Question Number 213467 by efronzo1 last updated on 06/Nov/24 $$\:\:\:\underbrace{\boldsymbol{{B}}} \\ $$ Commented by mr W last updated on 06/Nov/24 �� Commented by mr W…
Question Number 213463 by golsendro last updated on 06/Nov/24 $$\:\:\mathrm{For}\:\mathrm{p},\mathrm{q}\:\mathrm{and}\:\mathrm{r}\:\mathrm{prime}\:\mathrm{numbers}\: \\ $$$$\:\:\mathrm{satisfying}\:\begin{cases}{\mathrm{p}\left(\mathrm{q}+\mathrm{1}\right)\left(\mathrm{r}+\mathrm{1}\right)=\mathrm{1064}}\\{\mathrm{r}\left(\mathrm{p}+\mathrm{1}\right)\left(\mathrm{q}+\mathrm{1}\right)=\mathrm{1554}}\end{cases} \\ $$$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{p}\left(\mathrm{q}+\mathrm{1}\right)\mathrm{r}\: \\ $$ Answered by A5T last updated on 06/Nov/24 $${p}\mid\mathrm{1064}=\mathrm{2}^{\mathrm{3}} ×\mathrm{7}×\mathrm{19};\:{p}+\mathrm{1}\mid\mathrm{1554}=\mathrm{2}×\mathrm{3}×\mathrm{7}×\mathrm{37}…
Question Number 213459 by golsendro last updated on 06/Nov/24 $$\:\:\mathrm{Find}\:\mathrm{tupple}\:\mathrm{natural}\:\mathrm{numbers}\:\left(\mathrm{a},\mathrm{b},\mathrm{c}\right) \\ $$$$\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\begin{cases}{\mathrm{max}\left\{\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}+\frac{\mid\mathrm{a}−\mathrm{b}\mid}{\mathrm{2}}\:,\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}}+\frac{\mid\mathrm{b}−\mathrm{c}\mid}{\mathrm{2}}\:,\frac{\mathrm{c}+\mathrm{a}}{\mathrm{2}}+\frac{\mid\mathrm{c}−\mathrm{a}\mid}{\mathrm{2}}\right\}=\mathrm{a}}\\{\mathrm{min}\left\{\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}−\frac{\mid\mathrm{a}−\mathrm{b}\mid}{\mathrm{2}}\:,\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}}−\frac{\mid\mathrm{b}−\mathrm{c}\mid}{\mathrm{2}}\:,\:\frac{\mathrm{c}+\mathrm{a}}{\mathrm{2}}−\frac{\mid\mathrm{c}−\mathrm{a}\mid}{\mathrm{2}}\right\}=\mathrm{b}}\end{cases} \\ $$$$\:\:\mathrm{where}\:\mathrm{a}+\mathrm{b}+\mathrm{c}\:=\:\mathrm{10} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 213486 by a.lgnaoui last updated on 06/Nov/24 $$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{5}} +\mathrm{5}\boldsymbol{\mathrm{x}}−\frac{\mathrm{6}}{\boldsymbol{\mathrm{x}}}=\mathrm{0}\:\:\:\:\:\:\boldsymbol{\mathrm{x}}? \\ $$ Answered by Frix last updated on 06/Nov/24 $${x}^{\mathrm{6}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{6}}…
Question Number 213482 by efronzo1 last updated on 06/Nov/24 Answered by A5T last updated on 06/Nov/24 $$\mathrm{3}\mid{Y}\Rightarrow\mathrm{8}+{c}+\mathrm{2}+\mathrm{3}+{d}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)\Rightarrow{c}+{d}\equiv\mathrm{2}\left({mod}\:\mathrm{3}\right) \\ $$$$\mathrm{3}\mid{X}\Rightarrow\mathrm{9}\mid{Y}\::\: \\ $$$$\mathrm{3}+\mathrm{1}+{a}+\mathrm{5}+{b}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)\Rightarrow\mathrm{8}+{c}+\mathrm{2}+\mathrm{3}+{d}\equiv\mathrm{0}\left({mod}\:\mathrm{9}\right) \\ $$$$\Rightarrow{a}+{b}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)\Rightarrow{c}+{d}\equiv\mathrm{5}\left({mod}\:\mathrm{9}\right) \\ $$$$\mathrm{3}\mid{X}\Rightarrow{c}+{d}=\mathrm{5}\Rightarrow{a}+{b}=\mathrm{0},\mathrm{3},\mathrm{6},\mathrm{9}\:\wedge\:{c}+{d}=\mathrm{5}…
Question Number 213439 by Ari last updated on 05/Nov/24 Commented by BaliramKumar last updated on 06/Nov/24 $${l}\:{and}\:{b}\:=\:? \\ $$$$\mathrm{If}\:\mathrm{square}\:\:\:\:=\:\lceil\frac{\sqrt{\mathrm{2000}}}{\mathrm{7}}\rceil^{\mathrm{2}} =\:\mathrm{49} \\ $$$$ \\ $$ Commented…
Question Number 213430 by hardmath last updated on 05/Nov/24 $$\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{R} \\ $$$$\begin{cases}{\mathrm{x}\:+\:\left[\mathrm{y}\right]\:+\:\left\{\mathrm{z}\right\}\:=\:\mathrm{9},\mathrm{4}}\\{\left[\mathrm{x}\right]\:+\:\left\{\mathrm{y}\right\}\:+\:\mathrm{z}\:=\:\mathrm{11},\mathrm{3}}\\{\left\{\mathrm{x}\right\}\:+\:\mathrm{y}\:+\:\left[\mathrm{z}\right]\:=\:\mathrm{10},\mathrm{5}}\end{cases}\:\:\:\:\:\mathrm{find}:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by A5T last updated on 05/Nov/24 $$\left({i}\right)−\left({ii}\right):\:\left\{{x}\right\}+\left[{y}\right]−\left\{{y}\right\}−\left[{z}\right]=−\mathrm{1}.\mathrm{9}…\left({iv}\right) \\ $$$$\left({iv}\right)−\left({iii}\right)\Rightarrow\left[{y}\right]−\left\{{y}\right\}−{y}−\mathrm{2}\left[{z}\right]=−\mathrm{12}.\mathrm{4} \\…
Question Number 213417 by hardmath last updated on 04/Nov/24 $$\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{c}\:,\:\mathrm{d}\:\in\:\mathbb{N} \\ $$$$\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:+\:\mathrm{d}\:=\:\mathrm{63} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{maksimum}\left(\mathrm{ab}\:+\:\mathrm{bc}\:+\:\mathrm{cd}\right)\:=\:? \\ $$ Answered by Frix last updated on 05/Nov/24 $${d}=\mathrm{63}−{a}−{b}−{c} \\…
Question Number 213391 by York12 last updated on 04/Nov/24 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}\:\mathrm{where}\:{a},{b},{c}\geqslant\mathrm{0} \\ $$$${a}−\mathrm{2}{bc}={b}−\mathrm{2}{ac}={c}−\mathrm{2}{ab} \\ $$$${a}+{b}+{c}=\mathrm{2}\: \\ $$ Answered by Frix last updated on 04/Nov/24 $$\mathrm{Due}\:\mathrm{to}\:\mathrm{symmetry}\:{a}={b}={c}=\frac{\mathrm{2}}{\mathrm{3}} \\…
Question Number 213413 by hardmath last updated on 04/Nov/24 $$\mathrm{Find}: \\ $$$$\mathrm{A}=\:\left[\sqrt{\mathrm{1}}\right]\:+\:\left[\sqrt{\mathrm{2}}\right]\:+\:\left[\sqrt{\mathrm{3}}\:\right]+…+\:\left[\sqrt{\mathrm{323}}\right]\:=\:? \\ $$ Answered by mehdee7396 last updated on 04/Nov/24 $$=\left(\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]+\right)+\left(\left[\mathrm{4}\right]+\left[\mathrm{5}\right]+…+\left[\left[\sqrt{\mathrm{8}}\right]\right)\right. \\ $$$$+\left(\left[\sqrt{\mathrm{9}}\right]+\left[\sqrt{\mathrm{10}}\right]+…+\left[\sqrt{\mathrm{24}}\right]\right)+… \\…