Mock AIME 4 2006-2007 Problems/Problem 1

Problem

Albert starts to make a list, in increasing order, of the positive integers that have a first digit of 1. He writes $1, 10, 11, 12, \ldots$ but by the 1,000th digit he (finally) realizes that the list would contain an infinite number of elements. Find the three-digit number formed by the last three digits he wrote (the 998th, 999th, and 1000th digits, in that order).

Solution

It is clear that his list begins with 1 one-digit integer, 10 two-digits integers, and 100 three-digit integers, making a total of $321$ digits.

So he needs another $1000-321=679$ digits before he stops. He can accomplish this by writing 169 four-digit numbers for a total of $321+4(169)=997$ digits. The last of these 169 four-digit numbers is 1168, so the next three digits will be $116$.

See also

Mock AIME 4 2006-2007 (Problems, Source)
Preceded by
First question
Followed by
Problem 2
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