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Category: Arithmetic

P-and-Q-are-partners-in-a-venture-P-contributed-20-000-for-nine-month-and-Q-contributed-50-000-for-one-year-find-each-person-share-of-profit-of-6-300-

Question Number 6743 by Tawakalitu. last updated on 20/Jul/16 $${P}\:{and}\:{Q}\:{are}\:{partners}\:{in}\:{a}\:{venture},\:\:{P}\:\:{contributed}\:#\mathrm{20},\mathrm{000}\:{for} \\ $$$${nine}\:{month},\:{and}\:{Q}\:{contributed}\:#\mathrm{50},\mathrm{000}\:{for}\:{one}\:{year}.\:{find}\:{each} \\ $$$${person}\:{share}\:{of}\:\:{profit}\:{of}\:#\mathrm{6},\mathrm{300} \\ $$$$ \\ $$ Answered by Rasheed Soomro last updated on…

Question-6593

Question Number 6593 by Tawakalitu. last updated on 04/Jul/16 Commented by Yozzii last updated on 05/Jul/16 $${Given}\:{that}\:{p}\:{is}\:{prime}\:{and}\:{p}\leqslant\mathrm{100}, \\ $$$${show}\:{that}\:{there}\:{are}\:{solutions}\:\left({x},{y}\right),\: \\ $$$${x},{y}\in\mathbb{Z},\:{for}\:{which}\:{y}^{\mathrm{37}} \equiv{x}^{\mathrm{3}} +\mathrm{37}\:\left({mod}\:{p}\right). \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−…

Prove-and-show-why-k-1-n-H-k-n-1-H-n-1-1-Where-H-k-1-1-2-1-3-1-k-

Question Number 6487 by Temp last updated on 29/Jun/16 $$\mathrm{Prove}\:\mathrm{and}\:\mathrm{show}\:\mathrm{why}: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{H}_{{k}} =\left({n}+\mathrm{1}\right)\left({H}_{{n}+\mathrm{1}} −\mathrm{1}\right) \\ $$$$\mathrm{Where}: \\ $$$${H}_{{k}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{{k}} \\ $$ Commented by…

Solve-1-x-2-x-4-3-

Question Number 71925 by Henri Boucatchou last updated on 22/Oct/19 $$\:\boldsymbol{{Solve}}\:\:\sqrt{\mathrm{1}\:−\:\boldsymbol{{x}}^{\mathrm{2}} }\:=\:\boldsymbol{{x}}^{\mathrm{4}/\mathrm{3}} \\ $$ Answered by MJS last updated on 23/Oct/19 $${x}={t}^{\mathrm{3}/\mathrm{2}} \\ $$$$\sqrt{\mathrm{1}−{t}^{\mathrm{3}} }={t}^{\mathrm{2}}…

Question-6340

Question Number 6340 by sanusihammed last updated on 24/Jun/16 Commented by nburiburu last updated on 24/Jun/16 $${seems}\:{it}\:{need}\:{more}\:{info}.\:{For}\:{the}\:{moment}\:{being}\:{could}\:{be}\:{that}\:{moving}\:{one}\:{unit}\:{from}\:{right}\:{to}\:{left}\:{adds}\:\mathrm{2}\:{units}\:{up}\:{and}\:{cost}\:\mathrm{1}\:{in}\:{the}\:{centre}. \\ $$$${However}\:{many}\:{other}\:{posibilities}\:{could}\:{explain}\:{the}\:{numbers}\:{in}\:{the}\:{diagram}. \\ $$ Commented by prakash jain…

According-to-WolframAlpha-k-0-n-1-x-1-k-1-x-n-2-1-x-1-x-n-1-2-1-Can-anyone-work-out-how-

Question Number 6311 by FilupSmith last updated on 23/Jun/16 $$\mathrm{According}\:\mathrm{to}\:\mathrm{WolframAlpha}: \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\left(\mathrm{1}−{x}^{\left(−\mathrm{1}\right)^{{k}} } \right)=\left(\mathrm{1}−{x}\right)^{\lfloor\frac{{n}}{\mathrm{2}}\rfloor+\mathrm{1}} \left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\lfloor\frac{{n}−\mathrm{1}}{\mathrm{2}}\rfloor+\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{Can}\:\mathrm{anyone}\:\mathrm{work}\:\mathrm{out}\:\mathrm{how}? \\ $$ Commented by…