# Number having sexagesimal expansion end with infinitely many zeros?

I am looking for all the real numbers whose sexagesimal expansion (base $$60$$) ends in infinite tail of zeros. Does they really exist?

It seems absurd to me or mm thinking it in a wrong manner?

• You mean numbers like $1/60$ and so on? Or even like $17$ or $108$? – kimchi lover Oct 16 '18 at 15:26
• $1/60$ is just $1$. – Mittal G Oct 16 '18 at 15:36

$$\forall n,a \in N, 60 \nmid n \lor a = 0$$ $$\frac {n} {60^a}$$
Example: $$\frac {175371} {60^2}$$
A number $$N$$ has a finite sexagesimal expansion if it can be written as $$N=a_0+\frac{a_1}{60}+\frac{a_2}{60^2}+\dots+\frac{a_n}{60^n}$$ with $$a_0$$ any integer and $$0\le a_i<60$$ for $$i=1,2,\dots,n$$.
Then we can write $$N=\frac{m}{60^n}$$ The converse is obviously also true: take $$a_0=\lfloor N\rfloor$$ and write $$N-a_0=\frac{m'}{60^n}$$ which is true for a unique $$m'$$ with $$0\le b<60^n$$. Then write the base $$60$$ expansion of $$b$$ and you're done.
You can notice that base $$60$$ has nothing special. A number $$N$$ has a finite base $$b$$ expansion if and only if it has the form $$N=\frac{m}{b^n}$$ for some integer $$m$$ and nonnegative integer $$n$$.