Difficulty understanding how Babylonian reciprocals work

Question

According to the reciprocal tables for Babylonian's base $60$ system, dividing by $2$ is like multiplying like $30$. Dividing by $3$ is like multiplying by $20$. Dividing by $4$ is like multiplying by $15$. Dividing by $k$ is like multiplying by $60/k$.

I don't understand why the reciprocal works like this and not like $1/k$. You might say "It multiplies by $60/k$ because it's base $60$" but this doesn't make sense to me, it's not like our base-$10$ reciprocals look like $10/k$.

What's going on?

Answer 1

Our base-$10$ reciprocals do look like $10/k$! Dividing by $2$ is like multiplying by $5$ (and then shifting the decimal point) and vice versa. Indeed, I regularly divide by $5$ by doubling the number and shifting the decimal point.

$10$ has fewer divisors than $60$, so this "trick" (if you like) doesn't have as many applications—basically only this one.

Answer 2

Babylonians - at least in the tablets we have found - do not have the concept of order of magnitude: probably they calculated it in other ways. This means that - as far as the tablets are concerned - 1, 60, 3600=60*60, 1/60, 1/3600 are all represented as 1.

Therefore, if you have a number $n$ and you want to divide it by 4 (say), if the number is greater than 4 there is no problem; 9 / 4 = 2;15 (the semicolon has the same use as the comma in our system). For 3/4, they would have 0;45; but since they could not accept a 0 by itself, they multiplied the result for 60 obtaining 45, which is the same as 3*15. In this way, "divide by 4" is the same as "multiply by 15".

Answer 3

In our number system, $\dfrac 15 = \dfrac{2}{10}$. So,

to divide $678$ by $5$,

first divide by $10$, getting $67.8$,

and then multiply by $2$, getting $135.6$.

In a base $60$ number system, $\dfrac 15 = \dfrac{12}{60}$. So

to divide $[6,7,8]_{60}$ by $5$,

first divide by $[1,0]_{60}$, getting $[6,7 . 8]_{60}$

then multiply by $12_{60}$, getting $[72, 84. 96]_{60} = [72, 85. 36]_{60} =[73, 25. 36]_{60}==[1,13, 25. 36]_{60}$