Talk:Distribution of 0 digits in factorial series: Difference between revisions
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(added a section header for a new talk section concerning the number of trailing zeros in a factorial product.) |
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1,000,000! has 249,998 trailing zeroes. |
1,000,000! has 249,998 trailing zeroes. |
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True. So the number of terminal zeros actually goes up as the summation over [k from 1 to rounded up(log base 5 of n)] of (n / 5^k), ie n/5 + n/25 + n/125 .... Interesting, though the nonterminal part of n! still dominates the result eventually. |
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--[[User:Wherrera|Wherrera]] ([[User talk:Wherrera|talk]]) 18:27, 10 June 2021 (UTC) |