Caas wrote:

suithink,

So as far as I understood, any fraction that has a denom., which has 2 and 5 as the only prime factors, is a terminating decimal?

Is it a sufficient condition always?

The key point to be noted here is :

Terminating decimal --> A number having a 'fixed' number of decimal places...==> Can be expressed as N/

(10^n).....
=> Deno ..i.e 10 ^n= 2^n . 5^n.....

Then it means Yes...sufficient enough...

(Note the 5^n has disappaered when converting to N/10^n form a term will appear in num which will cancel out 5^n factor)

PS: BTW do a search in the forum ...this problem has been covered lot many times here...