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Category: Arithmetic

5-5-5-5-5-5-5-

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Question Number 103321 by bemath last updated on 14/Jul/20 $$\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+…}}}}}} \\ $$ Answered by Dwaipayan Shikari last updated on 14/Jul/20 $$\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}…}}}={p} \\ $$$$\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}…={p}^{\mathrm{2}} \\ $$$$\mathrm{5}+{p}=\left(\mathrm{5}−{p}^{\mathrm{2}}…

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Evaluate-r-0-2-r-1-

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Question Number 37712 by Rio Mike last updated on 16/Jun/18 $${Evaluate}\:\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}^{{r}−\mathrm{1}} \\ $$ Commented by prakash jain last updated on 17/Jun/18 $$−\mathrm{1}/\mathrm{2}\:\mathrm{is}\:\mathrm{also}\:\mathrm{a}\:\mathrm{valid}\:\mathrm{answer}\:\mathrm{using} \\…

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3-3-3-3-1-or-3-3-3-3-3-

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Question Number 168391 by Florian last updated on 09/Apr/22 $$\:\:\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{3}\sqrt{\mathrm{3}}}=\mathrm{1}\:\:\:\:{or}\:\:\mathrm{3}×\sqrt{\mathrm{3}}\boldsymbol{\div}\mathrm{3}×\sqrt{\mathrm{3}}\:=\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:??? \\ $$ Commented by Rasheed.Sindhi last updated on 09/Apr/22 $$\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{3}\sqrt{\mathrm{3}}}=\left(\mathrm{3}×\sqrt{\mathrm{3}}\:\right)\boldsymbol{\div}\left(\mathrm{3}×\sqrt{\mathrm{3}}\:\right) \\ $$$$\mathrm{Where}\:\mathrm{as} \\ $$$$\mathrm{3}×\sqrt{\mathrm{3}}\boldsymbol{\div}\mathrm{3}×\sqrt{\mathrm{3}}\:=\mathrm{3}×\left(\sqrt{\mathrm{3}}\boldsymbol{\div}\mathrm{3}\right)×\sqrt{\mathrm{3}} \\…

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During-a-sales-period-a-magazine-offers-a-t-discount-For-clients-in-possession-of-the-fidelity-card-an-extra-discount-of-t-5-is-offered-A-client-benefits-from-these-two-discounts-and-pays-

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Question Number 102773 by Ar Brandon last updated on 11/Jul/20 $$\:\:\:\mathrm{During}\:\mathrm{a}\:\mathrm{sales}\:\mathrm{period}\:\mathrm{a}\:\mathrm{magazine}\:\mathrm{offers}\:\mathrm{a}\:\mathrm{t\%}\:\mathrm{discount} \\ $$$$\mathrm{For}\:\mathrm{clients}\:\mathrm{in}\:\mathrm{possession}\:\mathrm{of}\:\mathrm{the}\:\mathrm{fidelity}\:\mathrm{card},\:\mathrm{an}\:\mathrm{extra}\:\mathrm{discount} \\ $$$$\mathrm{of}\:\left(\mathrm{t}+\mathrm{5}\right)\%\:\mathrm{is}\:\mathrm{offered}. \\ $$$$\:\:\:\mathrm{A}\:\mathrm{client}\:\mathrm{benefits}\:\mathrm{from}\:\mathrm{these}\:\mathrm{two}\:\mathrm{discounts}\:\mathrm{and}\:\mathrm{pays}\: \\ $$$$\mathrm{150}\:\epsilon\:\mathrm{for}\:\mathrm{an}\:\mathrm{article}\:\mathrm{whose}\:\mathrm{initial}\:\mathrm{price}\:\mathrm{is}\:\mathrm{250}\epsilon \\ $$$$\left({i}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{t}\:\mathrm{is}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{250}×\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{100}}\right)×\left(\mathrm{1}−\frac{\mathrm{t}+\mathrm{5}}{\mathrm{100}}\right)=\mathrm{150} \\ $$$$\left({ii}\right)\:\mathrm{Solve}\:\mathrm{this}\:\mathrm{equation}\:\mathrm{and}\:\mathrm{deduce}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{t}.…

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let-A-1-46-show-that-for-any-p-A-we-have-q-Z-such-as-p-q-1-47-please-i-need-your-help-

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Question Number 102549 by pticantor last updated on 09/Jul/20 $$ \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{A}}=\left[\mathrm{1},\mathrm{46}\right]\: \\ $$$$\boldsymbol{{show}}\:\boldsymbol{{that}}\:\boldsymbol{{for}}\:\boldsymbol{{any}}\:\boldsymbol{{p}}\in\boldsymbol{{A}}\:{we}\:\boldsymbol{{have}}\:\boldsymbol{{q}}\in\mathbb{Z} \\ $$$$\boldsymbol{{such}}\:\boldsymbol{{as}}\:\boldsymbol{{p}}×\boldsymbol{{q}}\equiv\mathrm{1}\left[\mathrm{47}\right] \\ $$$$\boldsymbol{{pleas}}{e}\:{i}\:\boldsymbol{{need}}\:\boldsymbol{{your}}\:\boldsymbol{{help}} \\ $$$$ \\ $$$$ \\ $$$$ \\…

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Lets-p-N-and-n-N-A-n-2-n-p-and-d-n-PGCD-A-n-A-n-1-1-show-that-d-n-2-n-2-determine-the-parity-of-A-n-as-a-function-of-that-of-p-3-determine-the-parity-of-d-n-as-a-function-of-that-of

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Question Number 102491 by pticantor last updated on 09/Jul/20 $$ \\ $$$$\boldsymbol{{L}}{e}\boldsymbol{{ts}}\:\boldsymbol{{p}}\:\in\mathbb{N}\:\boldsymbol{{and}}\:\boldsymbol{{n}}\in\mathbb{N}^{\ast} \\ $$$$\boldsymbol{{A}}_{{n}} =\mathrm{2}^{\boldsymbol{{n}}} +\boldsymbol{{p}}\:\boldsymbol{{and}}\:\boldsymbol{{d}}_{\boldsymbol{{n}}} =\boldsymbol{{P}}{GCD}\left(\boldsymbol{{A}}_{{n}} ,\boldsymbol{{A}}_{\boldsymbol{{n}}+\mathrm{1}} \right) \\ $$$$\left.\mathrm{1}\right)\:\boldsymbol{{show}}\:\boldsymbol{{that}}\:\:\boldsymbol{{d}}_{\boldsymbol{{n}}} /\mathrm{2}^{\boldsymbol{{n}}} \\ $$$$\left.\mathrm{2}\right)\boldsymbol{{determine}}\:\boldsymbol{{the}}\:\boldsymbol{{parity}}\:{of}\:\boldsymbol{{A}}_{{n}} \:{as}\:{a}\:{function}\:{of}\:{that}\:\boldsymbol{{of}}\:\boldsymbol{{p}}…

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