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back to boardShow all messages Hide all messagesit 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. |
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