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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 March 2014.

Problem 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}?

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

Problem 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),

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.)

Edition 2013-2 We received solutions from Leon van den Broek (Nijmegen), Alex Heinis (Amster- dam), Jos van Kan (Delft), Thijmen Krebs (Nootdorp), Javier S´anchez-Reyes (Castilla-La Mancha, Spain) and ´Angel Plaza (Las Palmas de Gran Canaria, Spain), and Robert van der Waall (Huizen).

Problem 2013-2A (based on a problem proposed by Gerard Renardel de Lavalette)

We have two hourglasses,Aforaseconds andBforbseconds, whereaandbare relatively prime integers and0< a < b. Lett₀be an integer witht₀≥ b + (¹₂a − 1)². Show thatAandB can be used to identify the timet = t0if the upper bulbs are empty att = 0.

Remark. The original problem received from the proposer was to prove a slightly stronger result.

Letmbe the remainder ofbupon division bya. The original problem was to prove that for any integert₀> b + m(a − m) − a, the timet = t₀can be identified usingAandB.

Solution We received only one correct solution, from Thijmen Krebs, who will receive the book token. The following solution is based on that solution.

Letmbe the remainder ofbupon division bya. For any integerTthat is a multiple ofaorb, we can use the following strategy:

− whilet < T, turn each hourglass whenever it is empty;

− whilet ≥ T, turn both hourglasses whenever at least one is empty.

If we apply this strategy toT = b, then we turn both hourglasses at the timest = b + kmfor k = 0, 1, 2, . . ..

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If we apply this strategy toT = a(1 +^b−m_a ) =b + (a − m), then we turn both hourglasses at the timest = b + k(a − m)fork = 1, 2, 3, . . ..

In particular, all elements of the following set are measurable times:

S = {b + km : 0 ≤ k < a − m} ∪ {b + k(a − m) : 0 < k ≤ m}.

Asaandbare coprime, so aremanda, henceScontains an element of each residue class moduloa. Moreover, the maximal element ofSisb + m(a − m).

Before starting the strategy above, we can measure any non-negative integer multiple ofa seconds usingA, while lettingBstay empty. In particular, we can measure any timet0 ≥ b + m(a − m) − a + 1.

Finally, notem(a − m) ≤ (¹₂a)², sob + m(a − m) − a + 1 ≤ b + (¹₂a)²− a + 1 = b + (¹₂a − 1)² and we can measure any timet0≥ b + (¹₂a − 1)².

Problem 2013-2B (folklore, communicated by Jeanine Daems)

In a two-player game, players take turns drawing a number of coins from a pile that starts withn coins. The first player takes at least one coin from the pile, but not all. In the subsequent turns, each player takes at least one coin, and at most twice the number of coins taken in the previous turn. The player who takes the last coin wins. For which numbersncan the first player win?

SolutionWe received correct solutions from Alex Heinis and Thijmen Krebs. The book token is awarded to Alex Heinis. The game is known as Fibonacci Nim, and the first player can win for those integersn > 1that are not a Fibonacci number.

Let(Fk)k≥1be the Fibonacci sequence: F1 = 1,F2 = 2andFk+2 =Fk+1+Fkfork ≥ 1. The proof uses Zeckendorf’s theorem: every positive integer can uniquely be written as the sum of non-consecutive Fibonacci numbers. Letzbe the function on the positive integers that assign tomthe smallest Fibonacci number occurring in the Zeckendorf decomposition ofm. E.g., we can write20 = 13 + 5 + 2 =F₆+F₄+F₂andz(20) = F₂= 2.

We define a position in this game to be a pair(m, d)wheremis number of coins left on the pile anddthe maximal number of coins that may be taken (by the player who is to move). The initial position is(n, n − 1)and the final positions are those of the form(0, d). Call a position (m, d)‘good’ if it is non-final andd ≥ z(m); call it ‘bad’ otherwise.

Lemma. Let(m, d)be a good position. There exists a move to a bad position.

Proof. Writem = Fk1+ · · · +Fk_r for the Zeckendorf decomposition, withki≥ ki+1+ 2. By assumption,d is at leastFk_r. Our move is to take exactlyFk_r coins. The new position is (m − F_kr, 2F_kr). This is a bad position: in the caser = 1it is even final, and otherwise it follows

from2Fk_r < Fk_{r −1}. 

Lemma. Let(m, d)be a non-final bad position. All moves lead to a good position.

Proof. WriteF_k=z(m). By assumption we haved < F_k. Suppose we takexcoins, for some x ∈ {1, . . . , d}. Lett ≥ 0be the even number such thatFk−t−2≤ x < Fk−t. Then

F_k− F_k−t < F_k− x ≤ F_k− F_k−t−2

hence

Fk−1+Fk−3+ · · · +Fk−t+1< Fk− x ≤ Fk−1+Fk−3+ · · · +Fk−t−1,

soz(F_k− x) ≤ F_k−t−1, which is smaller than2F_k−t−2 ≤ 2x. Note further thatz(F_k− x) =

z(m − x). Hence(m − x, 2x)is a good position. _

Together the lemmas show that the good positions are exactly the winning ones. The initial position(n, n − 1)is good if and only ifz(n) ≤ n − 1, i.e., if and only ifnis not Fibonacci.

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Problem 2013-2C (proposed by Bas Edixhoven and Maarten Derickx)

LetABCDbe a convex quadrilateral inside a planeU inR³. Suppose thatABCDis not a parallelogram. Show that there exist a planeVinR³and a point_{P ∈ R}³− (U ∪ V )such that if a light source is placed inP, then the shadow ofABCDonVis a square.

Solution We received solutions from Leon van den Broek, Alex Heinis, Jos van Kan, Javier S´anchez-Reyes and ´Angel Plaza, and Robert van der Waall. The book token goes to Jos van Kan.

The main idea of the solution is to pick the planeV and the pointP in such a way that the projection on V of the intersection of the linesAB andCD, and that ofBC andAD is ‘at infinity’. Some extra conditions onPrelated to the diagonals and consecutive edges will then ensure that the projection ofABCDonVis a square.

If the linesAB andCDintersect, denote their intersection byX₁. Similarly, ifBC andAD intersect, denote their intersection byX2.

We consider three cases, the first of which is the following.

Case 1. The linesABandCDintersect, and so doBCandAD. Moreover, both ofACandBD intersect the lineX1X2.

First note thatX1X2does not intersect the quadrilateralABCD, asABCDis convex. LetY1be the intersection ofACandX₁X₂, and likewise, letY₂be the intersection ofBDandX₁X₂. LetWbe a plane that has as intersection the lineX₁X₂withU. In particular,U 6= W. LetΓ1, Γ2 be the circles inW with the segmentsX1X2,Y1Y2as diameter, respectively. LetP be an intersection ofΓ1andΓ2, and letVbe any plane parallel toWsuch that the quadrilateralABCD lies betweenVandW. This intersection exists as one ofY1,Y2lies betweenX1andX2, and the other does not.

Then note thatPdoes not lie inU, as the pointsX₁, X₂, Y₁, Y₂are pairwise distinct, and thatP does not lie inV, asPlies inW, which is parallel toV. Hence_{P ∈ R}³− (U ∪ V ).

Now letA0, B0, C0, D0be the respective intersections ofAP , BP , CP , DPwithV. They exist, as the given lines intersect inPwithW, which is parallel toV. By construction ofV, and asX₁X₂ does not intersect the quadrilateralABCD, it now suffices to show thatA0B0C0D0is a square inV.

Letlbe a line inU, not equal toX₁X₂. Write^P(l)for the unique plane throughlandP, write I(l)for the intersection line ofWwith^P(l), and write^I0(l)for the intersection line ofVwith^P(l). (So for example,^L0(AB) = A₀B₀.) AsV andWare parallel, it follows that for all linesl,min U, the angle between^I(l)and^I(m)is equal to the one between^I0(l)and^I0(m). Note that a square is a quadrilateral such that

− every two successive edges are perpendicular, and

− the diagonals are perpendicular.

Therefore, to show thatA₀B₀C₀D₀is a square, it suffices to show that the^I(e)(withean edge or a diagonal of the quadrilateralABCD) satisfy the above properties.

Now we simply note that

I(AB) = I(CD) = P X₁, I(BC) = I(AD) = P X₂,

and that

I(AC) = P Y₁, I(BD) = P Y₂,

so by construction ofP, the quadrilateralA₀B₀C₀D₀is a square, as desired.

For the remaining cases, we will only state them, and the corresponding construction ofP, as the proof (and the construction ofV) is done in the same way.

Case 2. The linesABandCDintersect, and so doBCandAD. Moreover, at most one ofAC andBDintersects the lineX₁X₂.

Note here that at least one ofACandBDintersects the lineX1X2, asACandBDintersect, so exactly one of them intersectsX₁X₂. We assume without loss of generality thatACand

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X₁X₂ intersect, and letYbe their intersection. Note thatY lies betweenX₁andX₂, as the lineACintersects the segmentBD, which is parallel toX1X2. LetWbe any plane that has as intersection the lineX₁X₂withU, and letΓbe the circle inWwith diameterX₁X₂. Then we takePto be an intersection ofΓwith the line throughYperpendicular toX₁X₂.

Case 3.Exactly one of the pairs of lines(AB, CD)and(BC, AD)intersect.

We assume without loss of generality thatABandCDintersect, and letXdenote this inter- section. Letlbe the line throughXparallel toBC(hence also toAD). Then the linesACand BDboth intersectl, as they intersectBC. LetY1andY2be their respective intersections. Then Xlies betweenY1andY2, as forSthe intersection ofACandBD, the lineXSintersects the segmentsBCandAD, which are parallel tol. LetWbe any plane that haslas intersection with U, and let_Γbe the circle with diameterY1Y2. Then we takePto be an intersection of_Γwith the line throughXperpendicular tol.

References.This problem turned out to be rather well-known, as we received a lot of references.

Thanks to Leon van den Broek, Javier S´anchez-Reyes and ´Angel Plaza, and Robert van der Waall for these. The references given were, respectively,

− L. van den Broek, Welke schaduwbeelden, Euclides 64, nr. 3 (1988, in Dutch).

− Problem 72 of H. D¨orrie, 100 great problems of elementary mathematics, their history and solution (translation of Thriumph der Mathematik, 1932), reworked in 2010 by M. Woltermann http://www2.washjeff.edu/users/MWoltermann/Dorrie/72.pdf.

− E.J. Hopkins and J.S. Hails, An Introduction to Plane Projective Geometry, Clarendon Press (1953).