Background: A number is said to be (sexagesimally) regular if its reciprocal has a finite sexagesimal expansion (that is, a finite expansion when expressed as a radix fraction for base 60). With the exception of a single tablet in the Yale collection, all Babylonian tables of reciprocals contain only reciprocals of regular numbers. A Louvre tablet of about 300 B.C. contains a regular number of 7 sexagesimal places and its reciprocal of 17 sexagesimal places.
Problem: Show that a necessary and sufficient condition for $n$ to be regular is that $n = 2^a3^b5^c$, where $a,b,c$ are nonnegative integers. Then generalize this result to numbers having general base $b$.
Book solution: Suppose $n$ is regular. Then \begin{align*} \frac{1}{n} &= a_0 + \frac{a_1}{60} + \cdots + \frac{a_r}{60^r}\\[1em] &= \frac{a_060^r + a_160^{r-1}+\cdots+a_r}{60^r}\\[1em] &= \frac{m}{60^r},\; \textrm{say}. \end{align*} It follows that $mn = 60^r$ and $n$ can have no prime factors other than those of 60.
My work: Suppose $n \in \mathbb{N}$. Furthermore, suppose we have the following: \begin{align*} \frac{1}{n} &= \frac{a_1}{60}+\frac{a_2}{60}+\cdots+\frac{a^r}{60^r}\\[1em] &= \frac{a_160^{r-1}+\cdots+a_r}{60^r}\\[1em] &= \frac{\ell}{60^r}, \end{align*} where $\ell \in \mathbb{N}$. Hence, when the LHS and RHS are combined, we have that $\frac{1}{n} = \frac{\ell}{60^r} \leftrightarrow n\ell = 60^r$. Now consider the unique prime factorization of 60: $2^2\cdot 3\cdot 5 = 60$. Thus, we may write the following: $60^r = (2^2\cdot 3\cdot 5)^r = 2^{2r}3^r5^r = (2^a3^b5^c)(2^{2r-a}3^{r-b}5^{r-a})$. Here, without loss of generality, we let $n = 2^a3^b5^c$ and $\ell = 2^{2r-a}3^{r-b}5^{r-a}$. The important thing to note is that both $n$ and $\ell$ can have no prime factors other than those of $2,3,5$. Thus, we have shown that $n$ is regular iff $n = 2^a3^b5^c$, where $a,b,c \in \mathbb{N}\cup \{0\}$.
Note: Does this proof flow better than the one provided in the text? Given that we are dealing with a fractional part, it did not seem necessary to consider $a_0$ in the original expression of $\frac{1}{n}$ because $a_0 = 0$.
Generalized solution: The solution here largely imitates that from above but with some seemingly sticky abstraction concerning infinity. We proceed as before: \begin{align*} \frac{1}{n} &= \frac{a_1}{b}+\frac{a_2}{b^2}+\cdots+\frac{a_\Bbbk}{b^\Bbbk}\\[1em] &= \frac{a_1b^{\Bbbk-1}+\cdots+a_\Bbbk}{b^\Bbbk}\\[1em] &= \frac{\ell}{b^\Bbbk}. \end{align*} As in the solution above, consider the prime factorization of $b^\Bbbk = p_1^\alpha p_2^\beta\ldots p^\Omega_\wp$. We may write the following: $b^\Bbbk = (p_1^\alpha p_2^\beta\ldots p_\wp^Z)(p_1^{\alpha-a}p_2^{\beta-b}\ldots p_\wp^{\Omega-Z})$. Here, without loss of generality, we let $n = p_1^ap_2^b\cdots p_\wp^Z$ and $\ell = p_1^{\alpha-a}p_2^{\beta-b}\ldots p_\wp^{\Omega-Z}$. Again, the important thing to note is that both $n$ and $\ell$ can have no prime factorization other than those of $b$, namely $p_1, p_2, \ldots, p_\wp$. Thus, we have shown that $n$ is regular in base $b$ iff $n=p_1^ap_2^b\ldots p_\wp^Z$, where $a,b,\ldots,Z\in\mathbb{N}\cup\{0\}$. Note that the unique prime factorization can be as "long" as needed; that is, it is incorrect to think of $a,b,\ldots,Z$ as some sort of "finite restriction"--this notation was simply chosen for ease of demonstration, but the limited number of symbols available to use is the real issue.
Is this generalized solution/proof all that good or can it be cleaned up considerably? I was struggling with trying to communicate the proof clearly.
Suppose you have an expression with $n$ digits after the decimal point in base $b$. then the term is a sum of fractions whose denominator can be seen as $b^n$. Hence if a fraction has a finite representation in base $b$ it is of the form $\frac{k}{b^n}$ for some $n$.
Proving every number of the form $\frac{k}{b^n}$ has an expression is simple. Just write the integer $k$ in base $b$ and then place a decimal point $n$ places to the left of the last digit of $k$ in the base $b$ representation.