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…
Question Number 94025 by Rasheed.Sindhi last updated on 16/May/20 $$\mathrm{LCM}\left({a},\frac{\mathrm{3}}{\mathrm{5}}{a}\right)=\mathrm{3}{a}\:\wedge\:\mathrm{HCF}\left({a},\frac{\mathrm{3}}{\mathrm{5}}{a}\right)=\frac{\mathrm{1}}{\mathrm{5}}{a} \\ $$$${a}=? \\ $$ Commented by som(math1967) last updated on 16/May/20 $$\mathrm{sir}\:,\mathrm{if}\:\mathrm{a}=\mathrm{5}\:\mathrm{then}\:\:\mathrm{L}.\mathrm{C}.\mathrm{M}=\mathrm{15} \\ $$$$\mathrm{i}.\mathrm{e}\:\mathrm{3}×\mathrm{5} \\…
Question Number 93982 by Rasheed.Sindhi last updated on 16/May/20 $$\mathrm{LCM}\left({a},\frac{\mathrm{3}}{\mathrm{5}}{a}\right)=\mathrm{3}{a}\: \\ $$$${a}=? \\ $$ Commented by mr W last updated on 16/May/20 $${a}=\mathrm{5}{b} \\ $$$${LCM}\left(\mathrm{5}{b},\mathrm{3}{b}\right)=\mathrm{3}×\mathrm{5}{b}…
Question Number 159461 by Rasheed.Sindhi last updated on 17/Nov/21 $$\:\:\:\:\:\:\:\:\:\:\prec\mathbb{PRIME}-\mathcal{BIRTHDAYS}\succ \\ $$$$\mathrm{Do}\:\mathrm{you}\:\mathrm{know}\:'\mathrm{Prime1611}'?… \\ $$$$\mathrm{No},\mathrm{no}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{an}\:\mathrm{ID}\:\mathrm{of}\:\mathrm{the}\:\mathrm{forum}- \\ $$$$\mathrm{member}.\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{person}\:\mathrm{who}\:\mathrm{was}\:\mathrm{born} \\ $$$$\mathrm{on}\:\mathrm{November}\:\mathrm{16},\:\mathrm{0001}.\mathrm{On}\:\mathrm{his} \\ $$$$\mathrm{birthday}\:\mathrm{astrologers}\:\mathrm{formed}\:\mathrm{a}\:\mathrm{string} \\ $$$$\mathrm{from}\:\mathrm{his}\:\mathrm{birthdate}\:\mathrm{in}\:\mathrm{the}\:\mathrm{way}: \\ $$$$'\mathrm{ddmmyyyy}':\:'\mathrm{16110001}'\:\mathrm{The}\:\mathrm{astro}- \\…
Question Number 93911 by i jagooll last updated on 16/May/20 $$\mathrm{number}\:\mathrm{of}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{a}\:\mathrm{number}\: \\ $$$$\mathrm{2}^{\mathrm{2016}} \:\mathrm{and}\:\mathrm{5}^{\mathrm{2016}} \:\mathrm{is}? \\ $$ Commented by PRITHWISH SEN 2 last updated on…