Question Number 68234 by Mikael last updated on 07/Sep/19 Commented by kaivan.ahmadi last updated on 07/Sep/19 $$\mathrm{3}^{{x}+\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{{x}−\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{4}^{{x}} +\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} \Rightarrow \\ $$$$\mathrm{3}^{{x}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{3}+\mathrm{1}\right)=\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} \left(\mathrm{2}+\mathrm{1}\right)\Rightarrow…
Question Number 2699 by abcd last updated on 25/Nov/15 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when} \\ $$$$\mathrm{3}^{\mathrm{215}} \:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{43}. \\ $$ Answered by RasheedAhmad last updated on 25/Nov/15 $$ \\ $$$${Since}\:\left(\mathrm{3},\mathrm{43}\right)=\mathrm{1}\:{and}\:\mathrm{43}\:{is}\:{prime}…
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Question Number 133771 by bramlexs22 last updated on 24/Feb/21 $$\mathrm{A}\:\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{1}\:\mathrm{centered} \\ $$$$\mathrm{at}\:\left(\mathrm{0},\mathrm{0}\right)\:\mathrm{is}\:\mathrm{conjoined}\:\mathrm{to}\:\mathrm{a}\:\mathrm{circle} \\ $$$$\mathrm{of}\:\mathrm{radius}\:\mathrm{2}\:\mathrm{centered}\:\mathrm{at}\:\left(\mathrm{2},\mathrm{0}\right).\: \\ $$$$\mathrm{Forming}\:\mathrm{a}\:\mathrm{single}\:\mathrm{shape}\:.\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{shape}'\mathrm{s}\:\mathrm{area}? \\ $$ Commented by bramlexs22 last updated…
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Question Number 133765 by Salman_Abir last updated on 24/Feb/21 Answered by JDamian last updated on 24/Feb/21 $${R}_{{e}} =\infty \\ $$ Terms of Service Privacy Policy…
Question Number 133764 by liberty last updated on 24/Feb/21 Answered by EDWIN88 last updated on 24/Feb/21 $$\:\mathrm{let}\:\mathrm{x}\:\mathrm{be}\:\mathrm{a}\:\mathrm{page}\:\mathrm{number}\:\mathrm{was}\:\mathrm{counted}\:\mathrm{twice} \\ $$$$\mathrm{Assuming}\:\mathrm{the}\:\mathrm{pages}\:\mathrm{start}\:\mathrm{counting}\:\mathrm{at}\:\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{count}\:\mathrm{continously}\:\mathrm{up}\:\mathrm{we}\:\mathrm{use}\:\mathrm{formula} \\ $$$$\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{n}\right)+\mathrm{x}\:=\:\mathrm{1999}\: \\ $$$$\Rightarrow\mathrm{x}\:+\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\:=\:\mathrm{1999}\:;\:\mathrm{since}\:\mathrm{x}\:\mathrm{is}\:\mathrm{positive}\:…
Question Number 133761 by liberty last updated on 24/Feb/21 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{linear}\:\mathrm{congruence}\: \\ $$$$\mathrm{19}{x}\:\equiv\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{141}\:\right) \\ $$ Answered by bobhans last updated on 24/Feb/21 $${Using}\:{Euclidean}\:{algorithm}\: \\ $$$$\:\mathrm{141}\:=\:\mathrm{7}×\mathrm{19}+\mathrm{8} \\…
Question Number 68226 by Tinku Tara last updated on 07/Sep/19 $$\mathrm{We}\:\mathrm{are}\:\mathrm{working}\:\mathrm{on}\:\mathrm{problems} \\ $$$$\mathrm{reported}\:\mathrm{on}\:\mathrm{post}\:\mathrm{67927}. \\ $$$$ \\ $$$$\mathrm{We}\:\mathrm{will}\:\mathrm{update}\:\mathrm{on}\:\mathrm{the}\:\mathrm{resolution} \\ $$$$\mathrm{as}\:\mathrm{soon}\:\mathrm{as}\:\mathrm{possible}. \\ $$ Commented by Rasheed.Sindhi last…
Question Number 133757 by liberty last updated on 24/Feb/21 Answered by bobhans last updated on 24/Feb/21 $${Let}\:{x}\:{be}\:{the}\:{least}\:{number}\:{of}\:{marbles} \\ $$$${in}\:{the}\:{box}\:,\:{such}\:{that}\:\begin{cases}{{x}\equiv\mathrm{5}\:\left({mod}\:\mathrm{7}\right)}\\{{x}\equiv\mathrm{6}\:\left({mod}\:\mathrm{11}\right)}\\{{x}\equiv\:\mathrm{8}\:\left({mod}\:\mathrm{13}\right)}\end{cases} \\ $$$${We}\:{can}\:{use}\:{Chinese}\:{remainder}\:{theorem} \\ $$$${a}_{\mathrm{1}} =\mathrm{5}\:;\:{a}_{\mathrm{2}} =\mathrm{6}\:;\:{a}_{\mathrm{3}}…