Question Number 190 by 123456 last updated on 25/Jan/15 $$\mathrm{solve}\:\mathrm{30x}\equiv\mathrm{50}\left(\mathrm{mod}\:\mathrm{40}\right) \\ $$ Answered by mreddy last updated on 15/Dec/14 $$\mathrm{30}{x}+\mathrm{40}{y}=\mathrm{50} \\ $$$$\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{5} \\ $$$$\mathrm{general}\:\mathrm{solution} \\…
Question Number 185 by 123456 last updated on 25/Jan/15 $$\mathrm{solve}\:\mathrm{10x}\equiv\mathrm{25}\left(\mathrm{mod}\:\mathrm{15}\right) \\ $$ Answered by mreddy last updated on 14/Dec/14 $$\mathrm{gcd}\left(\mathrm{10},\mathrm{15}\right)=\mathrm{5}\:\mathrm{and}\:\mathrm{5}\:\mathrm{divided}\:\mathrm{25}\:\mathrm{so}\: \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{5}\:\mathrm{solutions} \\ $$$$\mathrm{Solutions}\:\mathrm{are}\:\mathrm{given}\:\mathrm{by}\:\mathrm{equation} \\…
Question Number 176 by 123456 last updated on 25/Jan/15 $$\mathrm{find}\:\mathrm{the}\:\mathrm{integer}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{3x}+\mathrm{4y}=\mathrm{5} \\ $$ Answered by prakash jain last updated on 14/Dec/14 $$\mathrm{gcd}\left(\mathrm{3},\mathrm{4}\right)=\mathrm{1}\:\mathrm{and}\:\mathrm{1}\:\mathrm{divides}\:\mathrm{5}\:\mathrm{so}\:\mathrm{it}\:\mathrm{is}\:\mathrm{solvable}. \\ $$$$\mathrm{Particular}\:\mathrm{solution}\:{x}=\mathrm{3},\:{y}=−\mathrm{1} \\ $$$$\mathrm{General}\:\mathrm{solution}…
Question Number 177 by 123456 last updated on 25/Jan/15 $$\mathrm{solve}\:\mathrm{12x}\equiv\mathrm{34}\left(\mathrm{mod}\:\mathrm{56}\right) \\ $$ Answered by prakash jain last updated on 14/Dec/14 $$\mathrm{gcd}\left(\mathrm{12},\mathrm{56}\right)=\mathrm{4}\: \\ $$$$\mathrm{4}\:\mathrm{does}\:\mathrm{not}\:\mathrm{divide}\:\mathrm{34}\:\mathrm{so}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solutions}. \\ $$…
Question Number 170 by 123456 last updated on 25/Jan/15 $$\mathrm{solve}\:\mathrm{12x}\equiv\mathrm{25}\left(\mathrm{mod}\:\mathrm{69}\right) \\ $$ Answered by prakash jain last updated on 14/Dec/14 $$\mathrm{gcd}\left(\mathrm{12},\mathrm{69}\right)=\mathrm{3} \\ $$$$\mathrm{3}\:\nmid\:\mathrm{25},\:\mathrm{3}\:\mathrm{does}\:\mathrm{not}\:\mathrm{divide}\:\mathrm{25}\:\mathrm{hence}\:\mathrm{there} \\ $$$$\mathrm{is}\:\mathrm{no}\:\mathrm{solution}.…
Question Number 139 by shaleen last updated on 25/Jan/15 $$\boldsymbol{{express}}\:\boldsymbol{{the}}\:\boldsymbol{{folowing}}\:\boldsymbol{{in}}\:\boldsymbol{{the}}\mathrm{0}.\mathrm{6}\left(\boldsymbol{{bar}}\right)= \\ $$$$\boldsymbol{{form}}\:\boldsymbol{{of}}\:\boldsymbol{{p}}/\boldsymbol{{q}}\:;{where}\:{p}\:{and}\:{q} \\ $$$${are}\:{integers}\:{and}\:{q}\:{is}\:{not}\:=\mathrm{0} \\ $$$$\mathrm{0}.\bar {\mathrm{6}} \\ $$ Answered by vkulkarni last updated on…
Question Number 131196 by john_santu last updated on 02/Feb/21 $$\:{The}\:{minimum}\:{value}\:{of}\:{the}\: \\ $$$${expression}\:{B}\:=\:\mid{z}\mid^{\mathrm{2}} +\mid{z}−\mathrm{3}\mid^{\mathrm{2}} +\mid{z}−\mathrm{6}{i}\mid^{\mathrm{2}} \\ $$$${is}\:{p}.\:{What}\:{the}\:{value}\:{of}\:\frac{{p}}{\mathrm{10}}\:.? \\ $$ Answered by liberty last updated on 02/Feb/21…
Question Number 12861 by Gauri last updated on 04/May/17 $${change}\mathrm{0}.\mathrm{35}\overset{−} {\mathrm{6}}\:{into}\:{p}/{q}\:{form} \\ $$$$ \\ $$ Answered by mrW1 last updated on 04/May/17 $${x}=\mathrm{0}.\mathrm{35}\bar {\mathrm{6}}=\mathrm{0}.\mathrm{356666666}…… \\…
Question Number 12843 by tawa last updated on 04/May/17 $$\mathrm{Let}\:\mathrm{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{cummutative}\:\mathrm{ring}\:\mathrm{with}\:\mathrm{1}.\:\mathrm{and}\:\mathrm{a},\mathrm{b}\:\:\mathrm{member}\:\mathrm{of}\:\mathrm{R}.\:\mathrm{Suppose}\:\mathrm{a}\:\mathrm{is} \\ $$$$\mathrm{invertible}\:\mathrm{and}\:\mathrm{b}\:\mathrm{is}\:\mathrm{nilpotent}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{a}\:+\:\mathrm{b}\:\mathrm{is}\:\mathrm{invertible}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 12794 by tawa last updated on 01/May/17 $$\mathrm{Let}\:\mathrm{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{cummutative}\:\mathrm{ring}\:\mathrm{with}\:\mathrm{1},\:\mathrm{and}\:\:\mathrm{a},\mathrm{b}\in\mathrm{R}.\:\mathrm{suppose}\:\mathrm{a}\:\mathrm{is}\:\mathrm{ivertible}\:\mathrm{and} \\ $$$$\mathrm{b}\:\mathrm{is}\:\mathrm{nilpotent}.\:\mathrm{Show}\:\mathrm{that}\:\:\mathrm{a}\:+\:\mathrm{b}\:\:\mathrm{is}\:\mathrm{ivertible}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com