Question Number 95768 by john santu last updated on 27/May/20 $$\mathrm{find}\:\mathrm{x}\:\mathrm{such}\:\mathrm{that}\: \\ $$$${x}\equiv\mathrm{3}\:\left(\mathrm{mod5}\right) \\ $$$${x}\equiv\mathrm{5}\:\left(\mathrm{mod7}\right) \\ $$$${x}\equiv\mathrm{7}\left(\mathrm{mod11}\right) \\ $$ Commented by PRITHWISH SEN 2 last…
Question Number 95697 by i jagooll last updated on 27/May/20 Answered by john santu last updated on 27/May/20 $${a}_{\mathrm{0}} =\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:\underset{−\pi} {\overset{\pi} {\int}}\:\mathrm{cos}\:{at}\:{dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}\pi{a}}\:\left[\mathrm{sin}\:{at}\:\right]_{−\pi} ^{\pi} \\ $$$$=\:\frac{\mathrm{sin}\:\pi{a}}{{a}\pi}\:.…
Question Number 30032 by Penguin last updated on 15/Feb/18 $$\left({x}+\mathrm{1}\right)^{{x}} −{x}^{\left({x}+\mathrm{1}\right)} =\mathrm{1} \\ $$$${x}=? \\ $$ Commented by Penguin last updated on 15/Feb/18 $$\mathrm{x}=\mathrm{2}\:{works} \\…
Question Number 94955 by Rasheed.Sindhi last updated on 22/May/20 $${Prove}\:{by}\:{set}\:{theory}\:{or}\:{otherwise} \\ $$$$\:\:\mathrm{lcm}\left(\:\mathrm{gcd}\left(\mathrm{x},\mathrm{y}\right),\mathrm{gcd}\left(\mathrm{y},\mathrm{z}\right),\mathrm{gcd}\left(\mathrm{z},\mathrm{x}\right)\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\:\mathrm{gcd}\left(\:\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right),\mathrm{lcm}\left(\mathrm{y},\mathrm{z}\right),\mathrm{lcm}\left(\mathrm{z},\mathrm{x}\right)\:\right) \\ $$$$\:\:{Or}\:{give}\:{a}\:{counter}\:{example}. \\ $$ Terms of Service Privacy Policy…
Question Number 160270 by alf123 last updated on 27/Nov/21 Answered by Rasheed.Sindhi last updated on 27/Nov/21 $$\left({x}^{\mathrm{2}} +{y}\right)\left({y}^{\mathrm{2}} +{x}\right)=\left({x}−{y}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\begin{cases}{\underset{\left({i}\right)} {\underbrace{{x}^{\mathrm{2}} +{y}={x}−{y}}}\:\wedge\underset{\left({ii}\right)} {\underbrace{\:{y}^{\mathrm{2}} +{x}=\left({x}−{y}\right)^{\mathrm{2}}…
Question Number 29140 by mathshooter last updated on 04/Feb/18 $${Find}\:{no}.\:{of}\:{rational}\:{roots}\:{of}\:{f}\left({x}\right)= \\ $$$$\mathrm{2}{x}^{\mathrm{98}} +\mathrm{3}{x}^{\mathrm{97}} +\mathrm{2}{x}^{\mathrm{96}} +…..+\mathrm{2}{x}+\mathrm{3}=\mathrm{0}. \\ $$ Commented by math solver last updated on 05/Feb/18…
Question Number 160061 by Rasheed.Sindhi last updated on 24/Nov/21 $$ \\ $$$${Find}\:{out}\:{some}\:{pairs}\:\left({a},{b}\right)\:{such}\:{that} \\ $$$${for}\:{some}\:{n}\geqslant\mathrm{1} \\ $$$${a}^{{n}} +{b}^{{n}} ,{a}^{\mathrm{2}{n}} +{b}^{\mathrm{2}{n}} ,{a}^{\mathrm{4}{n}} +{b}^{\mathrm{4}{n}} ,{a}^{\mathrm{8}{n}} +{b}^{\mathrm{8}{n}} \in\mathbb{P} \\…
Question Number 159925 by Rasheed.Sindhi last updated on 22/Nov/21 $$\mathrm{Find}\:\mathrm{out}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{numbers}\:\left(\mathrm{a},\mathrm{b}\right)\:\left(\mathrm{as}\right. \\ $$$$\left.\mathrm{many}\:\mathrm{as}\:\mathrm{you}\:\mathrm{can}\right)\:\mathrm{such}\:\mathrm{that}: \\ $$$$\sqrt{\mathrm{a}}\:+\sqrt{\mathrm{b}}\:,\:\mathrm{a}+\mathrm{b}\:,\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \:\in\:\mathbb{P} \\ $$ Answered by nikif99 last updated on 23/Nov/21…
Question Number 159775 by Ghaniy last updated on 22/Nov/21 $$\prod_{\mathrm{n}=\mathrm{1}} ^{\infty} \frac{\alpha^{\mathrm{3}} +\beta^{\mathrm{2}} }{\mathrm{3}^{\mathrm{n}} }=\:? \\ $$$$\mathrm{in}\:\mathrm{expanded}\:\mathrm{form} \\ $$ Answered by Canebulok last updated on…
Question Number 94080 by Rio Michael last updated on 16/May/20 $$\mathrm{3}.\left(\mathrm{a}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{number}\:{z}\:\mathrm{which}\:\mathrm{satisfy}\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:{z}^{\mathrm{3}} \:=\:\mathrm{8}{i}\:,\:\mathrm{giving}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{form}\:{a}\:+\:{bi}\:\mathrm{where}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{real}. \\ $$ Answered by mr W last updated on…