2011 AMC 12B Problems/Problem 17
Problem
Let , and for integers . What is the sum of the digits of ?
Solution
Proof by induction that :
For ,
Assume is true for n:
Therefore, if it is true for n, then it is true for n+1; since it is also true for n = 1, it is true for all positive integers n.
, which is the 2011-digit number 8888...8889
The sum of the digits is 8 times 2010 plus 9, or
Solution 2 (Quick, Non-Rigorous Trends)
As before, . Compute , , and to yield 9, 89, and 889. Notice how this trend will evidently repeat this trend (multiply by 10, subtract 1, repeat). As such, is just 2010 8's followed by a nine. .
~~BJHHar
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.