Last week, in Fundamental Understanding of Mathematics XXXIX, we took a look at the standard long division algorithm, to see how we could do divisions of large numbers using several divisions of simpler numbers. This week I'm going to wrap up the discussion of long division by showing how to work with multi-digit divisors.
Our earlier efforts have all used a single digit divisor, to make the exposition a bit more straightforward. It's a bit easier to explain things when all the example multiplications are in a standard multiplication table. But, not all real problems fit in a standard multiplication table. We will confront problems with multi-digit divisors, as well as multi-digit dividends.
Let's take a look at this one:
One often misunderstood aspect of the long division bracket is that we use the place value of the dividend (under that heavy black line) to keep track of the place value of the quotient. The digit "1" placed above the digit "4" stands for 1000. The digit "1" placed above the digit "6" stands for 10.
We teach students to not write in the zeros, since we are going to use that space later, and we have arranged the division algorithm so all the quotients of the simple divisions are single digit quotients. Students who have a weak grasp of place value often don't understand why the location of the answer above the line is significant, and just write the digits of the quotient any old place. Teachers sometimes have these students turn their lined paper sideways, so the lines form columns, but I'm not sure students see this as any more than an attempt to have their work come out neatly. I suggest emphasizing the place value of the digit by writing in the zeros in the multiplication product, and leaving the "bringing down" shortcut for later.
The first decision the problem solver must make is where to start. In effect, will the final quotient be a single digit number, a two digit number, a three digit number?
Division is the inverse of multiplication. Multiplication is repeated addition, so division is repeated subtraction. We subtract multiples of the divisor. In order to work as efficiently as possible, we want to take the largest multiples we can find, first.
So, we compare our divisor, 89, with the high place value numbers in our dividend, looking for at least a multiple of 1.
If we look at the ten thousands place, we see that we cannot subtract a multiple of 10,000 (which would be 890,000, a larger number than our dividend, 234,567.) This division will not have a five digit quotient. So we try the next smaller place, 1,000
In this case, we see that we can subtract at least one multiple of 1,000, and may be able to subtract more than one. But, since we already figured out that we can't subtract a multiple of 10,000, our highest possible multiple is 9. So, our quotient's highest digit will be in the thousands place.
But, what is this digit? We can't look it up in a multiplication table, they don't go up to 89. The proper way to do this is to estimate: round both numbers, ignore trailing zeros, and use the result from the single digit division: 234 rounds to 230, ignore the zero makes it 23. 89 rounds to 90, ignore the zero makes it 9. So our estimate is 23 divided by 9, or 2.
The division on the left is how this is usually written, ignoring the 567, and winding up with a remainder of 56. I'd recommend, at least at the beginning, emphasizing that the 2 is really a 2,000, and showing the complete product, including the zeros, and doing the complete subtraction.
To return momentarily to our "dealing cards" analogy: we started with 234,567 cards and 89 players. We just dealt 2,000 cards to each player, and we have 56,567 cards left to deal.
Our next cognitive leap is to recognize that, having done one simple division with a single digit quotient, we do it again. Using our long division notation, we are really looking at this new problem:
Although the way it is normally written is on the right. We estimate the hundreds digit in the quotient using 56 or 57 divided by 9, finding 9 x 6 = 54 to be the closest multiple, without going over.
Since we have a four digit quotient, we do two more simple divisions, to find our last two digits:
Estimating: 32 divided by 9 gives us 32 = 3 x 9 + 5
And, finally, 50 divided by 9 gives us 50 = 5 x 9 + 5
And there's our answer:
234567 = 2635 x 89 + 52
This is a very sophisticated algorithm, although, in principle, it uses the same "repeat a simple operation" strategy that our multi-digit multiplication, addition and subtraction algorithms use. The long division bracket is a very compact way of organizing all the multiplications and subtractions needed to do the repeated subtractions of smaller and smaller multiples of the divisor. To use the algorithm, the skill of rounding numbers and estimating should be well understood, the multiplication table should be memorized and the ability to work a multiplication problem when the multiplicands are in quite non-standard locations. It also helps to be able to regroup in one's head, since there really isn't any good place to note any carry over digits.
Given all the different skills that must come together to make working with this algorithm possible, students can be forgiven if they get the impression that it was designed by geeks to make them feel bad about their math skills.
Clever students may take issue with calling the intermediate quotients "single digit" quotients, especially when you explain that they really are 2000, or 600, or 30, obviously not single digit numbers. It would be tempting to tell them "the zeros don't count" which demotes zero from a real number to a place holder. It would be a better idea (these are your clever students, after all) to introduce the idea of "significant digits."
Have fun in the comments.