Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 1"
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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. | 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. | ||
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Revision as of 17:58, 13 February 2007
Problem
Albert starts to make a list, in increasing order, of the positive integers that have a first digit of 1. He writes 
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, 20 two digits integers, and 300 three digit integers. That's 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.
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