Question Number 3630 by prakash jain last updated on 16/Dec/15 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{prime}\:\mathrm{numbers}\:{p} \\ $$$$\mathrm{such}\:\mathrm{that}\:{n}\leqslant{p}\leqslant\mathrm{2}^{{k}} {n}\:\mathrm{is}\:\geqslant\mathrm{2}^{{k}} {n}. \\ $$$$\underset{{n}\leqslant{p}\leqslant\mathrm{2}^{{k}} {n}} {\sum}{p}\:\:\geqslant\:\mathrm{2}^{{k}} {n}\:\:\:\:\:\left({p}−{prime}\right) \\ $$ Terms of Service…
Question Number 3595 by Filup last updated on 16/Dec/15 $$\mathrm{I}\:\mathrm{just}\:\mathrm{thought}\:\mathrm{of}\:\mathrm{something}\:\mathrm{I}\:\mathrm{am}\:\mathrm{curious} \\ $$$$\mathrm{in}\:\mathrm{figuring}\:\mathrm{out}. \\ $$$$ \\ $$$$\mathrm{All}\:\mathrm{integer}\:\mathrm{numbers}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made}\:\mathrm{up}\:\mathrm{by} \\ $$$${prime}\:{factors}.\:\mathrm{That}\:\mathrm{is}: \\ $$$${n}={p}_{\mathrm{1}} ×{p}_{\mathrm{2}} ×…×{p}_{{i}} \\ $$$${n}\in\mathbb{Z}\:\:\:\:\:\:\:\:\:{p}_{{k}} \in\mathbb{P}…
Question Number 3519 by Rasheed Soomro last updated on 14/Dec/15 $${n}\:{is}\:{a}\:{number}\:{such}\:{that}\:{regular}\:\:{n}−{gon}\:{is} \\ $$$${possible}\:{with}\:{straightedge}\:{and}\:\:{compass}\:{only}. \\ $$$$\ast{Write}\:{first}\:{thirty}\:{values}\:{of}\:{n}. \\ $$$$\ast{What}\:{are}\:{other}\:{properties}\:{of}\:{such}\:{numbers}\:? \\ $$$$\ast{If}\:{values}\:{of}\:{n}\:{are}\:{arranged}\:{in}\:{order},\:{what}\:{is} \\ $$$${the}\:{formula}\:{for}\:{generating}\:{Nth}\:{number}? \\ $$ Commented by…
Question Number 134458 by bramlexs22 last updated on 04/Mar/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{of}\: \\ $$$$\:\mathrm{2025}^{\mathrm{2052}} \:+\:\mathrm{1392}^{\mathrm{1329}} \:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 134393 by liberty last updated on 03/Mar/21 $$\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{4}^{\mathrm{k}} \:\left(\mathrm{k}!\right)^{\mathrm{2}} }{\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} \:\left(\mathrm{2k}\right)!}\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 3296 by prakash jain last updated on 09/Dec/15 $$\mathrm{If}\:{n},{p},{q}\in\mathbb{Z}^{+} \:\mathrm{and}\:{p}\:\mathrm{and}\:{q}\:\mathrm{are}\:\mathrm{coprimes}, \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{HCF}\:\mathrm{of}\:\left({n}^{{p}} −\mathrm{1}\right)\:\mathrm{and}\:\left({n}^{{q}} −\mathrm{1}\right)\:\mathrm{is}\:\left({n}−\mathrm{1}\right). \\ $$$$\mathrm{Assume}\:{n}>\mathrm{1}. \\ $$ Commented by Rasheed Soomro last…
Question Number 134327 by mr W last updated on 02/Mar/21 Answered by aleks041103 last updated on 25/Dec/21 $${i}\Rightarrow\mid{z}_{{k}} ^{\mathrm{2}} \mid=\mathrm{1} \\ $$$$\Rightarrow{z}_{{k}} ={e}^{{it}_{{k}} } \\…
Question Number 3205 by Rasheed Soomro last updated on 07/Dec/15 $$\mathcal{S}{uggest}\:{minimum}\:{number}\:{of}\:\:{weights}\:,{two}\:{peices}\:{of}\:{each},\: \\ $$$${to}\:{weigh}\:{upto}\:{at}\:{least}\:\mathrm{60}\:{kg}\left({in}\:{whole}\:{kg}'{s}\right)\:{in}\:{a}\:{common} \\ $$$${balance}. \\ $$ Commented by prakash jain last updated on 07/Dec/15…
Question Number 134267 by bramlexs22 last updated on 01/Mar/21 $$\begin{cases}{\mathrm{x}\equiv\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{5}\right)}\\{\mathrm{x}\equiv\:\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{4}\:\right)}\end{cases} \\ $$ Answered by EDWIN88 last updated on 01/Mar/21 $$\begin{cases}{\mathrm{x}=\mathrm{5k}+\mathrm{4}}\\{\mathrm{x}=\mathrm{4}\ell+\mathrm{3}}\end{cases}\:\Leftrightarrow\:\mathrm{5k}+\mathrm{4}\:=\:\mathrm{4}\ell+\mathrm{3} \\ $$$$\ell\:=\:\frac{\mathrm{5k}+\mathrm{1}}{\mathrm{4}}\:;\:\left(\mathrm{k},\ell\right)\:=\:\left(\mathrm{3},\mathrm{4}\right),\:\left(\mathrm{7},\mathrm{9}\right),\left(\mathrm{11},\mathrm{14}\right),…, \\ $$$$\left(\mathrm{4}\lambda−\mathrm{1},\:\mathrm{5}\lambda−\mathrm{1}\right)\:\Rightarrow\mathrm{then}\:\mathrm{x}=\mathrm{5}\left(\mathrm{4}\lambda−\mathrm{1}\right)+\mathrm{4} \\…
Question Number 134252 by bramlexs22 last updated on 01/Mar/21 $$\:\mathrm{solve}\:\begin{cases}{\mathrm{x}\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{3}\right)}\\{\mathrm{x}\equiv\:\mathrm{5}\:\left(\mathrm{mod}\:\mathrm{7}\right)}\end{cases} \\ $$ Answered by EDWIN88 last updated on 01/Mar/21 $$\begin{cases}{\mathrm{x}\:=\:\mathrm{3k}+\mathrm{2}\:;\mathrm{k}\in\mathbb{Z}}\\{\mathrm{x}=\mathrm{7m}+\mathrm{5};\:\mathrm{m}\in\mathbb{Z}}\end{cases}\:\Rightarrow\:\mathrm{3k}+\mathrm{2}\:=\:\mathrm{7m}+\mathrm{5} \\ $$$$\Rightarrow\:\mathrm{m}=\frac{\mathrm{3k}−\mathrm{3}}{\mathrm{7}}\:;\:\left(\mathrm{k},\mathrm{m}\right)=\left(\mathrm{8},\mathrm{3}\right),\left(\mathrm{15},\mathrm{6}\right),\left(\mathrm{22},\mathrm{9}\right) \\ $$$$…,\:\left(\mathrm{7}\lambda+\mathrm{1},\:\mathrm{3}\lambda\right) \\…