The “Four Fours” is a very well-known little problem that encourages some creative thinking and use of the order of operations. It goes like this:

Using exactly four of the number 4, any of the operations +, -, *, / and as many brackets as you like, see if you can produce all the natural numbers from 1 to 20.

Taking these instructions at face value, there are some numbers you simply can’t make with exactly four fours, and it’s rather interesting to watch how people deal with this. Mostly they come up with creative approaches to subvert the instructions, by allowing you to concatenate 4’s to make bigger numbers, or allowing a decimal point (shamefully without a zero in front of it). It worries me that no-one takes a more systematic approach and asks if they could produce all the missing ones with just one of these innovations, or asks exactly how many outcomes there are with the original operations only.

But discussing what happens with the Four Fours is not the purpose of this post. The purpose of this post is to show you four fourfoursesque puzzles I’ve created which have encouraged some great learning.


Zero Zeros

The Zero Zeros problem goes like this:

Using all five of the numbers 10, 100, 1000, 10 000, 100 000,  any of the operations +, -, *, /, and as many brackets as you like,  make as many different numbers as you can whose digits contain no zeros.

I must confess I made this one up tonight in order to make my title have four alternatives rather than three. But it’s still, I think, a very interesting problem. It requires you to think about how zeros appear in the digits of a number when you do operations, and it is surprisingly difficult to come up with even one solution, letalone several (and yes there are several solutions).

UPDATE: We tried this at One Hundred Factorial today (the day after the post). Originally we used just the first four numbers 10, 100, 1000, 10 000, but every single person came up with the same first solution and then got stuck. Using five numbers seems to give simultaneously more scope for different answers and a more varied set of first answers. Either way, it was as interesting as predicted!

Only Ones

The Only Ones problem goes like this:

Using any of the operations of addition, subtraction, multiplication, division, and powers, as well as as many brackets as you want, and also as many of the number 1 as you need, make each of the numbers from 2 to 20. For example, here is a way to make 17: (1+1+1)^(1+1)*(1+1)-1 = 17.
What is the smallest number of the number 1 needed to produce each of the natural numbers from 2 to 20?
(Note: you can’t concatenate to make numbers like 11 — each 1 must stand alone as its own number.)

I find this one much more rewarding than the Four Fours. Firstly it allows for powers, which adds another operation to practice. Second, it requires a bit of working systematically in order to be sure you have the smallest number of 1’s possible. Thirdly it doesn’t create as much of a need to extend the operations allowed, because it’s quite interesting enough all by itself. Finally, it also has a nice extension problem which is to find out if there is a number whose digits are all 1, and can be calculated using the same number of 1’s as appear in it.

Here is a photo of Year 11 students working on this problem. When I suggested that perhaps they need to be more systematic, they decided to take a divide-and-conquer approach.


Pi on the Floor

The Pi on the Floor problem goes like this:

Using any of the operations of +, -, *, /, as many of the number π as you need, and as many of the floor function ⌊·⌋ as you need, make each of the whole numbers from 1 to 20. What is the least number of π’s required to make each number?

I made this one up on the spot at the One Hundred Factorial session on Pi Day (American Pi day anyway). I wanted something fourfoursesque that used the number pi, and it occurred to me that the floor function would allow us to create integers pretty easily. Little was I prepared for the interest it would generate or the clever maths that would be created by the students in response. I learned so very much about how the floor function worked, and about how close some multiples of pi were to various whole numbers.

Here is a photo of some people working on the problem. (I have blurred out the solutions though, so you can do them yourself!)


The i’s Have It

The i’s Have It problem goes like this:

Using any of the operations +, -, *, /, as many brackets as you want, and as many of the complex number number i as you need, make each of the numbers a+bi for a, b = -2, -1, 0, 1, 2. How does your answer change if you are not allowed to use the – symbol?


We did this one at One Hundred Factorial in January 2016 and it was a most interesting problem. You had to really use the fact that i is an ordinary number that does actually do ordinary operations, especially to produce the real number results. Amie Albrecht suggested the extension problem of not being allowed to use the – sign, which forces you to use the fundamental property of i that i*i = -1 and very much changes the answer! 0 in particular is much harder to make without using minus!


So there you have it. Four alternatives to the Four Fours. I like them all better than the Four Fours, mostly because none of them imply that you can do something that’s impossible, and some actually challenge you to find the most efficient solution. Plus, they all make you think about functions or numbers or properties of number that you don’t think about that often. I and various friends, students and strangers have had a lot of fun thinking about them, and have learned a lot. Try them out yourself and tell me how they go!


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When doing algebra and solving equations, there is this move we often make which is usually called “doing the same thing to both sides”. For many people it is their fundamental rule of algebra. (It’s not mine, but that’s a discussion for another day.) You use it when solving an equation like this:

4x-3 + 3 = 13 + 3
4x = 16
4x / 4 = 16 / 4
x = 4

On the second line we “added 3 to both sides” and on the fifth line we “divided both sides by 4″. The “sides” being referred to here are the two sides of the = sign.

Quite recently this phrase of “both sides” has begun to bother me.

It started when I was trying to explain to students how they needed to be careful when they were solving equations using matrices. In their question, the matrix A was invertible and they solved for X like this:

XA = B
X = A-1B

Imagine for a moment explaining to a student why this is not true. My first attempt was “When you’re solving this equation, you’re multiplying both sides by A-1. With matrices you have to be careful which side you multiply the matrix on, so you have to make sure you multiply on the same side of both sides.” Crap. I need a different word for the sides of the equation and the sides of the matrix!

It got worse when I was helping students manipulate inequalities. Consider this one:

4 < x + 2 < 5
4 – 2 < x + 3 – 2 < 5 – 2
2 < x < 3

How would you describe what I did on the second line there? I subtracted 2 from all three … all three what? Certainly not all three sides! What should I call those things? Parts?

And then there were all those times when students did this with their differential equations:

ln(y) = 2x + C
y = e2x + C

How do you explain to students that the C belongs up in the power of e? You could say, “You have to do the ‘e to the power of’ to this whole side.” And at least some will reply, “I need to do it to everything on both sides, right…”

ln(y) = 2x + C
y = e2x + eC

And now you have to dust off your emergency stash of extra patience.

Then suddenly it occurred to me. I should never call it a “side” because that’s not what it is. I’m not talking about locations relative to the = sign, I’m talking about two objects that the = sign is relating. The = sign or the < sign is telling me how things are related to each other. Those things are complete whole objects in their own right that may be written as some algebraic expression showing how they are constructed, but they really are single things. That is, it’s more (4x-3) = (13) rather than 4x-3 = 13.

So from now on I’m not saying “sides” of an equation or inequality. I will start calling them by what they are: “Divide both of these numbers by 4″, “Multiply both of these matrices by A-1 on their right”, “Subtract 2 from all three numbers”, “Raise e to the power of both expressions”. Failing that, I’m going to use the good old fashioned all-purpose word “thing”.

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Upon Amie and Cathy‘s request, I am writing a blog post about a problem we worked on at One Hundred Factorial recently. In fact, in order to do so I am creating a whole new category for the blog called One Hundred Factorial, so I can talk about the things that happen there. (Just so you know, One Hundred Factorial is the name of our regular puzzle-solving gathering of staff and students here at the Uni of Adelaide. I tweet about what we do under the hashtag #100factorial.)

Anyway, a few weeks ago we became interested in problems involving removing spots from dice. A good place to start is the following puzzle, which we created during our investigations:


In order to solve this, it might be useful to know the ways that we envisaged the dots on a die are arranged, which are shown below. We don’t really care about the orientation of each face, only the relative arrangement of the dots. Also, it might be very useful to know that on a standard die the faces opposite each other add to 7. (Go look at a real one and you’ll see it’s true.)


If you have a go at the puzzle, you’ll probably find yourself doing a bit of quite sophisticated reasoning using various if-then statements. I can imagine quite an interesting exercise of writing down all these statements and arranging them into a proper proof of what the original die must have looked like. If you’ve got a group of people, see if you can come up with multiple different proofs that follow different lines of reasoning.

But that’s just the starter! The real fun for us happened when I asked how I might change the question. In the problem above, we removed four of the spots and were still able to tell which face was which. Just how many of the spots can we actually remove?


This problem is very interesting indeed, especially because there are multiple ways to interpret the question. If you own the die and are familiar with how the faces were arranged relative to each other before you removed the spots, you might get a different answer than if you were a new user of the die who only knew it was a die with some spots removed. As you work through them, you can create some puzzles of your own like the intro puzzle (that’s how the intro puzzle was created in the first place, but I thought it’s a good way to get started with the concept).

There are several other questions you can ask that we didn’t even begin to delve into yet, but I would love to think about one day:

  • How many spots can you remove and still be able to tell what number was on each face? (Did this one already)
  • How many spots do you have to remove to make it so you can’t tell what some faces are?
  • How many spots can you add to the die and still be able to tell what number was on each face?
  • How many spots do you need to add to make it so you you can’t tell what some faces are?
  • How does all of this change if you don’t require/know that opposite faces add to 7? (thanks for this Fred Harwood.)
  • How would all of this change if the faces of the die were some other numbers from 0-9?

PS: I should say that the inspiration for this puzzle was from seeing Eric Harshbarger’s wonderful dice collection, though the versions of the puzzle here are all my own (with the help of some of the other puzzlers from One Hundred Factorial).

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If you google “fundamental trig identity” you will get many many images and handouts which all list the fundamental trig identity as:
sin2 t + cos2 t = 1
This is, of course in the wrong order and it should really have cos first then sin, like this:
(cos t)2 + (sin t)2 = 1
“But David,” you say, “it’s addition, […]

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In the Drop-In Centre, the majority of students visit to ask for help learning in a very small number of courses, mostly the first-year ones with “mathematics” in the title. Of course, any student from anywhere in the uni can visit to ask about maths relating to any course, and we do see them from […]

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After writing the previous blog post (Finding errors by asking how your answer is wrong) and rereading one I wrote three years ago (Who tells you if you’re correct?), I got to thinking about how students are supposed to learn how to check if they are right.
It occurred to me that, at least at university, […]

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One of the most common situations we face in the MLC is when a student says, “I’m wrong, but I don’t know why”. They’ve done a fairly long calculation and put their answer into MapleTA, only to get the dreaded red cross, and they have no idea why it’s wrong and how to fix it. […]

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Yesterday the Maths 1M students handed in an assignment question that asked them to prove a property of triangles using a vector-based argument. It’s not my job to help students do their assignment questions per se, but it is my job to help them learn skills to solve any future problem. This kind of problem, […]

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Did you know that cats have scent glands just inside their bottoms that are constantly being filled with liquid and are squeezed as their poos come out, and if their poos are too skinny the glands are not squeezed enough and get over-full making them very painful and inflamed? Neither did I, until my cat […]

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One of my friends and a past MLC staffer graduated from her PhD yesterday (congratulations Jo!). One of my strongest memories of Jo is when she told me something about my teaching that I never knew I was doing, but that she saw as an essential part of what I was trying to achieve at […]

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