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Stage 9 · Combinatorics

10 problems · IMO P2/P5; hard shortlist · mathsolympiadprep.com

The answer key prints on its own page at the end.

  1. Let nn be a positive integer. A Nordic square is an n×nn \times n board containing all the integers from 11 to n2n^2 so that each cell contains exactly one number. Two different cells are considered adjacent if they share a common side. Every cell that is adjacent only to cells containing larger numbers is called a valley. An uphill path is a sequence of one or more cells such that:

    (i) the first cell in the sequence is a valley,

    (ii) each subsequent cell in the sequence is adjacent to the previous cell, and

    (iii) the numbers written in the cells in the sequence are in increasing order.

    Find, as a function of nn, the smallest possible total number of uphill paths in a Nordic square.

    Author: Nikola Petrovi?

    International Mathematical Olympiad

    Combinatorics Solution and answer checking →

  2. Let nn and TT be positive integers. James has 4n4 n marbles with weights 1,2,…,4n1,2, \ldots, 4 n. He places them on a balance scale, so that both sides have equal weight. Andrew may move a marble from one side of the scale to the other, so that the absolute difference in weights of the two sides remains at most TT.
    Find, in terms of nn, the minimum positive integer TT such that Andrew may make a sequence of moves such that each marble ends up on the opposite side of the scale, regardless of how James initially placed the marbles.

    IMO Shortlist

    Combinatorics Solution and answer checking →

  3. A cake has the form of an n×nn \times n square composed of n2n^2 unit squares. Strawberries lie on some of the unit squares so that each row or column contains exactly one strawberry; call this arrangement A\mathcal{A}.
    Let B\mathcal{B} be another such arrangement. Suppose that every grid rectangle with one vertex at the top left corner of the cake contains no fewer strawberries of arrangement B\mathcal{B} than of arrangement A\mathcal{A}. Prove that arrangement B\mathcal{B} can be obtained from A\mathcal{A} by performing a number of switches, defined as follows:
    A switch consists in selecting a grid rectangle with only two strawberries, situated at its top right corner and bottom left corner, and moving these two strawberries to the other two corners of that rectangle.

    IMO Shortlist 2006

    Combinatorics Solution and answer checking →

  4. On a square table of 20112011 by 20112011 cells we place a finite number of napkins that each cover a square of 5252 by 5252 cells. In each cell we write the number of napkins covering it, and we record the maximal number kk of cells that all contain the same nonzero number. Considering all possible napkin configurations, what is the largest value of kk?

    IMO Shortlist 2011

    Combinatorics Solution and answer checking →

  5. Let nn be a positive integer. Consider sequences a0,a1,…,ana_0, a_1, \dots, a_n such that ai∈{1,2,…,n}a_i \in \{1, 2, \dots, n\} for each ii and an=a0a_n = a_0.

    a. Call such a sequence good if for all i=1,2,…,ni = 1, 2, \dots, n, ai−ai−1≢i(modn)a_i - a_{i-1} \not\equiv i \pmod n. Suppose that nn is odd. Find the number of good sequences.

    b. Call such a sequence great if for all i=1,2,…,ni = 1, 2, \dots, n, ai−ai−1≢i,2i(modn)a_i - a_{i-1} \not\equiv i, 2i \pmod n. Suppose that nn is an odd prime. Find the number of great sequences.

    USA IMO

    Combinatorics Solution and answer checking →

  6. Integers are written in the cells of a table 2010×20102010 \times 2010. Adding 11 to all the numbers in a row or in a column is called a move. We say that the table is equilibrium if one can obtain after finitely many moves a table in which all the numbers are equal.

    a) Find the largest positive integer nn, for which there exists an equilibrium table containing the numbers 20,21,…,2n2^0, 2^1, \dots, 2^n.

    b) For this nn, find the maximal number that may be contained in such a table.

    BMO Shortlist 2010

    Combinatorics Solution and answer checking →

  7. The kingdom of Anisotropy consists of nn cities. For every two cities there exists exactly one direct one-way road between them. We say that a path from XX to YY is a sequence of roads such that one can move from XX to YY along this sequence without returning to an already visited city. A collection of paths is called diverse if no road belongs to two or more paths in the collection.
    Let AA and BB be two distinct cities in Anisotropy. Let NABN_{A B} denote the maximal number of paths in a diverse collection of paths from AA to BB. Similarly, let NBAN_{B A} denote the maximal number of paths in a diverse collection of paths from BB to AA. Prove that the equality NAB=NBAN_{A B}=N_{B A} holds if and only if the number of roads going out from AA is the same as the number of roads going out from BB.

    IMO Shortlist 2021

    Combinatorics Solution and answer checking →

  8. Let n1,n2,…,n26n_1, n_2, \dots, n_{26} be pairwise distinct positive integers, satisfying:

    (1) In the decimal representation of each nin_i, each digit belongs to the set {1,2}\{1, 2\};

    (2) For any i,ji, j, njn_j cannot be obtained from nin_i by adding some digits on the right.

    Find the least possible value of ∑i=126S(ni)\sum_{i=1}^{26} S(n_i), where S(m)S(m) denotes the sum of all digits of mm in decimal representation.

    China National Team Selection Test

    Combinatorics Solution and answer checking →

  9. The columns and the rows of a 3n×3n3n \times 3n square board are numbered 1,2,…,3n1, 2, \ldots, 3n. Every square (x,y)(x, y) with 1≤x,y≤3n1 \leq x, y \leq 3n is colored asparagus, byzantium or citrine according as the modulo 33 remainder of x+yx+y is 00, 11 or 22 respectively. One token colored asparagus, byzantium or citrine is placed on each square, so that there are 3n23n^{2} tokens of each color.

    Suppose that one can permute the tokens so that each token is moved to a distance of at most dd from its original position, each asparagus token replaces a byzantium token, each byzantium token replaces a citrine token, and each citrine token replaces an asparagus token. Prove that it is possible to permute the tokens so that each token is moved to a distance of at most d+2d+2 from its original position, and each square contains a token with the same color as the square.

    IMO Shortlist

    Combinatorics Solution and answer checking →

  10. Let A0=(a1,…,an)A_{0} = (a_{1}, \ldots, a_{n}) be a finite sequence of real numbers. For each k≥0k \geq 0, from the sequence Ak=(x1,…,xn)A_{k} = (x_{1}, \ldots, x_{n}) we construct a new sequence Ak+1A_{k+1} in the following way.
    1. We choose a partition {1,…,n}=I∪J\{1, \ldots, n\} = I \cup J, where II and JJ are two disjoint sets, such that the expression
    ∣∑i∈Ixi−∑j∈Jxj∣ \left|\sum_{i \in I} x_{i} - \sum_{j \in J} x_{j}\right|
    attains the smallest possible value. (We allow the sets II or JJ to be empty; in this case the corresponding sum is 00.) If there are several such partitions, one is chosen arbitrarily.
    2. We set Ak+1=(y1,…,yn)A_{k+1} = (y_{1}, \ldots, y_{n}), where yi=xi+1y_{i} = x_{i} + 1 if i∈Ii \in I, and yi=xi−1y_{i} = x_{i} - 1 if i∈Ji \in J.
    Prove that for some kk, the sequence AkA_{k} contains an element xx such that ∣x∣≥n/2|x| \geq n / 2.

    IMO Shortlist 2007

    Combinatorics Solution and answer checking →

Answer key — Stage 9 · Combinatorics

Worked solutions for every problem are on the site, one page per problem.

  1. 2n(n−1)+12n(n - 1) + 1 open
  2. Prove it — see the worked solution open
  3. Prove it — see the worked solution open
  4. Prove it — see the worked solution open
  5. Prove it — see the worked solution open
  6. Prove it — see the worked solution open
  7. Prove it — see the worked solution open
  8. Prove it — see the worked solution open
  9. Prove it — see the worked solution open
  10. Prove it — see the worked solution open

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