Mock AIME 4 2006-2007 Problems/Problem 1

Revision as of 20:58, 13 February 2007 by Azjps (talk | contribs) (Solution: wikify, I believe its 1, 10, 100, etc.)

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 consists of one 1 digit integer, 10 two digits integers, and 100 three digit integers, making a total of $321$ digits.

So he needs another $1000-321=629$ digits before he stops. He can accomplish this by writing another 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