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Category: Arithmetic

1-S-n-a-a-d-a-d-a-2d-a-n-1-d-a-n-d-2-S-n-a-a-d-a-2d-a-d-a-2d-a-3d-a-n-1-d-a-n-d-a-n-1-d-

Question Number 202774 by BaliramKumar last updated on 03/Jan/24 $$\mathrm{1}.\:\:\:\:\:\:\mathrm{S}_{\mathrm{n}} \:=\:\mathrm{a}\left(\mathrm{a}+\mathrm{d}\right)+\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)+……+\left\{\mathrm{a}+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{d}\right\}\left\{\mathrm{a}+\left(\mathrm{n}\right)\mathrm{d}\right\} \\ $$$$\mathrm{2}.\:\:\:\:\:\:\mathrm{S}_{\mathrm{n}} \:=\:\mathrm{a}\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)+\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)\left(\mathrm{a}+\mathrm{3d}\right)+……+\left\{\mathrm{a}+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{d}\right\}\left\{\mathrm{a}+\left(\mathrm{n}\right)\mathrm{d}\right\}\left\{\mathrm{a}+\left(\mathrm{n}+\mathrm{1}\right)\mathrm{d}\right\} \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last updated on 03/Jan/24…

2-8-2-1-4-2-2-4-1-1-6-3-1-6-3-4-2-5-5-2-6-5-2-

Question Number 202796 by MathematicalUser2357 last updated on 03/Jan/24 $$\left[\mathrm{2}^{\mathrm{8}\boldsymbol{\div}\mathrm{2}−\mathrm{1}} +\mathrm{4}\boldsymbol{\div}\left\{\mathrm{2}+\mathrm{2}^{\mathrm{4}\boldsymbol{\div}\left(\mathrm{1}+\mathrm{1}\right)} \boldsymbol{\div}\left(\mathrm{6}\boldsymbol{\div}\mathrm{3}−\mathrm{1}\right)\right\}\right]\boldsymbol{\div}\left[\mathrm{6}\boldsymbol{\div}\left\{\left(\mathrm{3}^{\mathrm{4}\boldsymbol{\div}\mathrm{2}} +\mathrm{5}\boldsymbol{\div}\mathrm{5}\right)\boldsymbol{\div}\mathrm{2}−\mathrm{6}\boldsymbol{\div}\left(\mathrm{5}−\mathrm{2}\right)\right\}\right] \\ $$ Answered by a.lgnaoui last updated on 03/Jan/24 $$\mathrm{si}\:\left(\:\frac{.}{.}\right)\:\:\mathrm{est}\:\mathrm{une}\:\mathrm{operation}\:\mathrm{de}\left[\mathrm{division}\right. \\ $$$$\:\:\Rightarrow\:\mathrm{Expression}\:\mathrm{est}\:\mathrm{equivalente}\:\mathrm{a}…

The-sum-of-all-the-divisors-of-24-is-60-Find-the-sum-of-all-the-divisors-of-24-79-

Question Number 202672 by BaliramKumar last updated on 31/Dec/23 $$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{divisors}\:\mathrm{of}\:\mathrm{24}\:\mathrm{is}\:\mathrm{60}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{divisors}\:\mathrm{of}\:\mathrm{24}×\mathrm{79}. \\ $$ Answered by a.lgnaoui last updated on 31/Dec/23 $$\mathrm{60} \\ $$ Commented…

determine-le-reste-de-la-division-eucludienne-de-2023-2019-par-13-

Question Number 202698 by Bambamamoudou last updated on 31/Dec/23 $${determine}\:{le}\:{reste}\:{de}\:{la}\:{division}\:{eucludienne}\:{de}\:\mathrm{2023}^{\mathrm{2019}} {par}\:\mathrm{13} \\ $$ Answered by Rasheed.Sindhi last updated on 01/Jan/24 $$\because\:{gcd}\left(\mathrm{2023},\mathrm{13}\right)=\mathrm{1} \\ $$$$\therefore\:\:\:\:\mathrm{2023}^{\phi\left(\mathrm{13}\right)} \equiv\mathrm{1}\left({mod}\:\mathrm{13}\right) \\…

Question-202462

Question Number 202462 by pticantor last updated on 27/Dec/23 Answered by witcher3 last updated on 27/Dec/23 $$\mathrm{p}^{\mathrm{n}} −\mathrm{1}=\left(\mathrm{p}−\mathrm{1}\right)\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{p}^{\mathrm{k}} \right) \\ $$$$\Rightarrow\mathrm{p}.\left(\mathrm{p}^{\mathrm{n}} \right)+\left(\mathrm{p}−\mathrm{1}\right).\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}}…

1-1-2-3-1-2-3-4-1-3-4-5-1-n-n-1-n-2-

Question Number 202251 by BaliramKumar last updated on 23/Dec/23 $$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}×\mathrm{5}}\:+\:…………..\:+\:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Commented by BaliramKumar last updated on 23/Dec/23 $$\frac{\mathrm{1}}{\mathrm{a}\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)\left(\mathrm{a}+\mathrm{3d}\right)}\:+\:\:…….+\:\frac{\mathrm{1}}{\left\{\mathrm{a}+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{d}\right\}\left\{\mathrm{a}+\mathrm{nd}\right\}\left\{\mathrm{a}+\left(\mathrm{n}+\mathrm{1}\right)\mathrm{d}\right\}}\:=\:? \\ $$ Commented by MATHEMATICSAM…