Ok so im lost on this riddle... help?? Easy points for first right answer!

Question

Here it goes:

1. The code is made up of six unique digits and 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).

Any help??

Answer 1

This is whack. Lemme google it.

Aha!  903618

Answer 2

First, look at the last two digits.

the last one is 8, so the one before has to be odd:

18, 38, 58, 78, or 98

We have to pick a number that will have at least two different multiples of 99 or less, since the first two and middle two digits must be multiples of it...

The only one is 18. (Because 38 times 3 is greater than 99)

Let's list the multiples of 18...

36, 54, 72, 90

We can drop 54, since the difference of adjacent digits is not greater than one, leaving:

36, 72, 90

Try them out...

367218 -- 2 is next to 1

369018 -- 0 is next to 1

729018 -- 0 is next to 1

723618 -- 2 is next to 3

903618 -- this works!

907218 -- 2 is next to 1