The St. Ives problem is a riddle based on the well-known nursery rhyme, "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 kitts. Kitts, cats, sacks, wives, how many were going to St. Ives?" The usual trick interpretation counts only the narrator as going to St. Ives, taking the man and his party to be traveling in the opposite direction. However, the word "met" does not establish their direction of travel. The four categories named in the final question total 2800 items. Under the usual trick interpretation, the narrator is outside those categories and none of their members is going to St. Ives. Counting only the four categories then gives 0, whereas the customary answer 1 counts the narrator.
Should a diligent reader nevertheless wish to calculate the sum total 
of kitts, cats, sacks, wives, plus the man himself, the answer
is easily given by the geometric series
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(1)
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with 
and 
.
Therefore,
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(2)
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Computing the sum explicitly (but grouping cleverly),
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(3)
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(4)
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(5)
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(6)
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(7)
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(8)
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A similar question was given as problem 79 of the Rhind papyrus, dating from 1650 BC. This problem concerns 7 houses, each with 7 cats, each with 7 mice, each with 7 spelt, each with 7 hekat. The total number of items is then
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(9)
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(Wells 1986, p. 71). In turn, the problem of the Rhind papyrus is repeated in Fibonacci's Liber Abaci (Fibonacci 1202, 1228; Sigler 2002).
The riddle appears in the film Die Hard: With a Vengeance (1995), which uses the answer 1.
