Question Number 2381 by Yozzi last updated on 18/Nov/15 $${Of}\:{the}\:{numbers}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:…\:,\:\mathrm{6000}, \\ $$$${how}\:{many}\:{are}\:{not}\:{multiples}\:{of}\:\mathrm{2},\:\mathrm{3}\:{or}\:\mathrm{5}? \\ $$ Commented by 123456 last updated on 18/Nov/15 $$\mathrm{N}=\mathrm{N}_{\mathrm{2}} +\mathrm{N}_{\mathrm{3}} +\mathrm{N}_{\mathrm{5}} −\mathrm{N}_{\mathrm{2},\mathrm{3}}…
Question Number 133412 by EDWIN88 last updated on 22/Feb/21 $$\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{2}^{\mathrm{222}} −\mathrm{1}\:? \\ $$ Answered by liberty last updated on 22/Feb/21 $$\mathrm{2}^{\mathrm{10}} =\mathrm{1024}\equiv\mathrm{24}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{2}^{\mathrm{20}} \equiv\mathrm{24}^{\mathrm{2}}…
Question Number 133320 by bramlexs22 last updated on 21/Feb/21 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{n}\:\mathrm{for}\:\mathrm{which}\:\mathrm{n}^{\mathrm{2}} +\mathrm{2n}+\mathrm{4}\: \\ $$$$\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7}\: \\ $$ Answered by EDWIN88 last updated on 21/Feb/21 $$\mathrm{let}\:\mathrm{n}\:=\:\mathrm{7k}+\mathrm{r}\:\mathrm{then}\:\mathrm{n}^{\mathrm{2}} +\mathrm{2n}+\mathrm{4}\:=\:\left(\mathrm{7k}+\mathrm{r}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{7k}+\mathrm{r}\right)+\mathrm{4}…
Question Number 2148 by Yozzi last updated on 04/Nov/15 $${Is}\:\mathrm{3}^{\mathrm{2015}} −\mathrm{2}^{\mathrm{2015}} \:{prime}? \\ $$ Answered by sudhanshur last updated on 04/Nov/15 $$\left(\mathrm{3}^{\mathrm{5}} \right)^{\mathrm{403}} −\left(\mathrm{2}^{\mathrm{5}} \right)^{\mathrm{403}}…
Question Number 1942 by Yozzi last updated on 25/Oct/15 $${Let}\:{N}\:{be}\:{a}\:{positive}\:{integer}\:{with}\:{prime} \\ $$$${factorisation}\: \\ $$$$\:\:{N}={p}_{\mathrm{1}} ^{{m}_{\mathrm{1}} } {p}_{\mathrm{2}} ^{{m}_{\mathrm{2}} } {p}_{\mathrm{3}} ^{{m}_{\mathrm{3}} } ×…×{p}_{{n}−\mathrm{1}} ^{{m}_{{n}−\mathrm{1}} }…
Question Number 1895 by Yozzy last updated on 22/Oct/15 $${Let}\:{us}\:{generalise}\:{the}\:{result}\:{of}\:{taking}\:{the}\:{inverse}\:{tangent}\:{of}\:{a}\:{complex}\:{number} \\ $$$${to}\:{the}\:{form}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{tan}^{−\mathrm{1}} \left({c}+{id}\right)={a}+{ib} \\ $$$${where}\:{a},{b},{c},{d}\in\mathbb{R}\:{and}\:{i}=\sqrt{−\mathrm{1}}.\:{Determine}\:{a}\:{and}\:{b}\:{respectively}\:{in}\:{terms} \\ $$$${of}\:{c}\:{and}\:{d}.\: \\ $$ Commented by Rasheed Soomro…
Question Number 1776 by 123456 last updated on 20/Sep/15 $$\boldsymbol{\mathrm{P}}_{{k}} =\left\{{x}\in\mathbb{N},{n}\in\mathbb{N}:{x}>\mathrm{0},\underset{{n}\mid{x}} {\sum}\mathrm{1}={k}\right\} \\ $$$$\boldsymbol{\mathrm{C}}=\underset{{k}\geqslant\mathrm{3}} {\cup}\boldsymbol{\mathrm{P}}_{{k}} \\ $$$$\left\{\mathrm{0},\mathrm{1}\right\}\cup\boldsymbol{\mathrm{C}}\cup\mathbb{P}=\mathbb{N} \\ $$$$\mathrm{proof}\:\mathrm{or}\:\mathrm{give}\:\mathrm{a}\:\mathrm{counter}\:\mathrm{example}\:\mathrm{that} \\ $$$$\left({x},{y},{z}\right)\in\boldsymbol{\mathrm{C}}^{\mathrm{3}} ,{x}^{\mathrm{2}} \mid{yz}\Rightarrow{x}\mid{y}\vee{x}\mid{z} \\ $$…
Question Number 1700 by 112358 last updated on 01/Sep/15 $${Show}\:{that}\:\left(\mathrm{7}!\right)^{\frac{\mathrm{1}}{\mathrm{7}}} <\left(\mathrm{8}!\right)^{\frac{\mathrm{1}}{\mathrm{8}}} . \\ $$$${Also}\:{show}\:{that}\: \\ $$$$\sqrt{\mathrm{100001}}−\sqrt{\mathrm{100000}}<\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{100000}}}\:. \\ $$$$ \\ $$ Commented by 123456 last updated…
Question Number 67193 by Rasheed.Sindhi last updated on 23/Aug/19 Commented by Rasheed.Sindhi last updated on 24/Aug/19 $$\:\underset{−} {\:\:\:\:\:\:\:\:\:\:\mathbb{S}\mathrm{ome}\:\mathbb{C}\mathrm{ounting}\:\mathbb{P}\mathrm{roblems}\:\:\:\:\:\:\:\:\:\:} \\ $$$$\left({a}\right){How}\:{many}\:{rectangles}\:{are}\:{in}\:{the} \\ $$$$\:\:\:\:\:\:\:{above}\:{picture}? \\ $$$$\left({b}\right){How}\:{many}\:{rectangles}\:{contain} \\…
Question Number 132386 by liberty last updated on 13/Feb/21 $$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\begin{cases}{\mathrm{ln}\:{i}}\\{\mathrm{ln}\:\left(\mathrm{3}+\mathrm{4}{i}\right)}\end{cases} \\ $$ Answered by EDWIN88 last updated on 13/Feb/21 $$\left(\mathrm{1}\right)\mathrm{ln}\:{i}\:=\:\mathrm{ln}\:\mid{i}\mid\:+{i}\:\left(\mathrm{arg}\:\left({i}\right)+\mathrm{2}{n}\pi\right)=\mathrm{ln}\:\mathrm{1}+{i}\:\left(\frac{\pi}{\mathrm{2}}+\mathrm{2n}\pi\right) \\ $$$$\left(\mathrm{2}\right)\mathrm{ln}\:\left(\mathrm{3}+\mathrm{4}{i}\right)=\mathrm{ln}\:\mid\mathrm{3}+\mathrm{4}{i}\mid=\mathrm{ln}\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }\:+{i}\left(\mathrm{arctan}\:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)+\mathrm{2}{n}\pi\right) \\…