MathDB

2012 IMC

Part of IMC

Subcontests

(5)
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IMC 2012 Day 1, Problem 5

Let aa be a rational number and let nn be a positive integer. Prove that the polynomial X2n(X+a)2n+1X^{2^n}(X+a)^{2^n}+1 is irreducible in the ring Q[X]\mathbb{Q}[X] of polynomials with rational coefficients.
Proposed by Vincent Jugé, École Polytechnique, Paris.

Plunecke's Inequality

Let c≥1c \ge 1 be a real number. Let GG be an Abelian group and let A⊂GA \subset G be a finite set satisfying ∣A+A∣≤c∣A∣|A+A| \le c|A|, where X+Y:={x+y∣x∈X,y∈Y}X+Y:= \{x+y| x \in X, y \in Y\} and ∣Z∣|Z| denotes the cardinality of ZZ. Prove that ∣A+A+⋯+A⏟k∣≤ck∣A∣|\underbrace{A+A+\dots+A}_k| \le c^k |A| for every positive integer kk.
Proposed by Przemyslaw Mazur, Jagiellonian University.
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IMC 2012 Day 1, Problem 3

Given an integer n>1n>1, let SnS_n be the group of permutations of the numbers 1,  2,  3,  …,  n1,\;2,\;3,\;\ldots,\;n. Two players, A and B, play the following game. Taking turns, they select elements (one element at a time) from the group SnS_n. It is forbidden to select an element that has already been selected. The game ends when the selected elements generate the whole group SnS_n. The player who made the last move loses the game. The first move is made by A. Which player has a winning strategy?
Proposed by Fedor Petrov, St. Petersburg State University.

n!+1 divides (2012n)!

Is the set of positive integers nn such that n!+1n!+1 divides (2012n)!(2012n)! finite or infinite?
Proposed by Fedor Petrov, St. Petersburg State University.
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IMC 2012 Day 1, Problem 1

For every positive integer nn, let p(n)p(n) denote the number of ways to express nn as a sum of positive integers. For instance, p(4)=5p(4)=5 because
4=3+1=2+2=2+1+1=1+1+1.4=3+1=2+2=2+1+1=1+1+1.
Also define p(0)=1p(0)=1.
Prove that p(n)−p(n−1)p(n)-p(n-1) is the number of ways to express nn as a sum of integers each of which is strictly greater than 1.
Proposed by Fedor Duzhin, Nanyang Technological University.

Albert Einstein and Homer Simpson

Consider a polynomial f(x)=x2012+a2011x2011+⋯+a1x+a0.f(x)=x^{2012}+a_{2011}x^{2011}+\dots+a_1x+a_0. Albert Einstein and Homer Simpson are playing the following game. In turn, they choose one of the coefficients a0,a1,…,a2011a_0,a_1,\dots,a_{2011} and assign a real value to it. Albert has the first move. Once a value is assigned to a coefficient, it cannot be changed any more. The game ends after all the coefficients have been assigned values. Homer's goal is to make f(x)f(x) divisible by a fixed polynomial m(x)m(x) and Albert's goal is to prevent this. (a) Which of the players has a winning strategy if m(x)=x−2012m(x)=x-2012? (b) Which of the players has a winning strategy if m(x)=x2+1m(x)=x^2+1?
Proposed by Fedor Duzhin, Nanyang Technological University.