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Question Number 26473 by NECx last updated on 25/Dec/17
$${How}\:{many}\:{numbers}\:{less}\:{than} \\ $$$$\mathrm{1000}\:{and}\:{divisible}\:{by}\:\mathrm{5}\:{can}\:{be} \\ $$$${formed}\:{with}\:{the}\:{digits}\:\mathrm{0},\:\mathrm{1},\:\mathrm{2}\:,\mathrm{3}\:,\mathrm{4}\:,\mathrm{5}\: \\ $$$$\mathrm{6}\:,\mathrm{7}\:,\mathrm{8}\:,\mathrm{9},{each}\:{digit}\:{not}\:{occuring} \\ $$$${more}\:{than}\:{once}\:{in}\:{each}\:{number}? \\ $$
Answered by mrW1 last updated on 26/Dec/17
$${one}\:{digit}\:{numbers}: \\ $$$${N}_{\mathrm{1}} =\mathrm{1}\:\left({only}\:\mathrm{5},\:{without}\:{the}\:{number}\:\mathrm{0}\right) \\ $$$${two}\:{digits}\:{numbers}: \\ $$$${type}\:{X}\mathrm{0}:\:\Rightarrow\mathrm{9} \\ $$$${type}\:{X}\mathrm{5}:\:\Rightarrow\mathrm{8} \\ $$$$\Rightarrow{N}_{\mathrm{2}} =\mathrm{9}+\mathrm{8}=\mathrm{17} \\ $$$${three}\:{digits}\:{numbers}: \\ $$$${type}\:{XX}\mathrm{0}:\:\Rightarrow\mathrm{9}×\mathrm{8}=\mathrm{72} \\ $$$${type}\:{XX}\mathrm{5}:\:\Rightarrow\mathrm{8}×\mathrm{8}=\mathrm{64} \\ $$$$\Rightarrow{N}_{\mathrm{3}} =\mathrm{72}+\mathrm{64}=\mathrm{136} \\ $$$$ \\ $$$$\Rightarrow{N}=\mathrm{1}+\mathrm{17}+\mathrm{136}=\mathrm{154} \\ $$
Commented by NECx last updated on 26/Dec/17
$${ok}\:{sir} \\ $$
Commented by NECx last updated on 26/Dec/17
$${thanks}\:{so}\:{much}\:{sir}.... \\ $$$$ \\ $$$${I}\:{got}\:\mathrm{164}\:{but}\:{the}\:{book}\:{got}\:\mathrm{155} \\ $$
Commented by mrW1 last updated on 26/Dec/17
$${I}\:{have}\:{excluded}\:{the}\:{number}\:{zero}.\:{If}\:{zero} \\ $$$${is}\:{also}\:{taken}\:{into}\:{account}\:{then}\:{the} \\ $$$${result}\:{is}\:\mathrm{155}. \\ $$