Timeline for Calculating the number of zeroes at the end of a factorial
Current License: CC BY-SA 3.0
6 events
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| Oct 31, 2017 at 10:20 | history | edited | Mast♦ | CC BY-SA 3.0 |
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| Oct 19, 2017 at 12:44 | comment | added | Jim | After a night's sleep (the above was at 2am), I think you only need to count the times 5^N (5 to the power of N) divides into T, with a loop on N++, N>0. Stop the loop when 5^N > T. Why does this work - Since there are so many more 2 factors than 5 factors, any 5^N essentially becomes a number with N zeroes at the end (5x2=10, 25x4=100, 125x8=1000, etc.). Just up to 100!, there are 50 2-factors, but only 20 5-factors, giving us this surplus of 2s that make this work. | |
| Oct 19, 2017 at 5:47 | history | edited | Jim | CC BY-SA 3.0 |
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| Oct 19, 2017 at 5:34 | history | edited | Jim | CC BY-SA 3.0 |
Editted to add an example
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| Oct 19, 2017 at 5:29 | review | First posts | |||
| Oct 19, 2017 at 5:30 | |||||
| Oct 19, 2017 at 5:25 | history | answered | Jim | CC BY-SA 3.0 |