Determine all possible values of a 6-digit positive integer N having 6 different digits from 0 to 9, with the first digit being nonzero, such that:
(i) Either, the digits at the even positions are even and the digits at the odd positions are odd – or, the digits at the even positions are odd and the digits at the odd positions are even. Zero is regarded as an even digit.
(ii) The absolute difference between two adjacent digits is always greater than one.
(iii) The 2-digit number formed by the first and second digit as well as the 2-digit number formed by the third and fourth digit is divisible by the 2-digit number formed by the fifth and sixth digit.
(In reply to SPOILER
by Ady TZIDON)
KS asks for a 6-digit number, not an 8-digit one.
Also, MN don't account for 12 possibilities in your plan, as M must be 5 in order to be odd and differ from 8 by more than 1, and then N must be 2 for similar reason, so you'd be left with 90361852 as your 8-digit number.
But again, it's a 6-digit number that's sought, though of course 903618 is an example of one that works.
Edited on January 1, 2010, 1:14 pm
Posted by Charlie
on 2010-01-01 13:13:53