Question Number 150797 by mathdanisur last updated on 15/Aug/21 $$\mathrm{x};\mathrm{y}\in\mathbb{N} \\ $$$$\frac{\mathrm{18}^{\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:+\:\boldsymbol{\mathrm{y}}^{\mathrm{2}} }{\mathrm{2}}} }{\mathrm{9}^{\boldsymbol{\mathrm{xy}}} }\:=\:\mathrm{2592}\:\:\Rightarrow\:\mathrm{find}\:\:\boldsymbol{\mathrm{xy}}=? \\ $$ Answered by nimnim last updated on 15/Aug/21…
Question Number 150792 by mathdanisur last updated on 15/Aug/21 $$\mathrm{if}\:\:\:\frac{\mathrm{y}+\mathrm{z}}{\mathrm{x}}\:=\:\frac{\mathrm{10}}{\mathrm{3}}\:\:\mathrm{and}\:\:\frac{\mathrm{x}}{\mathrm{z}}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{find}\:\:\frac{\mathrm{x}}{\mathrm{y}}\:=\:? \\ $$ Answered by nimnim last updated on 15/Aug/21 $${y}=\frac{\mathrm{10}{x}−\mathrm{3}{z}}{\mathrm{3}}\:\:\:\:{and}\:\:\:\:\mathrm{3}{z}=\mathrm{4}{x} \\ $$$$\therefore\frac{{x}}{{y}}=\frac{{x}}{\frac{\mathrm{10}{x}−\mathrm{4}{x}}{\mathrm{3}}}\:=\frac{\mathrm{1}}{\mathrm{2}} \\…
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Question Number 150769 by mathdanisur last updated on 15/Aug/21 $$\frac{\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +…+\boldsymbol{\mathrm{x}}^{\mathrm{3}} }{\mathrm{1}\centerdot\mathrm{4}+\mathrm{2}\centerdot\mathrm{7}+\mathrm{3}\centerdot\mathrm{10}+…+\boldsymbol{\mathrm{x}}\left(\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{1}\right)}\:=\:\mathrm{2021} \\ $$$$\mathrm{find}\:\:\boldsymbol{\mathrm{x}}=? \\ $$ Answered by nimnim last updated on…
Question Number 19700 by Tinkutara last updated on 14/Aug/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of} \\ $$$${k}\:\mathrm{for}\:\mathrm{which}\:\mathrm{2013}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{sum}\:\mathrm{of}\:{k}\:\mathrm{consecutive}\:\mathrm{positive}\:\mathrm{integers}? \\ $$ Answered by mrW1 last updated on 15/Aug/17 $$\mathrm{keep}\:\mathrm{in}\:\mathrm{mind}:\:\mathrm{2013}=\mathrm{3}×\mathrm{11}×\mathrm{61} \\…
Question Number 19698 by Tinkutara last updated on 14/Aug/17 $$\mathrm{Let}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:−\:\mathrm{3}{x}\:+\:{b}\:\mathrm{and}\:{g}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:+ \\ $$$${bx}\:−\:\mathrm{3},\:\mathrm{where}\:{b}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}.\:\mathrm{What} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:{b}\:\mathrm{for} \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{equations}\:{f}\left({x}\right)\:=\:\mathrm{0}\:\mathrm{and}\:{g}\left({x}\right) \\ $$$$=\:\mathrm{0}\:\mathrm{have}\:\mathrm{a}\:\mathrm{common}\:\mathrm{root}? \\ $$ Answered by ajfour…
Question Number 19696 by Tinkutara last updated on 14/Aug/17 $$\mathrm{Let}\:{m}\:\mathrm{be}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{odd}\:\mathrm{positive} \\ $$$$\mathrm{integer}\:\mathrm{for}\:\mathrm{which}\:\mathrm{1}\:+\:\mathrm{2}\:+\:…\:+\:{m}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{square}\:\mathrm{of}\:\mathrm{an}\:\mathrm{integer}\:\mathrm{and}\:\mathrm{let}\:{n}\:\mathrm{be}\:\mathrm{the} \\ $$$$\mathrm{smallest}\:\mathrm{even}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{for} \\ $$$$\mathrm{which}\:\mathrm{1}\:+\:\mathrm{2}\:+\:…\:+\:{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{integer}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{m}\:+\:{n}? \\ $$ Answered by Rasheed.Sindhi…
Question Number 19690 by Tinkutara last updated on 14/Aug/17 $$\mathrm{If}\:\mid{z}\:−\:\frac{\mathrm{4}}{{z}}\mid\:=\:\mathrm{2},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mid{z}\mid. \\ $$ Answered by ajfour last updated on 15/Aug/17 $$\:\:\mid\left(\mid\mathrm{z}\mid−\frac{\mathrm{4}}{\mid\mathrm{z}\mid}\right)\mid\leqslant\mathrm{2} \\ $$$$\mathrm{let}\:\mid\mathrm{z}\mid=\mathrm{t} \\…
Question Number 85228 by mathocean1 last updated on 20/Mar/20 Commented by mathocean1 last updated on 20/Mar/20 $${please}\:{help}\:{me}… \\ $$ Commented by abdomathmax last updated on…