The book also discusses ‘Futurama’, and explains how 1,729 appears in several episodes, because the mathematician Ramanujan commented that it is the smallest natural number that is the sum of two cubes in two different ways.

1,729 = 10^{3} + 9^{3}

1,729 = 12^{3} + 1^{3}

Of course, there is link between Fermat near misses and Ramanujan’s 1,729, and therefore a link between the mathematics of ‘The Simpsons’ and the mathematics of ‘Futurama’, because:

**Fermat’s Last Theorem looks for solutions to x ^{n} + y^{n} = z^{n}**

**The two ways to form 1,729 can be matched as 10 ^{3} + 9^{3 }= 12^{3} + 1^{3}**

In other words (10^{3} + 9^{3 }= 12^{3}) is a near miss solution to Fermat’s Last Theorem, as it only misses by 1 (or 1^{3}).

I am grateful to Mike Hirschhorn, a mathematician at the University of New South Wales, who pointed out that Ramanujan identified a way to generate an infinite number of near misses of the form:

x^{3} + y^{3 }= z^{3} ± 1

If you want to find out more, then you can read Mike’s papers on Ramanujan and Fermat near misses on his website (39, 40, 107 and 128).

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DAVID x COHEN= 250266547

as

10651 x 23497 = 250266547

Kate added: “I’d love to win this, I sent my last copy of *Fermat’s Last Theorem* to a boy to impress him. It totally worked.”

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Thanks to everyone who entered the previous David X Cohen puzzle competition – there was such a great response that here is another one.

Once again, this cryptarithmetic puzzle substitutes each letter for a number in order to make the multiplication below valid. This puzzle is based on the name of the great David X. Cohen (a writer on *The Simpsons* and co-creator of *Futurama*), such that his middle initial represents a multiplication sign. The nine different letters in DAVID COHEN represent nine of the digits 0 to 9 (which means it is trickier than the previous puzzle, as it was restricted to the digits 1 to 9).

**DAVID x COHEN = 250266547
**

**Competition closed** – answer here.

Big thanks to James Grime (@jamesgrime), who invented this puzzle.

]]>DAVID x COHEN= 763860049

as

13241 X 57689 = 763860049

He summarised his method as: “Started from the lowest digit (there is only one assignment for D and N that works) and worked my way up.”

In other words, obtaining 9 as the final digit of the product means that the final two digits of DAVID & COHEN have to be 3 & 3, or 7 & 7, or 1 & 9, or 9 & 1. However, 3 & 3 and 7 & 7 cannot be correct, as the digits have to be different. Similarly, 9 & 1 cannot be correct, because if D=9 then it both ends and starts with a 9, and you that would force COHEN to be a 4-digit number, when it has to be a 5-digit number – does that make sense?

Once you have got as far as 1AVI1 x COHE9 = 763860049, then you can pick away at the rest of the number.

Alternatively, I know that some people used a brute force computer search approach. Moreover, some guessed that the solution might be a pair of primes, which made searching much easier.

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This cryptarithmetic puzzle substitutes each letter for a number in order to make the multiplication below valid. This puzzle is based on the name of the great David X. Cohen (a writer on *The Simpsons* and co-creator of *Futurama*), such that his middle initial represents a multiplication sign. The nine different letters in DAVID COHEN represent the digits 1 to 9.

**DAVID x COHEN = 763860049**

**Competition closed** – answer here.

Big thanks to James Grime (@jamesgrime), who invented this puzzle.

If you are new to cryptarithmetic puzzles, then here is a primer.

]]>The problem is that Ptolemy’s theory can be used to justify any set of orbits, because the epicycles can be adjusted to describe any path. Indeed, a sufficiently complex and honed set of epicycles can even describe a planetary path that draws an outline of Homer Simpson.

More about epicycles here and in loads of other places.

]]>was that everyone on the Math Team was co-captain, so that they all could put it on their college applications.

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Mathematics Team. As well as Mr. Kozikowski, who coached the team and

appears in the photograph, Reiss had many other mathematical mentors.

For example, Reiss’s geometry teacher was Mr. Bergstromm. In an episode

titled “Lisa’s Substitute” (1991), Reiss showed his gratitude by naming Lisa’s

inspirational substitute teacher Mr. Bergstromm.

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