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Category: Arithmetic

Question-22874

Question Number 22874 by selestian last updated on 23/Oct/17 Answered by $@ty@m last updated on 23/Oct/17 $${Solutionof}\:\mathrm{9}. \\ $$$${S}_{\mathrm{1}} =\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$${S}_{\mathrm{2}} =\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{6}} \\ $$$${S}_{\mathrm{3}}…

Question-153905

Question Number 153905 by liberty last updated on 12/Sep/21 Commented by MJS_new last updated on 12/Sep/21 $$\mathrm{3}^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}×\mathrm{5}}+\frac{\mathrm{1}}{\left(\mathrm{2}+\mathrm{3}\right)\left(\mathrm{5}+\mathrm{3}\right)}+\frac{\mathrm{1}}{\left(\mathrm{2}+\mathrm{2}×\mathrm{3}\right)\left(\mathrm{5}+\mathrm{2}×\mathrm{3}\right)}+…+\frac{\mathrm{1}}{\left(\mathrm{2}+\mathrm{11}×\mathrm{3}\right)\left(\mathrm{5}+\mathrm{11}×\mathrm{3}\right)}\right) \\ $$ Answered by MJS_new last updated…

let-d-n-denote-the-sum-of-the-digits-of-n-i-e-d-1000-1-d-999-27-find-minimum-k-such-that-d-d-k-times-d-5-10-100-10-

Question Number 153657 by talminator2856791 last updated on 09/Sep/21 $$\: \\ $$$$\:\mathrm{let}\:{d}\left({n}\right)\:\mathrm{denote}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\: \\ $$$$\:\mathrm{of}\:{n}. \\ $$$$\:\mathrm{i}.\mathrm{e}\:\:\:\:{d}\left(\mathrm{1000}\right)\:=\:\mathrm{1}\:,\:\:\:\:\:{d}\left(\mathrm{999}\right)\:=\:\mathrm{27} \\ $$$$\: \\ $$$$\:\mathrm{find}\:\mathrm{minimum}\:{k}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\underset{{k}\:\mathrm{times}} {\underbrace{{d}\left({d}\left(…..}{d}}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)…..\right)\:\ll\:\mathrm{10}\right.…

define-d-n-to-be-the-sum-of-the-digits-of-n-i-e-d-1000-1-d-999-27-find-d-d-d-d-d-5-10-100-

Question Number 153565 by talminator2856791 last updated on 08/Sep/21 $$\: \\ $$$$\:\mathrm{define}\:{d}\left({n}\right)\:\mathrm{to}\:\mathrm{be}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\: \\ $$$$\:\mathrm{of}\:{n}.\:\:\: \\ $$$$\:\mathrm{i}.\mathrm{e}\:\:{d}\left(\mathrm{1000}\right)\:=\:\mathrm{1}\:,\:\:\:\:{d}\left(\mathrm{999}\right)\:=\:\mathrm{27} \\ $$$$\: \\ $$$$\:\mathrm{find}\:\:{d}\left({d}\left({d}\left({d}\left({d}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)\right)\right)\right)\right) \\ $$$$\: \\…

n-1-H-n-2n-1-pi-2-2H-n-2-8H-2n-2-83-4-4-7log-2-3-8log-2-2-2-2-3-log-4-2-16-Li-4-1-2-where-H-n-m-1-1-2-m-1-n-m-represents-the-nth-gener

Question Number 88004 by M±th+et£s last updated on 07/Apr/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{\mathrm{2}{n}+\mathrm{1}}\left(\pi^{\mathrm{2}} +\mathrm{2}{H}_{{n}} ^{\left(\mathrm{2}\right)} −\mathrm{8}{H}_{\mathrm{2}{n}} ^{\left(\mathrm{2}\right)} \right) \\ $$$$=\frac{\mathrm{83}}{\mathrm{4}}\zeta\left(\mathrm{4}\right)−\mathrm{7}{log}\left(\mathrm{2}\right)\zeta\left(\mathrm{3}\right)−\mathrm{8}{log}^{\mathrm{2}} \left(\mathrm{2}\right)\zeta\left(\mathrm{2}\right)−\frac{\mathrm{2}}{\mathrm{3}}{log}^{\mathrm{4}} \left(\mathrm{2}\right)−\mathrm{16}\:{Li}_{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${where}\:{H}_{{n}}…

A-man-buys-a-hen-at-9-and-sells-it-at-11-He-buys-the-same-hen-later-on-at-12-and-sells-it-now-at-14-What-is-his-total-benefit-

Question Number 87935 by Ar Brandon last updated on 07/Apr/20 $${A}\:{man}\:{buys}\:{a}\:{hen}\:{at}\:\mathrm{9\$}\:{and}\:{sells}\:{it}\:{at}\:\mathrm{11\$}.\:{He}\:{buys} \\ $$$${the}\:{same}\:{hen}\:{later}\:{on}\:{at}\:\mathrm{12\$}\:{and}\:{sells}\:{it}\:{now} \\ $$$${at}\:\mathrm{14\$}.\:{What}\:{is}\:{his}\:{total}\:{benefit}\:? \\ $$ Answered by MJS last updated on 07/Apr/20 $$\mathrm{0}−\mathrm{9}+\mathrm{11}−\mathrm{12}+\mathrm{14}=\mathrm{4}…

x-y-7-x-y-11-find-x-y-

Question Number 87927 by john santu last updated on 07/Apr/20 $$\mathrm{x}+\sqrt{\mathrm{y}}\:=\:\mathrm{7} \\ $$$$\sqrt{\mathrm{x}}\:+\:\mathrm{y}\:=\:\mathrm{11} \\ $$$$\mathrm{find}\:\mathrm{x}+\mathrm{y}\: \\ $$ Commented by MJS last updated on 07/Apr/20 $$\mathrm{we}\:\mathrm{had}\:\mathrm{the}\:\mathrm{same}\:\mathrm{question}\:\mathrm{lately}…