Difference between revisions of "2006 AIME I Problems/Problem 13"
m (→Alternate Solution) |
|||
Line 11: | Line 11: | ||
First note that <math>g(k)=1</math> if <math>k</math> is odd and <math>2g(k/2)</math> if <math>k</math> is even. | First note that <math>g(k)=1</math> if <math>k</math> is odd and <math>2g(k/2)</math> if <math>k</math> is even. | ||
so <math> S_n=\sum_{k=1}^{2^{n-1}}g(2k). = \sum_{k=1}^{2^{n-1}}2g(k) = 2\sum_{k=1}^{2^{n-1}}g(k) = 2\sum_{k=1}^{2^{n-2}} g(2k-1)+g(2k).</math> | so <math> S_n=\sum_{k=1}^{2^{n-1}}g(2k). = \sum_{k=1}^{2^{n-1}}2g(k) = 2\sum_{k=1}^{2^{n-1}}g(k) = 2\sum_{k=1}^{2^{n-2}} g(2k-1)+g(2k).</math> | ||
− | <math>2k-1</math> must be odd so this reduces to <math>2\sum_{k=1}^{2^{n-2}}1+g(2k) = 2(2^{n-2}+\sum_{k=1}^{2^n-2}g(2k)).</math> Thus <math>S_n=2(2^{n-2}+S_{n-1})=2^{n-1}+2S_{n-1}.</math> Further noting that <math>S_0=1</math> we can see that <math>S_n=2^{n-1}\cdot (n-1)+2^n\cdot S_0=2^{n-1}\cdot (n-1)+2^{n-1}\cdot2=2^{n-1}\cdot (n+1). </math> which is the same as above. To | + | <math>2k-1</math> must be odd so this reduces to <math>2\sum_{k=1}^{2^{n-2}}1+g(2k) = 2(2^{n-2}+\sum_{k=1}^{2^n-2}g(2k)).</math> Thus <math>S_n=2(2^{n-2}+S_{n-1})=2^{n-1}+2S_{n-1}.</math> Further noting that <math>S_0=1</math> we can see that <math>S_n=2^{n-1}\cdot (n-1)+2^n\cdot S_0=2^{n-1}\cdot (n-1)+2^{n-1}\cdot2=2^{n-1}\cdot (n+1). </math> which is the same as above. To simplify the process of finding the largest square <math>S_n</math> we can note that if <math>n-1</math> is odd then <math>n+1</math> must be exactly divisible by an odd power of <math>2</math>. However, this means <math>n+1</math> is even but it cannot be. Thus <math>n-1</math> is even and <math>n+1</math> is a large even square. The largest even square <math>< 1001</math> is <math>900</math> so <math>n+1= 900 => n=899</math> |
− | |||
== See also == | == See also == |
Revision as of 14:50, 1 August 2013
Contents
[hide]Problem
For each even positive integer , let denote the greatest power of 2 that divides For example, and For each positive integer let Find the greatest integer less than 1000 such that is a perfect square.
Solution
Given , consider . Define . There are elements of that are divisible by , elements of that are divisible by but not by and elements of that are divisible by but not by .
Thus , so we need for $k \in \N$ (Error compiling LaTeX. Unknown error_msg). Now notice we also require , so if also (but ), then , so we have . If , then , so we have . Finally, could possibly be or 256. The maximum possible is thus .
Alternate Solution
First note that if is odd and if is even. so must be odd so this reduces to Thus Further noting that we can see that which is the same as above. To simplify the process of finding the largest square we can note that if is odd then must be exactly divisible by an odd power of . However, this means is even but it cannot be. Thus is even and is a large even square. The largest even square is so
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.