Question Number 216912 by ArshadS last updated on 26/Feb/25 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{three}-\mathrm{digit}\:\mathrm{numbers}\:{n}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{1}.\:{n}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{the}\:\mathrm{sum}\:\:\mathrm{of}\:\:\mathrm{its}\:\:\mathrm{digits}. \\ $$$$\mathrm{2}.\:{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 216913 by Bhutto last updated on 24/Feb/25 $$\mathrm{6}\frac{\mathrm{1}}{\mathrm{4}}\%\: \\ $$ Answered by MrGaster last updated on 24/Feb/25 $$\mathrm{6}\frac{\mathrm{1}}{\mathrm{4}}\%=\frac{\mathrm{25}}{\mathrm{4}}\%=\frac{\mathrm{25}}{\mathrm{400}}=\frac{\mathrm{1}}{\mathrm{16}} \\ $$ Terms of Service…
Question Number 216914 by hardmath last updated on 24/Feb/25 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{1}}{\mathrm{1001}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1002}}\:\:+\:\:…\:\:+\:\:\frac{\mathrm{1}}{\mathrm{2000}}\:\:>\:\:\frac{\mathrm{5}}{\mathrm{8}} \\ $$ Answered by MrGaster last updated on 24/Feb/25 $$\underset{{k}=\mathrm{1001}} {\overset{\mathrm{2000}} {\sum}}\frac{\mathrm{1}}{{k}}>\frac{\mathrm{5}}{\mathrm{8}} \\…
Question Number 216910 by Marzuk last updated on 24/Feb/25 $${Ed}:.\mathrm{06} \\ $$$${a}\:{function}\:\delta\left({x}\right)\:{is}\:{a}\:{composite}\:{function} \\ $$$${which}\:{is}\:{as}\:{follow}\: \\ $$$$\left[\left\{\left({f}\:\circ\:{g}\right)\left({x}\right)\right\}\:\circ\:\left\{\left({g}\:\circ\:{f}\right)\left({x}\right)\right\}\right]\:\circ\:\left[\left\{\left({f}\:'\:\circ\:{g}\:'\right)\left({x}\right)\right\}\:\circ\:\left\{\left({g}\:'\:\circ\:{f}\:'\right)\left({x}\right)\right\}\right] \\ $$$${where} \\ $$$${f}\left({x}\right)\:=\:\underset{{n}\:=\:\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{{nx}^{\mathrm{3}} \:−\:{nx}^{\mathrm{2}} \:−\:{nx}\:\:−{n}}{{n}^{\mathrm{3}} {x}\:−\:{n}^{\mathrm{2}}…
Question Number 216911 by ArshadS last updated on 24/Feb/25 $${Find}\:{all}\:{positive}\:{integer}\:\mathrm{x},\mathrm{y}\:{such}\:{that} \\ $$$$\mathrm{x}^{\mathrm{2}} +\:\mathrm{y}^{\mathrm{2}} +\:\mathrm{xy}\:=\:\mathrm{169} \\ $$ Answered by A5T last updated on 25/Feb/25 $$\mathrm{WLOG},\:\mathrm{let}\:\mathrm{x}\geqslant\mathrm{y} \\…
Question Number 216906 by MrGaster last updated on 24/Feb/25 $$\mathrm{Prove}:\Gamma\left({x}\right)=\frac{{x}^{{x}} }{\left(\mathrm{2}\pi\right)^{{x}−\mathrm{1}} }\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{{k}^{\mathrm{2}\left({x}−\mathrm{1}\right)} }{\underset{{i}=\mathrm{1}} {\overset{{x}−\mathrm{1}} {\prod}}\left[{k}^{\mathrm{2}} −\left(\frac{{i}}{\mathrm{2}}\right)^{\mathrm{2}} \right]} \\ $$ Terms of Service Privacy…
Question Number 216907 by MrGaster last updated on 24/Feb/25 $$\mathrm{Prove}:{n}!=\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{k}^{{n}} }{{e}^{{k}} }−\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{B}_{{k}} \mathrm{sin}\left(\pi{k}\right)\left({n}−{k}\right)!}{\pi{k}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 216900 by MathematicalUser2357 last updated on 24/Feb/25 Commented by MathematicalUser2357 last updated on 24/Feb/25 Is this right? (Part 2) - Complex number to the power of complex number Sorry for solving too complicated. But I was trying to solve clearly. Added more equivalent solution step from the last solution step and graph. The red one is the fixed one. Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 216861 by badiane last updated on 23/Feb/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 216859 by ajfour last updated on 23/Feb/25 Commented by ajfour last updated on 23/Feb/25 $${Radius}\:{of}\:{inner}\:{disc}\:{is}\:{R}.\:{As}\:{it}\:{rolls}\:{up} \\ $$$${the}\:{outer}\:{circular}\:{track}\:{of}\:{radius}\:\mathrm{2}{R},\:{find} \\ $$$${equation}\:{of}\:{trajectory}\:{of}\:{a}\:{point}\:\boldsymbol{{P}}\:{on}\:{the} \\ $$$${wheel}\:{until}\:{it}\:{comes}\:{into}\:{contact}\:{with} \\ $$$${the}\:{outer}\:{track}.…