Can someone please help me solve 2 math problems?

You need to know a couple of shortcuts.

1) any number is divisible by 4 if its last two digits are divisible by 4.

2) any number is divisible by 3 if the SUM of its digits is divisible by 3.

For "32m0n" to be divisible by 12, it has to be divisible by 4 and by 3.

For "32m0n" to be divisible by 4, "0n" has to be divisible by 4. Since we know n is less than 5, n can only be 0 or 4.

For "32m0n" to be divisible by 3, "3 + 2 + m + 0 + n" must be divisible by 3.

If n is 0, then m must be a number you can add to 5 that yields a sum divisible by 3: 1, 4, 7.

If n is 4, then m must be a number you can add to 9 that yields a sum divisible by 3: 0, 3, 6, 9.

So there are 7 possible values of m.

For "635ab" to be divisible by 4, "ab" must be divisible by 4. (Which means "b" must be even, since odd numbers are not divisible by 4.) And we know a-b = 4. Possibilities for a - b:

4 - 0 = 4; 40 is divisible by 4. GOOD.

6 - 2 = 4; 62 is not divisible by 4.

8 - 4 = 4; 84 is divisible by 4. GOOD.

For any higher value of b, a would not be a single-digit number. So there are two possibilities for "ab": "40" and "84". The sum of the possible values of a is therefore 4 + 8 = 12.

Done.

#2: If the set parameters are correct then there are only two possible answers, 63584 and 63540 being that those two numbers are both divisible by 4. meaning that those numbers are multiples of four.

hope it helps