Is there any way to find the number of zeroes at the end of n! for n≥5?

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seskew192atp · Accepted Answer

Step 1You have only counted the 5 of numbers like 25,125,250,1000 etc. once each. But these have more than one 5 in their factorisations. The correct method is as follows:Count one 5 from each of these numbers once:5,10,15,20,25,30,⋯⋯1000We get&amp;nbsp;⌊10005⌋=200&amp;nbsp;of these 5&#039;sNow count another 5 from each multiple of 25 (note that we have already counted one 5 from each of these!)25,50,75,100,⋯⋯1000We get&amp;nbsp;⌊100025⌋=40&amp;nbsp;of these 5&#039;sNow count another 5 from each multiple of 125 (we have already counted two 5s from each)125,250,375,⋯⋯1000We get&amp;nbsp;⌊1000125⌋=8&amp;nbsp;of these 5&#039;sFinally, count another 5 from each multiple of 625 (we have already counted three 5s from them) 625We get&amp;nbsp;⌊1000625⌋=1&amp;nbsp;of these 5&#039;sAdding all of these up, the maximum power of 5 in 1000! is&amp;nbsp;200+40+8+1=249.&amp;nbsp;Obviously, the maximum power of 2 is much more, so we won&#039;t bother about it. The maximum power of 10 is therefore equal to the maximum power of 5, that is 249.So 1000! has 249 zeroes at the end.

Is there any way to find the number

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