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Question Number 116091 by I want to learn more last updated on 30/Sep/20

$$\mathrm{Is}\mathrm{there}\mathrm{a}\mathrm{formular}\mathrm{to}\mathrm{tell}\mathrm{how}\mathrm{many}\mathrm{times}\mathrm{a}\mathrm{digit}\mathrm{occur}\mathrm{in}\mathrm{an}\mathrm{interval}.$$$$$$$$\mathrm{e}.\mathrm{g}.\mathrm{How}\mathrm{many}\mathrm{times}\mathrm{digits}2\mathrm{occur}\mathrm{between}1-100$$

Answered by 1549442205PVT last updated on 01/Oct/20

$$\mathrm{In}\mathrm{each}\mathrm{tens}\mathrm{the}\mathrm{digits}2\mathrm{appears}\mathrm{one}$$$$\mathrm{time},\mathrm{only}\mathrm{tens}\mathrm{from}20\mathrm{to}30\mathrm{digits}2$$$$\mathrm{appears}\mathrm{two}\mathrm{times}.\mathrm{Therefore},\mathrm{from}1$$$$\mathrm{to}100\mathrm{digts}2\mathrm{appears}11\mathrm{times}$$$$\mathrm{The}\mathrm{other}\mathrm{digits}\mathrm{are}\mathrm{the}\mathrm{same}$$

Commented by I want to learn more last updated on 01/Oct/20

$$\mathrm{Thanks}\mathrm{sir}$$

Commented by 1549442205PVT last updated on 01/Oct/20

$$\mathrm{You}\mathrm{are}\mathrm{welcome}$$

Commented by mr W last updated on 02/Oct/20

$$2\Rightarrow onetime2$$$$12,22,32,...,92\Rightarrow 10times2$$$$20,21,23,...,29\Rightarrow 9times2$$$$\Rightarrow totally1+10+9=20times2$$

Answered by mr W last updated on 01/Oct/20

$$ndigitnumbers:$$$${10}^{n-1}+(n-1)\times 9\times {10}^{n-2}$$$$=(9n+1){10}^{n-2}$$$$$$$$from1to100:$$$$1+19\times 1=20times$$

Commented by I want to learn more last updated on 06/Oct/20

$$\mathrm{I}\mathrm{just}\mathrm{see}\mathrm{this}\mathrm{now}\mathrm{sir},\mathrm{so}\mathrm{it}\mathrm{will}\mathrm{work}\mathrm{for}1\mathrm{to}50\mathrm{too}\mathrm{or}$$$$\mathrm{another}\mathrm{nth}\mathrm{derivative}$$

Commented by I want to learn more last updated on 06/Oct/20

$$\mathrm{And}\mathrm{sorry}\mathrm{for}\mathrm{bordering}\mathrm{you}\mathrm{sir},$$$$\mathrm{How}\mathrm{to}\mathrm{generalise}\mathrm{it}\mathrm{as}\mathrm{you}\mathrm{solve},\mathrm{so}\mathrm{that}\mathrm{i}\mathrm{can}\mathrm{think}\mathrm{too}\mathrm{on}\mathrm{my}\mathrm{own}.$$$$\mathrm{Thanks}\mathrm{sir}.$$

Commented by I want to learn more last updated on 06/Oct/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.$$

Commented by mr W last updated on 06/Oct/20

$${let}^{\prime }slookatnumberswithmdigits$$$$xxxx...x$$$$thedigit2(oranyothernon-zero$$$$digit)canbeatfirstposition$$$$2xxx..x$$$$thereare10\times 10\times ..\times 10={10}^{m-1}$$$$suchnumbers,i.e.digit2comes$$$${10}^{m-1}timesatthisposition.$$$$whenthedigit2isatotherpositions,$$$$x2xx..x(m-1possiblepositions)$$$$thereare9\times 10\times 10\times ...10=9\times {10}^{m-2}$$$$suchnumbers,i.e.digit2comes$$$$(m-1)\times 9\times {10}^{m-2}times.$$$$$$$$alltogether:$$$${10}^{m-1}+(m-1)\times 9\times {10}^{m-2}$$$$={10}^{m-2}[10+(m-1)\times 9]$$$$=(9m+1)\times {10}^{m-2}$$

Commented by I want to learn more last updated on 06/Oct/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

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