This section gives a worked example of solving a linear equation.
Chinese source text: Version A, Version B, Version C, Version D.
Unless noted otherwise, I follow the text from Version D, 《知不足齋叢書》本.
Source text | Target text | Notes |
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今有五等諸侯、共分橘子六十顆、人別加三顆。問五人各得幾何。 | Suppose there be [the] five ranks of all [the] marquesses, together sharing tangerines sixty, [and for each] person's distinction [there be] added three [unto his share]. [We] ask, how many getteth each of [the] five people? | |
答曰、公一十八顆、侯一十五顆、伯一十二顆、子九顆、男六顆。 | Answer saith: [the] lord eighteen [tangerines], [the] marquess fifteen [tangerines], [the] elder twelve [tangerines], [the] master nine [tangerines], [and the] male six [tangerines]. | |
術曰、先置人數別加三顆於下、次六顆、次九顆、次一十二顆、上十五顆。 | Method saith: first put [the] number of people [and the] distinctional addition of three [tangerines] below; next, six [tangerines]; next, nine [tangerines]; next, twelve [tangerines]; above, fifteen [tangerines]. |
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副并之、得四十五。 | Subsidiarily combining them, resulteth in forty-five. |
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以減六十顆、餘、人數除之、人得三顆。 | Subtract of [the] sixty [tangerines] by [this]: [the] remainder, dividing it [by the] number of people, [each] person getteth three [tangerines]. |
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各加不并者、上得一十八為公分、次得一十五為侯分、次得十二為伯分、次得九為子分、下得六為男分。 | Adding unto each of those not combined: above [there] resulteth eighteen as [the] lord's share; next [there] resulteth fifteen as [the] marquess's share; next [there] resulteth twelve as [the] elder's share; next [there] resulteth nine as [the] master's share; below [there] resulteth six as [the] male's share. |
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Conway (2023). "Sun Tzŭ's Computational Classic: Volume II §25". <https://yawnoc.github.io/sun-tzu/ii/25> Accessed yyyy-mm-dd.