Distribution of 0 digits in factorial series: Difference between revisions
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as the factorials become larger, the proportion of 0 digits in the factorial products shifts slowly |
as the factorials become larger, the proportion of 0 digits in the factorial products shifts slowly |
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from around 1/5 toward 1/10, since the number of terminating zeros in n! increases only in proportion |
from around 1/5 toward 1/10, since the number of terminating zeros in n! increases only in proportion |
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to n, whereas the number of digits of n in base 10 increases exponentially. |
to n, whereas the number of digits of n! in base 10 increases exponentially. |
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; The task: |
; The task: |
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Find the N in 10000 < N < 50000 where the mean of the proportions of 0 digits in the factorial products from 1 to N |
Find the N in 10000 < N < 50000 where the mean of the proportions of 0 digits in the factorial products from 1 to N |
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permanently falls below 0.16. |
permanently falls below 0.16. This task took many hours in the Python example, though I wonder if there is a faster |
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algorithm out there. |
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=={{header|Python}}== |
=={{header|Python}}== |
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