Since the first two digits are 12 in that order, there is only 1 way
to choose the first two digits.  That's 1 way to choose the first
two digits.

Since the third digit is bigger than 6, it can only be 7,8, or 9.

So that's 1×3 choices for the the first three digits.

Since the 4th digit is 3 times the 5th digit, the number can only end in 
31, 62, or 93

So for each of the 1×3 way to choose the first 3 digits, there are 3
ways to choose the last two digits.

That's a total of 1×3×3 = 9 possible numbers.  Here are all 9:

1.  12731
2.  12762
3.  12793
4.  12831
5.  12862
6.  12893
7.  12931
8.  12962
9.  12993

Edwin