Repeating decimals

by Willie Wong

Here’s something a bit random for Friday. (Presumably this actually is quite well-known; but having never taken a course in number theory…)

Question: Given a fraction m/n in lowest terms, and expand it in “decimal notation” in a number base b. What are the conditions on m, n, and b that guarantees that the expansion eventually consists of a single repeating “digit”? (Note, we can clearly assume 0 < m < n.)

To make the question more clear, let’s look at some examples in base b = 10. Obviously, if n = 1, then the fraction is in fact integral, and its expansion is m.\dot{0} has a repeated digit 0. If n = 2, the decimal expansion is either 0.\dot0 or 0.5\dot{0}. Similarly n = 4, 5, 8, 10 all have terminating decimals, so repeats the digit 0 eventually. n = 3,6,9 on the other hand, will lead to repeating digits other than 0, whereas n = 7 will lead to a repetition of a string of the 6 digits …142857142857…

Here’s the solution. Suppose m/n has an expansion of the prescribed form. Recall that a “decimal expansion”

0.a_1a_2a_3a_4\ldots in base b

is in fact a short hand for

\displaystyle 0 + \frac{a_1}{b} + \frac{a_2}{b^2} + \frac{a_3}{b^3} \ldots.

So the criterion specified in the question is equivalent to the condition that

There exists some integer K, a number c < b^K, and a digit d such that
\displaystyle \frac{m}{n} = \frac{c}{b^K} + \frac{d}{b^K} \sum_{j = 1}^{\infty} \frac{1}{b^j}
which corresponds to the decimal expansion
\displaystyle 0.\underbrace{XXXXXXXXXXXX}_{\mbox{the digits given by }c} ddddddddddd\ldots

The infinite sum on the far right can be solved: \sum_1^\infty b^{-j} = (b-1)^{-1}. Multiplying the expression through by b^K we have

\displaystyle \frac{m b^K}{n} = c + \frac{d}{b-1}

Now, by redefining c \to c\cdot b^k + d \sum_1^{k-1} b^j, we can replace K\to K+k. So we can set K arbitrarily large. Which means that by doing so, after setting the left hand side to lowest terms, we can “remove” from n any prime factors that also divides b. To be more precise: suppose p is a prime such that p^\ell | n and p^{\ell+1} does not divide n. (So that p goes into n exactly \ell times.) Now suppose p|b also, then the fraction b^\ell / n, when written in lowest terms, has a denominator that cannot be divided by p. Repeating this for all common prime factors of n and b we can get rid of all common prime factors from the denominator. Let us denote by n_0 the number n with all the prime divisors of b removed.

Our equation then implies that there exists some integer m' that is coprime with n_0 such that

\displaystyle \frac{m'}{n_0} = \frac{d}{b-1}

which means that n_0 must divide (b-1). That is

Answer: Let n_0 be the number n with all prime divisors of b removed. Then n_0 must divide (b-1).

For base b = 10, (b-1) = 9. This means that any n for which the decimal expansion is eventually repeating with period 1 must have the form

n = 2^s\cdot 5^t\cdot 3^r where s,t are non-negative integers, and r is one of 0, 1, or 2.