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Category: Algebra

Find-the-number-of-digit-in-2-50-

Question Number 62449 by Tawa1 last updated on 21/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{digit}\:\mathrm{in}\:\:\:\:\mathrm{2}^{\mathrm{50}} \\ $$ Commented by Tawa1 last updated on 21/Jun/19 $$\mathrm{And}\:\mathrm{can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{a}\:\mathrm{general}\:\mathrm{nth}\:\mathrm{number}\:\mathrm{of}\:\mathrm{term}\:\mathrm{in}\:\mathrm{any}\:\mathrm{number}\:\mathrm{and}\:\:\mathrm{powers} \\ $$ Answered by Rasheed.Sindhi…

Find-the-remainder-when-2014-is-divisible-by-2017-

Question Number 62452 by Tawa1 last updated on 21/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\:\:\mathrm{2014}!\:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\:\mathrm{2017} \\ $$ Answered by Rasheed.Sindhi last updated on 21/Jun/19 $${Wilson}'{s}\:{Theorm}: \\ $$$$\:\:\:\:\:\:\:\left({p}−\mathrm{1}\right)!\equiv−\mathrm{1}\left({mod}\:{p}\right)\::\:{p}\in\mathbb{P} \\ $$$$\:\:\:\because\:\:\:\:\mathrm{2017}\in\mathbb{P} \\…

How-many-x-R-satisfy-x-1-7-x-1-5-x-1-3-x-

Question Number 127968 by liberty last updated on 03/Jan/21 $$\:\mathrm{How}\:\mathrm{many}\:\mathrm{x}\epsilon\mathbb{R}\:\mathrm{satisfy}\:\sqrt[{\mathrm{7}}]{\mathrm{x}}\:−\sqrt[{\mathrm{5}}]{\mathrm{x}}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{x}}\:−\sqrt{\mathrm{x}}\: \\ $$ Answered by MJS_new last updated on 03/Jan/21 $${x}=\mathrm{0}\vee{x}\approx.\mathrm{0117154773}\vee{x}=\mathrm{1} \\ $$ Terms of Service…

Question-62424

Question Number 62424 by aliesam last updated on 21/Jun/19 Commented by mathmax by abdo last updated on 21/Jun/19 $$\mathrm{1}\:{is}\:{not}\:{root}\:{for}\:{this}\:{equatio}\: \\ $$$$\left({e}\right)\:\Leftrightarrow\frac{\mathrm{1}−{x}^{\mathrm{5}} }{\mathrm{1}−{x}}\:=\mathrm{0}\:\Leftrightarrow\:{x}^{\mathrm{5}} \:=\mathrm{1}\:\:{let}\:{x}\:={re}^{{i}\theta} \:\:\:\:\:\left(\:\:{we}\:{take}\:{x}\:{from}\:{C}\right) \\…

1-1-3-2-1-3-5-3-1-3-5-7-4-1-3-5-7-9-

Question Number 127908 by bramlexs22 last updated on 03/Jan/21 $$\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}+\frac{\mathrm{4}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}}+…=? \\ $$ Answered by liberty last updated on 03/Jan/21 $$\:\mathrm{Let}\:\lambda\:=\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{n}}{\mathrm{1}.\mathrm{3}.\mathrm{5}…\left(\mathrm{2n}+\mathrm{1}\right)}\: \\ $$$$\mathrm{consider}\:\frac{\mathrm{n}}{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2n}+\mathrm{1}\right)}\:=\:\frac{\mathrm{n}}{\mathrm{1}.\left(\mathrm{3}.\mathrm{5}…\left(\mathrm{2n}−\mathrm{1}\right)\right).\left(\mathrm{2n}+\mathrm{1}\right)} \\…