I have a four clue math problem. Someone help me find the answer?

Question

Clue #1) Six unit digits. The last number is eight

Clue #2) Odd and even integers alternate. (Zero is even)

Clue #3) The difference of adjacent digits is always bigger than one.

Clue #4) The first two digits (read as one number) as well as the two middle digits (as one number) are multiples of the last two digits (as one number).

Answer 1

903618

Starting with the first rule, you know the last digit is 8.  With the second rule, you find that the second last digit is odd, so the number must end in 18, 38, 58, or 78

Because of the third rule, you throw out 78 as a possible ending.  The number must end in 18, 38, or 58

Because of the second rule, you throw out 58 as a possible ending.  The only 2-digit number that is a multiple of 58 is 96, but you can't put 969658 because of rule 3. Likewise, you throw out 38 as a possible ending.  The only 2-digit number that is a multiple of 38 is 76, but you can't use 76 because of rule number 3.

That leaves you with 18 as the only possible ending.  Here are the multiples of 18:

36

54

72

90

Throw out 54 because it is odd-even instead of even-odd.  Rearrange the remaining 3 possibilities and the only one that doesn't break rule 3 is 903618.

Then again, since 18 is technically a multiple of 18, you could also use

183618 or

361818

_/