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Question Number 108430 by Rasikh last updated on 16/Aug/20

Answered by Sarah85 last updated on 17/Aug/20

$$42!\mathrm{has}\mathrm{got}\mathrm{more}2\mathrm{s}\mathrm{than}3\mathrm{s},\mathrm{so}\mathrm{we}\mathrm{have}\mathrm{to}$$$$\mathrm{count}\mathrm{the}3\mathrm{s}$$$$3,6,12,15,21,24,30,33,39,42/103\mathrm{s}$$$$9,18,36/63\mathrm{s}$$$$27/33\mathrm{s}$$$$\Rightarrow n=19$$

Commented by Rasikh last updated on 17/Aug/20

$$\mathrm{thans}\mathrm{a}\mathrm{lot}\mathrm{sir}$$

Commented by udaythool last updated on 17/Aug/20

$$\mathrm{There}\mathrm{are}14\mathrm{no}.\mathrm{s}\mathrm{which}\mathrm{are}$$$$\mathrm{multiple}\mathrm{of}3,4\mathrm{multiple}\mathrm{of}9\mathrm{and}$$$$1\mathrm{multiple}\mathrm{of}27.$$$$\mathrm{Therefore}42!\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}{3}^{14+4+1}=3^{19}.$$$$\mathrm{No}\mathrm{doubt}42!\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}$$$$\mathrm{more}\mathrm{than}19\mathrm{th}\mathrm{power}\mathrm{of}2.\mathrm{Thus}$$$$42!\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}$$$$19\mathrm{th}\mathrm{power}\mathrm{of}6,\mathrm{since}2\mathrm{and}3$$$$\mathrm{are}\mathrm{relatively}\mathrm{prime}.$$

Commented by Rasikh last updated on 18/Aug/20

$$\mathrm{thank}\mathrm{you}$$

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