Question Number 94573 by Ar Brandon last updated on 19/May/20 $$\mathrm{Given}\:\mathrm{a},\:\mathrm{b},\:\mathrm{and}\:\mathrm{c},\:\mathrm{3}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{which}\:\mathrm{satisfy} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\begin{cases}{\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{312}}\\{\mathrm{c}+\mathrm{a}=\mathrm{192}}\end{cases} \\ $$$$\mathrm{Find}\:\mathrm{these}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\:\mathrm{they}\:\mathrm{form} \\ $$$$\mathrm{3}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{Geometric}\:\mathrm{Progression}. \\ $$ Commented by mr W last updated…
Question Number 29035 by abdo imad last updated on 03/Feb/18 $${let}\:{give}\:{a}\:{prime}\:{number}\:{p}>\mathrm{2}\:\:{and}\:{a}\:/{D}\left({a},{p}\right)=\mathrm{1}\:{and}\: \\ $$$$\left.{suppose}\:{that}\:{the}\:{equation}\:{x}^{\mathrm{2}} \equiv\:{a}\left[{p}\right]{have}\:{a}\:{solution}\mathrm{1}\right) \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\:\:{a}^{\frac{{p}−\mathrm{1}}{\mathrm{2}}} \:\:\:\equiv\:\mathrm{1}\:\left[{p}\right] \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\:{x}^{\mathrm{2}} \equiv\:−\mathrm{1}\left[{p}\right]\:\Leftrightarrow\:\:\:{p}\equiv\:\mathrm{1}\:\left[\mathrm{4}\right] \\ $$ Terms of Service…
Question Number 29036 by abdo imad last updated on 03/Feb/18 $${p}=\mathrm{2}{m}+\mathrm{1}\:{is}\:{a}\:{prime}\:{number}\:{prove}\:{that} \\ $$$$\left.\mathrm{1}\right)\:\left({p}−\mathrm{1}\right)!\equiv\:−\mathrm{1}\left[{p}\right] \\ $$$$\left.\mathrm{2}\right)\:\left({m}!\right)^{\mathrm{2}} \equiv\:\left(−\mathrm{1}\right)^{{m}+\mathrm{1}} \:\left[{p}\right] \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 94553 by john santu last updated on 19/May/20 $$\mathrm{solve}\:\frac{\mid{x}+\mathrm{2}\mid−{x}}{{x}}\:<\:\mathrm{2}\: \\ $$ Answered by i jagooll last updated on 19/May/20 Terms of Service Privacy…
Question Number 160052 by mr W last updated on 24/Nov/21 $${find}\:\Phi\left({k}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{{k}} }{{n}!}\:{with}\:{k}\geqslant\mathrm{1}. \\ $$ Answered by Tokugami last updated on 24/Nov/21 $$\underset{{n}=\mathrm{0}} {\overset{\infty}…
Question Number 28932 by Rasheed.Sindhi last updated on 01/Feb/18 $$\mathrm{Determine}\:\mathrm{the}\:\mathrm{least}\:\mathrm{number}\:\mathrm{of}\:\mathrm{4}\:\mathrm{digits}, \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{perfect}\:\mathrm{square}. \\ $$$$\mathrm{Method}\:\mathrm{of}\:\mathrm{finding}\:\mathrm{is}\:\boldsymbol{\mathrm{required}}. \\ $$ Answered by mrW2 last updated on 01/Feb/18 $${x}={n}^{\mathrm{2}} \geqslant\mathrm{1000}…
Question Number 159891 by tounghoungko last updated on 22/Nov/21 $$\:\:\:\:{S}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2002}\:} {\sum}}\sqrt{\frac{{k}^{\mathrm{2}} +\mathrm{1}}{{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }}\:=? \\ $$ Answered by chhaythean last updated on 22/Nov/21 $$\mathrm{S}=\underset{\mathrm{k}=\mathrm{1}}…
Question Number 28805 by NECx last updated on 30/Jan/18 $${In}\:{a}\:{competition},\:{a}\:{school}\:{awarded} \\ $$$${medals}\:{in}\:{different}\:{categories}. \\ $$$$\mathrm{36}\:{medals}\:{in}\:{dance},\mathrm{12}\:{in}\:{dramatics} \\ $$$${and}\:\mathrm{18}\:{medals}\:{in}\:{music}.{If}\:{these} \\ $$$${medals}\:{went}\:{to}\:{total}\:\mathrm{45},{and}\:{only} \\ $$$$\mathrm{4}\:{persons}\:{got}\:{medals}\:{in}\:{all}\:{three} \\ $$$${catogories}.{Using}\:{set}\:{notations}, \\ $$$${how}\:{many}\:{received}\:{in}\:{exactly} \\…
Question Number 94298 by peter frank last updated on 17/May/20 Answered by Ar Brandon last updated on 18/May/20 $$\mathrm{log}_{\mathrm{2}} \left(\mathrm{1}×\mathrm{2}×\mathrm{3}×…×\mathrm{n}\right)=\mathrm{1994} \\ $$$$\mathrm{log}_{\mathrm{2}} \left(\mathrm{n}!\right)=\mathrm{1994}\Rightarrow\mathrm{n}!=\mathrm{2}^{\mathrm{1994}} \\ $$$$\mathrm{n}\approx\mathrm{295}…
Question Number 94297 by peter frank last updated on 17/May/20 Commented by mathmax by abdo last updated on 18/May/20 $${we}\:{know}\:{that}\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow{x}={y}={z}\:\:\:\left({for}\:{x},{y}\:,{z}\:{reals}\right) \\ $$$${so}\:\left({e}\right)\:\Rightarrow{cosx}\:={cos}\left(\mathrm{2}{x}\right)={cos}\left(\mathrm{3}{x}\right)=\mathrm{0}…