twixt wrote:

Linker,

I am not sure i understand your theory :

In case of your 2^n5^p denominator, what happens if numerator is a power of 3 ?? unterminated...

Yes. See examples below. I will take a common prime numerator, say 43.

43/32 -> terminating decimal -> Since it has only 2 as prime factor

43/625 -> terminating decimal -> Since it has only 5 as prime factor

43/50 -> terminating decimal -> Since it has only 2 and 5 as prime factors

43/150 -> not a terminating decimal -> Since it has 3 as a prime factor apart from 2 and 5

43/41 -> not a terminating decimal -> Since it does not have 2 or 5 as a prime factor.

I will rephrase my statement above: Any fraction with a denominator containing prime factors of only 2 and 5, will be have terminating decimal.