As I Was Going to St Ives
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"As I was going to St Ives" is a traditional nursery rhyme which is generally thought to be a riddle. The earliest known published version of it dates to around 1730, although a similar problem appears in the Rhind Mathematical Papyrus (Problem 79), dated to around 1650 BC. The words are, in one version, as follows:
- As I was going to St Ives
- I met a man with seven wives
- And every wife had seven sacks
- And every sack had seven cats
- And every cat had seven kits
- Kits, cats, sacks, wives
- How many were going to St Ives?
A second version is:
- As I was going to St Ives
- I met a man with seven wives
- Seven wives with seven sacks
- Seven sacks with seven cats
- Seven cats with seven kits
- Kits, cats, sacks, wives
- How many were going to St Ives?
There are a number of places called St Ives in England and elsewhere.
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[edit] Solution
The solution to the riddle depends in large part on how one interprets the ambiguous language used in it.
The number of people, sacks, and felines involved, regardless of how the riddle is answered, is 2,802, calculated as follows:
- Narrator--one
- Other man--one
- Wives--seven
- Sacks--49 (seven wives times seven sacks/wife)
- Adult cats--343 (49 sacks times seven cats/sack)
- Kittens--2,401 (343 cats times seven kittens/cat)
[edit] Traditional Answer
The traditional answer is only one person: the narrator.
This answer is based on the assumption that the narrator, while heading to St. Ives, met the large party going away from St. Ives (the most common assumption).
[edit] Alternate Interpretations, with same answer
- The narrator could have met the party crossing the road in front of him, neither heading to nor away from St. Ives, but heading elsewhere.
- The narrator could have met the party at a roadside establishment, but the party was not heading to St. Ives.
- The narrator could have met the party in front of their residence.
[edit] Alternate Answers
- Although it is usually assumed that the large party was traveling from St. Ives, it may well be true that they were going to St Ives as was the narrator. Obviously, a large party of eight adults, carrying 49 sacks containing 2,744 felines, will travel far slower than one narrator. But if one makes this assumption, the final question of "how many were going to St. Ives" still leaves room for ambiguity:
- If the term "how many" involves all people, felines, and inanimate objects, the answer is 2,802.
- If the term "how many" excludes inanimate objects, the answer is 2,753.
- If the term "how many" includes people only, the answer is eight.
- Another solution derives from the fact that the narrator only mentions that he met a man with seven wives (plus 49 sacks and 2,744 felines), but does not explicitly state that the wives are present, nor the sacks and felines. Then the answer is either one (if only the narrator is heading to St. Ives) or two (if both the narrator and the man are both heading to St. Ives).
- Yet another solution involves the use of the word had. Using this word could mean either had (in their immediate possession) or had (at one time, but not now):
- If the first meaning, it still depends on 1) the direction in which the man was travelling and 2) whether the entire party, or only the man, was met by the narrator.
- If the second meaning, there is no correct mathematical answer, in that anywhere from one to all seven sacks, cats, and kittens may not be presently with the party.
- One final odd interpretation has the narrator being one of the man's seven wives, heading to St. Ives while meeting her husband along the way. This leaves open numerous answers, depending on 1) if she was meeting her husband coming back from St. Ives or heading to it, 2) whether her complement of sacks, cats, and kittens was with her, her husband, or back at the house, 3) whether the other members of the party were with either her or her husband.
[edit] Rhind Mathematical Papyrus
A similar problem is found in the Rhind Mathematical Papyrus (Problem 79), dated to around 1650 BC. The papyrus is translated as follows [1]:
| houses | 7 | |||
| 1 | 2,801 | cats | 49 | |
| 2 | 5,602 | mice | 343 | |
| 4 | 11,204 | spelt | 2,301 [sic] | |
| hekat | 16,807 | |||
| Total | 19,607 | Total | 19,607 |
The problem appears to be an illustration of an algorithm for multiplying numbers. The sequence 7, 7 × 7, 7 × 7 × 7, ..., appears in the right-hand column, and the terms 2,801, 2 × 2,801, 4 × 2,801 appear in the left; the sum on the left is 7 × 2,801 = 19,607, the same as the sum of the terms on the right. Note that the author of the papyrus miscalculated the fourth power of 7; it should be 2,401, not 2,301. However, the sum of the powers (19,607) is correct.
The problem has been paraphrased by modern commentators as a story problem involving houses, cats, mice, and grain, although in the Rhind Mathematical Papyrus there is no discussion beyond the bare outline stated above. The hekat was 1/30 of a cubic cubit (approximately 4.8 litre).
[edit] Popular Culture
This "riddle" was used to test Bruce Willis and Samuel L. Jackson in Die Hard with a Vengeance. McClane seems to have not heard the rhyme before, while Zeus knows the riddles (along with some later ones), although it is a well-known children's poem. In this case, the villain applies the traditional "twist", in which the correct answer is 1.
The riddle also appeared as a Muppet skit in the first season of Sesame Street, in which a boy, holding a number 7, asks the riddle, in song, to a girl. As she is busy with the figures, he reveals the answer as 1 (just him), and then the disgruntled girl counters with the question as to how many were going the other way, which she answers, using the mathematical figures as described above in "Other Possible Solutions", the same result the section describes; 1 man, 7 wives, 49 sacks, 343 cats, 2,401 kittens, which comes to 2,801. Amazed, the boy says, "How about that?!"
