# What is “ terminating reciprocals”

I meet a problem "What kinds of numbers have terminating reciprocals in base 60?", but I don't know what is "terminating reciprocals". Please help me, thanks.

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When written out in base 60, do the expansions terminate. For example, in base 10, note 1/4 = 0.25 terminates, and is the reciprocal of 4. But 1/3 = 0.33333... does not terminate. – GEdgar Aug 30 '14 at 21:27
Those are reciprocals sent back in time to kill John (and/or Sarah) Connor. – Asaf Karagila Aug 30 '14 at 21:36

It means that all of the digits of the base-60 expansion of the reciprocal are $0$ beyond some finite number of nonzero digits.

For example $1/250$ has a "terminating reciprocal in base $10$ because $$\frac 1 {250} = 0.004\,0000000000000000\ldots$$ so after the $4$, all the digits are $0$, forever. By contrast, look at $$\frac 1 {37} = 0.027\,027\,027\,\ldots$$ where $027$ repeats forever.

One way to see that this happens is doing long division, dividing $1$ by $37$: \begin{array}{cccccccccccccccccccccccccccc} & & 0 & . & 0 & 2 & 7 \\ \\ 37 & ) & 1 & . & 0 & 0 & 0 & 0 & 0 \\ & & 1 & & 0 \\ \\ & & 1 & & 0 & 0 \\ & & & & 7 & 4 \\ \\ & & & & 2 & 6 & 0 \\ & & & & 2 & 5 & 9 \\ \\ & & & & & & 1 & 0 & & \longleftarrow\text{Back to where we started!} \end{array} The fact that we have returned to the problem we started with means we must get the same answer we got before, i.e. the digits repeat. With $1/250$ applying long division brings us to a point where we are dividing $0$ by $250$, so we get $0$, and then at the next step we're dividing $0$ by $250$ again, so the sequence must then procede as before and we will just keep getting $0$s.

It works the same way in base $60$, except the particular numbers whose reciprocals terminate are different. With base $10$, it is those whose only prime factor are $2$ and $5$, the prime factors of $10$.

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Thank you very much. – 6666 Aug 30 '14 at 23:51

If the decimal expansion of $\dfrac{1}{n}$ terminates in base-$60$, then we can write

$\dfrac{1}{n} = \dfrac{d_1}{60}+\dfrac{d_2}{60^2}+\cdots+\dfrac{d_m}{60^m}$ for some choice of $m$ integers $0 \le d_1,\ldots,d_m \le 59$.

Then, $\dfrac{60^m}{n} = d_1 \cdot 60^{m-1}+d_2 \cdot 60^{m-2} + \cdots + d_{m-1} \cdot 60 + d_m$ is an integer.

So, $n$ must divide $60^m = 2^{2m} \cdot 3^m \cdot 5^m$ for some integer $m$.

Therefore, if the decimal expansion of $\dfrac{1}{n}$ terminates in base-$60$, then the only primes which divide $n$ are $2$, $3$, and $5$, i.e. $n = 2^p3^q5^r$ for some non-negative integers $p,q,r$.

It is not hard to see that if $n = 2^p3^q5^r$ for some non-negative integers $p,q,r$, then the base-$60$ expansion of $\dfrac{1}{n}$ has exactly $\max\{\lceil\frac{p}{2}\rceil,q,r\}$ digits behind the decimal point.

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Thank you very much! – 6666 Aug 30 '14 at 23:51