Difference between revisions of "2011 AMC 12B Problems/Problem 17"
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Assume <math>h_{n}(x)=10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})</math> is true for n: | Assume <math>h_{n}(x)=10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})</math> is true for n: | ||
− | + | <cmath>\begin{align*} | |
h_{n+1}(x)&= h_{1}(h_{n}(x))\ | h_{n+1}(x)&= h_{1}(h_{n}(x))\ | ||
&=10 h_{n}(x) - 1\ | &=10 h_{n}(x) - 1\ | ||
Line 22: | Line 22: | ||
&= 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}) | &= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{(n+1)-1}) | ||
− | \end{align*} | + | \end{align*}</cmath> |
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. | 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. |
Revision as of 11:32, 11 October 2018
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
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 |
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