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Question Number 43787 by kkr1729 last updated on 15/Sep/18

$$\mathrm{The}\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{numbers}\mathrm{of}\mathrm{not}$$$$\mathrm{more}\mathrm{than}20\mathrm{digits}\mathrm{that}\mathrm{are}\mathrm{formed}$$$$\mathrm{by}\mathrm{using}\mathrm{the}\mathrm{digits}0,1,2,3\mathrm{and}4\mathrm{is}$$

Answered by MrW3 last updated on 17/Sep/18

$$letnumberofnumberswithndigits=S\left(n\right)$$$$S\left(1\right)=4$$$$S\left(2\right)=4\times 5=4\times 5^1$$$$S\left(3\right)=4\times 5\times 5=4\times 5^2$$$$S\left(4\right)=4\times 5\times 5\times 5=4\times 5^3$$$$S\left(n\right)=4\times 5^{n-1}$$$$S\left(20\right)=4\times 5^{19}$$$$\underset{n=1}{\sum ^{20}}S\left(n\right)=\frac{4\times (5^{20}-1)}{5-1}=5^{20}-1$$$$$$$$Ifnot5digits(0,1,2...4)but10digits$$$$(0,1,2....9)areused,weknowthereare$$$${10}^{20}-1numberswithmax.20digits:$$$$1,2,3,...,\underset{20times9}{9999....9}$$

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