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back to boardbig,big i mean huge problem 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 Re: big,big i mean huge problem Posted by Alex 28 Feb 2010 20:10 Input is not N but 2N - common length of ticket:) Thus they are ten: 00,11,22,33,44,55,66,77,88,99. Re: big,big i mean huge problem you mean than the lucky numbers are the mirror numbers ex: 123454321 1234554321 ???::? Re: big,big i mean huge problem Posted by panic 14 Apr 2010 01:57 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 Re: big,big i mean huge problem 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 Re: big,big i mean huge problem The problem states you are passed an even positive number, so you need not solve it for odd values. |
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