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Author: Tinku Tara

find-the-last-4-digits-of-2024-2023-

Question Number 203270 by MrGHK last updated on 13/Jan/24 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{last}}\:\mathrm{4}\:\boldsymbol{\mathrm{digits}}\:\boldsymbol{\mathrm{of}}\:\mathrm{2024}^{\mathrm{2023}} \\ $$ Answered by Frix last updated on 14/Jan/24 $$\mathrm{Last}\:\mathrm{4}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{2024}^{{n}} \\ $$$${n}=\mathrm{1}\:\mathrm{2024} \\ $$$$\mathrm{Then}\:\mathrm{a}\:\mathrm{loop}\:\mathrm{of}\:\mathrm{length}\:\mathrm{50} \\…

If-x-1-5-2-find-x-12-

Question Number 203264 by hardmath last updated on 13/Jan/24 $$\mathrm{If}\:\:\:\:\:\mathrm{x}\:=\:\frac{\mathrm{1}\:+\:\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:\mathrm{find}:\:\mathrm{x}^{\mathrm{12}} \:=\:? \\ $$ Answered by MM42 last updated on 13/Jan/24 $$\begin{cases}{{x}^{\mathrm{2}} =\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\\{{x}^{\mathrm{4}} =\frac{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}}\end{cases}\:\:\Rightarrow{x}^{\mathrm{6}} =\mathrm{9}+\mathrm{4}\sqrt{\mathrm{5}} \\…

Let-s-define-linear-Operator-L-as-L-0-e-st-L-W-t-W-t-is-inverse-function-of-y-t-te-t-t-1-e-

Question Number 203199 by MathedUp last updated on 12/Jan/24 $$\mathrm{Let}'{s}\:\mathrm{define}\:\mathrm{linear}\:\mathrm{Operator}\:\boldsymbol{\mathcal{L}}\:\mathrm{as}\:\boldsymbol{\mathcal{L}}=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{st}} \centerdot \\ $$$$\boldsymbol{\mathcal{L}}\left\{{W}\left({t}\right)\right\}=??? \\ $$$${W}\left({t}\right)\:\mathrm{is}\:\mathrm{inverse}\:\mathrm{function}\:\mathrm{of}\:{y}\left({t}\right)={te}^{{t}} \:,\:{t}\in\left[−\frac{\mathrm{1}}{{e}},\infty\right) \\ $$ Commented by shunmisaki007 last updated…

find-the-last-four-digits-of-2022-2023-2023-2022-

Question Number 203192 by MrGHK last updated on 12/Jan/24 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{last}}\:\boldsymbol{\mathrm{four}}\:\boldsymbol{\mathrm{digits}}\:\boldsymbol{\mathrm{of}}\:\mathrm{2022}^{\mathrm{2023}} +\mathrm{2023}^{\mathrm{2022}} \\ $$ Answered by AST last updated on 12/Jan/24 $${x}=\mathrm{2022}^{\mathrm{2023}} +\mathrm{2023}^{\mathrm{2022}} \overset{\mathrm{16}} {\equiv}\mathrm{6}^{\mathrm{2023}} +\mathrm{7}^{\mathrm{2022}}…