Answer: 4

The last zero in the first block of zeros is the the place of 4 +1;

The last zero in the second block of zeros is in the place of 4 + 1 + 4 + 2;

The last zero in the nth block of zeros is in the place of 4n + (1 + 2 + ... + n) 

Now, we need to choose the largest integer n such that

  4n + (1 + 2 + ... + n) <= 2550 

Hence, n = 67.

The last zero in the 67th block of zeros is in the place of 4*67 + 67*68/2 = 2546.

1 is in the 2547th place;

2 is in the 2548th place;

3 is in the 2549th place;

4 is in the 2550th place.

所有跟帖： 

• 谢谢!明白了. -studentdriver- ♀ (0 bytes) () 10/12/2012 postreply 21:22:53

• 我１分钟就得出了答案，但 -布兰雅- ♀ (9920 bytes) () 10/12/2012 postreply 21:42:14

• 谢谢! -studentdriver- ♀ (0 bytes) () 10/13/2012 postreply 11:47:33