Question Number 185270 by Mingma last updated on 19/Jan/23 Answered by mr W last updated on 20/Jan/23 $$\mathrm{2}{a}+\mathrm{1}\leqslant{d}\leqslant\mathrm{9}\:\Rightarrow{a}\leqslant\mathrm{4}\:\cap\:{a}\neq\mathrm{0} \\ $$$$\mathrm{2}{d}=\mathrm{10}+{a} \\ $$$${a}=\mathrm{4}:\:{d}=\mathrm{7}\:\Rightarrow\mathrm{2}×\mathrm{4}+\mathrm{1}=\mathrm{9}\:>\:\mathrm{7}\:\Rightarrow{rejected} \\ $$$${a}=\mathrm{2}:\:{d}=\mathrm{6}\:\Rightarrow\mathrm{2}×\mathrm{2}+\mathrm{1}=\mathrm{5}\:<\mathrm{6}\:\Rightarrow{ok} \\…
Question Number 185229 by Mingma last updated on 18/Jan/23 Answered by Rasheed.Sindhi last updated on 21/Jan/23 $${A}_{{S}} =\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{c}+\mathrm{2}\left({n}−\mathrm{1}\right)\right] \\ $$$${G}_{{S}} =\frac{{c}\left(\mathrm{1}−\mathrm{2}^{{n}} \right)}{\mathrm{1}−\mathrm{2}} \\ $$$$\mathrm{2}\centerdot{A}_{{S}} ={G}_{{S}}…
Question Number 185167 by Rupesh123 last updated on 18/Jan/23 Answered by aba last updated on 18/Jan/23 $$\mathrm{P}=\left(\mathrm{1}+\mathrm{2}+…+\mathrm{9}\right)\mathrm{N}=\mathrm{45N} \\ $$$$\mathrm{if}\:\mathrm{N}\:\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{integer},\:\mathrm{then}\:\mathrm{P}=\mathrm{45N}\:\mathrm{will}\:\mathrm{end}\:\mathrm{in}\:\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{possible}\:\mathrm{as}\:\mathrm{all}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{same} \\ $$$$\mathrm{if}\:\mathrm{N}\:\mathrm{is}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{integer},\:\mathrm{then}\:\mathrm{P}=\mathrm{45N}\:\mathrm{will}\:\mathrm{end}\:\mathrm{in}\:\mathrm{5} \\ $$$$\mathrm{P}=\mathrm{45N}=\mathrm{555}…\mathrm{55}…
Question Number 119575 by mathocean1 last updated on 25/Oct/20 $$\mathrm{a}\:\in\:\mathbb{N}.\:\mathrm{a}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{a}^{\mathrm{3}} \equiv−\mathrm{1}\left[\mathrm{9}\right]\:\mathrm{or}\:\mathrm{a}^{\mathrm{3}} \equiv\mathrm{1}\left[\mathrm{9}\right]. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Given}\:\mathrm{a};\:\mathrm{b};\:\mathrm{c}\:\in\:\mathbb{Z}. \\ $$$$\left.\mathrm{Deduct}\:\mathrm{from}\:\mathrm{1}\right)\:\mathrm{that}\:\mathrm{if}\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \equiv\mathrm{0}\left[\mathrm{9}\right]\:,\:\mathrm{then}\:\: \\ $$$$\mathrm{one}\:\mathrm{of}\:\mathrm{integers}\:\mathrm{a};\:\mathrm{b};\:\mathrm{c}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3}. \\ $$…
Question Number 185030 by Engr_Jidda last updated on 15/Jan/23 Commented by Engr_Jidda last updated on 15/Jan/23 $${help}\:{me}\:{solve}\:{these}\:{quantitative}\:{reasonings} \\ $$ Answered by manolex last updated on…
Question Number 53919 by byaw last updated on 27/Jan/19 Answered by $@ty@m last updated on 27/Jan/19 $${ATQ} \\ $$$$\mathrm{6}\:{oxen}\equiv\mathrm{8}\:{cows} \\ $$$$\Rightarrow\mathrm{1}\:{ox}\equiv\frac{\mathrm{8}}{\mathrm{6}}=\frac{\mathrm{4}}{\mathrm{3}}\:{cows} \\ $$$$\therefore\:\mathrm{9}\:{oxen}+\mathrm{2}\:{cows} \\ $$$$=\left(\mathrm{9}×\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{2}\right){cows}…
Question Number 119440 by mathocean1 last updated on 24/Oct/20 $$\mathrm{Determine}\:\mathrm{the}\:\mathrm{integer}\:\mathrm{which}\:\mathrm{can}\: \\ $$$$\mathrm{be}\:\mathrm{written}:\:\mathrm{N}=\overline {\mathrm{xyz}}\:^{\mathrm{7}} =\overline {\mathrm{zyx}}\:^{\mathrm{11}} \\ $$ Commented by mathocean1 last updated on 24/Oct/20 $$\mathrm{Please}\:\mathrm{can}\:\mathrm{you}\:\mathrm{detail}\:\mathrm{sir}…
Question Number 119395 by peter frank last updated on 24/Oct/20 Answered by mr W last updated on 24/Oct/20 Commented by mr W last updated on…
Question Number 119314 by peter frank last updated on 23/Oct/20 Commented by mr W last updated on 23/Oct/20 $${what}\:{is}\:{a}? \\ $$ Commented by mr W…
Question Number 119241 by harenderkumar last updated on 23/Oct/20 $$\mathrm{675}×\mathrm{54}/\mathrm{100} \\ $$ Answered by sahiljakhar04 last updated on 23/Oct/20 $$\mathrm{675}×\frac{\mathrm{54}}{\mathrm{100}}\:=\:\frac{\mathrm{675}×\mathrm{54}}{\mathrm{100}}\:=\:\frac{\mathrm{27}×\mathrm{27}}{\mathrm{4}}\:=\frac{\left(\mathrm{27}\right)^{\mathrm{2}} }{\left(\mathrm{2}\right)^{\mathrm{2}} }\:=\left(\frac{\mathrm{27}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\mathrm{13}.\mathrm{5}\right)^{\mathrm{2}} \:=\:\mathrm{182}.\mathrm{25}…