dari:1 s.d. 9 ada 9 bilangan10 s.d. 99 ada 90x2 = 180 bilangan

jumlahnya1 s.d 99 ada 189 bilangan

sisanya 2400-189=2211 bilangan lagikarena ada ratusan ada 3 bilangan, sisa 2211/3=737 suku lagi

737 suku lagi adalah 99+737 = 836

keseluruhannya1 s.d. 836 ada 2400 bilanganjadi, bilangan ke 2400 adalah 6

Lihat soal yang ini yah.. http://id.answers.yahoo.com/question/ind... terus kamu kurangi aja sendiri, kalau angka ke 2013 adalah digit terakhir dari 707, maka: suku ke 2012--> 0, 2011--> 7, 2010--> 6, 2009-->0, 2008-->7, 2007-->5, 2006-->0, 2005-->7, 2004--> 4 Jadi, angka yang terletak pada bilangan ke 2004 adalah 4I would use a multiple step approach, based on the number of digits of numbers. I give you the idea, but double-check he calculations themselves... So for numbers 1 to 9, the beginning of the number is at its rank (1--> 1st, ..., 9--> 9th). For two digit numbers: 10 begins at 10th, 11 begins at 12th, 12 begins at 13th... So for a two digit number n, the first digit is at rank 10+2(n-10). That leads 99 begins at 10+2(99-10)= 188. That means number 100 begins at 190th (188th is the first 9 of 99 and 189th the second 9 of 99). Using the same principle, for a 3 digit number n, the first digit of n is at rank 190+3(n-100) So 999 begins at rank 190+3(999-100)= 2887--in other words, rank 2887th is the first 9 of 999. 2887 is greater than 2013, so rank 2013th is part of a 3 digit number. So find the greatest n such as 190+3(n-100) is lower than or equal to 2013. You find n=707, and 707 begins at rank 2011.: 2011th: 7 (first 7 of 707), 2012th: 0 2013th: 7 (last digit of 707).

I did the calculation quickly, so you should double-check my numbers. But the strategy works :) Penilaian & komentar penanya

Thanks for your answer. That's true. Also thanks for the correction, DavidM, It's sequence, not a series. Sorry for my bad math english! :)