Difference between revisions of "2006 USAMO Problems"
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== Problem 3 == | == Problem 3 == | ||
− | For integral <math>m</math>, let <math>p(m)</math> be the greatest prime divisor of <math>m</math>. By convention, we set <math> p(\pm1)=1</math> and <math>p(0)=\infty</math>. Find all polynomials <math>f</math> with integer coefficients such that the sequence <math>\lbrace p(f(n^2))-2n | + | For integral <math>m</math>, let <math>p(m)</math> be the greatest prime divisor of <math>m</math>. By convention, we set <math> p(\pm1)=1</math> and <math>p(0)=\infty</math>. Find all polynomials <math>f</math> with integer coefficients such that the sequence <math>\lbrace p(f(n^2))-2n\rbrace_{n\ge0}</math> is bounded above. (In particular, this requires <math>f(n^2)\neq0</math> for <math>n\ge0</math>.) |
[[2006 USAMO Problems/Problem 3 | Solution]] | [[2006 USAMO Problems/Problem 3 | Solution]] |
Revision as of 14:23, 16 April 2009
Contents
[hide]Day 1
Problem 1
Let be a prime number and let be an integer with . Prove that there exist integers and with and if and only if is not a divisor of .
Note: For a real number, let denote the greatest integer less than or equal to , and let denote the fractional part of .
Problem 2
For a given positive integer find, in terms of , the minimum value of for which there is a set of distinct positive integers that has sum greater than but every subset of size has sum at most .
Problem 3
For integral , let be the greatest prime divisor of . By convention, we set and . Find all polynomials with integer coefficients such that the sequence is bounded above. (In particular, this requires for .)
Day 2
Problem 4
Find all positive integers such that there are positive rational numbers satisfying .
Problem 5
A mathematical frog jumps along the number line. The frog starts at 1, and jumps according to the following rule: if the frog is at integer , then it can jump either to or to where is the largest power of 2 that is a factor of . Show that if is a positive integer and is a nonnegative integer, then the minimum number of jumps needed to reach is greater than the minimum number of jumps needed to reach .
Problem 6
Let be a quadrilateral, and let and be points on sides and , respectively, such that . Ray meets rays and at and respectively. Prove that the circumcircles of triangles , , , and pass through a common point.
Resources
- USAMO Problems and Solutions
- USAMO 2006 Problems on the Resources Page
- USAMO 2006 Questions Document
- USAMO 2006 Solutions Document
2006 USAMO (Problems • Resources) | ||
Preceded by 2005 USAMO |
Followed by 2007 USAMO | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |