Difference between revisions of "2011 AMC 12B Problems/Problem 17"
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Assume <math>h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})</math> is true for n: | Assume <math>h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})</math> is true for n: | ||
| − | <math>h_{n+1}(x)\text{ = } h_{1}(h_{n}(x))\text{ = }10 h_{n}(x) - 1\text{ = 10 }(10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})) - 1 | + | <math>h_{n+1}(x)\text{ = } h_{1}(h_{n}(x))\text{ = }10 h_{n}(x) - 1\text{ = 10 }(10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})) - 1 |
\\= 10^{n+1} x - (10 + 10^2 + ... + 10^{n}) - 1 | \\= 10^{n+1} x - (10 + 10^2 + ... + 10^{n}) - 1 | ||
\\= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{(n+1)-1})</math> | \\= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{(n+1)-1})</math> | ||
Revision as of 13:46, 7 February 2012
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 $\boxed{16089\ \(\textbf{(B)}}$ (Error compiling LaTeX. Unknown error_msg)
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 | |
