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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

$$1time1$$$$2times2$$$$3times3$$$$...$$$$n-1timesn-1$$$$ntimesn$$$$numberofnumbersbeforethefirst$$$$numbernis1+2+3+...+(n-1),so$$$$thefirstnumbernisthe{x}^{th}number.$$$$x=1+2+3+...+(n-1)+1=\frac{n(n-1)}{2}+1$$$$\frac{n(n-1)}{2}+1⩾2005$$$$n(n-1)⩾4008\Rightarrow n⩾64$$$$i.e.the{2005}^{th}numberis64$$

Commented by Shrinava last updated on 06/Sep/22

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

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

$$thefirstnumbernisthe$$$${[1+2+3+...+(n-1)+1]}^{th}number.$$

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

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

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