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Question Number 160109 Answers: 0 Comments: 3
$$\mathrm{How}\:\mathrm{many}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{there}\:\mathrm{which} \\ $$$$\mathrm{contain}\:\mathrm{5}\:\mathrm{digits}\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{and}\: \\ $$$$\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{both}\:\mathrm{prime} \\ $$$$\mathrm{numbers}? \\ $$
Question Number 160102 Answers: 1 Comments: 1
Question Number 160092 Answers: 1 Comments: 0
$${x}\:\mathrm{and}\:{y}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{and}\: \\ $$$${x}\:×\:{x}\:−\:\mathrm{8}{y}\:=\:\mathrm{4}{x}\:\:\:. \\ $$$$\mathrm{If}\:{x}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{8},\:\mathrm{then}\:\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{for}\:{y}? \\ $$
Question Number 160091 Answers: 3 Comments: 1
$$\mathrm{51b2cd}\:\mathrm{is}\:\mathrm{a}\:\mathrm{six}-\mathrm{digit}\:\mathrm{perfect}\:\mathrm{square}\: \\ $$$$\mathrm{that}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{both}\:\mathrm{5}\:\mathrm{and}\:\mathrm{11}.\: \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{values}\: \\ $$$$\mathrm{of}\:\mathrm{it}? \\ $$
Question Number 160080 Answers: 1 Comments: 0
$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}−\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \mathrm{dt}}{\mathrm{x}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)}=? \\ $$
Question Number 160077 Answers: 1 Comments: 4
$$\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\mathrm{654}}{\mathrm{1}−\mathrm{x}^{\mathrm{654}} }−\frac{\mathrm{678}}{\mathrm{1}−\mathrm{x}^{\mathrm{678}} }\right)=? \\ $$
Question Number 160065 Answers: 1 Comments: 8
$$\mathrm{The}\:\mathrm{largest}\:\mathrm{value}\:\mathrm{of}\:\mathrm{non}-\mathrm{negative}\:\mathrm{integer}\:{a} \\ $$$$\mathrm{for}\:\mathrm{which}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left\{\frac{−{ax}+\mathrm{sin}\left({x}−\mathrm{1}\right)+{a}}{{x}+\mathrm{sin}\left({x}−\mathrm{1}\right)−\mathrm{1}}\right\}^{\frac{\mathrm{1}−{x}}{\:\mathrm{1}−\sqrt{{x}}}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{is}\:........? \\ $$
Question Number 160064 Answers: 0 Comments: 2
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{least}\:\mathrm{positive}\:\mathrm{integer}\:\:\boldsymbol{\mathrm{n}}\:\:\mathrm{for} \\ $$$$\mathrm{which}\:\:\mathrm{2}^{\boldsymbol{\mathrm{n}}} \:+\:\mathrm{5}^{\boldsymbol{\mathrm{n}}} \:-\:\boldsymbol{\mathrm{n}}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{1000} \\ $$
Question Number 160063 Answers: 0 Comments: 0
$$\mathrm{Find}: \\ $$$$\Omega\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{n}!}\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\left(\left(\mathrm{1}\:-\:\mathrm{x}\right)^{\boldsymbol{\mathrm{n}}} \:+\:\mathrm{cos}\boldsymbol{\mathrm{nx}}\right)\mathrm{e}^{\boldsymbol{\mathrm{x}}} \:\mathrm{dx} \\ $$
Question Number 160062 Answers: 0 Comments: 0
$$\mathrm{Find}: \\ $$$$\Omega\:=\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\mathrm{n}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}+\mathrm{1}}} }{\mathrm{2}}\right)\:=\:? \\ $$
Question Number 160061 Answers: 1 Comments: 0
$$ \\ $$$${Find}\:{out}\:{some}\:{pairs}\:\left({a},{b}\right)\:{such}\:{that} \\ $$$${for}\:{some}\:{n}\geqslant\mathrm{1} \\ $$$${a}^{{n}} +{b}^{{n}} ,{a}^{\mathrm{2}{n}} +{b}^{\mathrm{2}{n}} ,{a}^{\mathrm{4}{n}} +{b}^{\mathrm{4}{n}} ,{a}^{\mathrm{8}{n}} +{b}^{\mathrm{8}{n}} \in\mathbb{P} \\ $$$$ \\ $$
Question Number 160056 Answers: 0 Comments: 0
Question Number 160058 Answers: 1 Comments: 2
$${Find}\:{n}\:{so}\:{that}\:\frac{{a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} }\:{may}\:{be} \\ $$$${the}\:{arithmetic}\:{mean}\:{between}\:{a} \\ $$$${and}\:{b}. \\ $$
Question Number 160052 Answers: 1 Comments: 0
$${find}\:\Phi\left({k}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{{k}} }{{n}!}\:{with}\:{k}\geqslant\mathrm{1}. \\ $$
Question Number 160050 Answers: 1 Comments: 0
Question Number 160048 Answers: 0 Comments: 0
$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{x}\:\mathrm{sin}^{-\mathrm{1}} \:\mathrm{x}}{\mathrm{1}\:+\:\mathrm{sin}^{-\mathrm{1}} \:\mathrm{x}}\:\mathrm{dx}\:<\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Question Number 160045 Answers: 0 Comments: 0
Question Number 160036 Answers: 0 Comments: 0
Question Number 160035 Answers: 0 Comments: 0
Question Number 160025 Answers: 0 Comments: 1
Question Number 160023 Answers: 0 Comments: 0
Question Number 160014 Answers: 1 Comments: 2
Question Number 160013 Answers: 0 Comments: 1
$$\int\frac{\mathrm{1}}{\mathrm{4}{sin}\:{x}+\mathrm{3}{cos}\:{x}}{dx} \\ $$$${evaluate} \\ $$
Question Number 160009 Answers: 1 Comments: 0
$$\mathrm{Find}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt[{\boldsymbol{\mathrm{n}}}]{\mathrm{n}!}\:\centerdot\underset{\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:...\:+\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }} {\overset{\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}}} {\int}}\:\mathrm{e}^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\mathrm{dx}\right) \\ $$
Question Number 160008 Answers: 2 Comments: 2
$$\mathrm{x}_{\mathrm{1}} =\mathrm{3}\:;\:\mathrm{n}\left(\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +...+\mathrm{x}_{\boldsymbol{\mathrm{n}}} \right)=\mathrm{x}_{\boldsymbol{\mathrm{n}}} \:;\:\mathrm{n}\in\mathbb{N}\:;\:\mathrm{n}\geqslant\mathrm{1} \\ $$$$\mathrm{Find}: \\ $$$$\Omega\:=\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\left(-\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:\mathrm{x}_{\boldsymbol{\mathrm{n}}} \\ $$
Question Number 160007 Answers: 0 Comments: 0
$$\mathrm{Find}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{n}\left(\left(\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{n}}\right)^{\boldsymbol{\mathrm{n}}} -\:\mathrm{e}\:-\:\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} -\:\mathrm{e}^{-\:\frac{\mathrm{e}}{\mathrm{2}}} \right)\right) \\ $$$$ \\ $$
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