As in a related problem, let 3ⁿ and 3^m (n>m) have the same remainder when divided by 1000. Thus 3ⁿ - 3^m = 3^m(3^n-m - 1) is divisible by 1000. Since 1000 and 3^m have no common factors, 1000 is bound to divide the second factor (3^n-m - 1). This exactly means that 3^n-m ends with 001!