Question Number 126666 by bramlexs22 last updated on 23/Dec/20 $$\:\mathrm{3}+\mathrm{33}+\mathrm{333}+\mathrm{3333}+…+\underset{\mathrm{2020}\:{times}} {\underbrace{\mathrm{3333}…\mathrm{3}}}\: \\ $$$${divide}\:{by}\:\mathrm{2}.\:{find}\:{the}\:{remainder} \\ $$ Answered by JDamian last updated on 23/Dec/20 $${it}\:{is}\:{easy} \\ $$$$\mathrm{0}…
Question Number 126657 by adeyemiprof40 last updated on 23/Dec/20 Answered by Olaf last updated on 23/Dec/20 $${a}^{\mathrm{4}} +{a}^{\mathrm{3}} +{a}^{\mathrm{2}} +{a}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\Leftrightarrow\:\frac{{a}^{\mathrm{5}} −\mathrm{1}}{{a}−\mathrm{1}}\:=\:\mathrm{0},\:{a}\neq\mathrm{1} \\ $$$$\Leftrightarrow\:{a}^{\mathrm{5}}…
Question Number 191889 by cortano12 last updated on 03/May/23 $$\:\:\:\:\:\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{k}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{2n}}\:=? \\ $$ Answered by mahdipoor last updated on 03/May/23 $$\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}}…
Question Number 191862 by cortano12 last updated on 02/May/23 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{from}\: \\ $$$$\:\left(\mathrm{2}^{\mathrm{400}} −\mathrm{2}^{\mathrm{320}} \right)\left(\mathrm{2}^{\mathrm{200}} +\mathrm{2}^{\mathrm{160}} \right)\left(\mathrm{2}^{\mathrm{200}} −\mathrm{2}^{\mathrm{160}} \right) \\ $$ Commented by BaliramKumar last updated…
Question Number 191846 by malwan last updated on 01/May/23 $${find}\:{the}\:{last}\:{three}\:{digits} \\ $$$${of}\:\mathrm{4}^{\mathrm{2}^{\mathrm{42}} } \\ $$$${Mohammed}\:{Alwan} \\ $$ Answered by AST last updated on 01/May/23 $$\mathrm{4}^{\mathrm{2}^{\mathrm{42}}…
Question Number 126269 by liberty last updated on 19/Dec/20 $${If}\:{n}\:{is}\:{an}\:{odd}\:{integer}\:,\:{show}\:{that}\:\mathrm{32} \\ $$$${divides}\:\left({n}^{\mathrm{2}} +\mathrm{3}\right)\left({n}^{\mathrm{3}} +\mathrm{7}\right)\:?\: \\ $$ Answered by MJS_new last updated on 19/Dec/20 $$\mathrm{not}\:\mathrm{true}\:\mathrm{for}\:{n}=\mathrm{3} \\…
Question Number 191710 by pete last updated on 29/Apr/23 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{f}\:\mathrm{149}!\:\mathrm{when}\:\mathrm{divided} \\ $$$$\mathrm{by}\:\mathrm{139}? \\ $$ Commented by Rasheed.Sindhi last updated on 29/Apr/23 $$\mathrm{149}!=\mathrm{1}.\mathrm{2}.\mathrm{3}…\mathrm{138}.\mathrm{139}.\mathrm{140}…\mathrm{148}.\mathrm{149} \\ $$$$\therefore\:\mathrm{149}!\mathrm{mod139}=\mathrm{0} \\…
Question Number 191680 by pete last updated on 28/Apr/23 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{of}\:\mathrm{67}!\:\mathrm{when}\:\mathrm{divided} \\ $$$$\mathrm{by}\:\mathrm{7}! \\ $$ Answered by JDamian last updated on 28/Apr/23 $${zero} \\ $$$$ \\…
Question Number 125959 by bramlexs22 last updated on 15/Dec/20 Answered by liberty last updated on 15/Dec/20 $${Let}\:{q}\:{denote}\:{the}\:{numbers}\:{of}\:{eggs}\:{originally} \\ $$$${in}\:{the}\:{basket}.\:{We}\:{want}\:{to}\:{find}\:{the}\:{least}\:{number} \\ $$$${of}\:{eggs}\:{in}\:{the}\:{basket}.\: \\ $$$$\Leftrightarrow\:{q}\:\equiv\:\mathrm{1}\left({mod}\:\mathrm{2}\right)\:\equiv\:\mathrm{1}\left({mod}\:\mathrm{3}\right)\equiv\:\mathrm{1}\left({mod}\:\mathrm{5}\right)\:\equiv\:\mathrm{0}\:\left({mod}\:\mathrm{7}\right) \\ $$$${the}\:{first}\:{three}\:{congruences}\:{then}…
Question Number 191232 by BaliramKumar last updated on 21/Apr/23 $$\mathrm{If}\:{x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:=\:\mathrm{2023}^{\mathrm{2023}} \:\mathrm{then}\:\mathrm{how}\:\mathrm{many}\: \\ $$$$\mathrm{pair}\:\mathrm{of}\:{x},{y}\:{where}\:{x},\:{y}\:\in\:\mathrm{N} \\ $$ Answered by mr W last updated on 21/Apr/23…