**By Harold Reiter, posted February 27, 2017 —**

This blog was inspired by a message from Roger Howe. The earliest known published version of the riddle below comes from a manuscript dated to about 1730 (but differs in referring to nine rather than seven wives). The modern form was first printed circa 1825:

As I was going to St. Ives,

I met a man with 7 wives.

Each wife had 7 sacks.

Each sack had 7 cats.

Each cat had 7 kits.

Kits, cats, sacks and wives,

How many were going to St. Ives?

The rhyme is generally thought to refer to St. Ives, Cornwall, when it was a busy fishing port and had many cats to stop the rats and mice from destroying the fishing gear. Some people argue that it refers to St. Ives, Huntingdonshire, an ancient market town.

Because only the riddle’s poser is “on his way” to St. Ives, the trick answer is 1. But if we leave out the speaker, the intention is to ask the sum

7
+ 7^{2} +
7^{3} + 7^{4}.

Rather than simply crunching the numbers on a calculator, a teacher might write this:

Let
*S* denote the sum we seek. Then
compute 7*S*. Next consider 7*S* – *S*:

7*S*
– *S* = 7^{2} +
7^{3} +
7^{4} +
7^{5} – (7
+ 7^{2} +
7^{3} +
7^{4})

=
7^{5} + 7^{4} – 7^{4} +7^{3} – 7^{3} +7^{2} – 7^{2} – 7

=
7^{5} – 7 = 7(7^{4} – 1)

=
7(7^{2} – 1)(7^{2} + 1)

= 7 · 48 · 50 = 6 · 2800

So *S* = 2800. Adding the speaker, we get
2801 “things” going to St. Ives. Where is the algebra? So far, there isn’t any.
But replace the integer 7 with the letter *x*,
and consider *xS* – *S*:

*xS* – *S* = *x*^{2}
+ *x*^{3} + *x*^{4} + *x*^{5} – (*x* + *x*^{2}
+ *x*^{3} + *x*^{4})

= *x*^{5} + *x*^{4} – *x*^{4} + *x*^{3} – *x*^{3} + *x*^{2} – *x*^{2} – *x*

= *x*^{5} – *x*
= *x*(*x*^{4} – 1)

= *x*(*x*^{2} – 1)(*x*^{2} + 1)

This leads to the surprisingly lovely formula

Now
we have a solution to any of the possible variations of the riddle; we can
replace *x* with any value, say 9, as
in the original riddle. A teacher who is aware that students are eventually going
to need to understand the second general solution can pave the way by
discussing the specific case first.

Similarly, the finite geometric series

*S* = *a* + *ar*
+ *ar*^{2} +
... + *ar ^{n}*

leads to

*rS* – *S*
= *r*(*a* + *ar* + *ar*^{2} +
... + *ar ^{n}*) – (

*a*+

*ar*+

*ar*

^{2}+ ... +

*ar*)

^{n}

=
*ar ^{n}*

^{+1}–

*a*

and
*S* = (*ar ^{ n}*

^{+1}–

*a*)/(

*r*– 1).

In the case where Ι*r*Ι < 1, we have *r ^{n}*

^{+1}

**→**0 as

*n*→∞, so the infinite series converges, and we have

*S*=

*a*/(1 –

*r*).

As
another extension, consider the series for which the *n*th term is *n*/2* ^{n}*. Multiply the infinite sum

by 1/2 to get

and subtract to get

which
we recognize is just 2. Hence, our original sum is *S* = 4.

Even a third-grade teacher with a firm math background can present the arithmetic in a children’s poem in such a way that students later recognize the beautiful ideas associated with a geometric series.

Harold Reiter has taught mathematics for more than fifty-two years. In recent years, he has enjoyed teaching at summer camps, including Epsilon, MathPath, and MathZoom. His favorite current activity is teaching fourth and fifth graders two days each week.

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Sydney Long- 5/3/2019 6:44:38 PMI really like this riddle. After watching my students work through several simplistic arithmetic and geometric word problems I think this problem would have lead to a better mathematical discussion and a better understanding of why the equations work the way they do. Thank you for sharing I'll definitely be saving this one.

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Aaron Chaney- 4/28/2018 10:23:39 PMThis is a very interesting riddle that is a fun way of bringing math into any classroom where you have time to read a riddle and then give the students an opportunity to work it out. Like the article says even a third grade teacher with a strong math background could present this.

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