# How to Do Large Cube Roots in Your Head

Created | Updated Apr 22, 2012

You will have heard of those geniuses who can remember 100,000 digits of pi, who memorise whole books or who even manage to remember Auntie Betsy's birthday. You may not be one of them, but one thing you will certainly be able to do, to the amazement of your friends, is computing the cube roots of numbers up to one billion.

Doubts? Of course you do not have to calculate them the usual way. Instead, you will learn a much simpler method in this Entry.

### A Refresher Course in Cubes and Cube Roots

Just to remind you, a cube is when you take a number and multiply it by itself twice: that is, you take three copies of the number and multiply them together. For example, the cube of 4, or 4 cubed, is 4 × 4 × 4 which is 64. As a shorthand, we use a tiny '3' to denote a cube:

4^{3}= 4 × 4 × 4 = 64

A cube root is the opposite of a cube: given the cube, find the number which needs to be multiplied to produce it. This is the cube root. In the example above, the cube root of 64 is 4.

64 = 4 × 4 × 4

³√ 64 = 4

### Pick a Number, Any Number

Of course, most numbers have cube roots which are not nice round numbers like 4. The cube root of 100, for example, is approximately 4.641588834. The method used in this Entry will not work for such numbers. It only works for numbers which have a whole number as their cube root.

The best way to ensure this, when demonstrating your skills, is to get your friend to pick a three-digit number but to keep it secret from you. Get him to cube it (preferably on a calculator). He tells you the result, then you calculate the cube root in your head, giving him his original number. Gasps all round!

### Cube Roots of Numbers Up to Six Digits

Before tackling this, let's first take a look at numbers up to a million. Although a million already seems to be pretty large, its cube root is just 100. That means you could do the job by memorising 100 pairs of numbers. But the following eleven will also do:

0 | 0 |

1 | 1 |

2 | 8 |

3 | 27 |

4 | 64 |

5 | 125 |

6 | 216 |

7 | 343 |

8 | 512 |

9 | 729 |

10 | 1,000 |

The number in the right column is the cube of the number in the left column. Conversely, the number on the left is the cube root of the number on the right. Note that each of those numbers ends with a different digit. If you cube a number with two digits *ab*, only the second digit, *b*, will affect the last digit of the cube and can therefore be deduced easily. For example, if the cube ends in 3, the cube root must end in 7.

To find *a*, the first digit of the cube root, look at the powers of 10, 20, 30 and so on, which are those in the table above multiplied by 1,000. For example, 90^{3} is equal to 729,000. The biggest of them that is equal or less than *ab* cubed is assigned to *a* in the table. Problem solved!

#### Example

What is the cube root of 21,952? The cube lies between 8,000 and 27,000; so its root has to be between 20 and 30. The last digit equals 2 as in 512, so the second digit of the root must be 8. Therefore the whole root is 28.

### Up to Nine-Digit Cubes

Moreover, with this method we can tell the first and last digit of every natural cube root regardless of its size. With this in mind, we can proceed to cubes up to a billion.

A three-digit cube root *abc* bears only one problem, which is finding the middle digit, *b*. For this, we will need to add another column to our table.

We're going to use a divisibility rule. This is a simple rule which tells us when a number is divisible by another; that is, when it divides evenly into it without leaving any remainder. We will have to look at divisibility by eleven.

We calculate the alternating digit sum of a number by:

taking the last digit, subtracting the 2nd last digit, adding the 3rd last digit, subtracting the next digit and so on, always taking the digits from the right

If we get an answer which is below 0, we add 11 repeatedly until the answer is in the range 0 - 10.

If we get an answer which is 11 or greater, we subtract 11 repeatedly until the answer is in the range 0 - 10.

We're not going to prove it here, but the alternating digit sum gives us the remainder after the original number is divided by 11.

An example would be 184,758: its alternating digit sum equals 8-5+7-4+8-1 = 13. We take 11 from this to bring it into the range 0 - 10, giving us 2. This means that 184,758 is not divisible by 11, because it has a remainder of 2 when divided by 11.

The following table is the original one with an additional column: the remainder when the cube is divided by 11.

0 | 0 | 0 |

1 | 1 | 1 |

2 | 8 | 8 |

3 | 27 | 5 |

4 | 64 | 9 |

5 | 125 | 4 |

6 | 216 | 7 |

7 | 343 | 2 |

8 | 512 | 6 |

9 | 729 | 3 |

10 | 1,000 | 10 |

If you are given a cube between a million and a billion, determine its remainder when divided by eleven, by calculating its alternating digit sum. Then look up in the memorised table which remainder its root has: find the alternating digit sum in the third column, and read the remainder from the first column. For an alternating digit sum of two it would be seven, for example, or for eight it would be two. Do not mix up the columns!

With the remainder *r* and the easily computed first and last digit you can get *b* like this:

b = a + c - r

That is, add the first and second digit, then subtract the remainder to get the middle digit.

#### Example

Find the cube root of 122,023,936. The cube root's digits are determined as follows:

Our cube is more than 64,000,000 but less than 125,000,000, so the first digit of the cube root must be 4.

The last digit of the cube is 6 as in 216, so the last digit of the cube root must also be 6.

The alternating digit sum equals 6-3+9-3+2-0+2-2+1, which is 12. Subtract 11 from this to get it into the range 0 - 10, giving us 1. 1 in the third column corresponds to 1 in the first column, so r is 1. Add the first and third digits and subtract r to get 4+6-1 = 9, the second digit.

So the cube root must be 496 and we have 496^{3}=122,023,936.

After a little practice, you will soon be the centre of every party.