Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 1"
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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 begins with 1 one-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= | + | 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>. |
==See also== | ==See also== | ||
| − | + | {{Mock AIME box|year=2006-2007|n=4|before=First question|num-a=2|source=125025}} | |
| − | |||
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
Latest revision as of 22:41, 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 
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 
digits.
So he needs another 
digits before he stops. He can accomplish this by writing 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 
.
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 | ||
