Exploding Dots
1.6 Division
Lesson materials located below the video overview.
One of the many delights of the dots and boxes approach is that it explains long division with absolute conceptual ease!
Let’s first be clear on what division is.
In asking as division question such as \(3906 \div 3\), we’re asking for how many groups of three can we find in a picture of \(3906\) dots drawn on a page.
Cool Word: The division sign \( \div\) has an unusual name. It is called an obelus. Not many people seem to know this.
And without drawing the picture we can “see” that there will be 1000 groups of three among 3000 of the dots, 300 groups of three among 900 of the dots, and 2 groups of three among the remaining 6 dots.
The workings of this division problem are particularly easy to see in the \(1 \leftarrow 10\) system:
We are asking:
How many groups of \(3\) are to be found in \(3906\)?
Now, three dots looks like so we are really asking:
How many groups of can we see in the picture?
There is one group of three at the 1000s level, and three at the 100s level, none at the tens level, and two at the 1s level.
We have that 3 can be found in 3906 one thousand, three hundred and two times. That is,
\(3906 \div 3 = 1302\).
Question 27: Draw a dots and boxes picture of \(20486\). Use your picture to show why \(20486 \div 2\) equals \(10243\) . 
Let’s try a harder one. Consider:
\(402 \div 3\).
Here’s the picture:
and we are looking for groups of three dots: .
There is certainly one group at the 100s level.
and now it seems we are stuck – there are no more groups of three!
What can we do now? Are we really stuck?
Do you want a hint?
HINT: If our \(1 \leftarrow 10\) rule means that each group of \(10\) becomes one dot in the box to the left, what might happen if we move one dot to a box to the right?
EPIPHANY: We can UNEXPLODE dots!
Since each dot is worth ten dots in the box to the right we can write …
and find more groups of three:
There is still a troublesome extra dot. Let’s unexploded it too …
giving us more groups of three:
Finally we have the answer!
\(402 \div 3 = 134\).
Question 28: Compute \(62124 \div 3\) via the dots and boxes method. 
Question 29: Compute \(61230 \div 5\) via the dots and boxes method. 
Let’s go up another notch of difficulty. Consider \(156 \div 12\).
Here we are looking for groups of \(12\) in the picture:
What does \(12\) look like? It can be twelve dots in a single box:
We don’t see any of those twelve single dots, but twelve can also be the result of an explosion:
and we certainly see some of these in the picture. There is certainly one at the tens level …
(REMEMBER: With an unexplosion this would be twelve dots in the tens box.)
And there are three \(12\)s at the ones level. (With unexplosions those groups of twelve really do reside at the ones level.)
We see:
\(156 \div 12=13\).
Question 30: Compute \(13453 \div 11\) via the dots and boxes method. 
Recall that all solutions appear in the COMPANION GUIDE to this EXPLODING DOTS course.
Question 31: Compute \(4853 \div 23\) via this method. 
Here’s a bigger example:
What’s \(214506 \div 102\)?
STOP AND CONSIDER: Try this before continuing on. What do the \(102\) dots look like?
Here goes:
and we are looking for groups of \(102\) in this picture. These look like:
We can do it!
\(214506 \div 102 = 2103\).
Question 32: Compute the following using dots and boxes.
\(2781 \div 3\)\(31824 \div 102\)\(4061 \div 31\)\(2798 \div 11\)
The fourth one is interesting. What do you think your result means? 
Question 33: EXTRA PRACTICE IF YOU WANT ITCompute the following using dots and boxes.
\( 6412 \div 3\)\(44793 \div 21\)\(6182 \div 11\)\(99916131 \div 31\)\(637824 \div 302\)\(2125122 \div 1011\)

AHA MOMENT:
Question 34: Use dots and boxes to compute:
\(2130 \div 10\).Use dots and boxes to compute: \(41300 \div 100\).

WHAT IS TAUGHT IN THE TRADITIONAL CLASSROOM
We used dots and boxes to show …
\(402 \div 3 = 134\).
Some teachers have their students solve this division problem by using a diagram like the following:
At first glance this seems very mysterious, but it is really no different from the dots and boxes method. Here is what the table means.
To compute \(402 \div 3\) students first make a big estimation as to how many groups of 3 there are in 402. Let’s guess that there are 100 groups of three.
How much is left over after taking away 100 groups of 3?
How many groups of 3 are in 102? Let’s try 30:
How many are left? There are 12 left and there are four groups of 3 in 12.
The accounts for entire number 402. And where doe we find the final answer? Just add the total count of groups of three that we tallied:
\(402 \div 3 = 100 + 30 + 4 = 134\).
REFLECTION: Compare the table with the picture of the dots and boxes method. Do you see that the table is a shorthand summary, in some way, of what is happening with the dots?
WHY DO TEACHERS TEACH THEIR METHOD? Because it is quick, not too much to write down, and it works.
WHY DO WE LIKE THE DOTS AND BOXES METHOD? Because it easy to understand. (And drawing dots and boxes is kind of fun!) But once you understand what is going on, using any other shorthand method is, of course, fine and appropriate.
A COMMENT ON REMAINDERS
We saw that \(402\) is evenly divisible by \(3\):
\(402 \div 3 = 134\).
This means that \(403\), one more than \(402\), should not be divisible by three. It should be one dot too big. Do we see the extra dot if we try the dots and boxes method?
Yes we do! We have a remainder of one dot that can’t be divided. We say that we have a remainder of one and some people like to write:
\(403 \div 3 = 134 R 1\).
Let’s try another one:
\(263 \div 12\).
Here’s what we have …
and we are looking for …
Here goes!
Unexploding won’t help any further and we are indeed left with one remaining dot in the tens position and a dot in the ones position we cannot divide. This means we have a remainder of … eleven.
\(263 \div 12 = 21 R 11\).
Question 35: Use dots and boxes to show that \(5210 \div 4\) is \(1302\) with a remainder of \(2\). 
Question 36: Use dots and boxes to compute \(4857 \div 23\). 
Question 37: Use dots and boxes to show that \(31533 \div 101\) equals \(312\) with a remainder of \(21\). 
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