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Gabriele Dalla Torre Christophe Debry Jinbi Jin Marco Streng Wouter Zomervrucht Problemenrubriek NAW Mathematisch Instituut Universiteit Leiden Postbus 9512 2300 RA Leiden

problems@nieuwarchief.nl www.nieuwarchief.nl/problems

This Problem Section is open to everyone; everybody is encouraged to send in solutions and propose problems. Group contributions are welcome.

For each problem, the most elegant correct solution will be rewarded with a book token worth D20. At times there will be a Star Problem, to which the proposer does not know any solution.

For the first correct solution sent in within one year there is a prize of^D100.

When proposing a problem, please either include a complete solution or indicate that it is intended as a Star Problem. Electronic submissions of problems and solutions are preferred (problems@nieuwarchief.nl).

The deadline for solutions to the problems in this edition is 1 March 2015.

Problem A(proposed by Jan Turk)

Letk > 3be an integer. Determine the variance of the greatest common divisor ofkpositive integers. Here we mean the limit, asn → ∞, of the variance of the greatest common divisor of kintegers in{1, 2, . . . , n}with respect to the uniform distribution on{1, . . . , n}^k.

Problem B(folklore)

The evil Eve has locked Alice and Bob in a room without windows. Outside the room, there is a corridor with 64 doors. Eve puts a key behind one of the doors and a crocodile behind each of the others. Then she hangs up a light bulb above each of the doors, and for each light bulb, switches it on or off. Then Eve brings Alice into the corridor, tells her which door hides the key and tells her to choose one of the light bulbs and change the state of that chosen light bulb.

After Alice leaves, Eve brings Bob to the corridor, and tells him to open a door of his own choice.

Alice and Bob are allowed to discuss a strategy before Alice is shown where the key is, but not after.

a. Give a strategy that guarantees Bob to find the key.

b. For which positive integersndoes such a strategy exist if there arendoors?

Problem C(folklore)

LetX3be the collection of three-element subsets of{1, 2, . . . , 8}, and letX4be the collection of four-element subsets of{1, 2, . . . , 11}. Does there exist an injective mapφ : X3 → X₄with the following properties?

1. For all subsetsV ⊆ X3, we have#(T

v∈Vφ(v)) ≥ #(T

v∈Vv). 2. For allv, v^′∈ X₃, ifv ∩ v^′= ∅, thenφ(v) ∩ φ(v^′) = ∅.

Edition 2014-2 We received solutions from Rik Bos (Bunschoten), Pieter de Groen (Brussels), Alex Heinis (Amsterdam), Nicky Hekster (Amstelveen), Carlo Pagano (Leiden) and Rob van der Waall (Huizen).

Problem 2014-2/A (proposed by Wouter Zomervrucht)

Letnbe a positive integer. LetMbe ann × n-matrix with entries in{1, 2, . . . , n}. Letrbe the complex eigenvalue with the largest absolute value. Show thatn ≤ |r | ≤ n².

SolutionWe received correct solutions from Rik Bos, Pieter de Groen, Alex Heinis, Carlo Pagano and Rob van der Waall. The book token is awarded to Rob van der Waall.

Letk·kbe the Frobenius matrix norm (i.e. forM = (mij)ⁿ_i,j=1, we havekMk²=Pn i=1

Pn

j=1|m_ij|²).

Gelfand’s formula states

|r | = r = lim

k→∞kM^kk^1/k.

LetN be another matrix as in the problem, with largest eigenvalue s. Suppose M ≤ N,

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i.e. inequality on all entries, then alsoM^k ≤ N^k, sokM^kk ≤ kN^kk, andr ≤ s. LetM0be then × n-matrix with all entries equal to1. It has eigenvalues0andn. Note thatnM0has eigenvalues0andn². We haveM0≤ M ≤ nM₀, hencen ≤ r ≤ n²as desired.

Problem 2014-2/B (proposed by Hans Zwart)

LetXbe a unitalR-algebra with multiplicative unit1, and letk · kbe a submultiplicative norm onX, i.e. a mapk · k : X → Rsatisfying the following properties:

− k1k = 1;

− ifx ∈ Xsatisfieskxk = 0, thenx = 0;

− for alla ∈ R, x ∈ X, we havekaxk = |a| kxk;

− for allx, y ∈ X, we havekx + yk ≤ kxk + kykandkxyk ≤ kxk kyk. LetC : R → Xbe a map such thatC(0) = 1and such that for alls, t ∈ R, we have

2C(s)C(t) = C(s + t) + C(s − t).

Suppose that

sup

s∈R

kC(s) − 1k <³₂.

Show thatC = 1.

Solution We received no correct solutions. The following solution is based on that of the proposer, Hans Zwart.

Lett ∈ R. Then we have

C(2t) − 1 = 2C(t)²− C(0) − 1 = 2C(t)²− 2 = 2

C(t) − 12

+ 4

C(t) − 1 .

Hence for allt ∈ R, we have4 C(t)−1

=

C(2t)−1

−2

C(t)−12

, so ifρ = sup_s∈RkC(s)−1k, then for allt ∈ R, we have

4kC(t) − 1k =

C(2t) − 1

− 2

C(t) − 12

≤ kC(2t) − 1k + 2kC(t) − 1k²

≤ ρ + 2ρ²,

hence4ρ ≤ ρ + 2ρ², or equivalently,ρ(2ρ − 3) ≥ 0. As we assumed thatρ < ³₂, it follows that ρ = 0, henceC(t) = 1for allt ∈ R, as desired.

Problem 2014/2-C (proposed by Hendrik Lenstra)

LetGbe a finite group. Letnbe the number of automorphismsσofGsuch that for allx ∈ G, the elementσ (x)is conjugate tox. Show that every prime divisor ofndivides the order ofG.

SolutionWe received solutions from Rik Bos, Alex Heinis, Nicky Hekster, Carlo Pagano and Rob van der Waall. All their solutions were along the same lines, which we reproduce here. The book token goes to Alex Heinis.

LetAbe the set ofσ ∈Aut(G)such that for allx ∈ G, the elementσ (x)is conjugate tox, so n = #A. Letpbe a prime divisor ofn. Note thatAis a subgroup of Aut(G), hence by Cauchy’s theorem has an elementσof orderp. This elementσacts on each conjugacy class ofG, and the orbits have length1orp.

LetH = {x ∈ G : σ (x) = x}be the union of the orbits of length1, which is a proper subgroup ofG.

Lemma. Let H ⊂ Gbe a subgroup withH 6= G. Then the union of the conjugates of H is notG.

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Proof. The group H has at most [G : H] conjugates, becausegHg⁻¹ depends only on gH ∈ G/H. These conjugates all have#H elements and have the element1in common, hence the union of the conjugates has fewer than[G : H] · #H = #Gelements. 

By the lemma, we see that there is an element ofx ∈ Gthat is not conjugate to any element ofH. Its conjugacy classCis a union of orbits forhσ i, which all have lengthpasCis disjoint withH, so#Cis divisible byp.

By the orbit-stabilizer theorem, the order ofGis divisible by the length of the conjugation orbit

Cofx.