Question Number 175132 by mnjuly1970 last updated on 20/Aug/22
$$ \\ $$$$\:\:\:\:\:\:\prec\:\:\boldsymbol{{Question}}−\:\boldsymbol{{algebra}}\:\succ \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{Count}}\:\boldsymbol{{the}}\:\boldsymbol{{number}}\:\boldsymbol{{of}}\:\:\:''\:\boldsymbol{{zero}}\:\boldsymbol{{divisor}}\:'' \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{of}}\:\:\boldsymbol{{ring}}\:,\:\:\:\left(\:\mathbb{Z}_{\:\mathrm{45}} \:,\: \:,\:\oplus\:\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:\:\boldsymbol{{m}}.\boldsymbol{{n}}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−−−−−−−−\:\:\:\: \\ $$
Commented by kaivan.ahmadi last updated on 20/Aug/22
$${we}\:{must}\:{find}\:{the}\:{number} \\ $$$${of}\:{digits}\:{k}\:{such}\:{that} \\ $$$$\left({k},\mathrm{45}\right)\neq\mathrm{1}. \\ $$$$\emptyset\left(\mathrm{45}\right)=\phi\left(\mathrm{3}^{\mathrm{2}} ×\mathrm{5}\right)=\mathrm{45}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$$=\mathrm{45}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\left(\frac{\mathrm{4}}{\mathrm{5}}\right)=\mathrm{24} \\ $$$$\Rightarrow{number}\:{of}\:{zero}\:{divisor}= \\ $$$$\mathrm{44}−\mathrm{24}=\mathrm{20} \\ $$$$ \\ $$$$ \\ $$$${or}\:{K}=\left\{\mathrm{3},\mathrm{5},\mathrm{6},\mathrm{9},\mathrm{10},\mathrm{12},\mathrm{15},\mathrm{18},\mathrm{20},\mathrm{21}\right. \\ $$$$\left.,\mathrm{24},\mathrm{25},\mathrm{27},\mathrm{30},\mathrm{33},\mathrm{35},\mathrm{36},\mathrm{39},\mathrm{40},\mathrm{42}\right\} \\ $$$${for}\:{each}\:{k}\in{K}\:;\:\left(\mathrm{45},{k}\right)\neq\mathrm{1}\:{and} \\ $$$$\mid{K}\mid=\mathrm{20}. \\ $$
Commented by mnjuly1970 last updated on 21/Aug/22
![thank you so much.. in fact {[n − (ϕ (n) +1 )]=easy} =45−25=20](Q175145.png)
$${thank}\:{you}\:{so}\:{much}.. \\ $$$$\:\:{in}\:{fact} \\ $$$$\:\:\left\{\left[{n}\:−\:\left(\varphi\:\left({n}\right)\:+\mathrm{1}\:\right)\right]={easy}\right\} \\ $$$$\:\:\:\:\:\:\:=\mathrm{45}−\mathrm{25}=\mathrm{20} \\ $$$$\:\:\: \\ $$$$\: \\ $$
Commented by kaivan.ahmadi last updated on 20/Aug/22
$${yes}.\:{and}\:{so}\:{if}\:\mathbb{Z}_{{n}\:\:} {is}\:{a}\:{field} \\ $$$${it}\:{has}\:{no}\:{zero}\:{divisor}. \\ $$$${since}\:{n}={p}={prime}\:{and} \\ $$$${p}−\mathrm{1}−\phi\left({p}\right)= \\ $$$${p}−\mathrm{1}−{p}\left(\mathrm{1}−\frac{\mathrm{1}}{{p}}\right)={p}−\mathrm{1}−{p}+\mathrm{1}=\mathrm{0} \\ $$
Commented by mnjuly1970 last updated on 21/Aug/22
$$\:{zendeh}\:{bashid}\:... \\ $$