1. Terminating Decimal Proof

I am having troubles with this deceptively simple problem:

Show that for any integer n that $\displaystyle \frac{1}{n}$ will be a terminating sexagesimal iff n's prime factors only consist of 2,3 and 5.

In other words $\displaystyle \frac{1}{n}$ will terminate in base 60 iff $\displaystyle n = 2^a*3^b*5^c$

2. Define $\displaystyle n = 2^a3^b5^c$ and $\displaystyle m=max(a,b,c)$. Then $\displaystyle \frac1n=\frac{2^{m-a}3^{m-b}5^{m-c}}{60^m}$. Therefore $\displaystyle \frac1n$ terminates on the $\displaystyle m^{th}$ decimal place. QED

3. Hint for the other direction:
If $\displaystyle \frac{1}{n} = 0.a_1 a_2 a_3 ... a_n$ then look at $\displaystyle 60^n \frac{1}{n} = ?$

4. More generally, any integer, n, is equal to a product of powers of its prime divisors. Any terminating "decimal", base n, is equal to an integer times some power of n (the negative of the power of the last "decimal" place) and so is equal to that integer divided by that power of n. While some factors in denominator and numerator may cancel, that won't introduce new factors in the denominator. The only possible factors of the denominator are factors of n and so the only possible prime factors of the denominator are the prime factors of n.

5. Originally Posted by Media_Man
Define $\displaystyle n = 2^a3^b5^c$ and $\displaystyle m=max(a,b,c)$. Then $\displaystyle \frac1n=\frac{2^{m-a}3^{m-b}5^{m-c}}{60^m}$. Therefore $\displaystyle \frac1n$ terminates on the $\displaystyle m^{th}$ decimal place. QED
Why does that terminate?

6. The (Decimal) Terminator

Haven:

In base $\displaystyle 60$, by definition...

$\displaystyle \frac1{60}=.1$
$\displaystyle \frac1{60^2}=.01$
$\displaystyle \frac1{60^3}=.001$
...

Etcetera. So a decimal terminates in base $\displaystyle 60$ iff it is expressible in the form $\displaystyle \frac{N}{60^m}$, with $\displaystyle N,m$ integers.