Question Number 126795 by mathocean1 last updated on 24/Dec/20 $${N}\:{is}\:{a}\:{number}\:\in\:\mathbb{N}\:{which}\:{has}\:{three} \\ $$$${digits}\:{and}\:{written}\:{xyz}\:{in}\:{base}\:\mathrm{10}\: \\ $$$${such}\:{that}\:\begin{cases}{{xy}+{xz}+{yz}={xyz}}\\{\mathrm{0}<{x}<{y}<{z}}\end{cases} \\ $$$${Determinate}\:{N}. \\ $$ Commented by JDamian last updated on 24/Dec/20…
Question Number 192331 by Mingma last updated on 14/May/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 192266 by Shlock last updated on 13/May/23 Answered by Frix last updated on 13/May/23 $$\mathrm{Since}\:{a}\in\mathbb{N}\wedge\mathrm{0}\leqslant{a}\leqslant\mathrm{9}\:\mathrm{it}'\mathrm{s}\:\mathrm{best}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{for} \\ $$$${n}\:\mathrm{and}\:\mathrm{try}: \\ $$$$\mathrm{2}{n}^{\mathrm{2}} +\mathrm{14}{n}+\mathrm{83}=\mathrm{1111}{a} \\ $$$${n}=\frac{−\mathrm{7}+\sqrt{\mathrm{2222}{a}−\mathrm{117}}}{\mathrm{2}} \\…
Question Number 192208 by Shlock last updated on 11/May/23 Answered by witcher3 last updated on 11/May/23 $$\Rightarrow\forall\left(\mathrm{7k}+\mathrm{r}\right)\in\mathbb{N}\Rightarrow\exists\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)\in\left[\mathrm{7k}+\mathrm{r},\mathrm{7k}+\mathrm{r}+\mathrm{6}\right] \\ $$$$\mathrm{r}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\right\} \\ $$$$\mathrm{a}\equiv\mathrm{r}+\mathrm{1}\left[\mathrm{7}\right];\mathrm{a}=\mathrm{7k}+\mathrm{r}+\mathrm{1} \\ $$$$\mathrm{b}\equiv\mathrm{r}+\mathrm{2}\left[\mathrm{7}\right];\mathrm{b}=\mathrm{7k}+\mathrm{r}+\mathrm{2} \\ $$$$\mathrm{c}\equiv\mathrm{r}+\mathrm{4}\left[\mathrm{7}\right];\mathrm{c}=\mathrm{7k}+\mathrm{r}+\mathrm{4}…
Question Number 192187 by alcohol last updated on 10/May/23 $$\begin{cases}{{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:=\:{a}}\\{{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \:=\:{b}}\\{{z}_{\mathrm{3}} \:+\:{z}_{\mathrm{4}} \:=\:{c}}\\{{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{4}} \:=\:{d}}\end{cases} \\ $$$${solve}\:{using}\:{gaussian}\:{elimination} \\ $$ Terms of Service…
Question Number 61096 by Arthur El-bomart last updated on 29/May/19 $$\forall\:{a},\:{n}\:\in\:{N}\::\:\mid{a}−{n}\mid=\mathrm{1}\:{pour}\:{a},\:{n}\:\geqslant\mathrm{3} \\ $$$${a}^{{m}} \equiv\mathrm{1}{modn}\:\left(\ast\right) \\ $$$${posons}\::\:{m}={n}−\mathrm{1}\:\left(\ast'\right) \\ $$$${subtituons}\:{cette}\:{valeur}\:{dans}\:\left(\ast\right). \\ $$$${on}\:{a}:\:{a}^{{n}−\mathrm{1}} \equiv\mathrm{1}{modn}.\:{Mais}\:{n}\:{n}'{est}\:{pas}\:{forcement}\:{premier}. \\ $$$${Test}\:{de}\:{primalite} \\ $$$$\forall\:{n}\:\in\:{N},\:{n}\:\geqslant\mathrm{3}.…
Question Number 191966 by Shlock last updated on 04/May/23 Answered by mr W last updated on 04/May/23 $${y}={x}^{{x}^{…} } ={x}^{{y}} \\ $$$${y}={e}^{{y}\mathrm{ln}\:{x}} \\ $$$$\left(−{y}\mathrm{ln}\:{x}\right){e}^{−{y}\mathrm{ln}\:{x}} =−\mathrm{ln}\:{x}…
Question Number 60853 by Tony Lin last updated on 26/May/19 $$\sqrt{\mathrm{5}−\mathrm{12}{i}}+\sqrt{\mathrm{5}+\mathrm{12}{i}}=? \\ $$ Answered by tanmay last updated on 26/May/19 $$\sqrt{\mathrm{9}−\mathrm{4}−\mathrm{12}{i}}\:+\sqrt{\mathrm{9}−\mathrm{4}+\mathrm{12}{i}}\: \\ $$$$\sqrt{\left(\mathrm{3}−\mathrm{2}{i}\right)^{\mathrm{2}} }\:+\sqrt{\left(\mathrm{3}+\mathrm{2}{i}\right)^{\mathrm{2}} }\:…
Question Number 191831 by Shlock last updated on 01/May/23 Answered by mehdee42 last updated on 02/May/23 $${suppose}\:,\:\:{n}=<{abcdefghij}>\:,{is}\:{an}\:<{i}.{n}> \\ $$$${a}+{b}+{c}+…+{i}+{j}\overset{\mathrm{9}} {\equiv}\mathrm{0}\Rightarrow\mathrm{9}\mid{n} \\ $$$$\mathrm{9}\mid{n}\:,\:\mathrm{11111}\mid{n}\:\:,\:\left(\mathrm{9},\mathrm{11111}\right)=\mathrm{1}\Rightarrow\mathrm{99999}\mid{n} \\ $$$${let}\:\:{x}=<{abcde}>\:\:\&\:\:{y}={f}<{fghij}> \\…
Question Number 191821 by mehdee42 last updated on 01/May/23 $$\:{Q}\:\blacktriangleright\:{Show}\:{that}: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} \frac{\mathrm{1}}{{i}}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{i}+{n}} \\ $$ Answered by mehdee42 last updated on…