Number of element in \(\displaystyle{Z}{18}×{Z}{15}\) is \(\displaystyle{18}×{15}={270}.\)

So, 270x will be the zero element of \(\displaystyle{Z}{18}×{Z}{15}={\left(\begin{array}{cc} {0}&{0}\end{array}\right)}{Z}={\left(\begin{array}{cc} {0}&{0}\end{array}\right)}\) of for all \(\displaystyle{x}\in{Z}{18}×{Z}{15}.\) Let nn is charactiristic of Z18×Z15 then nn must divide 270 and again let \(\displaystyle{\left({1},{1}\right)}\in{Z}{18}×{Z}{15}{\left({1},{1}\right)}\in{Z}\) then \(\displaystyle{n}{\left({1},{1}\right)}={\left({n},{n}\right)}≡{\left({0},{0}\right)}\in{Z}{18}×{Z}{15}{Z}\), that is nn must be divisible by 18 and 15, infact nn must be least common multiple of 15 and 18. So, charactiristic of \(\displaystyle{Z}{18}×{Z}{15}={l}.{c}.{m}{\left({18},{15}\right)}={90}.\)

So, 270x will be the zero element of \(\displaystyle{Z}{18}×{Z}{15}={\left(\begin{array}{cc} {0}&{0}\end{array}\right)}{Z}={\left(\begin{array}{cc} {0}&{0}\end{array}\right)}\) of for all \(\displaystyle{x}\in{Z}{18}×{Z}{15}.\) Let nn is charactiristic of Z18×Z15 then nn must divide 270 and again let \(\displaystyle{\left({1},{1}\right)}\in{Z}{18}×{Z}{15}{\left({1},{1}\right)}\in{Z}\) then \(\displaystyle{n}{\left({1},{1}\right)}={\left({n},{n}\right)}≡{\left({0},{0}\right)}\in{Z}{18}×{Z}{15}{Z}\), that is nn must be divisible by 18 and 15, infact nn must be least common multiple of 15 and 18. So, charactiristic of \(\displaystyle{Z}{18}×{Z}{15}={l}.{c}.{m}{\left({18},{15}\right)}={90}.\)