Question Number 203566 by BaliramKumar last updated on 22/Jan/24 $$'\mathrm{P}'\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:\left(\mathrm{P}>\mathrm{1000}\right).\: \\ $$$$\mathrm{If}\:\:\:\mathrm{P}\:\equiv\:\mathrm{r}\:\left(\mathrm{mod}\:\mathrm{1000}\right).\:\mathrm{how}\:\mathrm{many}\:\mathrm{value}\:\mathrm{of}\:'\mathrm{r}'. \\ $$ Answered by MM42 last updated on 22/Jan/24 $${p}=\mathrm{1000}{k}+{r}\:\:\&\:\:\mathrm{0}<{r}<\mathrm{1000}\: \\ $$$$\Rightarrow“{p}''\:{is}\:{prime}\:{number}\: \\…
Question Number 203051 by ajfour last updated on 08/Jan/24 $${A}\:{two}\:{digit}\:{number}\:\:\left({AB}\right)_{\mathrm{10}} \\ $$$$\left({AB}\right)_{\mathrm{10}} −{A}^{{B}} =\left({BA}\right)_{\mathrm{10}} \\ $$$${Find}\:{the}\:{number}.\:{There}\:{is}\:{a}\:{poem}. \\ $$$${Having}\:{arrived}\:{at}\:{the}\:{age}\:{of}\:\left({BA}\right)_{\mathrm{10}} . \\ $$ Answered by Rasheed.Sindhi last…
Question Number 202780 by BaliramKumar last updated on 03/Jan/24 $$\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{7}+\mathrm{11}+\mathrm{16}+…………\mathrm{up}\:\mathrm{to}\:\mathrm{25}\:\mathrm{terms}\:=\:? \\ $$ Answered by mr W last updated on 03/Jan/24 $${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} ={a}_{\mathrm{1}}…
Question Number 202774 by BaliramKumar last updated on 03/Jan/24 $$\mathrm{1}.\:\:\:\:\:\:\mathrm{S}_{\mathrm{n}} \:=\:\mathrm{a}\left(\mathrm{a}+\mathrm{d}\right)+\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)+……+\left\{\mathrm{a}+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{d}\right\}\left\{\mathrm{a}+\left(\mathrm{n}\right)\mathrm{d}\right\} \\ $$$$\mathrm{2}.\:\:\:\:\:\:\mathrm{S}_{\mathrm{n}} \:=\:\mathrm{a}\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)+\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)\left(\mathrm{a}+\mathrm{3d}\right)+……+\left\{\mathrm{a}+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{d}\right\}\left\{\mathrm{a}+\left(\mathrm{n}\right)\mathrm{d}\right\}\left\{\mathrm{a}+\left(\mathrm{n}+\mathrm{1}\right)\mathrm{d}\right\} \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last updated on 03/Jan/24…
Question Number 202796 by MathematicalUser2357 last updated on 03/Jan/24 $$\left[\mathrm{2}^{\mathrm{8}\boldsymbol{\div}\mathrm{2}−\mathrm{1}} +\mathrm{4}\boldsymbol{\div}\left\{\mathrm{2}+\mathrm{2}^{\mathrm{4}\boldsymbol{\div}\left(\mathrm{1}+\mathrm{1}\right)} \boldsymbol{\div}\left(\mathrm{6}\boldsymbol{\div}\mathrm{3}−\mathrm{1}\right)\right\}\right]\boldsymbol{\div}\left[\mathrm{6}\boldsymbol{\div}\left\{\left(\mathrm{3}^{\mathrm{4}\boldsymbol{\div}\mathrm{2}} +\mathrm{5}\boldsymbol{\div}\mathrm{5}\right)\boldsymbol{\div}\mathrm{2}−\mathrm{6}\boldsymbol{\div}\left(\mathrm{5}−\mathrm{2}\right)\right\}\right] \\ $$ Answered by a.lgnaoui last updated on 03/Jan/24 $$\mathrm{si}\:\left(\:\frac{.}{.}\right)\:\:\mathrm{est}\:\mathrm{une}\:\mathrm{operation}\:\mathrm{de}\left[\mathrm{division}\right. \\ $$$$\:\:\Rightarrow\:\mathrm{Expression}\:\mathrm{est}\:\mathrm{equivalente}\:\mathrm{a}…
Question Number 202677 by BaliramKumar last updated on 31/Dec/23 $$ \\ $$$$\mathrm{If}\:{x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:=\:\mathrm{2023}^{\mathrm{2024}} \:\&\:{x},\:{y}\:\in\:\boldsymbol{\mathrm{N}} \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{set}\left\{{x},\:{y}\right\} \\ $$ Answered by AST last updated on…
Question Number 202672 by BaliramKumar last updated on 31/Dec/23 $$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{divisors}\:\mathrm{of}\:\mathrm{24}\:\mathrm{is}\:\mathrm{60}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{divisors}\:\mathrm{of}\:\mathrm{24}×\mathrm{79}. \\ $$ Answered by a.lgnaoui last updated on 31/Dec/23 $$\mathrm{60} \\ $$ Commented…
Question Number 202698 by Bambamamoudou last updated on 31/Dec/23 $${determine}\:{le}\:{reste}\:{de}\:{la}\:{division}\:{eucludienne}\:{de}\:\mathrm{2023}^{\mathrm{2019}} {par}\:\mathrm{13} \\ $$ Answered by Rasheed.Sindhi last updated on 01/Jan/24 $$\because\:{gcd}\left(\mathrm{2023},\mathrm{13}\right)=\mathrm{1} \\ $$$$\therefore\:\:\:\:\mathrm{2023}^{\phi\left(\mathrm{13}\right)} \equiv\mathrm{1}\left({mod}\:\mathrm{13}\right) \\…
Question Number 202589 by mnjuly1970 last updated on 30/Dec/23 $$ \\ $$$$\:\:\Omega\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{k}}\:+\:\mathrm{2}{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{k}}\right)\right)=?\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$ Commented by mnjuly1970 last updated…
Question Number 202462 by pticantor last updated on 27/Dec/23 Answered by witcher3 last updated on 27/Dec/23 $$\mathrm{p}^{\mathrm{n}} −\mathrm{1}=\left(\mathrm{p}−\mathrm{1}\right)\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{p}^{\mathrm{k}} \right) \\ $$$$\Rightarrow\mathrm{p}.\left(\mathrm{p}^{\mathrm{n}} \right)+\left(\mathrm{p}−\mathrm{1}\right).\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}}…