Difference between revisions of "1991 USAMO Problems/Problem 3"
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Note that for all integers <math>b</math>, the sequence <math>2^0, 2^1, 2^2, \dotsc</math> is eventually becomes cyclic mod <math>b</math>. Let <math>k</math> be the period of this cycle. Since there are <math>k-1</math> nonzero residues mod <math>b</math>. <math>1 \le k\le b-1 < b</math>. Since | Note that for all integers <math>b</math>, the sequence <math>2^0, 2^1, 2^2, \dotsc</math> is eventually becomes cyclic mod <math>b</math>. Let <math>k</math> be the period of this cycle. Since there are <math>k-1</math> nonzero residues mod <math>b</math>. <math>1 \le k\le b-1 < b</math>. Since | ||
<cmath> 2, 2^2, 2^{2^2}, 2^{2^{2^2}}, \dotsc </cmath> | <cmath> 2, 2^2, 2^{2^2}, 2^{2^{2^2}}, \dotsc </cmath> | ||
− | does not become constant mod <math> | + | does not become constant mod <math>b</math>, it follows the sequence of exponents of these terms, i.e., the sequence |
<cmath> 1, 2, 2^2, 2^{2^{2}}, \dotsc </cmath> | <cmath> 1, 2, 2^2, 2^{2^{2}}, \dotsc </cmath> | ||
does not become constant mod <math>k</math>. Then the problem statement is false for <math>n=k</math>. Since <math>k<b</math>, this is a contradiction. Therefore the problem statement is true. <math>\blacksquare</math> | does not become constant mod <math>k</math>. Then the problem statement is false for <math>n=k</math>. Since <math>k<b</math>, this is a contradiction. Therefore the problem statement is true. <math>\blacksquare</math> |
Revision as of 18:33, 21 January 2013
Problem
Show that, for any fixed integer the sequence is eventually constant.
[The tower of exponents is defined by . Also means the remainder which results from dividing by .]
Solution
Suppose that the problem statement is false for some integer . Then there is a least , which we call , for which the statement is false.
Since all integers are equivalent mod 1, .
Note that for all integers , the sequence is eventually becomes cyclic mod . Let be the period of this cycle. Since there are nonzero residues mod . . Since does not become constant mod , it follows the sequence of exponents of these terms, i.e., the sequence does not become constant mod . Then the problem statement is false for . Since , this is a contradiction. Therefore the problem statement is true.
Note that we may replace 2 with any other positive integer, and both the problem and this solution remain valid.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
1991 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |