The Cube Root Trick

02 Apr, 2024

The Trick

This is a mathemagic stunt I love: if you tell me any cube number up to $100^{3}$ , I can instantly tell you its cube root. $103, 823$ ? That's $47^{3}$ . It's a fairly well-known trick that I first learned from this cheesy old video aimed at amateur magicians. Here's the basic procedure:

Look at the digits before the comma. Find the next perfect cube below that, in this case $64$ (since $125$ is too large). The cube root of $64$ will be our first digit.
Look at the last digit of the cube.
- If it's a $1, 4, 5, 6, 9,$ or $0,$ that's also the last digit of your cube root.
- Otherwise, subtract from $10$ : $2$ becomes $8$ and vice versa, and $3$ becomes $7$ and vice versa.
- In this case, the last digit was $3$ so our cube root will end in a $7$ .

And voilà, the cube root of $103, 823$ is $47$ . Just memorize all of your one-digit cubes, and you too can perform this minor miracle. If that's all you wanted to know, you can stop reading now. The rest of this article will be about how the trick can be used for teaching.

Underlying Concepts

I haven't seen anyone use the cube root trick for its full educational potential. I've never seen it implemented in a classroom, and aside from encouraging someone to memorize the one-digit cubes, I've never seen it used in extracurricular materials either. But it relies on a host of theorems and principles, and it would be great for teaching any one of them. I've sometimes challenged students to explain the trick, and they impress me with their insights every time.

Determining the first digit

Let's say we're trying to find the cube root of $103, 823$ . We're going to figure out the first digit by just looking at the first half of the cube: $103$ . That's between $4^{3} = 64$ and $5^{3} = 125$ , so our first digit is a $4$ . Why does that work?

Monotonicity

The function $f (x) = x^{3}$ is monotonic. That means that for any two inputs, the bigger input will always give the bigger output. Without actually checking, I automatically know that $196^{3}$ is bigger than $194^{3}$ , just because $196 > 194$ . Formally, we can state that a function is monotonic if $x_{1} > x_{2} ⟹ f (x_{1}) > f (x_{2})$ . To check if a function is monotonic, we can look at the graph and see that it always seems to slope upward. If we want to prove it formally, we'll usually need a bit of calculus, although algebra is enough this time around: you'll need to show that $x^{3} < (x + ε)^{3}$ . I'll leave that as an exercise for the reader.

With this, we can see that we just need to pin down two cubes that $103, 823$ is between, and that will give us a range of what numbers we may have cubed to get $103, 823$ . In this case, I can tell that $103, 823$ is between $40^{3}$ and $50^{3}$ , so its cube root is between $40$ and $50$ . Next: how did I know the cubes of $40$ and $50$ ?

Powers of products

I can find $40^{3}$ pretty quickly: it's $64, 000$ . If you're experienced with math, that was probably trivial for you too. But let's not overlook that for a middle or even high school student, this might be a bit of magic. We can cube the $4$ and $10$ separately?? Let's break it down:

\begin{matrix} 40^{3} & = 40 \cdot 40 \cdot 40 \\ = 4 \cdot 10 \cdot 4 \cdot 10 \cdot 4 \cdot 10 \\ = 4 \cdot 4 \cdot 4 \cdot 10 \cdot 10 \cdot 10 \\ = 4^{3} \cdot 10^{3} \end{matrix}

We relied on the commutativity of multiplication to reorder the factors, and also on associativity when I ignored the possibility of adding parentheses. With that, knowing $60^{3}$ is as easy as knowing $6^{3}$ and tacking on three zeros.

Determining the second digit

We have a sort of table that let's us look at the last digit of $103, 823$ and figure out the last digit of its cube root. It looks like this:

Cube ends in:	Cube Root ends in:
0	0
1	1
2	8
3	7
4	4
5	5
6	6
7	3
8	2
9	9

Again, why will that work?

Multiplication mod $10$

The first insight here is that when we cube a number, the last digit of the cube only depends on the last digit of the number we cubed. I can tell that $212^{3}$ ends in an $8$ just because $2^{3}$ ends in an $8$ . The principle at work here is modular arithmetic: the last digit of any number is its remainder after dividing by $10$ . For example, $314 = 31 \cdot 10 + 4$ , so $4$ is the remainder.

Generally, multiplying numbers and then looking at the remainder is the same as first finding the remainders and then multiplying. Consider two numbers, $A$ and $B$ . Let's rewrite them in terms of their remainders, like so:

A = 10 a + r B = 10 b + q

We can multiply these using FOIL (or the box method) and get

A \cdot B = 100 a b + 10 a q + 10 b r + p q

When we divide by $10$ , the first three terms will all be evenly divisible. Only the last bit, $p q$ , will have any effect on the remainder. Thus, when we multiply two numbers, the remainder (last digit) only depends on the original remainders (original last digits).

Of course, younger students can use long multiplication to find out the same thing just by seeing that the first step is the only one that contributes to the product's final digit. Formal isn't always better, and I usually choose long multiplication over the formal way, even with adults.

Injective functions

This method won't work for some powers, like squares. Why? Because $1^{2}$ and $9^{2}$ both end in a $1$ . If we see a $1$ at the end of a square, we can immediately tell that its square root ended in either a $1$ or $9$ , but we don't know which of those digits it was. The problem is that the squaring function mod $10$ has overlaps: two different inputs could give you the same output.

Cubing mod $10$ , however, doesn't have this problem! It is injective, meaning that two different inputs will always give two different outputs. It just so happens that the cubes of one-digit numbers all have different values mod $10$ (that is, different last digits). This allows us to work backwards and find an inverse function, taking us from a cube to its root.

Extensions

Brown & Walter (2005) suggest the "what if" and "what if not" methods for generating interesting problems. I've used their strategies to come up with extensions — but be warned, there are some tough ones here that I haven't even solved yet! You can ponder these questions to deepen your (or your students') understanding:

We used cubes. Does the trick work for other powers?
- Which other powers will it work for in general?
- How might you modify it to work for powers where it fails?
We used base $10$ . Would the trick look different in other base systems?
How could you trip up someone using this trick?
- If someone tried to trip you up like that, could you catch it? How?
Could you reverse the trick to quickly estimate the cube of a number?
- How precise could that estimate be?

#math #teaching