Question Number 136792 by EDWIN88 last updated on 26/Mar/21 $$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{diophantine}\:\mathrm{equation} \\ $$$$\left(\mathrm{1}\right)\mathrm{3xy}\:+\mathrm{2x}\:+\mathrm{y}\:=\:\mathrm{12}\:? \\ $$$$\left(\mathrm{2}\right)\:\mathrm{x}^{\mathrm{3}} =\:\mathrm{4y}^{\mathrm{2}} +\mathrm{4y}−\mathrm{3}\:? \\ $$ Answered by mindispower last updated on 26/Mar/21…
Question Number 5216 by Yozzii last updated on 01/May/16 $${Let}\:{p}_{{j}} \:{represent}\:{the}\:{j}−{th}\:{prime}\:{number}. \\ $$$${Now},\:{define}\:{the}\:{number}\:{n}\:{whose} \\ $$$${decimal}\:{representation}\:{is}\:{written}\:{out} \\ $$$${in}\:{terms}\:{of}\:{p}_{{j}} \:\left({j}\in\mathbb{N}\right)\:{in}\:{the}\:{following} \\ $$$${way}: \\ $$$${n}=\mathrm{0}.{p}_{\mathrm{1}} {p}_{\mathrm{2}} {p}_{\mathrm{3}} {p}_{\mathrm{4}}…
Question Number 5185 by Yozzii last updated on 28/Apr/16 $${If}\:\:\mathrm{2}^{{x}\:} \:{and}\:\:\mathrm{3}^{{x}} \:{are}\:{integers}\:{for}\:{some} \\ $$$${x}\in\mathbb{R}^{+} ,\:{must}\:{x}\:{be}\:{an}\:{integer}?\: \\ $$ Answered by 123456 last updated on 28/Apr/16 $$\mathrm{2}^{{x}}…
Question Number 5165 by 1771727373 last updated on 24/Apr/16 $${what}\:{is} \\ $$$$\sqrt{{i}+\mathrm{1}}\:\:\:\:? \\ $$ Commented by FilupSmith last updated on 24/Apr/16 $$\mathrm{assuming}\:{i}^{\mathrm{2}} =−\mathrm{1}, \\ $$$$\mathrm{according}\:\mathrm{to}\:\mathrm{WolframAlpha}:…
Question Number 5168 by Yozzii last updated on 24/Apr/16 $${Find}\:{the}\:{value}\:{of}\:\mathrm{2023}!\:\left({mod}\:\mathrm{2027}\right). \\ $$ Commented by prakash jain last updated on 27/Apr/16 $$\mathrm{If}\:\mathrm{you}\:\mathrm{find}\:\mathrm{an}\:\mathrm{answer},\:\mathrm{please}\:\mathrm{do}\:\mathrm{post}\:\mathrm{it}.\:\mathrm{I} \\ $$$$\mathrm{have}\:\mathrm{not}\:\mathrm{yet}\:\mathrm{been}\:\mathrm{able}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}. \\ $$…
Question Number 5158 by 1771727373 last updated on 24/Apr/16 $${why} \\ $$$$\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{8}}\:+\:………..\:=\:\mathrm{2} \\ $$$$ \\ $$ Answered by FilupSmith last updated on 24/Apr/16 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+…={S} \\…
Question Number 5152 by 1771727373 last updated on 23/Apr/16 $${there}\:{is}\:{an}\:{ineterger}\:{a},{b},{c} \\ $$$${can}\:{there}\:{be}\:{an}\:{interger}\:{as}\: \\ $$$${a}^{{n}} +{b}^{{n}+\mathrm{1}} ={c}^{{n}+\mathrm{2}} \:\:\:\:\:\:\left({n}\:{is}\:{a}\:{interger}\right) \\ $$$$ \\ $$ Commented by Yozzii last…
Question Number 5144 by Yozzii last updated on 19/Apr/16 $${Let}\:{n},{j},{q}\in\left(\mathbb{Z}^{+} −\left\{\mathrm{1}\right\}\right).\:{Are}\:{there}\: \\ $$$${triples}\:\left({n},{j},{q}\right)\:{such}\:{that}\:{the}\:{following} \\ $$$${conditions}\:{are}\:{satisfied}\:{altogether}? \\ $$$$\left({i}\right)\:{n}={j}^{{q}} \:\:\:\: \\ $$$$\left({ii}\right){n}^{\mathrm{2}} ={j}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$$$−−−−−−−−−−−−−−−−−−−−−− \\…
Question Number 5125 by Rojaye Shegz last updated on 16/Apr/16 Commented by Rojaye Shegz last updated on 16/Apr/16 $$\mathrm{Sorry},\:\mathrm{that}\:\mathrm{was}\:\mathrm{an}\:\mathrm{error}. \\ $$$$\frac{\mathrm{p}}{\mathrm{q}}\:\mathrm{is}\:\mathrm{the}\:\left\{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{p}+\mathrm{q}−\mathrm{1}\right)\left(\mathrm{p}+\mathrm{q}−\mathrm{2}\right)+\mathrm{p}\right\}\mathrm{th}\:\mathrm{term} \\ $$ Commented by…
Question Number 5082 by sandhapadarbinda@gmail.com last updated on 10/Apr/16 $${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =? \\ $$ Answered by FilupSmith last updated on 11/Apr/16 $$\mathrm{difference}\:\mathrm{of}\:\mathrm{two}\:\mathrm{squares} \\ $$$$ \\…