Question Number 34186 by Joel578 last updated on 02/May/18 $$\mathrm{Let}\:{x}_{\mathrm{1}} \:=\:\mathrm{0},\:{x}_{\mathrm{2}} \:=\:\mathrm{1}\:\mathrm{and}\:{x}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}_{{n}−\mathrm{1}} \:+\:{x}_{{n}−\mathrm{2}} \right) \\ $$$$\mathrm{Show}\:\mathrm{that}\: \\ $$$${x}_{{n}} \:=\:\frac{\mathrm{2}^{{n}−\mathrm{1}} \:+\:\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}\:.\:\mathrm{2}^{{n}−\mathrm{2}} } \\ $$…
Question Number 99495 by bobhans last updated on 21/Jun/20 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{x},\mathrm{y}\:\in\:\mathbb{N}\: \\ $$$$\mathrm{7}^{\mathrm{y}} +\mathrm{2}\:=\:\mathrm{3}^{\mathrm{x}} \: \\ $$ Commented by PRITHWISH SEN 2 last updated on 21/Jun/20…
Question Number 164940 by amin96 last updated on 23/Jan/22 $$−−−−−−−−− \\ $$$$\mathrm{1}!−\mathrm{2}!+\mathrm{3}!−\mathrm{4}!+\mathrm{5}!−\ldots−\mathrm{14}!+\mathrm{15}!=? \\ $$$$ \\ $$$$−−−−−−−−−−\boldsymbol{{by}}\:\boldsymbol{{M}}.\boldsymbol{{A}} \\ $$ Commented by MJS_new last updated on 24/Jan/22…
Question Number 99322 by bramlex last updated on 20/Jun/20 $${what}\:{is}\:{remainder}\:{of}\:\mathrm{2}^{\mathrm{7}^{\mathrm{2002}} } \:{divided} \\ $$$${by}\:\mathrm{352}\: \\ $$ Answered by john santu last updated on 20/Jun/20 Commented…
Question Number 33463 by NECx last updated on 17/Apr/18 $${Find}\:{the}\:{pricipal}\:{and}\:{ordinary} \\ $$$${argument}\:{of}\:{z}=\frac{{i}}{−\mathrm{2}−\mathrm{2}{i}} \\ $$ Commented by abdo imad last updated on 17/Apr/18 $${z}\:=\:\frac{−{i}}{\mathrm{2}\left(\mathrm{1}+{i}\right)}\:=\:\frac{−{i}\left(\mathrm{1}−{i}\right)}{\mathrm{4}}\:=\:\frac{\mathrm{1}−{i}}{\mathrm{4}}\:\Rightarrow\mid{z}\mid\:=\frac{\mathrm{1}}{\mathrm{4}}\mid\mathrm{1}−{i}\mid\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${z}\:=\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{{i}}{\mathrm{4}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}}…
Question Number 33369 by Rasheed.Sindhi last updated on 15/Apr/18 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\mathrm{gcd}\left(\:\:\mathrm{gcd}\left(\mathrm{A},\mathrm{B}\right),\mathrm{gcd}\left(\mathrm{B},\mathrm{C}\right),\mathrm{gcd}\left(\mathrm{C},\mathrm{A}\right)\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{gcd}\left(\mathrm{A},\mathrm{B},\mathrm{C}\right) \\ $$ Answered by MJS last updated on 15/Apr/18 $$\mathrm{set}\:\mathrm{theory} \\…
Question Number 33323 by Rasheed.Sindhi last updated on 14/Apr/18 $$\mathrm{Question}\:\mathrm{related}\:\mathrm{to}\:\mathrm{Q}#\mathrm{33217} \\ $$$$\mathrm{If}\:\mathrm{A}_{\mathrm{1}} ,\mathrm{A}_{\mathrm{2}} ,…\mathrm{A}_{\mathrm{n}} \:\mathrm{are}\:\mathrm{n}\:\mathrm{points}\:\mathrm{with}\:\mathrm{integer} \\ $$$$\mathrm{coordinates}\:\mathrm{of}\:\mathrm{a}\:\mathrm{plane}\:\mathrm{such}\:\mathrm{that}\:\mathrm{every}\:\mathrm{triangle} \\ $$$$\mathrm{whose}\:\mathrm{vertices}\:\mathrm{are}\:\mathrm{any}\:\mathrm{three}\:\mathrm{of}\:\mathrm{the}\:\mathrm{above} \\ $$$$\mathrm{points}\:\mathrm{has}\:\mathrm{its}\:\mathrm{centroid}\:\mathrm{with}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one} \\ $$$$\mathrm{non}-\mathrm{integer}\:\mathrm{coordinate}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\mathrm{possible}\:\mathrm{n}.\:\:\:…
Question Number 164279 by amin96 last updated on 15/Jan/22 $${x}^{{n}} ={n}^{{x}} \:\:\:{x}=? \\ $$ Answered by Saiki last updated on 15/Jan/22 $${lnx}^{{n}} ={lnn}^{{x}} \\ $$$${nlnx}={xlnn}…
Question Number 98602 by bemath last updated on 15/Jun/20 $$\mathrm{find}\:\mathrm{integral}\:\mathrm{solution} \\ $$$$\mathrm{of}\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{x}^{\mathrm{3}} +\mathrm{1}\: \\ $$ Commented by bemath last updated on 15/Jun/20 $$\mathrm{oo}\:\mathrm{yes}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\…
Question Number 98596 by bemath last updated on 15/Jun/20 Commented by bobhans last updated on 15/Jun/20 $$\Rightarrow\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{2}\right)+\left({x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}\right){i}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:{real}\:{parts}\:\Rightarrow{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{2}\:=\:\mathrm{0} \\…