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Einige Berechnungen am spirituellen Maya-Kalender (Modulo-Rechnungen/Kongruenzen).

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Der spirituelle Kalender der Maya hieß \"Tzolk'in\" und verwendet die Tageszahlen 1 - 13 alternierend mit 20 Tagesnamen, so dass sich dieselben Kombinationen (Datumsangaben) nach $13 \\cdot 20 = 260$ Tagen ($\\approx$ ein Mondjahr) wiederholen (s. Tabelle unten für die ersten 20 Tage im Jahr).

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Erste 20 Datumsangaben im Tzolk'in\"maja\"/
1 Imix11 Chuwen
2 Ik'12 Eb'
3 Ak'bal13 Ben
4 K'an1 Ix
5 Chikchan2 Men
6 Kimi'3 Kib'
7 Manik'4 Kab'an
8 Lamat5 Etz'nab'
9 Muluk6 Kawak
10 Ok7 Ajaw
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Hinweis: Für diese Aufgabe sollten Sie etwas Modulo-Rechnung beherrschen.

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a) Da dies der 260. Tag ist, liegt sowohl ein ganzzahliges Vielfaches von 20, als auch von 13 vor. Der Tag ist also der 13 Ajaw.

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b) Alle Tage, für die sich die Datumszahl {maya_date(forb)} ergibt, liegen in der Restklasse $\\var{mod(maya_date(forb),13)}\\pmod{13}$. Alle, für die sich der Monat {maya_name(forb)} ergibt, liegen in der Restklasse $\\var{mod(forb+1,20)}\\pmod{20}$. Für den gesuchten Tag muss gelten:

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\\[13\\cdot x+\\var{mod(maya_date(forb),13)}=20\\cdot y+\\var{mod(forb+1,20)}\\Leftrightarrow13\\cdot x-20\\cdot y= \\var{mod(forb+1,20)-maya_date(forb)}\\] Diese Gleichung kann man mit dem erweiterten Euklidischen Algorithmus (oder durch geschicktes Probieren) lösen und erhält $x=\\var{(forb-mod(maya_date(forb),13)+1)/13},y=\\var{(forb-mod(forb,20))/20}$, also den $13\\cdot\\var{(forb-mod(maya_date(forb),13)+1)/13}+\\var{mod(maya_date(forb),13)}=20\\cdot\\var{(forb-mod(forb,20))/20}+\\var{mod(forb+1,20)}=\\var{forb+1}$-ten Tag des Maya-Kalenders.

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c) Es ist $40\\pmod{20}=0$ und $40\\pmod{13}=1$ (weil $3\\cdot 13=39$ ), d. h. alle 40 Tage ändert sich der Datumsname nicht, die Datumszahl rückt genau eins weiter. Im Beispiel rückt die Datumszahl genau $\\var{maya_date(forc3)-maya_date(forc1)}$ weiter, also sind $\\var{maya_date(forc3)-maya_date(forc1)}\\cdot 40=\\var{forc4}$ Tage vergangen. Alternativ bestimmt man nach der Methode aus b) das größere der beiden Daten und zieht das kleinere Datum (das eh in der Tabelle steht) davon ab ($\\var{forc3+1}-\\var{forc1+1}=\\var{forc4}$).

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Der letzte Tag des Jahres hat die Datumsangabe (erste Lücke: Zahl, zweite Lücke: Name): [[0]] [[1]]

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Der Tag {maya_date(forb)} {maya_name(forb)} ist der [[0]]-te Tag des Maya-Kalenders.

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Der Tag {maya_date(forc3)} {maya_name(forc3)} ist der [[0]]-te Tag nach dem Tag {maya_date(forc1)} {maya_name(forc1)}.

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