Question Number 193485 by Socracious last updated on 15/Jun/23 $$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{Show}}\:\boldsymbol{\mathrm{that}}\:\mathrm{2}^{\boldsymbol{\mathrm{n}}} −\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} \:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{divisible}}\:\boldsymbol{\mathrm{by}} \\ $$$$\:\:\:\:\:\:\:\mathrm{3}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{integers}}\:\boldsymbol{\mathrm{n}}. \\ $$ Answered by MM42 last updated on 15/Jun/23…
Question Number 193286 by BaliramKumar last updated on 09/Jun/23 $$\mathrm{Let}\:'\mathrm{P}'\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:\left(\mathrm{P}\:>\:\mathrm{1000}\right). \\ $$$$\mathrm{If}\:\:'\mathrm{P}'\:\mathrm{devided}\:\mathrm{by}\:\mathrm{1000},\:\mathrm{then}\:\mathrm{remainder}\:\mathrm{is}\:'\mathrm{r}'. \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{value}\:\mathrm{of}\:\:'\mathrm{r}'\:? \\ $$ Commented by Tinku Tara last updated on 09/Jun/23 $$\mathrm{All}\:\mathrm{number}\:\mathrm{other}\:\mathrm{than}\:\mathrm{multiple}…
Question Number 130727 by EDWIN88 last updated on 28/Jan/21 Commented by MJS_new last updated on 28/Jan/21 $$\mathrm{6} \\ $$ Commented by EDWIN88 last updated on…
Question Number 130109 by Adel last updated on 22/Jan/21 $$\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{6}}{\mathrm{5}^{\mathrm{3}} }+\centerdot\centerdot\centerdot\centerdot\centerdot\centerdot\centerdot\centerdot\centerdot=? \\ $$ Answered by Olaf last updated on 22/Jan/21 $$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty}…
Question Number 63917 by raj last updated on 11/Jul/19 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}−\sqrt{{x}}\right) \\ $$ Commented by kaivan.ahmadi last updated on 11/Jul/19 $$×\frac{\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}+\sqrt{{x}}}{\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}+\sqrt{{x}}}}={lim}_{{x}\rightarrow\infty} \frac{\sqrt{{x}+\sqrt{{x}}}}{\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}=\mathrm{1} \\ $$$$ \\…
Question Number 63674 by Tawa1 last updated on 07/Jul/19 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{number}\:\:\mathrm{122}^{\mathrm{n}} \:−\:\mathrm{102}^{\mathrm{n}} \:−\:\mathrm{21}^{\mathrm{n}} \:\:\mathrm{is}\:\mathrm{always}\:\mathrm{one}\:\mathrm{less}\:\mathrm{than}\:\mathrm{a} \\ $$$$\:\mathrm{multiple}\:\mathrm{of}\:\:\mathrm{2020}.\:\:\mathrm{For}\:\mathrm{every}\:\mathrm{positive}\:\mathrm{integer}\:\:\mathrm{n}. \\ $$ Commented by Prithwish sen last updated on 07/Jul/19…
Question Number 128740 by bemath last updated on 10/Jan/21 $$\:\:\underset{\mathrm{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{3}\right)\mathrm{n}!}\:=?\: \\ $$ Answered by liberty last updated on 10/Jan/21 $$\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{n}+\mathrm{2}}{\left(\mathrm{n}+\mathrm{3}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{1}\right)\mathrm{n}!}\:=\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty}…
Question Number 62554 by Rasheed.Sindhi last updated on 22/Jun/19 $${If}\:{a}\:\&\:{b}\:{are}\:{two}\:{natural}\:{numbers} \\ $$$${then}\:{the}\:{following}\:{relation}\:{holds} \\ $$$${always}. \\ $$$$\:\:\:\:\:{a}×{b}=\mathrm{gcd}\left({a},{b}\right)×\mathrm{lcm}\left({a},{b}\right) \\ $$$$\:^{\bullet} {If}\:{a},{b}\:\&\:{c}\:\:{are}\:{three}\:{natural}\:{numbers} \\ $$$$\:\:\:{what}\:{relation}\left({anologous}\:{to}\:{the}\:{above}\right) \\ $$$$\:\:\:\:{holds}\:{between}\:{numbers}\:{and}\:{their} \\ $$$$\:\:\:\mathrm{gcd}\:\&\:\mathrm{lcm}?…
Question Number 127979 by bobhans last updated on 03/Jan/21 $$\:\:\left(\mathrm{1}+{i}\right)^{\mathrm{2020}} \:=? \\ $$ Answered by liberty last updated on 03/Jan/21 $$\:\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{2020}} \:=\:\left(\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\:+\:\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)\right)^{\mathrm{2020}} \\ $$$$\:=\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2020}} \:\left(\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}\right)+\mathrm{i}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}\right)\right)^{\mathrm{2020}}…
Question Number 127940 by bramlexs22 last updated on 03/Jan/21 $$\:\begin{cases}{\mathrm{3x}=\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{4}\right)}\\{\mathrm{4x}=\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{5}\right)\:}\\{\mathrm{5x}=\mathrm{7}\:\left(\mathrm{mod}\:\mathrm{11}\right)}\end{cases} \\ $$ Answered by floor(10²Eta[1]) last updated on 03/Jan/21 $$\mathrm{3x}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{4}\right)\Rightarrow\mathrm{x}\equiv\mathrm{3}\left(\mathrm{mod}\:\mathrm{4}\right) \\ $$$$\mathrm{x}=\mathrm{4a}+\mathrm{3} \\ $$$$\mathrm{4}\left(\mathrm{4a}+\mathrm{3}\right)\equiv\mathrm{3}\left(\mathrm{mod}\:\mathrm{5}\right) \\…