Question Number 55482 by pooja24 last updated on 25/Feb/19
$${A}\:{group}\:{of}\:{n}\:{students}\:{are}\:{numbered} \\ $$$${continously}\:{from}\:{first}\:{sdtudent}\:{as}\:\mathrm{1},\mathrm{2},\mathrm{3},...... \\ $$$${If}\:\mathrm{1101}\:{digits}\:{had}\:{to}\:{be}\:{used}\:{in}\:{all},{what}\:{is}\:{the}\: \\ $$$${number}\:{of}\:{students}\:{in}\:{the}\:{group} \\ $$
Answered by mr W last updated on 25/Feb/19
$$\mathrm{1},\mathrm{2},\mathrm{3}...,\mathrm{9}\Rightarrow\Sigma=\mathrm{9}×\mathrm{1}=\mathrm{9}\:{digits} \\ $$$$\mathrm{10},\mathrm{11},\mathrm{12},...,\mathrm{99}\Rightarrow\Sigma=\mathrm{90}×\mathrm{2}=\mathrm{180}\:{digits} \\ $$$$\mathrm{100},\mathrm{101},\mathrm{102},...,\mathrm{999}\Rightarrow\Sigma=\mathrm{900}×\mathrm{3}=\mathrm{2700}\:{digits} \\ $$$${since}\:{totally}\:\mathrm{1101}\:{digits}\:{are}\:{used}, \\ $$$${n}\:{must}\:{be}\:{a}\:\mathrm{3}−{digit}\:{number} \\ $$$$\mathrm{9}+\mathrm{180}+\left({n}−\mathrm{99}\right)×\mathrm{3}=\mathrm{1101} \\ $$$$\Rightarrow{n}=\mathrm{403} \\ $$$${i}.{e}.\:{there}\:{are}\:\mathrm{403}\:{students}\:{in}\:{the}\:{group}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 25/Feb/19
$${excellent}\:{sir}... \\ $$
Commented by mr W last updated on 25/Feb/19
$${thanks}\:{sir}! \\ $$$${if}\:{m}={f}\left({n}\right)\:{is}\:{the}\:{total}\:{number}\:{of}\:{digits} \\ $$$${from}\:\mathrm{1}\:{to}\:{n}.\:{is}\:{it}\:{possible}\:{to}\:{find}\:{a} \\ $$$${formula}\:{for}\:{f}\left({n}\right)\:{in}\:{terms}\:{of}\:{n}? \\ $$$${furthermore}\:{find}\:{n}={f}^{−\mathrm{1}} \left({m}\right)\:{if}\: \\ $$$${possible}. \\ $$