Question Number 175750 by Shrinava last updated on 06/Sep/22 | ||
Answered by mr W last updated on 06/Sep/22 | ||
$$\mathrm{1}\:{time}\:\mathrm{1} \\ $$$$\mathrm{2}\:{times}\:\mathrm{2} \\ $$$$\mathrm{3}\:{times}\:\mathrm{3} \\ $$$$... \\ $$$${n}−\mathrm{1}\:{times}\:{n}−\mathrm{1} \\ $$$${n}\:{times}\:{n} \\ $$$${number}\:{of}\:{numbers}\:{before}\:{the}\:{first} \\ $$$${number}\:{n}\:{is}\:\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\left({n}−\mathrm{1}\right),\:{so} \\ $$$${the}\:{first}\:{number}\:{n}\:{is}\:{the}\:{x}^{{th}} \:{number}.\: \\ $$$${x}=\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\left({n}−\mathrm{1}\right)+\mathrm{1}=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}+\mathrm{1} \\ $$$$\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}+\mathrm{1}\geqslant\mathrm{2005} \\ $$$${n}\left({n}−\mathrm{1}\right)\geqslant\mathrm{4008}\:\Rightarrow{n}\geqslant\mathrm{64} \\ $$$${i}.{e}.\:{the}\:\mathrm{2005}^{{th}} \:{number}\:{is}\:\mathrm{64} \\ $$ | ||
Commented by Shrinava last updated on 06/Sep/22 | ||
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{professor} \\ $$$$\mathrm{but}\:\mathrm{how}\:\mathrm{did}\:\mathrm{it}\:\mathrm{happen}\:\mathrm{without} \\ $$$$\mathrm{mentioning}\:+\mathrm{1} \\ $$ | ||
Commented by mr W last updated on 06/Sep/22 | ||
$${the}\:{first}\:{number}\:{n}\:{is}\:{the} \\ $$$$\left[\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\left({n}−\mathrm{1}\right)+\mathrm{1}\right]^{{th}} \:{number}. \\ $$ | ||
Commented by Shrinava last updated on 06/Sep/22 | ||
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$ | ||
Commented by Tawa11 last updated on 15/Sep/22 | ||
$$\mathrm{Great}\:\mathrm{sir}. \\ $$ | ||