Congratulations! I’ll post the solution for the 2b:
[spoiler]
First, a number is divisible by 3 if the sum of its digits is. EX. 297 is divisible because 2+9+7=18, 4568 is not becuase 4+5+6+8=23.
Secondly, a number is divisible by 4 is the last 2 digits are. IE. 12736 is divisible because 36 is.
Therefore, we can study the problem digit by digit.
So, let’s study the last 2 digits. The possibilities are 00, 04, 08, 12, … 88, 92, 96, and that makes 25 possibilities, without the 00, 04, 08, 20, 40, 60, 80, so that makes 18.
All other digits can range from 1 to 9, so that would make 9*9=81 possibilities for the 3rd and the 4th, then the 5th must be left for the divisibility by 3. You see, having only numbers from 1 to 9 to choose, it is much easier, as they are grouped 3 by 3 for the disision rest by 3: 3,6,9; 1,4,7; 2,5,8 are 3 groups of rest 0, 1, 2. This way, whatever the rest of the division by 3 of the sum of the other digits will be, we will always have the possibility to choose any of 3 digits IE. X3712 has rest 1 so far (like 3+7+1+2=13 has), so we can use either 2, 5 or 8; X8328 has rest 0 (like 8+3+2+8=21), so we can use 3, 6 or 9.
So, all numbers responding to all criteria are 18 * 81 * 3 = 4374 numbers.[/spoiler]
Feel free to add any problems, I won’t solve them and I probably won’t put up any until mid January.