Division on the Abacus

Now we’re going to try something challenging on the abacus: *division*. Like multiplication, abacus division is close to the way you’d do it on paper. But just like doing paper division is trickier than paper multiplication, abacus division is tricker than abacus multiplication. But the technique that is used to do division on the abacus is an important fundamental one: it’s what makes it possible to use the abacus for more advanced operations, like roots.

Before going into the algorithm, there’s one important new technique that we need, called *partitioning* on the abacus. The idea is that we’re going to pick some column on the abacus, which we’ll call the *reference column*; and *in our minds*, we’re going to split the abacus so that the reference column and everything to its left is one abacus, and everything to its right is a second abacus.

The way that we’re going to use this is that we’re going to put the *dividend* into the section to the right of the reference column, and we’re going to accumulate the *quotient* to the left.

So, let’s start by reviewing the standard paper method.

1. Find the starting column for the quotient. This will be the *first* position *n* where the number formed in columns 1 through *n* of the dividend is greater than or equal to the divisor. Columns *1* through *n* are called the *working digits*; column *n* is called the *current quotient column*.
2. Using approximation, figure out the *largest* number *i* such that *i* times the divisor is *less than or equal* to the number formed by the working digits.
3. Write *i* in the current quotient column, and subtract *i* times the divisor from the working digits. The result *should be* a number *smaller* than the divisor. This is the *working remainder*.
4. Copy digits to the right of the working digits, and append them to the working remainder from step 3, until you get a number *greater than or equal to* the divisor. The working remainder + the copied digits become the new working digits. The last column that you copied is the new *current quotient column*. If there are any blank spaces between the old and new current quotient columns, fill them with zeros.
5. Go back to step 2, using the new working digits and current quotient column, until either the working remainder is zero, or you’re bored and don’t want to keep going.

As usual, it’s hard to follow something like that without an example. Let’s divide 4582 by 17.

* Find the starting column. It will be column 2, because 4<17, but 45>17.
* Find the largest multiple of 17 that’s smaller than 45. That will be 2, and it will be the first digit of our answer. Subtract 2*17=34 from 45, leaving a working remainder of 11. We can pull down “8″ from the dividend and append it, giving us new working digits 118; and the new current quotient column will be just one digit to the right of the old.

        2
     +---------------
     17 | 4582
          34
         ----
          118

* Find the largest multiple of 17 ≤ 118. That would be 6. 6*17=102, 118-102=16. The working remainder is 16. So we pull down a digit; 2. That gives us new working digits 162,

       26
     +---------------
    17 | 4582.0
         34
        ----
         118 
         102
        -----
          162

* Find the largest multiple of 17 ≤ 162: 9. 9*17=153; 162-153=9. We’ll stop here, so our answer is 269 with a remainder of 9.

       269  remainder=9
     +---------------
    17 | 4582
         34
        ----
         118 
         102
        -----
          162
          153
         -----
            9

The way we do it on the abacus is very similar to the way we did it on paper. The one difference is that on paper, we did the multiplication and subtraction as two steps; on the abacus, we’re going to combine them.

1. With the abacus, we start by setting the reference column. The way that we do that is by deciding *how many digits* we want in our answer. For our example, we can say three digits – because we can easily see that the integer part of dividing 4582 by 17 will be three digits (even if we hadn’t already worked out the answer). So, we’ll use the leftmost three columns for building the quotient. We zero out the quotient area, and put out divisor into the rightmost columns. The initial setup looks like figure 0 in the image.

2. With the abacus set up, we start following the same basic algorithm that we used on paper. What’s the first digit of the quotient? It’s two. So we put a 2 in the *leftmost* column of the quotient area; and then we start to multiply 2×17; but instead of *adding* the results the way we did in multiplication, we’re going to *subtract them* from the dividend. So 17×2=34; we subtract 5 from the third column, and 4 from the fourth column. The abacus now looks like figure 1 in the image.

3. Now, we look at the dividend section of the abacus, and find the rightmost column where the digits up to that column are larger than 17. That’s one column to the right of where we started working last time: the second column from the right. So our working digits are 118. The largest multiple of 17 smaller than 118 is 6×17. So, we subtract 6×7=42 from columns 2 and 3, which requires borrowing 1 from column 4. So after this part of the subtraction, our abacus looks like figure 2. Then we subtract 6×1=6 from column 3. Finally, we add this new digit of the quotient to the second column from the left of the quotient area. The abacus now looks like figure 3.

4. Finally, we’re on the third digit of the quotient. The working remainder is the three remaining digits (because the working remainder from the last digit is 16, we shift right, and out working remainder becomes the last three digits, 162.) 162 is very close to 10×17, so 9 must be the next digit of the quotient. We multiply 9×7=63, and subtract that from columns one and two. We need to borrow one from column 2 to subtract 3 from column one; and then we need to borrow one from column 3 to subtract 6 from column 3. The result of that is shown in fig 4. Then we subtract 9×1=9 from column 2; and we put 9 in column three of the quotient. So the answer is 269, with a remainder of 9, as shown in fig 5.

—–

You’re welcome to call me crazy, but I find division on the abacus to be really interesting. There’s just something about doing division on such a trivial addition-based device that’s really impressive. What’s even more impressive is just how *fast* you can do it. Just an afternoon practicing, and you can probably do most divisions on an abacus much faster than you could on paper.

As I said back in the first post about the abacus: it’s like a *better* pencil and paper. If you can do it on paper, you can do it on the abacus, only faster. The real trick to doing complicated things is exactly what we saw in division: partitioning the abacus into regions. That’s the reason why when you see a Chinese abacus, it’s usually so long: 13 column abacuses are very common, and seeing a 20 digit abacus isn’t terribly unusual. It’s not because anyone is expecting to be actually doing math on 13 digit numbers, but because they want the room to partition.

Good Math, Bad Math