22 Division Algorithms

You will need: Base Blocks (Material Cards 3-15)

There are two main algorithms for division; one based on partitioning by subsets and one based on repeated subtraction. These algorithms are just the number version of what is done with base blocks.

Partitioning by Subsets Model and Long Division (Standard) Algorithm

Exploration Activity

Pull out your base ten blocks. We will solve the base ten division problem [latex]151\div 4[/latex] using partitioning into subsets.

a. Lay out 151 in your base ten blocks.

b. We need to partition these into 4 subsets, so draw 4 empty circles.

c. Divide the blocks as evenly as possible between the 4 circles. (You should have 3 unit pieces leftover.)

We know that the answer to partitioning into subsets is the number of items in a subset. Each subset has 3 longs and 7 units and there are 3 units that were not put into any subset. In other words, each group has a value of 37 with 3 leftover. The answer is 37 remainder 3.

Using blocks to solve [latex]151 \div 4[/latex] is reasonable, but it is not reasonable to use blocks to solve things like [latex]146,573\div 5[/latex]. We need to take the process we do with the blocks and translate that into an algorithm using numbers.

Understanding Long Division

Action with Base Ten Blocks	Matching Action in Long Division
Let’s start by looking at how you distributed your blocks in the exploration activity. The one flat cannot be evenly distributed in its current form, so we need to break it into 10 longs. This gives us a total of 15 longs and one unit.	In terms of long division, we say “4 cannot go into 1” so we look at the number 150 instead of 100. Basically we are saying one thing of size 100 does not work for partitioning the pieces, so we consider 15 things of size 10.
Next we distribute 12 of the 15 longs into the 4 circles.	Because [latex]4\times 4[/latex] is more than 15, and [latex]4\times 3[/latex] is less than 15, we put a 3 in the tens place. This represents the 3 longs that were put into each circle. Then we subtract [latex]4\times 3=12[/latex] from 15 to show that we have now used 12 of the 15 longs and have 3 longs remaining.
We have three unused longs that cannot be distributed evenly into the 4 circles in their current form, so we need to break them into 30 units, meaning we have 31 total unit pieces.	In long division we say you “bring down the 1.” This represents  moving from 3 pieces of size ten to 31 pieces of size one.
Now we distribute the units as evenly as possible, placing 7 pieces in each circle, using a total of 28 unit pieces.	Because [latex]4\times 8[/latex] is more than 31, and [latex]4\times 7[/latex] is less than 31, we put a 7 in the ones place. This represents the 7 units that were put into each circle. Then we subtract [latex]4\times 7=28[/latex] from 31 to show that we have now used 28 of the 31 unit pieces and have 3 unit pieces leftover.
We have 3 units leftover, which is smaller than 4. We are done! Our quotient is what is inside of the circle and our remainder is 3.	We have 3 leftover, which is smaller than 4, so this is our remainder. Our quotient is what is above the line (what is in a circle) with remainder 3.
Answer: 37 remainder 3	Answer: 37 remainder 3

Long Division in Other Number Bases

This same process can be done with different number bases. If you can confidently perform long division calculations in other number bases, you will be better prepared to help your future students understand long division in base ten.

Be aware that your multiplication facts look different in each base. For example, in base ten, [latex]2\times 5=10[/latex], but in base six [latex]2\times 3=10_{\text{six}}[/latex]. Let’s consider [latex]345_{\text{six}}\div4_{\text{six}}[/latex] by partitioning with blocks and using long division. Before starting our long division, it is helpful to write out our multiplication facts for 4 in base six. We only need the facts for each digit possible in base six: 0, 1, 2, 3, 4, and 5.

Action with Base Six Blocks	Matching Action in Long Division
First, we need to trade the three flats for 30[latex]_{\text{six}}[/latex] (which is 18) longs.	Since “4 cannot go into 3,” we look at the number 340[latex]_{\text{six}}[/latex] instead of 300[latex]_{\text{six}}[/latex].
Next we distribute twenty (32[latex]_{\text{six}}[/latex]) of the twenty-two (34[latex]_{\text{six}}[/latex]) longs into the 4 circles.	Because 4[latex]_{\text{six}}\times[/latex]5[latex]_{\text{six}}[/latex] is close to without going over 34[latex]_{\text{six}}[/latex], we put a 5 in the sixes place. This represents the 5 longs that were put into each circle. Then we subtract 4[latex]_{\text{six}}\times[/latex]5[latex]_{\text{six}}=[/latex]32[latex]_{\text{six}}[/latex] from 34[latex]_{\text{six}}[/latex] to show that we have now used 32[latex]_{\text{six}}[/latex] (which is twenty) of the 34[latex]_{\text{six}}[/latex] (which is twenty-two) longs and have 2 longs remaining.
We have two unused longs that cannot be distributed evenly into the 4 circles in their current form, so we need to break them into twelve (20[latex]_{\text{six}}[/latex]) units, meaning we have seventeen (25[latex]_{\text{six}}[/latex]) total unit pieces.	“Bring down the 5” to show that we are converting the 2 in the sixes place to unit pieces, leaving us with 25[latex]_{\text{six}}[/latex] unit pieces (i.e. seventeen unit pieces).
Now we distribute the units as evenly as possible, placing 4 pieces in each circle, using a total of sixteen (24[latex]_{\text{six}}[/latex]) unit pieces.	Because 4[latex]_{\text{six}}\times[/latex]5[latex]_{\text{six}}[/latex] is more than 25[latex]_{\text{six}}[/latex], and 4[latex]_{\text{six}}\times[/latex]4[latex]_{\text{six}}[/latex] is less than 25[latex]_{\text{six}}[/latex], we put a 4 in the ones place. This represents the 4 units that were put into each circle. Then we subtract 4[latex]_{\text{six}}\times[/latex]4[latex]_{\text{six}}=[/latex]24[latex]_{\text{six}}[/latex] from 25[latex]_{\text{six}}[/latex] to show that we have now used 24[latex]_{\text{six}}[/latex] of the 25[latex]_{\text{six}}[/latex] unit pieces and have 1 unit piece leftover.
We have 1 unit leftover, which is smaller than 4. We are done! Our quotient is what is inside of the circle and our remainder is 1.	We have 1 leftover, which is smaller than 4, so this is our remainder. Our quotient is what is above the line (what is in a circle) with remainder 3.
Answer:  [latex]54_{\text{six}}[/latex] remainder [latex]1_{\text{six}}[/latex]	Answer:  [latex]54_{\text{six}}[/latex] remainder [latex]1_{\text{six}}[/latex]

There are several ways to figure out the multiples for a number in a given base. One way is to use repeated addition. If you are trying to solve [latex]4301_{\text{five}}\div 13_{\text{five}}[/latex], you need the multiples of [latex]13_{\text{five}}[/latex] up to 4. We know [latex]13_{\text{five}}\times 1_{\text{five}}=13_{\text{five}}[/latex], so we can start there and repeatedly add [latex]13_{\text{five}}[/latex] until we have our first few multiples.

[latex]13_{\text{five}}\times 1_{\text{five}}=\color{blue}13_{\text{five}}[/latex]

[latex]13_{\text{five}}\times 2_{\text{five}}={\color{blue}13_{\text{five}}}+13_{\text{five}}={\color{red}31_{\text{five}}}[/latex]

[latex]13_{\text{five}}\times 3_{\text{five}}={\color{red}31_{\text{five}}}+13_{\text{five}}={\color{cyan}44_{\text{five}}}[/latex]

[latex]13_{\text{five}}\times 4_{\text{five}}={\color{cyan}44_{\text{five}}}+13_{\text{five}}=112_{\text{five}}[/latex]

Repeated Subtraction Model and Algorithm

Long division is the standard algorithm for division because it keeps work condensed nicely, but it is not the most intuitive algorithm. Repeated subtraction, while more lengthy to write out, feels more intuitive to most people. Again, we will start with blocks and move into numbers.

Exploration Activity

Pull out your base ten blocks. We will solve the base ten division problem [latex]86\div 12[/latex] using repeated subtraction.

a. Lay out 86 in your base ten blocks.

b. We need to make groups of size 12. Separate out as many groups of size 12 as you can without trading any blocks.

c. Trade one long for 10 units and continue making groups of size 12. (You should have 2 unit pieces leftover.)

We know that the answer when using repeated subtraction is the number of subsets/groups made. There are 7 subsets of size 12 and there are 2 units that were not put into any subset. The answer is 7 remainder 2.

As with partitioning into subsets, it is not practical to use blocks for larger numbers, so we need to develop an algorithm that matches the blocks.

Action with Base Ten Blocks	Matching Action with Repeated Subtraction Algorithm
We need to make groups of size 12, which is one long and two units each. We can separate out three groups of this size. (Picture)	Begin subtracting 12 repeatedly. We can subtract 12 three times before we do not have enough ones to continue subtracting in the current form. Notice, we are able to subtract 12 three times, matching the three groups we were able to make with the blocks.
In order to continue making groups of size 12, we need more unit pieces, so break one long into 10 unit pieces, leaving 4 longs and 10 units. (Picture)	Take one from the tens place and convert it to ones, leaving 4 tens, and 10 ones. This matches converting one long into unit pieces.
Now we can make 4 more groups of size 12 and that will leave 2 unit pieces left unused. (Picture)	Continue to subtract 12 repeatedly. We can subtract 12 four more times before we have too few pieces to subtract any more. Each [latex]"-12"[/latex] matches with creating another group with the blocks.
Total we have created 7 groups/subsets of size 12 and have 2 unit pieces leftover.	Total we have subtracted 12 seven times and have 2 leftover after all of the subtractions.
Answer: [latex]7[/latex] remainder [latex]2[/latex]	Answer: [latex]7[/latex] remainder [latex]2[/latex]

Repeated Subtraction in Other Number Bases

As with long division, this same process can be done with other number bases. Pay close attention when subtracting to be sure you do not use base ten values! For example, [latex]10-9=1[/latex] in base ten, but [latex]10_{\text{twelve}}-9_{\text{twelve}}=3_{\text{twelve}}[/latex]. Let’s consider [latex]3\text{E}_{\text{twelve}}\div \text{T}_{\text{twelve}}[/latex]. (Recall that “E” is the symbol for eleven and “T” is the symbol for ten.) Since we will be subtracting [latex]\text{T}_{\text{twelve}}[/latex], it will be useful to list our addition facts for [latex]\text{T}_{\text{twelve}}[/latex]. We need all of the digits possible in base twelve: 0, 1, 2, 4, 5, 6, 7, 8, 9, T, and E. Remember that [latex]10_{\text{twelve}}[/latex] is equivalent to the base ten number, twelve.

Before beginning a repeated subtraction problem in an unusual number base, it is useful to list out your addition facts for all digits appearing in the divisor. For example, for the problem [latex]244_{\text{five}} \div 32_{\text{five}}[/latex] you would need the addition facts for both the digits 2 and 3.

(Fast) Repeated Subtraction

For the problems above, repeated subtraction wasn’t too tedious, but when facing a division problem like [latex]4025\div 3[/latex] it is not practical to repeatedly subtract 3. It would take over 1,000 steps to finish the algorithm! Thus, we must alter our repeated subtraction algorithm a little bit in order to make the process a feasible number of steps. The key idea is to subtract multiple groups of size 3 at one time instead of individually. We will accomplish this by using easy multiples of 3.

Thought Process	Written Work (Scaffold)
4023 is larger than 3000, which is 1000 groups of size 3. So if we subtract 3000, we are measuring out 1000 groups in one step. This leaves 1025 leftover. In our work we show the number of pieces on the left and the number of groups on the right.
Now we need to find a multiple of 3 close to 1025. We can use any multiple, so let’s use 900. This is 300 groups of 3. Subtracting 900 from 1025 leaves 125. We have now made 1300 groups of size 3.
Continue the process of finding multiples of 3 close to our leftover number. the number 120 is 40[latex]\times[/latex]3, which is close to 125. Subtracting 125 from 120 leaves 5. We have how made 1340 groups of size 3.
Finally, we can subtract one more group 3, to see we have a remainder of 2. The right hand side of our work, shows the total number of groups of size 3 we used.   Totaling up the number of groups created we have [latex]1000+300+40+1=1341[/latex] groups of size 3.   Answer: [latex]1341[/latex] remainder [latex]2[/latex]

Reconsider exercise 7, part b.  You may have had a difficult time choosing efficient powers of ten that worked. Using doubles can sometimes make the problem much more efficient. To use doubles, take your divisor, double it, then double that answer, and keep repeating the process until you have a large number close to your desired number. Suppose you start the problem with subtracting 12,000. Now if you cannot think of a nice multiple of 12 close to 3,437, you can list out doubles to see if any of them are close.

Notice that [latex]12\times 256=3072[/latex] is close to the desired 3,437, so we can use that in our scaffold and continue. This is probably fewer steps than you used when solving exercise 7, part b. In different situations, different approaches will be more efficient. Some people see doubles very easily and some do not. As a teacher, you will need to keep yourself open to the variety of approaches students might take.

The moral of the story is that you can use any multiples you are comfortable using. The closer you can get to the total, the fewer steps you will have.

(Fast) Repeated Subtraction in Other Number Bases

As with our other algorithms, we should extend smart repeated subtraction to other number bases. Consider [latex]4324_{\text{five}}\div 4_{\text{five}}[/latex]. Since we do not have them memorized, we should write out  our multiplication facts for 4 to make it easier.

Thought Process	Written Work (Scaffold)
Note that [latex]4000_{\text{five}}=4\times 1000_{\text{five}}[/latex] is close to [latex]4324_{\text{five}}[/latex]. We can subtract [latex]4000_{\text{five}}[/latex] from [latex]4324_{\text{five}}[/latex] to get [latex]1000_{\text{five}}[/latex] groups of size 4 with [latex]324_{\text{five}}[/latex] unused so far.
Now we need to find a multiple of 4 close to [latex]324_{\text{five}}[/latex]. Notice that [latex]4\times 4=31_{\text{five}}[/latex], so that means [latex]4\times 40=310_{\text{five}}[/latex]. Subtracting [latex]310_{\text{five}}[/latex] from [latex]324_{\text{five}}[/latex] leaves [latex]14_{\text{five}}[/latex]. Since [latex]310_{\text{five}}[/latex] was [latex]40_{\text{five}}[/latex] groups of size 4, we have now made [latex]1040_{\text{five}}[/latex] groups of size 4.
The closest multiple of 4 to [latex]14_{\text{five}}[/latex] is [latex]4\times 2=13_{\text{five}}[/latex]. Thus we can subtract two more groups of size 4. Solving [latex]14_{\text{five}}-13_{\text{five}}[/latex] yields a remainder of 1. Totaling up the number of groups created we have [latex]1000_{\text{five}}+40_{\text{five}}+2_{\text{five}}=1042_{\text{five}}[/latex] groups of size 4 with 1 leftover.   Answer: [latex]1042_{\text{five}}[/latex] remainder [latex]1_{\text{five}}[/latex]