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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?

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  • $\begingroup$ You mean numbers like $1/60$ and so on? Or even like $17$ or $108$? $\endgroup$ – kimchi lover Oct 16 '18 at 15:26
  • $\begingroup$ $1/60$ is just $1$. $\endgroup$ – Mittal G Oct 16 '18 at 15:36
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$$ \forall n,a \in N, 60 \nmid n \lor a = 0 $$ $$ \frac {n} {60^a} $$

Example: $$ \frac {175371} {60^2} $$

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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 $n$.

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