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When $n$ limited to odd numbers, for $\frac{1}{n}$ to be a terminal decimal$,\quad n$ must be in the form of $5^k$ ($k$ is positive integer). Why is $5$ so unique in this case?

Guess it is related to decimal system? I checked https://en.m.wikipedia.org/wiki/Repeating_decimal, where there is a statement “A fraction in lowest terms with a prime denominator other than $2$ or $5$ (i.e. coprime to $10$) always produces a repeating decimal” without further expectations. Just curious if there is a simple proof for this observation out there. Thanks!

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  • $\begingroup$ Why limit this to odd numbers? That's pretty arbitrary and also, incidentally, makes you miss half the pattern at work here. It's easier to spot the reason if you allow yourself to see all of it. $\endgroup$
    – Arthur
    Commented Mar 5, 2021 at 7:06
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    $\begingroup$ $5$ is the only odd number greater than $1$ that is a factor of $10$. $\endgroup$
    – user2661923
    Commented Mar 5, 2021 at 7:44

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If $\frac1n$ terminates after $m$ decimals, then $10^m\cdot\frac 1n$ is an integer, so $n$ is a divisor of $10^m$. All these divisors are of the form $2^a5^b$, but as you wanted $n$ odd, we must have $a=0$.

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