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Arithmettik & Algebra in der Renaissance

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Division durch eine einstellige Zahl gemäß dem 2. Rechenbuch

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Nach Adam Ries sei die Aufgabe $\\var{c}:\\var{a}$ schriftlich zu berechnen.

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Erster Schritt: Wir schreiben unter die 2. Stelle den Divisor, also $\\var{a}$. Wir rechnen $\\var{(c1)*10+c2}:\\var{a}=\\var{b_digits[0]}$, als Rest erhalten wir $\\var{carry1}$, den wir über die 2. Ziffer notieren.

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{carry1}
{c1}{c2}{c3}{c4}|{b_digits[0]}
{a}\u200b\u200b
\n

Zweiter Schritt: Wir schreiben unter die 3. Stelle den Divisor, also wieder $\\var{a}$. Wir rechnen $\\var{(carry1)*10+c3}:\\var{a}=\\var{b_digits[1]}$, als Rest erhalten wir $\\var{carry2}$, den wir über die 3. Ziffer notieren.

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{carry1}{carry2}
{c1}{c2}{c3}{c4}|{b_digits[0]}{b_digits[1]}
{a}{a}\u200b\u200b
\n

Dritter Schritt: Wir schreiben unter die letzte Stelle wieder den Divisor, also $\\var{a}$. Wir rechnen $\\var{(carry1*10+floor(mod(c,100)/10)-b_digits[1]*a)*10+c4}:\\var{a}=\\var{b_digits[2]}$. Das geht sich ohne Rest aus, also haben wir die Lösung $\\var{b}$ gefunden.

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{carry1}{carry2}
{c1}{c2}{c3}{c4}|{b_digits[0]}{b_digits[1]}{b_digits[2]}
{a}{a}\u200b{a}\u200b
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Tragen Sie unten die fehlenden Ziffern ein.

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In jede Lücke bitte nur genau eine Ziffer eintragen.

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[[3]][[4]]
\n

{c1}

\n
\n

{c2}

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\n

{c3}

\n
\n

{c4}

\n
\n

|

\n
[[5]][[6]][[7]]
[[0]][[1]]\u200b[[2]]\u200b
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Adaptiert von: Katz, V. J. (2009). A history of mathematics: An introduction (3. ed.). Boston: Addison-Wesley, S. 418.

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Das folgende Problem stammt (bis auf die Zahlenwerte)  von Piero della Francesca (1410/20-1492).

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Drei Männer gehen eine Geschäftspartnerschaft ein. Der erste investiert {a} Dukaten in das Geschäft, der zweite investiert {b} Dukaten und wir wissen nicht, wie viel der dritte investiert hat.

\n

Als Erlös erhalten die drei Partner zusammen {sum} Dukaten, von denen der erste Partner {a2} Dukaten erhält.

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Es ist davon auszugehen, dass der Erlös aus dem Geschäft proportional zur in die Partnerschaft investierten Summe ist.

\n

Der erste erhält bei einer Investition von $\\var{a}$ Dukaten einen Erlös von $\\var{a2}$ Dukaten.

\n

Dann muss der zweite bei einer Investition von {b} Dukaten einen Erlös von $\\var{b}\\cdot\\frac{\\var{a2}}{\\var{a}}=\\var{b}\\cdot\\var{a2/a}=\\var{b2}$ Dukaten erhalten.

\n

Für den dritten Partner bleiben dann vom Erlös noch $\\var{sum}-\\var{a2}-\\var{b2}=\\var{c2}$ Dukaten übrig. Er hat demnach $\\var{c2}:\\frac{\\var{a2}}{\\var{a}}=\\var{c2}:\\var{a2/a}=\\var{c}$ Dukaten investiert gehabt.

\n

Im Original hat das Problem eine nicht-ganzzahlige Lösung, hier wurde als Erlös zur Vereinfachung ein ganzzahliges Vielfaches der Investitionssume gewählt.

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Welche Anteile vom Erlös erhalten der zweite und der dritte Partner und wie viel hatte der dritte investiert?

\n

Der zweite Mann erhält [[0]] Dukaten vom Erlös.

\n

Der dritte Mann erhält [[1]] Dukaten vom Erlös.

\n

Der dritte Mann hat [[2]] Dukaten investiert gehabt.

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