Question Number 144468 by mathdanisur last updated on 25/Jun/21 Commented by Rasheed.Sindhi last updated on 26/Jun/21 $${It}'{s}\:{sufficient}\:{to}\:{say}\:'{natural}\:{numbers}' \\ $$$${instead}\:{of}\:'{positive}\:{natural}\:{numbers}'. \\ $$ Terms of Service Privacy…
Question Number 144470 by mathdanisur last updated on 25/Jun/21 $${Solve}\:{for}\:{real}\:{positive}\:{numbers}\:{the} \\ $$$${equation}: \\ $$$$\boldsymbol{{z}}^{\boldsymbol{{log}}\left(\mathrm{3}\right)} \:+\:\boldsymbol{{z}}^{\boldsymbol{{log}}\left(\mathrm{4}\right)} \:+\:\boldsymbol{{z}}^{\boldsymbol{{log}}\left(\mathrm{5}\right)} \:=\:\boldsymbol{{z}}^{\boldsymbol{{log}}\left(\mathrm{6}\right)} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 78931 by mr W last updated on 21/Jan/20 $${if}\:{x}+\frac{\mathrm{1}}{{x}}={a} \\ $$$${find}\:{x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }=? \\ $$ Commented by mathmax by abdo last updated on…
Question Number 13395 by Tinkutara last updated on 19/May/17 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:{n}\:\mathrm{for}\:\mathrm{which} \\ $$$${n}^{\mathrm{2}} \:+\:\mathrm{96}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$ Answered by ajfour last updated on 19/May/17 $$\left({n}+{m}\right)^{\mathrm{2}} ={n}^{\mathrm{2}} +\mathrm{96}…
Question Number 13391 by Tinkutara last updated on 19/May/17 $$\mathrm{A}\:\mathrm{four}\:\mathrm{digit}\:\mathrm{number}\:\mathrm{has}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{properties}: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{first}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{equal} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{The}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{equal}. \\ $$$$\mathrm{Find}\:\mathrm{all}\:\mathrm{such}\:\mathrm{numbers}. \\ $$ Answered by RasheedSindhi…
Question Number 13394 by Tinkutara last updated on 19/May/17 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{any}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{centre}\:\left(\sqrt{\mathrm{2}},\:\sqrt{\mathrm{3}}\right) \\ $$$$\mathrm{cannot}\:\mathrm{pass}\:\mathrm{through}\:\mathrm{more}\:\mathrm{than}\:\mathrm{one} \\ $$$$\mathrm{lattice}\:\mathrm{point}.\:\left[\mathrm{Lattice}\:\mathrm{points}\:\mathrm{are}\:\mathrm{points}\right. \\ $$$$\mathrm{in}\:\mathrm{cartesian}\:\mathrm{plane},\:\mathrm{whose}\:\mathrm{abscissa}\:\mathrm{and} \\ $$$$\left.\mathrm{ordinate}\:\mathrm{both}\:\mathrm{are}\:\mathrm{integers}.\right] \\ $$ Answered by mrW1 last updated…
Question Number 13388 by Tinkutara last updated on 19/May/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{all}\:\mathrm{rational}\:\mathrm{numbers} \\ $$$$\frac{{m}}{{n}}\:\mathrm{such}\:\mathrm{that}\:\left(\mathrm{i}\right)\:\mathrm{0}\:<\:\frac{{m}}{{n}}\:<\:\mathrm{1},\:\left(\mathrm{ii}\right)\:{m}\:\mathrm{and} \\ $$$${n}\:\mathrm{are}\:\mathrm{relatively}\:\mathrm{prime}\:\mathrm{and}\:\left(\mathrm{iii}\right)\:{mn}\:=\:\mathrm{25}!. \\ $$ Commented by RasheedSindhi last updated on 19/May/17 $$\mathrm{There}\:\mathrm{is}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{such}\:\mathrm{number}: \\…
Question Number 13389 by Tinkutara last updated on 19/May/17 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\left[{x}\right]\:+\:\left[\mathrm{2}{x}\right]\:+\:\left[\mathrm{4}{x}\right]\:+\:\left[\mathrm{8}{x}\right]\:+\:\left[\mathrm{16}{x}\right]\:+\:\left[\mathrm{32}{x}\right]\:=\:\mathrm{12345} \\ $$$$\mathrm{has}\:\mathrm{no}\:\mathrm{solution}. \\ $$ Commented by prakash jain last updated on 20/May/17 $${x}={n}+{y}\:,\:{n}\in\mathbb{Z}^{+}…
Question Number 13377 by tawa tawa last updated on 19/May/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17 $$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{4}\Rightarrow{x}−\frac{\mathrm{1}}{{x}}=\pm{i}\sqrt{\mathrm{3}} \\ $$$$\begin{cases}{{x}−\frac{\mathrm{1}}{{x}}=\pm{i}\sqrt{\mathrm{3}}}\\{{x}+\frac{\mathrm{1}}{{x}}=−\mathrm{1}}\end{cases}\Rightarrow{x}=\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} =−{x}−\mathrm{1}…
Question Number 144439 by mathdanisur last updated on 25/Jun/21 Commented by Rasheed.Sindhi last updated on 25/Jun/21 $${gcd}\left({m}^{\mathrm{2}} −{m}+\mathrm{1},{m}+\mathrm{1}\right)=\mathrm{1}? \\ $$ Commented by mathdanisur last updated…