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Repeat Length Again
I found the last two pieces of the repeat length puzzle in the book Dead Reckoning, and now the whole thing is perfectly clear to me. If you want to know the repeat length of some fraction m/n, here's what you should do.
 Make sure the fraction is in lowest terms. After that, the number m is irrelevant.
 Factor the number n and organize it into primepower components p^{k}. As often happens in number theory, the components work independently.
 Throw out the components with p = 2 and 5, or, more generally, with primes that are factors of the number base. (If you're interested, the number of nonrepeating digits is equal to the number of powers of the number base that it takes to cover these components. In base 10, with components 2^{a} and 5^{b}, the number is max(a,b), but in other bases the formula can be more complex. In base 12, for example, the fraction 1/40 has two nonrepeating digits.)
 Use the following rule to reduce the prime powers to simple primes.
len(p^{k}) = p^{k1} len(p)
The rule does have exceptions, but they're unbelievably rare; the only one you need to remember is that for p = 3, the first power of p gets lost on the way through the len sign, so that the exponent is k2 instead of k1.
 Fill in known values for len(p). As I explained in Repeat Length, len(p) always divides evenly into p1; and as I'll explain later, different divisors have different probabilities; but the actual values are unpredictable, you just have to know them, or work them out.
 The repeat length of m/n is the least common multiple of the lengths for the various prime powers.
Now let me tell you the thing that boggled my mind and guaranteed I'd write an essay on the subject. Just how rare are those unbelievably rare exceptions? Well, the first exception after 3 is … 487! That one was mentioned in Dead Reckoning; to find the next I had to resort to the OnLine Encyclopedia of Integer Sequences, specifically A045616. And, the next exception after 487 is … 56598313! And that's all the exceptions anyone has ever found; the next one must be at least 2×10^{11}.
So why 3? Why 487? I eventually figured out an answer that I found satisfactory, but before I can explain it, I need to digress and tell you all about multiplication in base 7. After that, the rest will be easy.
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See Also
Dead Reckoning
Duodecimal
Exceptions Explained, The
Favorite Things
@ April (2006)
