Please Help With This Math Riddle? "What Number Follows This Sequence?"?

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ssccrab

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This comes from a video game. In order to progress through a level, one needs an access code, which is figured out through this math riddle:

What Number Follows This Sequence?

#1: The door code is made up of six unique digits. the last digit is 8
#2: Odd and even digits alternate. (0 is even)
#3: The difference of adjacent digits is always bigger than one.
#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)

I've tried number combinations, but just can't figure it out. Your help is appreciated! Thank you!
 
The answer is 903618.

Here's why:

The possibilities for the last two digits are what?

X = 18, 38, 58, 78, 98.

If we need two unique multiples of this to be in the other two spots, then X = 18. Why? If X=38 (or bigger), then the other two spots are at least 2×38 and 3×38=114.

Thus the second to last digit is 1.

Now that we have 18, we need to find multiples of 18 that are two digits only. We can multiply 18 to get:

18×2=36
18×3=54
18×4=72
18×5=90

We cannot use 54 since it has two digits that are 1 apart.
We are left with 90, 36, 72 to pick from. We cannot use 36 and 72 together. Why? Because either the 2 and 3 are together, or the 6 and 7 are together:

367218
723618

If we use the 72 and 90 we get a problem because either the 0 or the 2 is next to the 1. So 72 cannot be used at all.

Finally, we know it is 36 and 90, and since 0 can't be next to 1, we have to have 36 next to 18. That is it:

903618

All other answers are wrong:

Datta has 08, which is two evens next to each other.
Pyz01 has 32, which are digits 1 apart next to each other.
 
The 5th digit could not be 7 or 9 because of rule #3
It has to be odd, therefore it could be 1, 3 or 5

If 5th digit of 5, last 2 digit is 58
Multiples of 58 which are 2 digits is only 58
But since the six digits should be unique, 58 could not be the last 2 digits.

If 5th digit is 3, last 2 digits = 38
Multiple of 38 which 2 digits are 76,
But since the six digits should be unique, 38 could not be the last 2 digits.

If 5th digit is 1, last 2 digits = 18
Multiple of 18 which 2 digits are 36, 72, 90
72 and 90 could not be the 3rd and fourth digit because their last digit has a difference of only 1 from 1.
Therefore the 3rd and 4th digits are 36
The digits known are now ??3618
72 could not be the first 2 digits because the difference of 3 and 2 is only 1.
Therefore the 1st 2 digits are 90
The access code is 903618
 
OK

I disagree with the first person. He is close, but 0 is an even number. I get

72 36 18 , notice how all rules are met:

1. Last digit is 8 - check
2. Odd and even alternate - check
3. The difference of adjacent digits is always bigger than one 7-2, 6-3 and 8-1 - check
4. First two digits and second set of two are multiples of the last 2 - 72/18 =4 and 36/18 = 2, - check

Hope that helps.
 
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