Category: Arithmetic

Question Number 209167 by lmcp1203 last updated on 02/Jul/24 $${please}\:{convert}\:\:\mathrm{2531}_{\left(\mathrm{5000}\right)\:} {to}\:\:{base}\:\mathrm{5002}.\:\:{thanks}.\:\: \\ $$ Answered by mr W last updated on 03/Jul/24 $$\mathrm{2531}_{\left(\mathrm{5000}\right)} =\mathrm{2}×\mathrm{5000}^{\mathrm{3}} +\mathrm{5}×\mathrm{5000}^{\mathrm{2}} +\mathrm{3}×\mathrm{5000}+\mathrm{1}…

Question Number 209123 by Spillover last updated on 02/Jul/24 Answered by A5T last updated on 02/Jul/24 $${max}\left(\mathrm{1000}{x}+\mathrm{600}{c}\right)\:{given}\:{c}\geqslant\mathrm{4}{x}\:{and}\:{x}+{c}\leqslant\mathrm{200} \\ $$$$\mathrm{4}{x}−{c}\leqslant\mathrm{0}\:{and}\:{x}+{c}\leqslant\mathrm{200}\Rightarrow{x}\leqslant\mathrm{40} \\ $$$$\Rightarrow{max}\left(\mathrm{1000}{x}+\mathrm{600}{c}\right)=\mathrm{1000}×\mathrm{40}+\mathrm{600}×\mathrm{160} \\ $$$$=\mathrm{136000}. \\ $$…

Question Number 208980 by lmcp1203 last updated on 30/Jun/24 $${please}\:.\:\:\:\:\:{find}\:\:\mathrm{2}^{\mathrm{11001}^{\mathrm{666}} } {mod}\:\mathrm{23}\:\:\:\:\:\:\:\:{thanks}. \\ $$ Answered by A5T last updated on 30/Jun/24 $$\mathrm{2}^{\mathrm{11001}^{\mathrm{666}} } {mod}\left(\mathrm{23}\right)\equiv\mathrm{2}^{\mathrm{11001}^{\mathrm{666}} \left[{mod}\:\phi\left(\mathrm{23}\right)\right]}…

Question Number 208445 by alcohol last updated on 16/Jun/24 $$\left.{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} −{u}_{{n}} ^{\mathrm{3}} ,\:{u}_{\mathrm{0}} \in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$${v}_{{n}} \:=\:\frac{\mathrm{1}}{{u}_{{n}+\mathrm{1}} ^{\mathrm{2}} }−\frac{\mathrm{1}}{{u}_{{n}} ^{\mathrm{2}} }\:=\:{f}\left({u}_{{n}} ^{\mathrm{2}} \right)\:;\:{f}\left({x}\right)\:=\:\frac{\mathrm{2}−{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }…

Question Number 207979 by Tawa11 last updated on 01/Jun/24 generate nth term for the sequence: 1, 1, 1, 2, 3, 5, 9, 18, 35, 75…