Difference between revisions of "Talk:1935: 2018"
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This is easy! Don't factor it - just subtract 4 repeatedly. If you end up at 0, it's divisible. If you end up at 1, 2, or 3, it's not. -- 17:55, 29 December 2017 (UTC) | This is easy! Don't factor it - just subtract 4 repeatedly. If you end up at 0, it's divisible. If you end up at 1, 2, or 3, it's not. -- 17:55, 29 December 2017 (UTC) | ||
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+ | The calculation of Christmas is trivial{{Citation needed}} it's December 25th. Where as the calculation of Easter is complex ([https://en.wikipedia.org/wiki/Computus]). [[Special:Contributions/172.68.133.18|172.68.133.18]] 18:03, 29 December 2017 (UTC) |
Revision as of 18:03, 29 December 2017
This is easy! Don't factor it - just multiply by 25 and if that ends in two zeros, but not four zeros then it's a leap year, at least most of the time.....17:25, 29 December 2017 (UTC)
This is easy! Don’t factor it - just convert it into a binary and look at the 2 least significant bits. If they are 00 the number is multiple of four. —172.69.33.35 17:37, 29 December 2017 (UTC)
This is easy! Don't factor it - just subtract 4 repeatedly. If you end up at 0, it's divisible. If you end up at 1, 2, or 3, it's not. -- 17:55, 29 December 2017 (UTC)
The calculation of Christmas is trivial[citation needed] it's December 25th. Where as the calculation of Easter is complex ([1]). 172.68.133.18 18:03, 29 December 2017 (UTC)