HMMT 二月 2007 · 冲刺赛 · 第 4 题

HMMT February 2007 — Guts Round — Problem 4

[ 6 ] A sequence consists of the digits 122333444455555 . . . such that the each positive integer n is repeated n times, in increasing order. Find the sum of the 4501st and 4052nd digits of this sequence. 2007!

解析

[ 6 ] A sequence consists of the digits 122333444455555 . . . such that the each positive integer n is repeated n times, in increasing order. Find the sum of the 4501st and 4052nd digits of this sequence. Answer: 13 . Note that n contributes n · d ( n ) digits, where d ( n ) is the number of digits of n . Then because 1 + · · · + 99 = 4950, we know that the digits of interest appear amongst copies of two digit numbers. Now for 10 ≤ n ≤ 99, the number of digits in the subsequence up to the last copy of n is 2 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. 2007!