The next to last digit must be odd, because odd and even digits
must alternate and the last digit is 8.
Therefore the last two digits must either be 18, 38, 58, 78, or 98
But since the first two digits and the middle two digits
are unique multiples of the last two digits, the last two digits
must have at least two different 2-digit multiples.
38 has only one 2-digit multiple, 76. So it won't do. 58, 78,
and 98 have no 2-digit multiples. So they won't do.
So the only possibility for the last two digits is 18, because 18 has
four 2-digit multiples, 36, 54, 72, and 90
So we have:
_ _ _ _ 1 8
The first two digits and the middle two digits must both be
multiples of 18. We cannot use 54 because its digits differ
by 1, so they must be two of these: 36, 72 or 90
We cannot put 72 or 90 in the middle before the 18, because the
second digit of either one and the first digit of 18 would only
differ by 1 in either case.
Therefore the middle two digits can only be 36.
So we have
_ _ 3 6 1 8
So that leaves 72 or 90 for the first two digits.
We cannot put 72 there for its second digit 2
would differ from the third digit 3, by only 1.
That leaves only 90 for the first two digits.
So the answer is
903618
Edwin
