答曰、六千五百六十一。
Answer saith: six thousand five hundred [and] sixty-one.
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In modern notation:
\begin{aligned} 9 \times 9 &= 81 \\ 81 \times 81 &= 6561. \end{aligned}
- See §8.
- Version B has 收上頭位 for 收上位.
-
In animated form (crappy source code here):
This section gives a fully worked example of the rod calculus multiplication algorithm described in §8.
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 |
|---|---|---|
| 九九八十一、自相乘、得幾何。 答曰、六千五百六十一。 |
Nine nines [are] eighty-one, [which],
multiplied with itself, resulteth in how much? Answer saith: six thousand five hundred [and] sixty-one. |
|
| 術曰、重置其位。 | Method saith: put layered [the] places. |
|
| 以上八呼下八、八八六十四、即下六千四百於中位。 | Calling [the] lower eight with [the] upper eight: eight eights [are] sixty-four, that is, put down six thousand four hundred upon [the] middle places. | |
| 以上八呼下一、一八如八、即於中位下八十。 | Calling [the] lower one with [the] upper eight: one eight [is] as eight, that is, upon [the] middle places put down eighty. | |
| 退下位一等、收上位八十。 | Retreat [the] lower places [by] one rank, [and] take back eighty [the] upper place. |
|
| 以上位一呼下八、一八如八、即於中位下八十。 | Calling [the] lower eight with one [the] upper place: one eight [is] as eight, that is, upon [the] middle places put down eighty. | |
| 以上一呼下一、一一如一、即於中位下一。 | Calling [the] lower one with [the] upper one: one one [is] as one, that is, upon [the] middle places put down one. | |
| 上下位俱收、中位即得六千五百六十一。 | [The] upper [and the] lower places [now] all taken back, [the] middle places do result in six thousand five hundred [and] sixty-one. |
|
Conway (2023). "Sun Tzŭ's Computational Classic: Volume I §13". <https://yawnoc.github.io/sun-tzu/i/13> Accessed yyyy-mm-dd.