Exploding and Collapsing Boxes


  1. Naming Conventions for boxes
  2. Getting Started
  3. Big Sums
  4. Square and Triangular numbers

Authors: Brian, Deepu, Fiona, Jamison, Johann, Levi, Lexi, Matthew, Sam, Travis, Zoe


1 ← 10 Procedure: Let's say that you are at school and your teacher asks you about the 1 ← 10 rule. Do you freak out? Or do you think about it? Lets say the number 100. How are you going to fit it in three boxes? So 100 divided by 10 = 10 With 0 left in the third box. Then 10 divided by ten is 0 in the second box and 1 in the First box. How about 555? 555 divided by 10. 555 divided by 10 is 55 with 5 left behind. So 5 will be left in the third box and 55 goes to the 2nd box. 55 divided by 10 = 5 in the second box and 5 in the First box.


Exploration One Q: 11

1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37+39+41+43+45+47+49 = 625

I figured out that the middle number times itself equals the answer. Plus count the numbers to get the answer. Also multiply the vertical number of dots by the horizontal number of dots to get the total dots.

Exploration Three-Q:1
In exploration three the code for the number ten is: 1010


Naming Conventions for boxes

Do you know your place values? In the base 10's system you have 1's, 10's 100's, 1000's and on and on. So the 1's place is 10^0, or ten to the zero power. Then that means the 10's place is 10^1, or ten to the first power. So the 100's place is 10^2, or ten to the second power. For the last place value that I am going to tell you is the 1000's place which is 10^3, or ten to the third power. Now I am going to show you a naming convention that is not wrong but will have troubles later on.

This is what they did, they only had three boxes and they named the first one on the left the "first" box then the one next to it the "second" box and the last one the "third" box. This is the problem, if they had a large number like one thousand five hundred thirty three, they would not have enough boxes to fit the dots in after they explode. Another problem would be the naming because it would be hard to name the boxes to the left.

Here is another naming convention that has problems. I am going to start naming the boxes from right to left, calling the farthest box to the right the "first" box, the next box to the left the "second" box, and so on. A good thing is that it can have as many boxes as you wish. The problem with it is that the powers will get all confusing because the exponent on the "first" box is zero, the exponent on the "second" box is one, and all of the other names are off by one. The last naming convention I am going to tell you about is when you can have as many boxes as you want and the box farthest to the right is called the 0th box then the next box to the left is called the 1st box and on and on and on if you know your numbers.

100 × 100 = 10,000


1 ← 2 Procedure: When we first started out, we had four boxes and a certain number of dots in the box the farthest to the right. Then we had a challenge that Amanda gave us, and it was pretty easy. It was to take a pair of dots from the box, and make it into one in the next box over. You keep doing it until there are none left in the box farthest to the right and the same number divided by two in the 1st box over. If the number in the 0th box is odd, for example, 7, you would explode the dots (divide them by two) until there is only one left. Then you move onto the 1st one over. There should be half of the number you originally had in the 0th box if that number was even. If it was odd, then there should be 1 dot left over. Once there are 1 or 0 dots in the 0th box, then move onto the 1st. Then you keep exploding pairs of dots, and doing the same thing that you did in the 1st one until you have all four boxes filled up with either 1 or 0 dots.

Example: If you are doing base two and the problem was six, you would put six dots in the 0th box and you group them into twos and then you would put three dots in the 1st box, leaving none in the 0th. Then you would leave 1 dot in the 1st box, and have 1 in the 2nd, and that would explode into 0 in the 3rd box. So in the 1 ← 2 theory, the code for six is no dots, one dot, one dot, no dots, or 0110.

Theorem: The 1 ← 2 theory is the same as binary. It is just another way to figure out the code for binary. It is important to remember this.

Dark Brian

I am going to tell you how to use the 1-2 code. Like I said back up there.

Try the number 4. 4 divided by 2 equals 0 in the third box 2 in the second box, and then 2 divided by 2 equals 0 in the in the second box and 1 in the first box. So it should look like this (1) (0) (0). Another example is using the number "6". 6 divided by 2 equals 3 in the 2 box and 0 third box. 3 divided 2 equals 1 in the 2 box and 1 in the 1 box.

1 ← 3 Procedure: The 1 ← 3 procedure is a bit like the 1 ← 2, but instead of exploding 2 dots, you explode 3. It's as simple as that!

Example: This 1 ← 3 example is as follows: 9 divided by 3 is 3, so you put three in the next box over. 3 can be divided by 3 to make 1, so you put a 1 in the next box over, and a zero in the last box because it ends on 1. So 9 dots becomes 100.

Theorem: The 1 ← 3 theory is the same as base 3. Just like how the 1 ← 2 theory is base two, the 1 ← 3 theory is the same as base 3.

Now here's something to think about: If there were a 1 ← 4 theory, what would it be like? Even with out having studied it, I can assume that it it means dividing by 4, and leaving 1 to 3 dots in the boxes. That would be base 4. Once you have done one or two of the the theories, then you start to get the idea. Now we can move on to fractional bases.

Actually, before we move on, I will explain what the pattern is: Divide by the number of the base you are trying to write in. If the number was, say, 16, and your base was 4, then your ending number woud be four, because 16 divided by four is 4.


Some of us have not learned base four but we are looking forward in learning it this Sunday.


We are also going to try to ask Amanda to teach us fractional bases.


Two Sundays ago Zoe taught me a method that was hard when I first learned it but then it got easier when I understood it well, so you can teach yourself many other ways to do bases.


I taught Fiona and another girl working on it, Mia, how to write in base two without having to do the dots and boxes. I would explain the formula here, but even though it is quite simple I can not, since it requires writing straight horizontal lines that this computer can not do. It is a matter of memorizing the places of the blank spots...which is a lot easier than it sounds.


I have explained to you about many different bases, but they were all positive and whole. I wonder what it would be like in base 0, 1/3, or -2. Once I show Amanda what we have written, maybe she will teach us those, though I have a vauge idea what they would be like.

Dr. Duck Sam

A fast way of doing 1 ← 5 is round to the to the lowest 5 then divide by 5. Then add the left overs to the box you are working on.


For any system b ← ab, we can determine the number (given the amount in the boxes) by doing the following:

  1. Truncate the sequence of boxes ck, ck-1, ..., c1, c0 by removing c0 and rewriting indices to obtain ck, ck-1, ..., c2, c1
  2. Divide the number in each box by b to get c'k, c'k-1, ..., c'2, c'1
  3. Calculate a value m based on the 1 ← a machine

    m = \sum_{i = 1}^k c'_i a^{i-1}

  4. Multiply m by ab and add c0 to get n, the original number that went into the system
  5. This can be written in one step as

    n = ab (\sum_{i = 1}^k \frac{c_i}{b}\cdot a^{i-1}) + c_0\]

  6. Which simplifies to

    n = \sum_{i = 0}^k c_ia^i

For example: If we have 3 ← 6 and the numbers in the boxes are 3, 0, 3, 5

a = 2, b = 3
we have 3·23+0·22+3·21+5·20, which is
24+0+6+5 = 35

Alternately, we can reverse this process by starting with n and doing the following:

  1. Have c0 = n mod ab and m = (n - c0)/ab
  2. Run m through the 1 ← a system to get c'k, c'k-1, ..., c'2, c'1. This is the same as assigning

    c'\''_i = \frac{(m \mod a^i) - (m \mod a^{i-1})}{a^{i-1}}

    for i ≥ 1
  3. Multiply each box by ab to get ck, ck-1, ..., c2, c1
  4. Attatch c0 at the end to get ck, ck-1, ..., c1, c0
Using the same example above, we have:

c_0 = 35 mod 6 = 5
m = 5
c'_1 = (5 mod 2^1 - 5 mod 2^0)/(2^0) = (1 - 0)/(1) = 1
c'_2 = (5 mod 2^2 - 5 mod 2^1)/(2^1) = (1 - 1)/(2) = 0
c'_3 = (5 mod 2^3 - 5 mod 2^2)/(2^2) = (5 - 1)/(4) = 1

then we multiply each one by 3 and add the last to get 3, 0, 3, 5

The next step is to discover the properties of a system which "explodes" a given number of boxes into a smaller number which has no common factors with the first. The smallest example of this is the 2 ← 3 machine. If we write the number of items in each box as 0, 1, or 2, we have the following simple situations for any digit after the first when it is incremented:

0 → 0 2
1 → 2 0
2 → 2 1

If, instad of numbers, we use letters, we can write

a → ac → cb    → cba
b → ca → cc    → cbb
c → cb → cba  → cbc

Thus, after three iterations for any digit, cb is appended to the front. Now we are interested in the behavior of the "cb" system. For our purposes, cb will be written as d_

d_ = cb → cbc = d_c
d_c = cbc → cbaa = d_aa
d_aa = cbaa → cbcb = d_d_
d_d_ → d(d_c) → d(d_aa) = (d+d)aa = d_caa
d_caa → d_cd_
d_cd_ → (d_c)(d_c) → (d_c)(d_aa) = (d_c+d)aa = daaaa

So, after 2, aa is appended. After 3, d changes to d_c. after 3, another aa is appended. so, the rule appears to be:

three iterations → d appears/incremented
three of above → two to left appears/incremented
this continues on forever

From this, we can conclude a pattern, for 3n will cause d to appear (n-1)·2+1 to the left. Thus, if we are interested in the kth digit, we obtain k' by subtracting two. Then, k' = (n-1)·2+3 (d_ adds two) So, k-2 (to be continued)


Getting Started

The counting numbers are the normal numbers that that we use everyday, such as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, and so on. They are called natural numbers, also. Is there a biggest counting number? You may say infinity, but you can't have infinity hot dogs, or infinity computers, can you? Infinity isn't a definite number. So you might say Googleplex, but then there is always Googleplex and one. There is not a biggest counting number.

0 is not a counting number. It is nothing, and there can't be nothing telephones, or keys, so it is not a counting number.

Rule: ab=ba.
We will be working with dots to explain this. 4×5 would be four dots up and 5 over. A bit like a graph. A boy looks at this, and sees 5 vertical rows and 4 horizontal one
Point: The 0 property of Addition
We will use O for the number 0 when working with dots.
The picture for 7 plus 0 would be
. . . . . . . + O = . . . . . . .
(7 dots + 0 = 7 dots)
Point: The 0 property of Multiplication
The picture for 5 × 0 would be blank, because it would be 5 dots down, but 0 across, so consequently 0 dota down, so that must mean 5 × 0 = 0.
Point: The Identity law of Multiplication
The picture for 1 × 4 would be
because it is 4 dots down, and 1 dots across. 1 × 4 = 4.


I learned how to count grids diagonally. For example, If I had a 4 × 4 grid could count diagonally, 1,2,3,4,3,2,1. The middle number tells me I need to take.


Big Sums

On the first problem there were six numbers 1,2,3,4,5, and 6 and the problem was asking Diagonal Dan looks at this picture and instead of seeing six rows of six.

What I learned in explorations one is that to get number 100 you have to go 1+2+3+4+5+6+7+8+9+10+9+8+7+6+5+4+3+2+1=100 because to get that number what is the number in the middle × itself

take the middle number and × it by itself: 10 × 10 = 100


Square and Triangular numbers

The square numbers are numbers that can be represented as a square chart in dots. 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 are the first 10 of them.

Triangular numbers are numbers that can be represented as a triangle chart in dots. 1, 3, 6, 10, 15, 21, 28, 36, 45, and 55 are 10 of them.

The 6th square number and the 8th triangular number, 36, is the same. That is considered a Squangular number. The Squangular numbers are numbers that are both square and triangular.

Rule: Add any two consecutive triangle numbers and the result will always be a square number.
Becuase, if you think about it, if you put two triangles together, it makes the shape of a square.
Rule: Take any triangular number, double it and add its matching square number. The result is always another triangular number.
Because if you take that square that you already made, and add another triangle to any one of its edges, it will make one, much larger, triangle.
Rule: Take any three consecutive triangle numbers, and take six copies of the middle number, and then add them to the other two numbers, and the result will be a square number.
Why? I have an explanation, but there are probably many more. What is yours?


What's next? ????


6 + 6 + 6 + 6 + 6 + 6 = 6 × 6 = 36


The 1 ← 5 machine (Base 5): Every 5 dots in a box "explode" and make one dot in the box on the left.

Example: If there are 42 dots...

  1. 40 of the dots "explode," creating 40 / 5 or 8 dots in the box to the left. The remaining 2 dots stay in the same place.
  2. 5 of the eight dots "explode," creating 1 dot in the box to the left. The remaining 3 dots stay in the same place.
  3. The 1 dot is less than 5, so it cannot explode and stays in the same place.
  4. So 42 in base 10 = 132 in base 5.

The 1 ← 5 machine is the same as base 5.

Diagram of 42 exploded into 3 base-5 boxes

(1 × 25) + (3 × 5) + (2 × 1)
= 25 + 15 + 2
= 42


Alright. So, I've talked about the 1 ← 2 theory and I said that was base 2, right? Well, there's an easier and faster way to do it. There are many base two place values. What I mean by that is.....

In base ten, there are place values, right? 1, 10, 100, 1000, 10,000.... Get it?

In base two, there are place values just like that. 1, 2, 4, 8, 16, 32, and 64, and so on. Each one is the double of the last. Times two, right?

So, say you had the number 11101. Reading from right to left, that would mean you have 1 1s, 0 2s, 1 4s, 1 8s, and 1 16s. So, add those all together and you get 29. 29 is the base 10 answer to 11101, a base 2 number. Try to figure out what 1101 is in binary. Trying to figure out 29 insted of 11101 is called reverse.

I'm wondering what one of the harder bases would be like. This is a really hard one..... what about 1/4 ← 5/4? That would be so hard! an easier one could be 1 ← 1/2. Still, that's pretty hard....

That would be taking half of a dot and moving it over. In the dots and boxes theory. I don't even know what it would be like in place value version.

This is the kind of thing you need to think about. Imagine the unimaginable. Think the unthinkable. Try to make the irrational rational. Change logic, and think about what it would be like if these things happen. Like if another color was created. What would it look like? You can't picture it, since it dosen't look like any color you already know. If you get to where you just can't picture what ever this new thing is like, stop and go back to reality.


You have heard of 1 ← # rules, but what about 29 ← 30? Here is a chart.

90  29,28,27,0

How did I get 29,28,27,0? It is 29,28,0 + 29,0 = 29,28,27,0

+ 0,29,0


Base 3/2, or, the 1 ← 3/2 machine In this section, I will refer to the box furthest to the right as Box A, the box second to the right as Box B, and so on.

An example of the 1 ← 3/2 machine:

Place values in base 3/2:

How to convert from base 3/2 to base 10:

Base 10 equivalent of a number in base 3/2 = (1 × Number of dots in Box A) + (3/2 × Number of dots in Box B) + (2 1/4 × Number of dots in Box C) etc. etc.

Let's try it with the example at the beginning of this section:

(1 × 0) + (3/2 × 0) + (2 1/4 × 1) + (3 3/8 × 1/2) + (5 1/16 × 1) = 0 + 0 + 2 1/4 + 1 11/16 + 5 1/16 = 9

Why does this work? If there are three dots in Box B, that is the same as three groups of 3/2, or 3 × 3/2. If there are three dots in Box C, that is the same as three groups of 2 1/4, or 3 × 2 1/4. The same applies with all the other boxes. Another way of writing the conversion method above is:

[(Number of dots in Box A) groups of 1] + [(Number of dots in Box B) groups of 3/2] + [(Number of dots in Box C) groups of 2 1/4] etc. etc.

An observation: The example at the beginning showed that 9 in base 10 is the same as 1 1/2 1 0 0 in base 3/2. The 1/2 refers to a single digit. So, in base 3/2, it would be ideal to have a symbol for the digit "1/2".

The 1/2 ← 3/2 machine: Picture a machine where 3/2 dots in one box collapse into 1/2 of a dot in the box to the left. (Note: As in the section above, I will refer to the box furthest to the right as Box A, the second box to the right as Box B, and so on.)

An example of the 1/2 ← 3/2 machine:


I would like to say that I have had a very good time here at the riverbend math community research project. It has been educational and fun. I would like to come back next year, but if I can't, I recommend this math center to any kids that like math, or possibly don't.


You have 4 boxes. Say the number is 8. Put the number 8 into dots in the very last box. Then explode the dots into 2. So circle the 2 dots. then put the 2 dots you circled put it into 1 dot and keep on doing that untill you have the answer wich is 1000 for this problem.


The 2 ← 3 machine: In the 2 ← 3 machine, 3 dots in the furthest box to the right (or Box A) collapse to make 2 dots in the second box to the right (or Box B). If there are 12 dots in Box A:

Place values in the 2 ← 3 machine: I have not found a way to calculate place values in 2 ← 3, nor have I found a way to convert from base 3 to base 10. I did make some assumptions about the place values, which I later found were wrong. My incorrect assumptions went like this:

Converting from 2 ← 3 to base 10: Since I thought I knew the first four place values, I tried to use them to convert from the 2 ← 3 machine to base 10. Here's how I tried to do it.

Base 102 ← 3
25  210111
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