I am having troubles with this deceptively simple problem:

Show that for any integer n that will be a terminating sexagesimal iff n's prime factors only consist of 2,3 and 5.

In other words will terminate in base 60 iff

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- Jan 15th 2010, 11:38 AMHavenTerminating Decimal Proof
I am having troubles with this deceptively simple problem:

Show that for any integer n that will be a terminating sexagesimal iff n's prime factors only consist of 2,3 and 5.

In other words will terminate in base 60 iff - Jan 15th 2010, 12:05 PMMedia_Man
Define and . Then . Therefore terminates on the decimal place. QED

- Jan 16th 2010, 01:23 AMUnbeatable0
Hint for the other direction:

If then look at - Jan 16th 2010, 05:51 AMHallsofIvy
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. - Jan 18th 2010, 06:20 PMHaven
- Jan 18th 2010, 06:52 PMMedia_ManThe (Decimal) Terminator
Haven:

In base , by definition...

...

Etcetera. So a decimal terminates in base iff it is expressible in the form , with integers.