Päevaülesanne

Juhuslik

Harjutuskomplekt

Balti Tee 2017 · Ülesanne 9

Kombinatoorika

A positive integer nn is Danish if a regular hexagon can be partitioned into nn congruent polygons. Prove that there are infinitely many positive integers nn such that both nn and 2n+n2^{n}+n are Danish.

Muuda valikut

Kui oled valmis

Ülevaatematerjal muutub kättesaadavaks järgmise päevaülesannete komplektiga.

Ülevaade

Teemad

Dirichlet’ printsiip ja ekstremaalargumendid

Lahendused

Lahendus

At first we note that n=3kn=3 k is danish for any positive integer kk, because a hexagon can be cut in 3 equal parallelograms each of which can afterwards be cut in kk equal parallelograms (Fig 1).

Furthermore a hexagon can be cut into two equal trapezoids (Fig. 2) each of which can afterwards be cut into 4 equal trapezoids of the same shape (Fig. 3) and so on. Therefore any number of the form n=2⋅4kn=2 \cdot 4^{k} is also danish.

Official solution diagram for Baltic Way 2017 Problem 9 (Figure 1).

Figure 1

Official solution diagram for Baltic Way 2017 Problem 9 (Figure 2).

Figure 2

Official solution diagram for Baltic Way 2017 Problem 9 (Figure 3).

Figure 3

If we take any danish number n=2⋅4kn=2 \cdot 4^{k} of the second type, then

2n+n=22⋅4k+2⋅4k≡1+2≡0( mod 3)2^{n}+n=2^{2 \cdot 4^{k}}+2 \cdot 4^{k} \equiv 1+2 \equiv 0(\bmod 3)

showing that 2n+n2^{n}+n is also a danish number.

Võistluse kontekst

Balti Tee tulemused 2017

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