8

\(\displaystyle{\left|{a}\right|}={\left|{3}\right|}=\frac{{12}}{{{\gcd{{\left({3},{12}\right)}}}}}=\frac{{12}}{{3}}={4}\)

\(\displaystyle{\left|{b}\right|}={\left|{8}\right|}=\frac{{12}}{{{\gcd{{\left({8},{12}\right)}}}}}=\frac{{12}}{{4}}={3}\)

\(\displaystyle{\left|{a}+{b}\right|}={\left|{11}\right|}=\frac{{12}}{{{\gcd{{\left({11},{12}\right)}}}}}=\frac{{12}}{{1}}={12}\)