Question Number 67481 by Mohamed Amine Bouguezzoul last updated on 27/Aug/19 $${p}\:{is}\:{a}\:{prime}\:{number}\:{such}\:{that}\:\left(\mathrm{1}+{p}\right)^{{p}} \equiv\mathrm{2}\left[\mathrm{7}\right] \\ $$$${find}\:{all}\:{k}\:{such}\:{that}\:{p}\equiv{k}\left[\mathrm{42}\right] \\ $$ Commented by Rasheed.Sindhi last updated on 30/Aug/19 $$\boldsymbol{{Some}}\:{values}\:{of}\:{k}…
Question Number 1831 by Filup last updated on 10/Oct/15 $$\mathrm{Solve}: \\ $$$$\mathrm{1}+\left(\mathrm{1}+\mathrm{2}\right)+\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)+…+\left(\mathrm{1}+\mathrm{2}+…+{n}\right) \\ $$ Answered by Rasheed Soomro last updated on 10/Oct/15 $$\mathrm{1}+\left(\mathrm{1}+\mathrm{2}\right)+\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)+…+\left(\mathrm{1}+\mathrm{2}+…+{n}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\right)\left(\mathrm{1}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\right)\left(\mathrm{2}+\mathrm{1}\right)+…+\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right)…
Question Number 66778 by Tony Lin last updated on 19/Aug/19 $$\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{2}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{2}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{11}}−\frac{\mathrm{2}}{\mathrm{12}}+\centerdot\centerdot\centerdot= \\ $$ Commented by Prithwish sen last updated on 19/Aug/19 $$\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+……\right)−\left[\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\right)+\left(\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{2}}{\mathrm{6}}\right)+\left(\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{2}}{\mathrm{9}}\right)+\left(\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{2}}{\mathrm{12}}\right)+…\right] \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty}…
Question Number 1181 by 22 last updated on 11/Jul/15 $$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}=? \\ $$ Answered by Rasheed Ahmad last updated on 24/Jul/15 $${In}\:{certain}\:{cases}\:\sqrt{{a}+{b}\sqrt{{c}}\:}\:{can}\:{be} \\ $$$${simplified}\:{into}\:{p}+{q}\sqrt{{c}}\:{form}\left({not}\right. \\ $$$$\left.{in}\:{all}\:{cases}\right).\:{The}\:{procedure}\:{is}\:{as}…
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Question Number 1112 by malwaan last updated on 14/Jun/15 $${compare}\:\frac{\mathrm{1}}{{log}_{\mathrm{2}} \pi}\:+\:\frac{\mathrm{1}}{{log}_{\mathrm{5}} \pi}\:{and}\:\mathrm{2} \\ $$ Answered by prakash jain last updated on 15/Jun/15 $$\mathrm{log}_{\pi} \mathrm{2}+\mathrm{log}_{\pi} \mathrm{5}=\mathrm{log}_{\pi}…
Question Number 1100 by malwaan last updated on 13/Jun/15 $${compare}\:{log}_{\mathrm{2}} \mathrm{3}\:{and}\:{log}_{\mathrm{3}} \mathrm{5} \\ $$ Answered by 123456 last updated on 13/Jun/15 $$\mathrm{log}_{\mathrm{2}} \mathrm{3}={x}\Leftrightarrow\mathrm{2}^{{x}} =\mathrm{3}\: \\…
Question Number 1083 by Vishal last updated on 08/Jun/15 $${Let}\:{a},{b},{c},{p}\:{be}\:{rational}\:{numbers}\:{such}\:{that}\:{p}\:{is}\:{not}\:{a}\:{perfect}\:{cube}. \\ $$$${If}\:{a}+{bp}^{\frac{\mathrm{1}}{\mathrm{3}}} +{cp}^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{0},\:{then}\:{prove}\:{that}\:{a}={b}={c}=\mathrm{0}. \\ $$ Answered by prakash jain last updated on 08/Jun/15 $${p}^{\mathrm{1}/\mathrm{3}}…
Question Number 1077 by Vishal last updated on 07/Jun/15 $${Find}\:{the}\:{smallest}\:{number}\:{which}\:{leaves}\:{remainders}\:\mathrm{8}\:{and}\:\mathrm{12}\: \\ $$$${when}\:{divided}\:{by}\:\mathrm{28}\:{and}\:\mathrm{32}\:{respectively}. \\ $$ Commented by prakash jain last updated on 07/Jun/15 $$\mathrm{I}\:\mathrm{assume}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{smallest}\:+\mathrm{ve}\:\mathrm{integer}. \\ $$…
Question Number 1063 by lops55 last updated on 26/May/15 $${rational}\:{number}\:{which}\:{is}\:{neither}\:{negetive}\:{nor}\:{positive} \\ $$ Answered by 123456 last updated on 01/Jun/15 $$\mathrm{0}? \\ $$ Terms of Service…