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

m (→‎Solution: wikify, I believe its 1, 10, 100, etc.)
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==Problem==
 
==Problem==
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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==
 +
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]], 10 two digits integers, and 100 three digit integers, making a total of <math>321</math> digits.
+
So he needs another <math>1000-321=629</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 <math>1000-321=629</math> digits before he stops. He can accomplish this by writing another 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>.
 
  
 
==See also==
 
==See also==
 
*[[Mock AIME 4 2006-2007 Problems/Problem 2| Next Problem]]
 
*[[Mock AIME 4 2006-2007 Problems/Problem 2| Next Problem]]
 
*[[Mock AIME 4 2006-2007 Problems]]
 
*[[Mock AIME 4 2006-2007 Problems]]
 +
 +
[[Category:Intermediate Combinatorics Problems]]

Revision as of 21:10, 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 $1, 10, 11, 12, \ldots$ 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 $1000-321=629$ 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

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