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
has a repeated digit 0. If n = 2, the decimal expansion is either
or
. 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”
in base
is in fact a short hand for
.
So the criterion specified in the question is equivalent to the condition that
There exists some integer K, a number
, and a digit d such that

which corresponds to the decimal expansion
The infinite sum on the far right can be solved:
. Multiplying the expression through by
we have
Now, by redefining
, we can replace
. So we can set
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
. To be more precise: suppose
is a prime such that
and
does not divide
. (So that p goes into n exactly
times.) Now suppose
also, then the fraction
, when written in lowest terms, has a denominator that cannot be divided by
. 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
the number
with all the prime divisors of
removed.
Our equation then implies that there exists some integer
that is coprime with
such that
which means that
must divide
. That is
Answer: Let
be the number
with all prime divisors of
removed. Then
must divide
.
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
where
are non-negative integers, and
is one of 0, 1, or 2.