INMO 2003 q6

a) 222 222 222 Green (given)

b) 122 222 222 Orange (given)

c) 311 311 311 White (a and b)

d) 333 333 333 White (a and b)

e) 133 133 133 Orange (a and c)

f) 211 211 211 Green (d and e)

g) 123 123 123 Orange (c and f)

The solution is here (JUST DO IT)

333333333 has no digits in common with ticket 122222222 or 222222222, so it must be White. Likewise 111111111 is Orange. Note that ticket 211111111 must be Green because it is different from 333333333 and 122222222. Call a ticket that starts with 3, followed by a bunch of 1's and 3's ticket A. Then A must be White because it is different from 122222222 and 222222222, which are Orange and Green.

Now consider the ticket starting with a 3, followed by a bunch of 3's and 2's. This is different from 211111111 and 111111111, so it must be White. We can carefully arrange the string of 3's and 2's so no digit matches a digit in 123123123, so we know 123123123 cannot be White.

Now consider the ticket starting with a 2, followed by a bunch of 3's and 1's. This is different from 122222222 and we can carefully place the digits to also be different from ticket A, so this ticket is neither Orange nor White, so it must be Green. But this ticket can be engineered to be different from 123123123 in every digit place, so 123123123 cannot be Green.

Hence 123123123 must be Orange .

A diagram that summarizes the above explanation: