[math-fun] Mayan Calendar and the Chinese Remainder Theorem
13 May
2007
13 May
'07
7:50 a.m.
playing with the Mayan calendar, cfr http://hermetic.nofadz.com/cal_stud/maya/chap1.htm I noticed that some web pages allow conversion of the Gregorian date to the Mayan Tzolkin-Haab date, but never give the inverse. Calculation of this inverse needs the Chinese Remainder Theorem: given integers u, v and w, find n so that mod(n,20)=u, mod(n,13)=v, mod(n,365)=w (with the added complication that 20 and 365 share a common factor of 5). Would the Mayan priests have had a trick to solve this (in Roman times)? for Mathematica fans: http://users.pandora.be/Wouter.Meeussen/mayan_calendar.nb enjoy, Wouter.
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wouter meeussen