7.2: Division Algorithms

As with other operations, there are many ways of performing division (not just many perspectives). You are likely aware of the fact that we can directly model (if the numbers are fairly small and usually integers), repeatedly subtract the divisor, or use long division. But there are other ways of performing division. Let's start with a division algorithm that is "between" repeated subtraction of the divisor and long division.

Scaffold Division

Let's consider \(\frac{17}{3}\). We will do this three different ways using what's called a scaffold. To make a scaffold, write the dividend, then draw a line starting to the left of the dividend and going up and around to the right of the dividend, extending as far downward as needed (it will not be clear how far this is until we are finished, so start short and simply extend downward as you continue through the problem). Write the divisor to the left of the part of the scaffold that goes around the dividend. See below.

In the first example below, we subtract 3 repeatedly (no short cuts!). We keep track of how many threes we subtract at each step to the right of the scaffold. Since 3 is subtracted one at a time, and a total of five 3's are subtracted before we get to 2 (any more and we would get a negative value as a result, so we stop), we have 5 ones to the right of the scaffold.

In the second one, first 2 threes are subtracted at once, then 2 more, and then 1 more. Notice that the numbers 2, 2, and 1 are written to the right of the scaffold. We stop because the last difference is 2, which means we cannot take away another 3. That means, again, we subtracted a total of 5 threes.

In the third one, first 4 threes are subtracted at once (and 4 is written to the right of the scaffold), and then only 1 more (and 1 is written to the right of the scaffold). Again, we stop because the difference is 2. And again this means we subtracted a total of 5 threes.

In each case, we got that \(17 \div 3 = 5\) with a remainder of 2.

Long division with large numbers is often difficult because you have to multiply in your head or use a lot of space for scratch work to find the largest possible factor for each multiplication step. It's easy to make mistakes! But, by using scaffold division, you can keep the scratch work organized, and you do not have to find the largest number times one number "goes into" another number. Try the following problem on your own using scaffold division: \(361 \div 53\).

Now that you have tried it, let's do it together. The way we will do this might look different than what you did, but that does not mean your work is incorrect! Scaffold division lets the user choose what works best for them.

Before we do division, let's make a quick partial multiplication table up for 53, by using repeated addition! This table is your prep work for division. A simple way to make up a partial multiplication table for 53 without ever multiplying entails doubling numbers. We know that \(1 \times 53 = 53\). To figure out \(2 \times 53\), just add two copies of 53 together. Two copies of 53 added to two more copies of 53 would equal four copies of 53, or \(4 \times 53\). Four copies of 53 added to four more copies of 53 would equal eight copies of 53, or \(8 \times 53\). Okay, so how do you show this? Start with one 53 and double it to find \(2 \times 53\). Double that number to find \(4 \times 53\). Double that number to find \(8 \times 53\). Look in the left part of the box shown below. There is a 1 pointing to 53 because that is the result of \(1 \times 53\). If you double the 53, (53 + 53), then you now know \(2 \times 53 = 106\). So a 2 points to 106. If you double that (106 + 106), you now know \(4 \times 53 = 212\). So a 4 points to 212. And if you double this number (212 + 212), then you now know \(8 \times 53 = 424\). So an 8 points to 424. You now know \(1 \times 53, 2 \times 53, 4 \times 53,\), and \(8 \times 53\).

Instead of guessing how many times 53 "goes into" 361, we note how many copies of 53 can be subtracted from 361 where underestimating is okay. Since \(8 \times 53 = 424\) and \(424 > 361\), we don't bother guessing 8. The biggest number in our list products we already found that can be subtracted from 361 is 212. Subtract 212 from 361 and write 4 on the right because you just subtracted 4 copies of 53. The resulting difference is 149. The biggest number you can subtract from 149 is 106. If you do that, you've subtracted 2 more copies of 53. So, write a 2 on the right. After subtracting, 43 is left. Since \(43 < 53\), no more copies of 53 can be subtracted. Therefore, 43 is the remainder. Add the numbers on the right to see how many copies of 53 you subtracted. Since 6 copies of 53 were subtracted, we have \(361 \div 53 = 6 + \frac{43}{53}\) or \(361 \div 53 = 6\) remainder \(3\).

To check, multiply \(53 \times 6\) and add 43. You should have gotten 361.

Here are some more examples for you to study. In each example, a partial multiplication table is made up first using the doubling technique. This is not necessary, and you will see more examples in a bit that do not do this. This prep work which involves a few additions makes the division much easier. Then, the problem is done using a scaffold. After adding up how many times the divisor was subtracted from the dividend (the numbers to the right of the scaffold), the answer (quotient and remainder) is written in a box up above the problem. Finally, the check is shown. Also, check to make sure the remainder is less than the divisor.

Example \(\PageIndex{1}\)

Divide \(\frac{279}{37}\) using a scaffold.

Solution

Example \(\PageIndex{1}\)

Divide \(\frac{200}{24}\) using a scaffold.

Solution

Example \(\PageIndex{1}\)

Divide \(\frac{887}{231}\) using a scaffold.

Solution

In the examples so far, all of the quotients were less than 10. It's possible to continue doubling the divisor in the partial multiplication table, so that you can find the divisor times any power of 2 — 1, 2, 4, 8, 16, 32, 64, etc. You would stop doubling if you notice that doubling once more gives a number bigger than the quotient. We do not need to use this doubling method, however. Let's see how two children perform the division in the example below.

Example \(\PageIndex{1}\)

Divide \(\frac{561}{29}\) using a scaffold.

Solution

Child A:

This child used a multiplication table but did not use only doubling. Notice that they crossed out the two times when the result was bigger than what they wanted. This tells us that they tried doubling 174 after they had already subtracted \(561-290=271\) because \(348 > 271\). The child never wrote their answer, but if they interpreted their work correctly, they would have that \(\frac{561}{29} = 19 + \frac{10}{29}\) or \(\frac{561}{29} = 19\) remainder \(10\).

Child B:

This child did not use a multiplication table at all. Their work is a little more cluttered because they did the multiplication inside the scaffold (note the crossed out 4 and the 3 above and to the left of the divisor). They also wrote their answer, 19 remainder 10, above the scaffold.

In each case, the child should have checked their work. If they did, they would get that \(19 \times 29 + 10 = 551+10 = 561\).

It's time for you to try using the scaffold method.

Exercises

Solve each problem twice. First by doing prep work with the doubling method to make a partial multiplication table first before drawing the scaffold. Then, redo it either with a different multiplication table or without using a multiplication table at all. Check your answers.

A. \(415 \div 72\) = _______________

B. \(1235 \div 214\) = _____________

C. \(3128 \div 321\) = _____________