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Übungen zur Ethnomathematik

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Dechiffrierung einer einfachen Zahlenschnur

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Unten ist ein Ausschnitt aus einem Quipu, einer Zahlenschnur der Inka, dargestellt. Der obere Teil mit dem Ergebnis fehlt.

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", "advice": "

Man liest die Schnüre von außen nach innen (also unten nach oben für die unteren Schnüre), jede Schnur stellt (in diesem Fall eine dreistellige) Zahl dar:

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  1. Links: {einer[O[0]]} meint {O[0]}, {zehner[T[0]]} meint {T[0]*10}, {zehner[H[0]]} meint {H[0]*100}, ergibt {a}.
  2. \n
  3. Mitte: {einer[O[1]]} meint {O[1]}, {zehner[T[1]]} meint {T[1]*10}, {zehner[H[1]]} meint {H[1]*100}, ergibt {b}.
  4. \n
  5. Rechts: {einer[O[2]]} meint {O[2]}, {zehner[T[2]]} meint {T[2]*10}, {zehner[H[2]]} meint {H[2]*100}, ergibt {c}.
  6. \n
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Als Lösung (dargestellt auf der fehlenden oberen Schnur) ergibt sich $\\var{a}+\\var{b}+\\var{c}=\\var{a+b+c}$.

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Welche Rechnung ist durch die drei Schnüre dargestellt?

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[[0]]+[[1]]+[[2]]=[[3]]

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Dechiffrierung von Maya-Zahlen

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Unten sei ein Auszug aus einem Codex der Maya dargestellt, auf dem drei Zahlenangaben zu finden sind.

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", "advice": "

a) Die Zahlen werden von links nach rechts und unten nach oben gelesen. Die Stellenwerte sind Einer (Unten), 20er (Mitte) und 360er (Oben).

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Erste Zahl (Links): $\\qquad\\var{H[0]}\\cdot 360+\\var{T[0]}\\cdot 20+\\var{O[0]}=\\var{a}$

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Zweite Zahl (Mitte): $\\quad\\var{H[1]}\\cdot 360+\\var{T[1]}\\cdot 20+\\var{O[1]}=\\var{b}$

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Dritte Zahl (Rechts): $\\quad\\var{Hc}\\cdot 360+\\var{Tc}\\cdot 20+\\var{Oc}=\\var{c}$

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b) Es gilt {bop[switch]}, es liegt also eine {cstring[switch]}-Aufgabe vor.

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Welche drei Zahlen sind hier dargestellt?

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1. Zahl: [[0]] 2. Zahl: [[1]] 3. Zahl: [[2]]

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Bei der Rechnung handelt es sich um eine [[0]]-Aufgabe.

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

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

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.

", "advice": "

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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\n

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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Aktivität zum Erstellen eines Kolam

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Ein Kolam ist eine in einem einzigen Linienzug durchführbare, regelgeleitet erstellte Kreidezeichnung, wie sie sich noch heute bei den südindischen Tamilen als ethnomathematische Praxis auffinden lässt.
Eine vereinfachte Variante für 8 Arme mit 5 Punkten sieht etwa so aus:

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\"kolam-demo\"/

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Unten ist ein Punktmuster zur Erstellung eines solchen Kolam mit 4 Armen zu je 3 Punkten abgebildet (bitte etwas Geduld, bis das Applet lädt).

", "advice": "

a) Es ergibt sich:

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{geogebra_applet('https://www.geogebra.org/m/fwc7k2nw',defs)}

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b) Angenommen, ein Kolam hat $a$ Arme mit je $b$ Punkten.

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i) Für $a,b$ muss dann zwingend $\\operatorname{ggT}(a,b)=1$ gelten, damit wir einen Eulergraphen haben.

\n

ii) Der Graph hat dann genau $a\\cdot b$ Kanten.

\n

\n

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Linie fehlt\"));\n if(exists(app,\"r\")=true,if(definition_string(app,\"r\")=r_correct,add_credit(1/12,\"12. Linie ok. \"),negative_feedback(\"12. Linie falsch\")),negative_feedback(\"12. Linie fehlt\"))\n \ninterpreted_answer:\n f", "extendBaseMarkingAlgorithm": false, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

{app}

\n

Erstellen Sie ein Kolam gemäß der Punktfolge {kolam_string}. Beginnen Sie dabei mit dem Zählen der Punkte beim innersten Punkt des nach oben zeigenden Arms (der gemeinsame Punkt aller vier Arme in der Mitte sei 0) und schreiten Sie im Uhrzeigersinn (rechts herum) fort.

\n

Sie können das Icon \"line\" zum Zeichnen von Linien verwenden, das Icon \"bin\" um fehlerhafte Linien wieder zu löschen. Mit \"reset\" können Sie die ganze Zeichnung wieder zurücksetzen, mit den beiden Pfeilen oben rechts einen Schritt rückgängig machen bzw. wiederholen.

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Angenommen, ein Kolam hat $a$ Arme mit je $b$ Punkten.

\n

i) Für $a,b$ muss dann zwingend $\\operatorname{ggT}(a,b)=$[[0]] gelten, damit wir einen Eulergraphen haben.

\n

ii) Der Graph hat dann genau [[1]] Kanten.

\n

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