Question-149439

Question Number 149439 by SLVR last updated on 05/Aug/21

Commented by SLVR last updated on 05/Aug/21

$${solution}\:{please} \\ $$

Answered by mr W last updated on 05/Aug/21

m(n)=number of digits of number n m(n)=1 for 1≤n≤9 m(n)=2 for 10≤n≤99 m(n)=3 for 100≤n≤999 m(n)=4 for 1000≤n≤9999 S(k)=Σ_(n=1) ^k m(n) S(9)=9×1=9 S(99)=9×1+90×2=189 S(999)=9×1+90×2+900×3=2889>2021 S(2021)=2889+(2021−999)×4=6977 ⇒the resulting number has 6977 digits. ⌊((2021−189)/3)⌋=610 99+610=709 189+610×3=2019 the 2021^(st) digit is the second digit from the number 710, i.e. 1. ⇒answer B)

$${m}\left({n}\right)={number}\:{of}\:{digits}\:{of}\:{number}\:{n} \\ $$$${m}\left({n}\right)=\mathrm{1}\:{for}\:\mathrm{1}\leqslant{n}\leqslant\mathrm{9} \\ $$$${m}\left({n}\right)=\mathrm{2}\:{for}\:\mathrm{10}\leqslant{n}\leqslant\mathrm{99} \\ $$$${m}\left({n}\right)=\mathrm{3}\:{for}\:\mathrm{100}\leqslant{n}\leqslant\mathrm{999} \\ $$$${m}\left({n}\right)=\mathrm{4}\:{for}\:\mathrm{1000}\leqslant{n}\leqslant\mathrm{9999} \\ $$$${S}\left({k}\right)=\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}{m}\left({n}\right) \\ $$$${S}\left(\mathrm{9}\right)=\mathrm{9}×\mathrm{1}=\mathrm{9} \\ $$$${S}\left(\mathrm{99}\right)=\mathrm{9}×\mathrm{1}+\mathrm{90}×\mathrm{2}=\mathrm{189} \\ $$$${S}\left(\mathrm{999}\right)=\mathrm{9}×\mathrm{1}+\mathrm{90}×\mathrm{2}+\mathrm{900}×\mathrm{3}=\mathrm{2889}>\mathrm{2021} \\ $$$${S}\left(\mathrm{2021}\right)=\mathrm{2889}+\left(\mathrm{2021}−\mathrm{999}\right)×\mathrm{4}=\mathrm{6977} \\ $$$$\Rightarrow{the}\:{resulting}\:{number}\:{has}\:\mathrm{6977} \\ $$$${digits}. \\ $$$$\lfloor\frac{\mathrm{2021}−\mathrm{189}}{\mathrm{3}}\rfloor=\mathrm{610} \\ $$$$\mathrm{99}+\mathrm{610}=\mathrm{709} \\ $$$$\mathrm{189}+\mathrm{610}×\mathrm{3}=\mathrm{2019} \\ $$$${the}\:\mathrm{2021}^{{st}} \:{digit}\:{is}\:{the}\:{second}\:{digit} \\ $$$${from}\:{the}\:{number}\:\mathrm{710},\:{i}.{e}.\:\mathrm{1}. \\ $$$$\left.\Rightarrow{answer}\:{B}\right) \\ $$

Commented by Rasheed.Sindhi last updated on 05/Aug/21