See A018247 - OEIS and A018248 - OEIS for the numbers given by values of N up to about 100. I don't think there is any simple formula for the general case.
hi, given N, how can one find all such N digit natural numbers such that the N least significant digits of any integral power of that number is the number itself?
E.g if N =2
then the required numbers are 25 and 76.
because 25^2 = 625 and the 2 least significant digits of 625 are 25 which is the number itself , similary for any integral power of 25 and 76.
Here N can be as large as 500 digits.
Thanks.
See A018247 - OEIS and A018248 - OEIS for the numbers given by values of N up to about 100. I don't think there is any simple formula for the general case.