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Problemen NAW 5/12 nr. 4 december 2011

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Johan Bosman Gabriele Dalla Torre Jinbi Jin Ronald van Luijk Lenny Taelman 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 20 Euro. 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 100 Euro.

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 March 1st, 2011.

Problem A(folklore)

Let_Γbe a finite undirected graph (without loops or multiple edges). Denote the set of vertices byV. Assume that there are a function_{f : V → Z}and a positive integernsuch that

X

v

X

w

(f (v) − f (w))     

= 2n,

wherevruns over all the vertices of_Γandwover all the neighbours ofv. Show that there are anm ≤ nand a collection ofmedges such that the graph obtained from_Γby removing those edges is not connected.

Problem B(folklore)

Letbe a positive real number. Show that there is a finite groupGthat is not a2-group, but in which the proportion of elements that have2-power order is at least1 −.

Problem C(proposed by Hendrik Lenstra)

LetBbe a commutative ring andAa subring ofB. Assume that the additive group ofAhas finite index inB. Show that the unit group ofAhas finite index in the unit group ofB.

Edition 2011-2 We have received correct solutions from Pieter de Groen (Brussel), Alex Heinis (Hoofddorp), Wim Hesselink (Groningen), Alexander van Hoorn (Abcoude), Thijmen Krebs (Noot- dorp), Tejaswi Navilarekallu (Amsterdam), Michiel Smid (Ottawa), Rob van der Waall (Huizen) and Martijn Weterings (Wageningen).

Problem 2011-2/A LetSbe an open subset ofR>0that contains arbitrarily small elements.

Prove that every positive real number can be written as a sum of finitely many elements ofS.

Solution We have received solutions from Pieter de Groen, Alex Heinis, Wim Hesselink, Alexan- der van Hoorn, Thijmen Krebs, Tejaswi Navilarekallu, Michiel Smid and Martijn Weterings. The book token goes to Wim Hesselink.

Letxbe a positive real number. By assumption there existy < xand0<  < x − ysuch that the open interval(y − , y + )is contained inS. Also choosez ∈ Swithz < 2. Then the interval(x − y − , x − y + )contains a multiplenzofz, withn ≥ 0an integer. Now w = x − nzlies in(y − , y + ) ⊂ Sandx = w + nzis a finite sum of elements ofS.

Problem 2011-2/B Letnbe a positive integer. Show that every sequence ofnelements of{0, . . . , 9}occurs as a sequence of consecutive digits in the last2ndigits of the decimal representation of some power of2. Also, determine all_{α ∈ R>0}for which the statement still holds if we replace2nbydαne.

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Solution We have received solutions from Alex Heinis, Tijmen Krebs and Tejaswi Navilarekallu.

The book token goes to Alex Heinis. The following is partly based on the solution of Tejaswi Navilarekallu.

Note that for every positive integern,2is a primitive root mod5ⁿ. We will use this fact later on.

We prove that the statement holds if and only ifα ≥ ^{log 10}_{log 5}. First note that forα < 1, the statement cannot hold, asdαne < nfor sufficiently largen.

So suppose1 ≤α < ^{log 10}_{log 5}. Note that there are at most4 · 5^dαne−1residue classes modulo 5^dαnethat are residues of powers of2. Also note that modulo2^dαne, there aredαneresidue classes that are residues of powers of2, and that fork ≥ dαne, we have2^k≡ 0 (mod 2^dαne). Hence, by the Chinese remainder theorem, there are at mostAn= 4 · 5^dαne−1+ dαneresidue classes modulo10^dαnethat are residues of powers of2.

Each of theseAnresidue classes gives rise to at mostdαne − n + 1sequences ofndigits in the lastdαnedigits of a power of 2. So if we can show that there exists annwith(dαne−n+1)An<

10ⁿ, then we are done. To this end, note that the left-hand side can be bounded from above by a constant multiple ofn5^αn. Since5^α< 10, for any_{c ∈ R}>0, we have for sufficiently largen thatcn5^αn< 10ⁿ. Hence also(dαne − n + 1)An< 10ⁿfor sufficiently largen.

Now supposeα ≥ ^{log 10}_{log 5}. Letn ≥ 1be an integer. Supposea0, . . . , an−1∈ {0, 1, . . . , 9}and setA =Pn−1

k=010^kak. Note that forr = dαne − n, we have2^n+r < 10^r. Therefore, there is a multipleYof2^n+rwithA · 10^r< Y < (A + 1) · 10^r. So lets ≤ rbe the smallest positive integer for which there exists a multipleYof2^n+swithA · 10^s< Y < (A + 1) · 10^s.

IfYwere divisible by10, then it would follow thats ≥ 2, as there are no multiples of 10 between 10Aand10(A + 1). ThenA · 10^s−1 < ₁₀^Y < (A + 1) · 10^s−1, and ₁₀^Y is divisible by2^n+s−1, contradicting the minimality ofs. HenceY is coprime to5. Since2is a primitive root mod 5ⁿ, it follows that there exists at ≥ n + swith2^t≡Y (mod 5^n+s). We also have2^t≡ 0 ≡Y (mod 2^n+s), so by the Chinese remainder theorem, it follows that2^t≡Y (mod 10^n+s). In other words, the lastn + sdigits of2^tare the same as those ofY. In particular,2^tcontainsAas a sequence of digits in its lastdαnedigits.

Problem 2011-2/C Letpbe a prime number. Determine the smallest integerdfor which there is a monic polynomialfof degreedwith integer coefficients such thatp^p+1dividesf (n)for all integersn.

Solution We have received solutions from Pieter de Groen, Alex Heinis, Thijmen Krebs, Tejaswi Navilarekallu, Michiel Smid, Rob van der Waall and Martijn Weterings. The book token goes to Tejaswi Navilarekallu, whose solution was as follows.

We claim that the smallest degree isp².

For a polynomialf (X) ∈ Z[X]we denote byf⁽¹⁾the polynomialf (X + 1) − f (X)and byf^(m) the polynomial obtained by repeating this operationmtimes. Now assume thatf is monic of degreed, and thatp^p+1dividesf (n)for alln. Then alsop^p+1dividesf^(d)(n)for alln. But we havef^(d)(X) = d!, so we conclude thatp^p+1dividesd!and hence thatdis at leastp². To see that the minimal degree is exactlyp², it suffices to observe that the polynomial

f (X) = (X − 1)(X − 2) · · · (X − p²)

satisfies the required condition.