it says than "the sum of the first three digits....is equal with the sum of nthe last three".
In that case 45 first there 4+5+(null)=9
                last three 5+4+(null)=9
=>45 is a lucky number!!!
that means than there are (at nr 2) 11,12,13,...,99(100-1-10)=89 lucky numbers...
Can anyone explain me why they say yhere are only 10 numbers at example 2??
THANX

Input is not N but 2N - common length of ticket:) Thus they are ten: 00,11,22,33,44,55,66,77,88,99.

you mean than the lucky numbers are the mirror numbers
ex: 123454321
    1234554321
???::?

yes but not only mirror numbers are the lucky ones,
also for ex.: 1234532145  the point is that the sum of digits on the left must be equal to sum of the right side to meet this demand of being lucky

qs:If we have the 3 case...wich is the middle?
for eg 123 the middle is 12 or only 1(23 or only 3)

Edited by author 04.06.2010 21:41

The problem states you are passed an even positive number, so you need not solve it for odd values.