Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

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

Commented by SLVR last updated on 05/Aug/21

solution please

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

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

Wonderful  mr W sir!

$$\mathbb{W}\mathrm{onderful}\:\:\mathrm{mr}\:\mathbb{W}\:\boldsymbol{\mathrm{sir}}! \\ $$

Commented by SLVR last updated on 05/Aug/21

Thanks..Mr.W..so kind of sir..  extreemly nice ..

$${Thanks}..{Mr}.{W}..{so}\:{kind}\:{of}\:{sir}.. \\ $$$${extreemly}\:{nice}\:.. \\ $$$$ \\ $$$$ \\ $$

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

thanks for reviewing sirs!

$${thanks}\:{for}\:{reviewing}\:{sirs}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com