Question Number 137005 by greg_ed last updated on 28/Mar/21 $$\boldsymbol{\mathrm{Hi}},\:\boldsymbol{\mathrm{guyz}}\:! \\ $$$$\boldsymbol{\mathrm{For}}\:\boldsymbol{\mathrm{R}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{5}}{\mathrm{6}}×…×\frac{\mathrm{223}}{\mathrm{224}}\:\:\:\boldsymbol{\mathrm{and}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{6}}{\mathrm{7}}×…×\frac{\mathrm{224}}{\mathrm{225}}\:. \\ $$$$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}}\::\:\:\boldsymbol{\mathrm{R}}\:<\:\frac{\mathrm{1}}{\mathrm{15}}\:<\:\boldsymbol{\mathrm{S}}. \\ $$ Commented by greg_ed last updated on 01/May/21…
Question Number 136987 by Ousmane last updated on 28/Mar/21 $$\Rightarrow\mathrm{2}{x} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 5851 by sanusihammed last updated on 01/Jun/16 $${Solve}\:{simultaneously} \\ $$$$ \\ $$$$\mathrm{2}{x}\:+\:{y}\:−\:{z}\:=\:\mathrm{8}\:\:\:\:………\:\left({i}\right) \\ $$$${x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:+\:\mathrm{2}{z}^{\mathrm{2}} \:=\:\mathrm{14}\:\:\:\:………\:\left({ii}\right) \\ $$$$\mathrm{3}{x}^{\mathrm{3}} \:+\:\mathrm{4}{y}^{\mathrm{3}} \:+\:{z}^{\mathrm{3}} \:=\:\mathrm{195}\:\:\:\:………..\:\left({iii}\right) \\…
Question Number 5834 by sanusihammed last updated on 31/May/16 $${Find}\:{all}\:{positive}\:{integers}\:{n}\:{for}\:{which}\:{there}\:{exist} \\ $$$${non}−{negative}\:{integer}\:.\:{a}_{\mathrm{1}\:} {a}_{\mathrm{2}} \:{a}_{\mathrm{3}} \:…….\:{a}_{{n}} \:.\:{Such}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{{a}_{\mathrm{1}} } }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{{a}_{\mathrm{2}} } }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{{a}_{\mathrm{3}} } }\:+\:….\:+\:\frac{\mathrm{1}}{\mathrm{2}^{{a}_{{n}} }…
Question Number 71327 by TawaTawa last updated on 13/Oct/19 Answered by mind is power last updated on 13/Oct/19 $$\mathrm{ferma}\:\mathrm{lemma}\: \\ $$$$\mathrm{x}^{\mathrm{5}} =\mathrm{x}\left(\mathrm{5}\right) \\ $$$$\mathrm{x}^{\mathrm{4}} =\mathrm{1}\left(\mathrm{5}\right)\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:\mathrm{x}\neq\mathrm{0}…
Question Number 71294 by Mohamed Amine Bouguezzoul last updated on 13/Oct/19 $${solve}\:{in}\:\mathbb{Z}\:\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{1}}{{p}}\:{with}\:{p}\in\mathbb{P} \\ $$ Commented by Rasheed.Sindhi last updated on 13/Oct/19 $$\frac{{x}+{y}}{{xy}}=\frac{\mathrm{1}}{{p}} \\ $$$$\frac{{xy}}{{x}+{y}}={p} \\…
Question Number 5738 by FilupSmith last updated on 26/May/16 $$\mathrm{I}'\mathrm{m}\:\mathrm{having}\:\mathrm{trouble}\:\mathrm{understanding}\:\mathrm{why}: \\ $$$$\mathrm{2}^{−\mathrm{1}} \equiv\mathrm{3}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$ \\ $$$$\mathrm{Can}\:\mathrm{someone}\:\mathrm{explain},\:\mathrm{please}?\:\mathrm{Thank}\:\mathrm{you}! \\ $$ Commented by Rasheed Soomro last updated…
Question Number 136739 by Ar Brandon last updated on 25/Mar/21 $$\mathrm{Given}\:\mathrm{0}<\mathrm{a}<\mathrm{b},\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} }{\mathrm{8b}}\leqslant\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}−\sqrt{\mathrm{ab}}\leqslant\frac{\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} }{\mathrm{8a}} \\ $$ Answered by snipers237 last updated on 26/Mar/21 $$\frac{{a}+{b}}{\mathrm{2}}−\sqrt{{ab}\:}=\:\frac{\left(\sqrt{{a}}−\sqrt{{b}}\right)^{\mathrm{2}}…
Question Number 5609 by FilupSmith last updated on 22/May/16 $$\lfloor{x}\rfloor\:=\:\mathrm{floor}\:\mathrm{function} \\ $$$$\lceil{x}\rceil\:=\:\mathrm{ceiling}\:\mathrm{function} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{seen}\:\mathrm{somewhere}\:\mathrm{either}: \\ $$$$\lfloor{x}\rceil\:\:\mathrm{or}\:\:\lceil{x}\rfloor\:\:\:\left(\mathrm{i}\:\mathrm{dont}\:\mathrm{remember}\:\mathrm{which}\right) \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{it}/\mathrm{are}\:\mathrm{they}? \\ $$ Terms of Service…
Question Number 71115 by TawaTawa last updated on 11/Oct/19 Answered by mind is power last updated on 11/Oct/19 $$\mathrm{A}.\left(\mathrm{999994}\right)……\left(\mathrm{1000002}\right)=\mathrm{1}\left(\mathrm{1000003}\right) \\ $$$$\mathrm{used}\:\mathrm{wilson}\:\mathrm{theorem}\:\left(\mathrm{p}−\mathrm{1}\right)!=\mathrm{1}\left(\mathrm{p}\right) \\ $$$$\Rightarrow \\ $$$$\Rightarrow\mathrm{A}\left(−\mathrm{1}\right).\left(−\mathrm{2}\right)……\left(−\mathrm{9}\right)=\mathrm{1}\left(\mathrm{1000003}\right)…