Difference between revisions of "2016 USAMO Problems"

(Problem 2)
(Problem 2)
Line 7: Line 7:
 
{{problem}}
 
{{problem}}
 
Prove that for any positive integer <math>k,</math>
 
Prove that for any positive integer <math>k,</math>
<cmath>\left(k^2\right)!\cdot\product_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}</cmath>
+
<cmath>\left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}</cmath>
 
is an integer.
 
is an integer.
  

Revision as of 00:48, 27 April 2016

Day 1

Problem 1

Let $X_1, X_2, \ldots, X_{100}$ be a sequence of mutually distinct nonempty subsets of a set $S$. Any two sets $X_i$ and $X_{i+1}$ are disjoint and their union is not the whole set $S$, that is, $X_i\cap X_{i+1}=\emptyset$ and $X_i\cup X_{i+1}\neq S$, for all $i\in\{1, \ldots, 99\}$. Find the smallest possible number of elements in $S$.

Solution

Problem 2

This problem has not been edited in. Help us out by adding it. Prove that for any positive integer $k,$ \[\left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}\] is an integer.

Problem 3

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Day 2

Problem 4

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$, \[(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.\]

Solution

Problem 5

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Problem 6

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These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

2016 USAMO (ProblemsResources)
Preceded by
2015 USAMO
Followed by
2017 USAMO
1 2 3 4 5 6
All USAMO Problems and Solutions