Question Number 21154 by youssoufab last updated on 14/Sep/17 $${prove}:\:\forall{n}\in\mathbb{N}^{\ast} ,\forall\left({a},{b}\right)\in\mathbb{C}^{\mathrm{2}} ,\:{a}^{\mathrm{2}{n}+\mathrm{1}} +{b}^{\mathrm{2}{n}+\mathrm{1}} = \\ $$$$\underset{{k}={o}} {\overset{\mathrm{2}{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} {a}^{{k}} {b}^{\mathrm{2}{n}−{k}} \\ $$ Commented by alex041103…
Question Number 86672 by M±th+et£s last updated on 30/Mar/20 $${show}\:{proofs}\:{by}\:{induction},{that} \\ $$$$\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +….+{x}_{{n}} }{{n}}\geqslant\left({x}_{\mathrm{1}} {x}_{\mathrm{2}} ….{x}_{{n}} \right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$\forall{n}=\mathrm{2}^{{k}} ,{k}>\mathrm{1}\:{and}\:\left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,{x}_{\mathrm{3}} ,…..{x}_{{n}} \right)>\mathrm{0}.…
Question Number 152139 by Ar Brandon last updated on 26/Aug/21 $$\mathrm{How}\:\mathrm{many}\:\mathrm{numbers}\:{x}\:\mathrm{are}\:\mathrm{found}\:\mathrm{between}\:\mathrm{9000}\:\mathrm{and}\:\mathrm{11000} \\ $$$$\mathrm{and}\:\mathrm{verify}\:{x}=\mathrm{12}\left[\mathrm{19}\right],\:{x}=\mathrm{5}\left[\mathrm{13}\right],\:{x}=\mathrm{9}\left[\mathrm{11}\right]\:? \\ $$ Commented by Rasheed.Sindhi last updated on 26/Aug/21 $$\mathrm{One}\:\left({x}=\mathrm{10899}\right) \\ $$…
Question Number 21067 by youssoufab last updated on 11/Sep/17 $$\forall{n}\in\mathbb{N},\:{prove}\:\mathrm{9}\mid\left[{n}^{\mathrm{3}} +\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\left({n}+\mathrm{2}\right)^{\mathrm{3}} \right] \\ $$ Answered by dioph last updated on 12/Sep/17 $${n}^{\mathrm{3}} +\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\left({n}+\mathrm{2}\right)^{\mathrm{3}}…
Question Number 86592 by M±th+et£s last updated on 29/Mar/20 $${prove}\:{that} \\ $$$$\mathrm{2}\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{{x}} \:\left({x}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{x}\right)!\:\left({x}\right)} \\ $$ Commented by M±th+et£s last updated on 29/Mar/20…
Question Number 20927 by amaresh pradhan last updated on 08/Sep/17 $$\left({a}+{b}\right)×\left({a}+{b}\right) \\ $$ Answered by Joel577 last updated on 08/Sep/17 $$=\:\left({a}\:+\:{b}\right)^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:+\:\mathrm{2}{ab}\:+\:{b}^{\mathrm{2}} \\ $$…
Question Number 20907 by DKumar last updated on 07/Sep/17 $${How}\:{many}\:{zeroes}\:\left(\mathrm{0}\right)\:{there}\:{in}\:\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}……….\mathrm{99}×\mathrm{100}\: \\ $$ Answered by dioph last updated on 07/Sep/17 $$\mathrm{zeros}\:\mathrm{in}\:\mathrm{number}\:=\:\mathrm{maximum}\:\mathrm{integer} \\ $$$${n}\:\mathrm{such}\:\mathrm{that}\:\mathrm{10}^{{n}} =\mathrm{2}^{{n}} \mathrm{5}^{{n}} \:\mathrm{divides}\:\mathrm{the}\:\mathrm{number}.…
Question Number 20827 by New Mathematics last updated on 04/Sep/17 $${find}\:{cube}\:{root}\:{of}\:\mathrm{12167}\:{by}\:{divison}\:{method}. \\ $$ Answered by Joel577 last updated on 04/Sep/17 $${I}\:{dont}\:{know}\:{about}\:{division}\:{method},\:{but} \\ $$$${I}\:{use}\:{this}\:{method} \\ $$$$\mathrm{12}.\mathrm{167}…
Question Number 20810 by Tinkutara last updated on 03/Sep/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{triples}\:\left({a},\:{b},\:{c}\right)\:\mathrm{of} \\ $$$$\mathrm{positive}\:\mathrm{integers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left(\mathrm{i}\right)\:{a}\:<\:{b}\:<\:{c}\:<\:\mathrm{10}\:\mathrm{and}\:\left(\mathrm{ii}\right)\:{a},\:{b},\:{c},\:\mathrm{10}\:\mathrm{form} \\ $$$$\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{quadrilateral}? \\ $$ Commented by Tinkutara last updated on 04/Sep/17…
Question Number 86343 by jagoll last updated on 28/Mar/20 $$\mathrm{Dear}\:\mathrm{mr}\:\mathrm{w}.\:\mathrm{i}\:\mathrm{want}\: \\ $$$$\mathrm{discuss}\:\mathrm{for}\:\mathrm{equation} \\ $$$$\mathrm{find}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{xy}\:+\mathrm{2}\:\mathrm{with}\:\mathrm{constraint} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}\:} =\:\mathrm{6}. \\ $$$$\mathrm{my}\:\mathrm{way}\:\left(\mathrm{short}\:\mathrm{cut}\right) \\ $$$$\Rightarrow\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{y}^{\mathrm{2}}…