19th IMC

IMC 2012

Blagoevgrad, Bulgaria · papers 28 July & 29 July 2012 · 10 problems across 2 papers

Official result files →

Day 1

28 July 2012 · 5 problems
  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+1.4 = 3 + 1 = 2 + 2 = 2 + 1 + 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 11.

  2. Problem 2

    Let nn be a fixed positive integer. Determine the smallest possible rank of an n×nn \times n matrix that has zeros along the main diagonal and strictly positive real numbers off the main diagonal.

  3. Problem 3

    Given an integer n>1n > 1, let SnS_n be the group of permutations of the numbers 1,2,…,n1, 2, \dots, n. Two players, AA and BB, 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 AA. Which player has a winning strategy?

  4. Problem 4

    Let f:R→Rf : \mathbb{R} \to \mathbb{R} be a continuously differentiable function that satisfies f′(t)>f(f(t))f'(t) > f(f(t)) for all t∈Rt \in \mathbb{R}. Prove that f(f(f(t)))≤0f(f(f(t))) \le 0 for all t≥0t \ge 0.

  5. 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.

Day 2

29 July 2012 · 5 problems
  1. Problem 1

    Consider a polynomial

    f(x)=x2012+a2011x2011+⋯+a1x+a0.f(x) = x^{2012} + a_{2011}x^{2011} + \dots + a_1 x + a_0.

    Albert Einstein and Homer Simpson are playing the following game. In turn, they choose one of the coefficients a0,…,a2011a_0, \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?

  2. Problem 2

    Define the sequence a0,a1,…a_0, a_1, \dots inductively by a0=1a_0 = 1, a1=12a_1 = \frac{1}{2} and

    an+1=nan21+(n+1)anfor n≥1.a_{n+1} = \frac{na_n^2}{1 + (n+1)a_n} \quad \text{for } n \ge 1.

    Show that the series ∑k=0∞ak+1ak\sum\limits_{k=0}^{\infty} \dfrac{a_{k+1}}{a_k} converges and determine its value.

  3. Problem 3

    Is the set of positive integers nn such that n!+1n! + 1 divides (2012n)!(2012n)! finite or infinite?

  4. Problem 4

    Let n≥2n \ge 2 be an integer. Find all real numbers aa such that there exist real numbers x1,…,xnx_1, \dots, x_n satisfying

    x1(1−x2)=x2(1−x3)=⋯=xn−1(1−xn)=xn(1−x1)=a.(1)x_1(1 - x_2) = x_2(1 - x_3) = \dots = x_{n-1}(1 - x_n) = x_n(1 - x_1) = a. \tag{1}
  5. Problem 5

    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 \mid x \in X, y \in Y\} and ∣Z∣|Z| denotes the cardinality of ZZ. Prove that

    ∣A+A+⋯+A⏟k times∣≤ck∣A∣|\underbrace{A + A + \dots + A}_{k \text{ times}}| \le c^k |A|

    for every positive integer kk. (Plünnecke's inequality)