Päevaülesanne

Juhuslik

Harjutuskomplekt

Balti Tee 1997 · Ülesanne 18

Kombinatoorika

a) Prove the existence of two infinite sets AA and BB, not necessarily disjoint, of non-negative integers such that each non-negative integer nn is uniquely representable in the form n=a+bn=a+b with a∈A,b∈Ba \in A, b \in B.

b) Prove that for each such pair (A,B)(A, B), either AA or BB contains only multiples of some integer k>1k>1.

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Teemad

Värvimised ja konfiguratsioonid · Dirichlet’ printsiip ja ekstremaalargumendid

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Lahendus

Solution: a) Let AA be the set of non-negative integers whose only non-zero decimal digits are in even positions counted from the right, and BB the set of non-negative integers whose only non-zero decimal digits are in odd positions counted from the right. It is obvious that AA and BB have the required property.

b) Since the only possible representation of 00 is 0+00+0, we have 0∈A∩B0 \in A \cap B. The only possible representations of 11 are 1+01+0 and 0+10+1. Hence 11 must belong to at least one of the sets AA and BB. Let 1∈A1 \in A, and let kk be the smallest positive integer such that k∉Ak \notin A. Then k>1k>1. If any number bb with 0<b<k0<b<k belonged to BB, it would have the two representations b+0b+0 and 0+b0+b. Hence no such number belongs to BB. Also, in k=a+bk=a+b with a∈Aa \in A and b∈Bb \in B the number bb cannot be 00 since then a=ka=k, contradicting the assumption that k∉Ak \notin A. Hence b=kb=k, and k∈Bk \in B.

Consider the decomposition of AA into the union A1∪A2∪⋯A_{1} \cup A_{2} \cup \cdots of its maximal subsets A1,A2,…A_{1}, A_{2}, \ldots of consecutive numbers, where each element of A1A_{1} is less than each element of A2A_{2} etc. In particular, A1={0,1,…,k−1}A_{1}=\{0,1, \ldots, k-1\}. By our assumption the set of all non-negative integers is the union of non-intersecting sets An+b={a+b∣a∈An}A_{n}+b=\left\{a+b \mid a \in A_{n}\right\} with n∈Nn \in \mathbb{N} and b∈Bb \in B, each of these consisting of some number of consecutive integers. We will show that each subset AnA_{n} has exactly kk elements. Indeed, suppose mm is the smallest index for which the number ll of elements in AmA_{m} is different from kk, then l<kl<k since Am+0A_{m}+0 and Am+kA_{m}+k do not overlap. Denoting by cc the smallest element of AmA_{m}, we have c+k−1∉Ac+k-1 \notin A, so c+k−1=a+bc+k-1=a+b with a∈Aa \in A and 0≠b∈B0 \neq b \in B. Hence, b⩾kb \geqslant k and a<ca<c. Suppose a∈Ana \in A_{n}, then n<mn<m and the subset AnA_{n} has kk elements. But then An+bA_{n}+b overlaps with either Am+0A_{m}+0 or Am+kA_{m}+k, a contradiction.

Hence, the set of non-negative integers is the union of non-intersecting sets An+bA_{n}+b with n∈Nn \in \mathbb{N} and b∈Bb \in B, each of which consists of kk consecutive integers. The smallest element of each of these subsets is a multiple of kk. Since each integer b∈Bb \in B is the smallest element of A1+bA_{1}+b, it follows that each b∈Bb \in B is a multiple of kk.

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Balti Tee tulemused 1997

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