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Question Number 175750 by Shrinava last updated on 06/Sep/22

Answered by mr W last updated on 06/Sep/22

1 time 1  2 times 2  3 times 3  ...  n−1 times n−1  n times n  number of numbers before the first  number n is 1+2+3+...+(n−1), so  the first number n is the x^(th)  number.   x=1+2+3+...+(n−1)+1=((n(n−1))/2)+1  ((n(n−1))/2)+1≥2005  n(n−1)≥4008 ⇒n≥64  i.e. the 2005^(th)  number is 64

$$\mathrm{1}\:{time}\:\mathrm{1} \\ $$$$\mathrm{2}\:{times}\:\mathrm{2} \\ $$$$\mathrm{3}\:{times}\:\mathrm{3} \\ $$$$... \\ $$$${n}−\mathrm{1}\:{times}\:{n}−\mathrm{1} \\ $$$${n}\:{times}\:{n} \\ $$$${number}\:{of}\:{numbers}\:{before}\:{the}\:{first} \\ $$$${number}\:{n}\:{is}\:\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\left({n}−\mathrm{1}\right),\:{so} \\ $$$${the}\:{first}\:{number}\:{n}\:{is}\:{the}\:{x}^{{th}} \:{number}.\: \\ $$$${x}=\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\left({n}−\mathrm{1}\right)+\mathrm{1}=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}+\mathrm{1} \\ $$$$\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}+\mathrm{1}\geqslant\mathrm{2005} \\ $$$${n}\left({n}−\mathrm{1}\right)\geqslant\mathrm{4008}\:\Rightarrow{n}\geqslant\mathrm{64} \\ $$$${i}.{e}.\:{the}\:\mathrm{2005}^{{th}} \:{number}\:{is}\:\mathrm{64} \\ $$

Commented by Shrinava last updated on 06/Sep/22

thank you so much dear professor  but how did it happen without  mentioning +1

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{professor} \\ $$$$\mathrm{but}\:\mathrm{how}\:\mathrm{did}\:\mathrm{it}\:\mathrm{happen}\:\mathrm{without} \\ $$$$\mathrm{mentioning}\:+\mathrm{1} \\ $$

Commented by mr W last updated on 06/Sep/22

the first number n is the  [1+2+3+...+(n−1)+1]^(th)  number.

$${the}\:{first}\:{number}\:{n}\:{is}\:{the} \\ $$$$\left[\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\left({n}−\mathrm{1}\right)+\mathrm{1}\right]^{{th}} \:{number}. \\ $$

Commented by Shrinava last updated on 06/Sep/22

thank you dear professor

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$

Commented by Tawa11 last updated on 15/Sep/22

Great sir.

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

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