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標題:
發問:
Given that n!, in decimal notation, has exactly 57 ending zeros, find the sum of all possible values of n.
最佳解答:
Giventhat n!，in decimal notation，has exactly 57 ending zeros，find the sum of all possible values of n. Sol [240/5]+[240/25]+[240/12]=48+10+2=60>57 [239/5]+[239/25]+[239/125]=47+9+1=57 [238/5]+[238/25]+[238/125]=47+9+1=57 [237/5]+[237/25]+[237/125]=47+9+1=57 [236/5]+[236/25]+[236/125]=47+9+1=57 [235/5]+[235/25]+[235/125]=47+9+1=57 [234/5]+[234/25]+[234/125]=46+9+1=56<57 So the sum of all possible values of n. =239+239+237+236+235 =(239+235)*5/2 =1185
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Factorial發問:
Given that n!, in decimal notation, has exactly 57 ending zeros, find the sum of all possible values of n.
最佳解答:
Giventhat n!，in decimal notation，has exactly 57 ending zeros，find the sum of all possible values of n. Sol [240/5]+[240/25]+[240/12]=48+10+2=60>57 [239/5]+[239/25]+[239/125]=47+9+1=57 [238/5]+[238/25]+[238/125]=47+9+1=57 [237/5]+[237/25]+[237/125]=47+9+1=57 [236/5]+[236/25]+[236/125]=47+9+1=57 [235/5]+[235/25]+[235/125]=47+9+1=57 [234/5]+[234/25]+[234/125]=46+9+1=56<57 So the sum of all possible values of n. =239+239+237+236+235 =(239+235)*5/2 =1185
其他解答:
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