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因為n!中2的約數遠多於5的約數,因此只需考慮5的約數.f(n)=[n/5]+[n/25]+[n/125]+..31n/125~53125x53/31=213f(213)=42+8+1=51f(215)=43+8+1=52f(220)=44+8+1=53f(224)=44+8+1=53f(225)=45+8+1=54n最大為224....
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