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Johan Bosman 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 September 2014.

Problem 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.

Problem 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. Let_{C : R → X}be a map such thatC(0) = 1and such that for all_{s, 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.

Problem 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.

Edition 2013-4 We received solutions from Alexandros Efthymiadis (St. Andrews), Alex Heinis (Amsterdam), Richard Kraaij (Delft), Thijmen Krebs (Nootdorp), Matth´e van der Lee (Amsterdam), Tejaswi Navilarekallu, Rohith Varma (Chennai, India), Traian Viteam (Cape Town, South Africa) and Hans Zwart (Enschede).

Problem 2013-4/A (folklore, communicated by Jaap Top)

Does there exist an integern > 1such that the set of leading digits of2ⁿ, 3ⁿ, . . . , 9ⁿis equal to {2, 3, . . . , 9}?

Solution The answer to the question is ‘no’. Let ld(k)denote the leading digit ofk. The more general question which patterns occur in the table

ld(2ⁿ),ld(3ⁿ), . . . ,ld(9ⁿ)

n

is known as Gelfand’s question. See, e.g., mathworld.wolfram.com/GelfandsQuestion.html.

We received correct solutions from Alexandros Efthymiadis, Alex Heinis and Thijmen Krebs. The book token goes to Thijmen Krebs, and the following solution is based on his.

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Problemen NAW 5/15 nr. 2 juni 2014

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Supposen > 1as in the problem exists and letσd=ld(dⁿ), so we have a permutationd 7→ σd

of{2, 3, . . . , 9}. Then for everyd, there exists somem_d∈ Zwith

σ_d10^md ≤ dⁿ< (σ_d+ 1)10^md. (1)

In particular, we have

σ2σ510^m2+m5≤ 10ⁿ< (σ2+ 1)(σ5+ 1)10^m2+m5, (2)

son = m2+m5+ 1andσ2σ5∈ {6, 8, 10}.

In caseσ2σ5= 10, we have equality in the left hand side of (2), hence5ⁿ=σ510^m5, but the left hand side has no factors2, som₅= 0andn = 1, contradiction. This leaves the three cases σ2= 2, 3, 4.

Ifσ2= 4, thenσ5= 2, but alsoσ4= 2because of

σ₂²10^2m2≤ 4ⁿ< (σ₂+ 1)²10^2m2. (3)

A contradiction sinceσ₂6= σ₄.

Ifσ2= 3, thenσ5= 2andσ4= 9because of (3). But thenσ8∈ {2, 3}because of

σ2σ410^m2+m4≤ 8ⁿ< (σ2+ 1)(σ4+ 1)10^m2+m4. (4)

This is a contradiction asσ2, σ5, σ8∈ {2, 3}are distinct.

Finally, ifσ₂ = 2, thenσ₅ ∈ {3, 4}andσ₄ ≤ 4by 4 (otherwiseσ₈ = 2 =σ₂). Similarly, the inequality

σ2σ310^m2+m3≤ 6ⁿ< (σ2+ 1)(σ3+ 1)10^m2+m3 (5)

impliesσ3≤ 4. Nowσ2, σ3, σ4, σ5∈ {2, 3, 4}is another contradiction.

This proves thatndoes not exist and solves the problem.

Wolfram MathWorld (mathworld.wolfram.com/GelfandsQuestion.html) calls the question whether there is annwith ld(dⁿ) =dfor alld ∈ {2, 3, . . . , 9}an open question. The solu- tion above implies that the answer to this question is ‘no’ as well.

Problem 2013-4/B (proposed by Bart de Smit and Hendrik Lenstra)

Rings are unital, and morphisms of rings send1to1. LetAandBbe commutative rings.

Suppose that there exists a ringCsuch that there are injective morphismsA → CandB → Cof rings. Show that there exists a commutative such ring.

Solution We received solutions from Matth´e van der Lee and Rohith Varma. The book token is awarded to Matth´e van der Lee.

Letf : A → Candg : B → Cbe the given injections. The map

A × B → C, (a, b) 7→ f (a)g(b)

is Z-bilinear, so induces a ring homomorphismh : A ⊗_ZB → C. Leti : A → A ⊗_ZBand j : B → A ⊗_ZBbe the canonical maps. The compositionsf = h ◦ iandg = h ◦ jare injective by assumption, henceiandjare injective. The ringA ⊗_ZBis commutative, so we are done.

Problem 2013-4/C (proposed by Jinbi Jin)

LetC(R, R)denote the set of continuous maps fromRto itself. A (not necessarily continuous) mapf : C(R, R) → C(R, R)is called good if it satisfies, for alls, t ∈ C(R, R), the identity

f (s ◦ t) = f (s)f (t),

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where the product on the right hand side is the point-wise multiplication of maps.

− Find a non-constant good mapf : C(R, R) → C(R, R).

− Show thatf (exp) = 0for all non-constant good mapsf : C(R, R) → C(R, R). (Here,expis given byx 7→ e^x.)

Solution We received solutions from Alex Heinis, Richard Kraaij, Thijmen Krebs, Tejaswi Navi- larekallu, Traian Viteam and Hans Zwart. The following solution (of the second part) is based on that of Tejaswi Navilarekallu, to whom we also award the book token for this problem.

Giving a non-constant good map Letf : C(R, R) → C(R, R)be given by

f (s) =







1 s is bijective, 0 otherwise.

Showing thatfis indeed a good map amounts to proving the following lemma.

Lemma 1. Lets, t : R → Rbe continuous maps such thats ◦ tis bijective. Then bothsand tare bijective.

Proof. First note thattands ◦ tare injective, so by the intermediate value theorem, they are either strictly increasing, or strictly decreasing. Assume without loss of generality thattand s ◦ tis strictly increasing.

Suppose for a contradiction that the image oft is bounded from above, and let M be its supremum inR. Let_{x ∈ R}be arbitrary, and consider the intervalI = [x, ∞). Astis increasing, it follows that the image ofIundertlies in[t(x), M]. Ass ◦ tis increasing and bijective, it follows thats ◦ t(I) = [s ◦ t(x), ∞). But on the other hand, as[t(x), M]is closed and bounded, s ◦ t(I)must be a closed and bounded interval, and this is a contradiction. Hence the image oftis not bounded from above, and similarly, we can show that it is not bounded from below either. Astis a strictly increasing continuous map, it follows thattis a bijection, and hence

thatsis a bijection as well, as desired. 

Showing thatf (exp) = 0forfnon-constant and good

Letfbe a non-constant good map. In what follows below we denote by0,1and−1the constant maps taking the corresponding values.

Let

i(x) = x, j(x) = −x, u(x) =







x ifx > 0, 0 otherwise.

For anys ∈ C(R, R)such thats ◦ s = swe havef (s) = f (s ◦ s) = f (s)f (s). Sincef (s)is a continuous function, it follows thatf (s) = 1orf (s) = 0.

Note thati ◦ i = i, sof (i) = 1orf (i) = 0. Iff (i) = 0, thenf (s) = f (s)f (i) = 0for alls ∈ C(R, R) and thusfis constant, contrary to our assumption. Thereforef (i) = 1. Also note that0 ◦ 0 = 0, sof (0) = 1orf (0) = 0. Iff (0) = 1, thenf (s) = f (0 ◦ s) = 1for alls ∈ C(R, R), sof is again constant, contrary to our assumption. Hencef (0) = 0. Finally, note thatf (i) = f (j ◦ j) = f (j)² and hencef (j) = 1orf (j) = −1.

Now

f (u)f (j) = f (u ◦ j)

=f (u ◦ u ◦ j) = f (u)f (u ◦ j) = f (u ◦ j)f (u) = f (u ◦ j ◦ u)

=f (0) = 0.

Sincef (j) = ±1, it follows that f (u) = 0. Therefore, asu ◦ exp = exp, we havef (exp) = f (u ◦ exp) = f (u)f (exp) = 0.