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Question Number 116360 by bemath last updated on 03/Oct/20

$$\mathrm{A}\mathrm{five}\mathrm{digits}\mathrm{number}\mathrm{divisible}\mathrm{by}3$$$$\mathrm{is}\mathrm{to}\mathrm{be}\mathrm{formed}\mathrm{using}\mathrm{the}\mathrm{number}$$$$0,1,2,3,4\mathrm{and}5\mathrm{without}\mathrm{repetition}$$$$\mathrm{The}\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{this}\mathrm{can}$$$$\mathrm{be}\mathrm{done}\mathrm{is}\mathrm{\_}\mathrm{\_}$$

Answered by mr W last updated on 03/Oct/20

$$0+1+2+3+4=10\not\equiv 0mod3$$$$0+1+2+3+5=11\not\equiv 0mod3$$$$0+1+2+4+5=12\equiv 0mod3✓$$$$0+1+3+4+5=13\not\equiv 0mod3$$$$0+2+3+4+5=14\not\equiv 0mod3$$$$1+2+3+4+5=15\equiv 0mod3✓$$$$with0,1,2,4,5wecanform$$$$4\times 4\times 3\times 2\times 1=96numbers$$$$with1,2,3,4,5wecanform$$$$5!=120numbers$$$$$$$$\Rightarrow total96+120=216numbers$$

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