Question Number 80645 by mr W last updated on 05/Feb/20 $${if} \\ $$$${a}_{\mathrm{1}} =\mathrm{2} \\ $$$${a}_{{n}+\mathrm{1}} ={a}_{{n}} ^{\mathrm{2}} −\mathrm{1} \\ $$$${find}\:{a}_{{n}} =? \\ $$ Commented…
Question Number 80643 by mr W last updated on 05/Feb/20 $${Prove}\:{that}\:{there}\:{is}\:{no}\:{solution}\:{for} \\ $$$$\mathrm{7}^{{n}} \equiv\mathrm{1}\:{mod}\left(\mathrm{35}\right)\:{with}\:{n}\in\mathbb{N}. \\ $$ Answered by MJS last updated on 05/Feb/20 $$\forall{k}\in\mathbb{N}:\:\mathrm{7}^{\mathrm{20}{k}} \equiv\mathrm{1mod}\:\mathrm{55}…
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Question Number 15051 by tawa tawa last updated on 07/Jun/17 $$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$ \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:…………\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} \:=\:\mathrm{92}\:\:\:\:\:\:\:\:\:……….\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{x}\:−\:\mathrm{y}\:=\:\mathrm{z}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:………..\:\mathrm{equation}\:\left(\mathrm{iii}\right) \\ $$ Answered…
Question Number 80580 by mr W last updated on 04/Feb/20 $${Find}\:{general}\:{solution}\:{for}\:{k}\:{such}\:{that} \\ $$$$\mathrm{7}^{{k}} \equiv\mathrm{1}\:{mod}\:\left(\mathrm{35}\right) \\ $$ Answered by Rio Michael last updated on 04/Feb/20 $$\:{k}\:=\:\mathrm{8}{n}\:,\:{n}\:\in\:\mathbb{N}…
Question Number 146096 by mathdanisur last updated on 10/Jul/21 Answered by mindispower last updated on 10/Jul/21 $${tg}\left(\mathrm{2}{a}\right)={tg}\left({a}+{b}+{a}−{b}\right)=\frac{{tg}\left({a}+{b}\right)+{tg}\left({a}−{b}\right)}{\mathrm{1}−{tg}\left({a}+{b}\right){tg}\left({a}−{b}\right)}=\frac{\frac{\mathrm{3}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\mathrm{3}\Rightarrow\mathrm{2}{tg}\left(\mathrm{2}{a}\right)=\mathrm{6} \\ $$ Commented by mathdanisur last…
Question Number 80543 by Power last updated on 04/Feb/20 Commented by mr W last updated on 04/Feb/20 $$\mathrm{1010} \\ $$ Commented by Power last updated…
Question Number 14988 by 433 last updated on 06/Jun/17 $${Solve}\:{on}\:\mathbb{Z}_{\mathrm{4}} \: \\ $$$${ax}+{b}=\left[\mathrm{0}\right]_{\mathrm{4}} \:\:{a},{b}\in\mathbb{Z}_{\mathrm{4}} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\left[\mathrm{0}\right]_{\mathrm{4}} \:\:{a},{b},{c}\in\mathbb{Z}_{\mathrm{4}} \\ $$ Commented by prakash jain last…