olymmath

CIIM 2014

Iberoamerican Interuniversity Mathematics Competition 2014.

Day 1

Problem 1

  • Day 1

Let g:[2013,2014]→Rg:[2013,2014]\to\mathbb{R} a function that satisfy the following two conditions:

i) g(2013)=g(2014)=0,g(2013)=g(2014) = 0, ii) for any a,b∈[2013,2014]a,b \in [2013,2014] it hold that g(a+b2)≤g(a)+g(b).g\left(\frac{a+b}{2}\right) \leq g(a) + g(b). Prove that gg has zeros in any open subinterval (c,d)⊂[2013,2014].(c,d) \subset[2013,2014].

Answer, solutions and source

Problem 2

  • Day 1

Let nn be an integer and pp a prime greater than 2. Show that: (p−1)nn!∣(pn−1)(pn−p)(pn−p2)⋯(pn−pn−1).(p-1)^nn!|(p^n-1)(p^n-p)(p^n-p^2)\cdots(p^n-p^{n-1}).

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Problem 3

  • Day 1

Given n≥2n\geq2 , let A\mathcal{A} be a family of subsets of the set {1,2,…,n}\{1,2,\dots,n\} such that, for any A1,A2,A3,A4∈AA_1,A_2,A_3,A_4 \in \mathcal{A} , it holds that ∣A1∪A2∪A3∪A4∣≤n−2|A_1 \cup A_2 \cup A_3 \cup A_4| \leq n -2 . Prove that ∣A∣≤2n−2.|\mathcal{A}| \leq 2^{n-2}.

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Day 2

Problem 4

  • Day 2

Let {ai}\{a_i\} be a strictly increasing sequence of positive integers. Define the sequence {sk}\{s_k\} as sk=∑i=1k1[ai,ai+1],s_k = \sum_{i=1}^{k}\frac{1}{[a_i,a_{i+1}]}, where [ai,ai+1][a_i,a_{i+1}] is the least commun multiple of aia_i and ai+1a_{i+1} . Show that the sequence {sk}\{s_k\} is convergent.

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Problem 5

  • Day 2

A analityc function f:C→Cf:\mathbb{C}\to\mathbb{C} is call interesting if f(z)f(z) is real along the parabola Re(z)=(Im(z))2Re (z) = (Im (z))^2 .

a) Find an example of a non constant interesting function. b) Show that every interesting function ff satisfy that f′(−3/4)=0.f'(-3/4) = 0.

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Problem 6

  • Day 2

a) Let {xn}\{x_n\} be a sequence with xn∈[0,1]x_n \in [0,1] for any nn . Prove that there exists C>0C > 0 such that for every positive integer rr , there exists m≥1m \geq 1 and n>m+rn > m + r that satisfy (n−m)∣xn−xm∣≤C(n-m)|x_n-x_m| \leq C . b) Prove that for every C>0C > 0 , there exists a sequence {xn}\{x_n\} with xn∈[0,1]x_n \in [0,1] for all nn and an integer rr such that, if m≥1m \geq 1 and n>m+rn > m+r , then (n−m)∣xn−xm∣>C.(n-m)|x_n-x_m| > C.

Answer, solutions and source