Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | It is clear that his list consists of one 1 digit integer, | + | 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 1000-321=629 digits before he stops. He can accomplish this by writing another 169 | + | 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>. |
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| + | ==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]] | ||
Revision as of 20:58, 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, 10 two digits integers, and 100 three digit integers, making a total of 
digits.
So he needs another 
digits before he stops. He can accomplish this by writing another 169 four digit numbers for a total of 
digits. The last of these 169 four digit numbers is 1168, so the next three digits will be 
.
