# St. Ives Problem

## Contents

## Classic Problem

*As I was going to St. Ives**I met a man with seven wives**Every wife had seven sacks**Every sack had seven cats**Every cat had seven kits**Kits, cats, sacks and wives**How many were going to St. Ives?*

### Rhind Papyrus Variant

Problem $79$ of the *Rhind Papyrus*, written by Ahmes some time around $1650$ BCE, concerns:

- $7$ houses, each with:
- $7$ cats, each with:
- $7$ mice, each with:
- $7$ spelt, each with:
- $7$ hekat

## Solution 1

This is a trick question, to which the trick answer is:

- $1$.

The narrator is travelling to St. Ives.

The man he meets, with the colossal entourage, is, by assumption of the definition of the word **meet**, travelling in the opposite direction, **from** St. Ives.

Technically, as it is not specified whether the narrator is travelling alone or not, the answer should actually be:

- at least $1$.

## Solution 2

Trick question aside, one can suppose that in order for the narrator to find out all about the unusual nature of this man's ménage, he would have had to spend some considerable time in his company, more than just a tip of the hat as they passed each other.

Hence it can be assumed that they met at a meeting of the ways, for example, when the narrator's path merged with that of the man and his family, and they travelled to St. Ives together.

So the number of entities mentioned in the riddle are counted as follows:

\(\displaystyle 1\) | \(\) | \(\displaystyle \text{man}\) | |||||||||||

\(\displaystyle 7\) | \(\) | \(\displaystyle \text{wives}\) | |||||||||||

\(\displaystyle 7 \times 7\) | \(\) | \(\displaystyle \text{sacks}\) | |||||||||||

\(\displaystyle 7 \times 7 \times 7\) | \(\) | \(\displaystyle \text{cats}\) | |||||||||||

\(\displaystyle 7 \times 7 \times 7 \times 7\) | \(\) | \(\displaystyle \text{kits}\) |

This is a geometric sequence:

\(\displaystyle 7^0 + 7^1 + 7^2 + 7^3 + 7^4\) | \(=\) | \(\displaystyle \dfrac {7^5 - 1} {7 - 1}\) | Sum of Geometric Sequence | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2 \, 801\) |

... from which we subtract the number of sacks (that is, if it is tacitly understood that the count is to be of the sentient beings):

- $2 \, 801 - 49 = 2 \, 752$

... and don't forget the narrator (a surprising number of the treatments of this subject do just that):

- $2 \, 753$

Quite a crowd.

$\blacksquare$

## Historical Note

The similarities between the St. Ives Problem and the version found in the *Rhind Papyrus* are too close to be coincidental.

Leonardo Fibonacci himself included a version in his *Liber Abaci* of $1202$ and $1228$.

Charles Sanders Peirce suggested a link between this problem and the nursery rhyme *This Is the House that Jack Built*, and went on to note that Fibonacci uses the same numbers as Ahmes and performs the calculation in the same way.

It may of course be suggested that the St. Ives Problem and the Rhind papyrus problem are in fact the same problem, having been passed down virtually unchanged for over $36$ centuries.

Hence it may be considered one of the oldest mathematical problems ever.

## Linguistic Note

Prime.mover remembers this riddle as the first in his book of nursery rhymes which he had when he was somewhat younger.

In that edition, the cats, as documented here, each had $7$ **kits**.

The word is an archaic or poetic one for the more usual **kittens**, which can also be seen in some renditions of this riddle.

Some sources use the word **kitts** in order to reduce the possibility of the word being confused with the word **kit** meaning **pack of useful tools and materials**.

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $7$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $7$

- Weisstein, Eric W. "St. Ives Problem." From
*MathWorld*--A Wolfram Web Resource. http://mathworld.wolfram.com/StIvesProblem.html