Quote:
Originally Posted by Citrix
Given a natural number N is there a fast way to test if the number is of the form k*b^n+c where b^n contributes to 99% size of N; c<m*n where m is small.

Yes.
The first number to match the criteria is 100 = 1*99^1+1
Then 200 = 1*198^1+2
Then x*100 = 1*(99x)^1+x
etc.
So all you have to do is check that the number is divisible by 100.