Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 1"

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==Problem==
 
==Problem==
Albert starts to make a list, in increasing order, of the positive integers that have a first digit of 1. He writes <math>1, 10, 11, 12, \ldots</math> 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).
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Albert starts to make a list, in [[increasing sequence | increasing]] order, of the [[positive integer]]s that have a first [[digit]] of 1. He writes <math>1, 10, 11, 12, \ldots</math> 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==
 
==Solution==
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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 <math>321</math> digits.
  
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.
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So he needs another <math>1000-321=679</math> digits before he stops.  He can accomplish this by writing 169 four-digit numbers for a total of <math>321+4(169)=997</math> digits.  The last of these 169 four-digit numbers is 1168, so the next three digits will be <math>116</math>.
  
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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==See also==
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{{Mock AIME box|year=2006-2007|n=4|before=First question|num-a=2|source=125025}}
  
drunner2007
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[[Category:Intermediate Combinatorics Problems]]
{{solution}}
 
 
 
 
 
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*[[Mock AIME 4 2006-2007 Problems/Problem 2| Next Problem]]
 
*[[Mock AIME 4 2006-2007 Problems]]
 

Latest revision as of 23:41, 22 April 2010

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15