the no. of times of the digits 3 will be written when listing the integers from 1 to 1000

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Any} \mathrm{number} \mathrm{between} 1 \mathrm{and} 999 \mathrm{can} \mathrm{be} \mathrm{expressed} \mathrm{in} \mathrm{the} \mathrm{form} \mathrm{of} \mathrm{abc} \mathrm{where} 0<\mathrm{a},\mathrm{b},\mathrm{c}<9 \\ \mathbf{Case}\mathbf{ }\mathbf{1}\mathbf{:}\mathbf{ }\mathrm{The} \mathrm{numbers} \mathrm{in} \mathrm{which} 3 \mathrm{occurs} \mathrm{only} \mathrm{once}. \mathrm{This} \mathrm{means} \mathrm{that} 3 \mathrm{is} \mathrm{one} \mathrm{of} \mathrm{the} \mathrm{digits} \mathrm{and} \mathrm{the} \\ \mathrm{remaining} \mathrm{two} \mathrm{digits} \mathrm{will} \mathrm{be} \mathrm{any} \mathrm{of} \mathrm{the} \mathrm{other} 9 \mathrm{digits}. \\ \mathrm{We} \mathrm{have} 1\times 9\times 9 = 81 \mathrm{such} \mathrm{numbers}. \\ \mathrm{However}, 3 \mathrm{could} \mathrm{appear} \mathrm{as} \mathrm{the} \mathrm{first} \mathrm{or} \mathrm{the} \mathrm{second} \mathrm{or} \mathrm{the} \mathrm{third} \mathrm{digit}. \\ \mathrm{Therefore}, \mathrm{there} \mathrm{will} \mathrm{be} 3\times 81 = 243 \mathrm{numbers} \left(1 \mathrm{digit}, 2 \mathrm{digits} \mathrm{and} 3 \mathrm{digits}\right) \mathrm{in} \\ \mathrm{which} 3 \mathrm{will} \mathrm{appear} \mathrm{only} \mathrm{once}. \\ \mathbf{Case}\mathbf{ }\mathbf{2}\mathbf{:} \mathrm{The} \mathrm{numbers} \mathrm{in} \mathrm{which} 3 \mathrm{will} \mathrm{appear} \mathrm{twice}. \mathrm{In} \mathrm{these} \mathrm{numbers}, \mathrm{one} \mathrm{of} \mathrm{the} \mathrm{digits} \mathrm{is} \mathrm{not} 3 \\ \mathrm{and} \mathrm{it} \mathrm{can} \mathrm{be} \mathrm{any} \mathrm{of} \mathrm{the} 9 \mathrm{digits}. \mathrm{There} \mathrm{will} \mathrm{be} 9 \mathrm{such} \mathrm{numbers}. \mathrm{However}, \mathrm{this} \mathrm{digit} \mathrm{which} \mathrm{is} \mathrm{not} 3 \\ \mathrm{can} \mathrm{appear} \mathrm{in} \mathrm{the} \mathrm{first} \mathrm{or} \mathrm{second} \mathrm{or} \mathrm{the} \mathrm{third} \mathrm{place}. \mathrm{So} \mathrm{there} \mathrm{are} 3\times 9=27 \mathrm{such} \mathrm{numbers}. \\ \mathrm{In} \mathrm{each} \mathrm{of} \mathrm{these} 27 \mathrm{numbers}, \mathrm{the} \mathrm{digit} 3 \mathrm{is} \mathrm{written} \mathrm{twice}. \mathrm{Therefore}, 3 \mathrm{is} \mathrm{written} 54 \mathrm{times}. \\ \mathbf{Case}\mathbf{ }\mathbf{3}\mathbf{:} \mathrm{The} \mathrm{number} \mathrm{in} \mathrm{which} 3 \mathrm{appears} \mathrm{thrice} \mathrm{is} 333 \mathrm{i}.\mathrm{e}. 1 \mathrm{number}. 3 \mathrm{is} \mathrm{written} \mathrm{thrice} \mathrm{in} \mathrm{it}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{times} \mathrm{the} \mathrm{digit} 3 \mathrm{is} \mathrm{written} \mathrm{between} 1 \mathrm{and} 999 \mathrm{is} \\ 243 + 54 + 3 = \mathbf{300}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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