HMMT 二月 2007 · 冲刺赛 · 第 4 题

英文解析

4. [ 6 ] A sequence consists of the digits 122333444455555 . . . such that the each positive integer n isrepeated n times, in increasing order. Find the sum of the 4501 st and 4052 nd digits of this sequence.
   Answer: 13 . Note that n contributes n · d ( n ) digits, where d ( n ) is the number of digits of n . Thenbecause 1 + · · · + 99 = 4950, we know that the digits of interest appear amongst copies of two digitnumbers. Now for 10 ≤ n ≤ 99, the number of digits in the subsequence up to the last copy of n is
   1 + 2 + 3 + · · · + 9 + 2 · (10 + · · · + n ) = 2 · (1 + · · · + n ) − 45 = n + n − 45 .2
   Since 67 + 67 − 45 = 4511, the two digits are 6 and 7 in some order, so have sum 13.2
   2007!