id stringlengths 10 15 | question stringlengths 63 2.3k | solutions stringlengths 20 28.5k |
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IMOSL-2020-G8 | Let \(\Gamma\) and \(I\) be the circumcircle and the incenter of an acute- angled triangle \(ABC\) . Two circles \(\omega_{B}\) and \(\omega_{C}\) passing through \(B\) and \(C\) , respectively, are tangent at \(I\) . Let \(\omega_{B}\) meet the shorter arc \(AB\) of \(\Gamma\) and segment \(AB\) again at \(P\) and \(M... | Solution 1. Let \(AI\) , \(BI\) , and \(CI\) meet \(\Gamma\) again at \(D\) , \(E\) , and \(F\) , respectively. Let \(\ell\) be the common tangent to \(\omega_{B}\) and \(\omega_{C}\) at \(I\) . We always denote by \(\hat{x} (p, q)\) the directed angle from a line \(p\) to a line \(q\) , taken modulo \(180^{\circ}\) . ... |
IMOSL-2020-G9 | Prove that there exists a positive constant \(c\) such that the following statement is true:
Assume that \(n\) is an integer with \(n \geqslant 2\) , and let \(\mathcal{S}\) be a set of \(n\) points in the plane such that the distance between any two distinct points in \(\mathcal{S}\) is at least 1. Then there is a ... | Solution. We prove that the desired statement is true with \(c = \frac{1}{8}\) . Set \(\delta = \frac{1}{8} n^{- 1 / 3}\) . For any line \(\ell\) and any point \(X\) , let \(X_{\ell}\) denote the projection of \(X\) to \(\ell\) ; a similar notation applies to sets of points.
Suppose that, for some line \(\ell\) , th... |
IMOSL-2020-N1 | Given a positive integer \(k\) , show that there exists a prime \(p\) such that one can choose distinct integers \(a_{1},a_{2},\ldots ,a_{k + 3}\in \{1,2,\ldots ,p - 1\}\) such that \(p\) divides \(a_{i}a_{i + 1}a_{i + 2}a_{i + 3} - i\) for all \(i = 1,2,\ldots ,k\) . | Solution. First we choose distinct positive rational numbers \(r_{1},\ldots ,r_{k + 3}\) such that
\[r_{i}r_{i + 1}r_{i + 2}r_{i + 3} = i\quad \mathrm{for}1\leqslant i\leqslant k.\]
Let \(r_{1} = x\) \(r_{2} = y\) \(r_{3} = z\) be some distinct primes greater than \(k\) ; the remaining terms satisfy \(r_{4} = \fr... |
IMOSL-2020-N2 | For each prime \(p\) , there is a kingdom of \(p\) - Landia consisting of \(p\) islands numbered 1, 2, ..., \(p\) . Two distinct islands numbered \(n\) and \(m\) are connected by a bridge if and only if \(p\) divides \((n^{2} - m + 1)(m^{2} - n + 1)\) . The bridges may pass over each other, but cannot cross. Prove that... | Solution 1. We prove that for each prime \(p > 3\) dividing a number of the form \(x^{2} - x + 1\) with integer \(x\) there are two unconnected islands in \(p\) - Landia.
For brevity's sake, when a bridge connects the islands numbered \(m\) and \(n\) , we shall speak simply that it connects \(m\) and \(n\) .
A br... |
IMOSL-2020-N3 | Let \(n\) be an integer with \(n \geqslant 2\) . Does there exist a sequence \((a_{1}, \ldots , a_{n})\) of positive integers with not all terms being equal such that the arithmetic mean of every two terms is equal to the geometric mean of some (one or more) terms in this sequence? | Answer: No such sequence exists.
Solution 1. Suppose that \(a_{1},\ldots ,a_{n}\) satisfy the required properties. Let \(d = \gcd (a_{1}\ldots ,a_{n})\) If \(d > 1\) then replace the numbers \(a_{1},\ldots ,a_{n}\) by \(\frac{a_{1}}{d},\ldots ,\frac{a_{n}}{d}\) ; all arithmetic and all geometric means will be divide... |
IMOSL-2020-N4 | For any odd prime \(p\) and any integer \(n\) , let \(d_{p}(n) \in \{0,1,\ldots ,p - 1\}\) denote the remainder when \(n\) is divided by \(p\) . We say that \((a_{0},a_{1},a_{2},\ldots)\) is a \(p\) - sequence, if \(a_{0}\) is a positive integer coprime to \(p\) , and \(a_{n + 1} = a_{n} + d_{p}(a_{n})\) for \(n\geqsla... | Answer: Yes, for both parts.
Solution. Fix some odd prime \(p\) , and let \(T\) be the smallest positive integer such that \(p\mid 2^{T} - 1\) in other words, \(T\) is the multiplicative order of 2 modulo \(p\) .
Consider any \(p\) - sequence \((x_{n}) = (x_{0},x_{1},x_{2},\ldots)\) . Obviously, \(x_{n + 1}\equiv... |
IMOSL-2020-N5 | Determine all functions \(f\) defined on the set of all positive integers and taking non-negative integer values, satisfying the three conditions:
\((i)\) \(f(n)\neq 0\) for at least one \(n\) .
\((ii)\) \(f(xy) = f(x) + f(y)\) for every positive integers \(x\) and \(y\) ;
\((iii)\) there are infinitely many p... | Answer: The sought functions are those of the form \(f(n) = c\cdot \nu_{p}(n)\) , where \(p\) is some prime, \(c\) is a nonnegative integer, and \(\nu_{p}(n)\) denotes the exponent of \(p\) in the prime decomposition of \(n\) .
Solution 1. If a number \(n\) is a product of primes, \(n = p_{1}p_{2}\cdot \cdot \cdot \... |
IMOSL-2020-N6 | For a positive integer \(n\) , let \(d(n)\) be the number of positive divisors of \(n\) , and let \(\phi (n)\) be the number of positive integers not exceeding \(n\) which are coprime to \(n\) . Does there exist a constant \(C\) such that
\[\frac{\phi(d(n))}{d(\phi(n))}\leqslant C\]
for all \(n\geqslant 1\) ? | Answer: No, such constant does not exist.
Solution 1. Fix \(N > 1\) , let \(p_{1},\ldots ,p_{k}\) be all primes between 1 and \(N\) and \(p_{k + 1},\ldots ,p_{k + s}\) be all primes between \(N + 1\) and \(2N\) . Since for \(j\leqslant k + s\) all prime divisors of \(p_{j} - 1\) do not exceed \(N\) , we have
\[\p... |
IMOSL-2020-N7 | Let \(\mathcal{S}\) be a set consisting of \(n \geqslant 3\) positive integers, none of which is a sum of two other distinct members of \(\mathcal{S}\) . Prove that the elements of \(\mathcal{S}\) may be ordered as \(a_{1}, a_{2}, \ldots , a_{n}\) so that \(a_{i}\) does not divide \(a_{i - 1} + a_{i + 1}\) for all \(i ... | Common remarks. In all solutions, we call a set \(\mathcal{S}\) of positive integers good if no its element is a sum of two other distinct members of \(\mathcal{S}\) . We will use the following simple observation.
Observation A. If \(a\) , \(b\) , and \(c\) are three distinct elements of a good set \(\mathcal{S}\) w... |
IMOSL-2021-A1 | Let \(n\) be an integer, and let \(A\) be a subset of \(\{0,1,2,3,\ldots ,5^{n}\}\) consisting of \(4n + 2\) numbers. Prove that there exist \(a,b,c\in A\) such that \(a< b< c\) and \(c + 2a > 3b\) . | Solution 1. (By contradiction) Suppose that there exist \(4n + 2\) non- negative integers \(x_{0}<\) \(x_{1}< \dots < x_{4n + 1}\) that violate the problem statement. Then in particular \(x_{4n + 1} + 2x_{i}\leqslant 3x_{i + 1}\) for all \(i = 0,\ldots ,4n - 1\) , which gives
\[x_{4n + 1} - x_{i}\geqslant \frac{3}{2... |
IMOSL-2021-A2 | For every integer \(n \geq 1\) consider the \(n \times n\) table with entry \(\left\lfloor \frac{ij}{n + 1} \right\rfloor\) at the intersection of row \(i\) and column \(j\) , for every \(i = 1, \ldots , n\) and \(j = 1, \ldots , n\) . Determine all integers \(n \geq 1\) for which the sum of the \(n^2\) entries in the ... | Answer: All integers \(n\) for which \(n + 1\) is a prime.
Solution 1. First, observe that every pair \(x, y\) of real numbers for which the sum \(x + y\) is integer satisfies
\[\left|x\right| + \left|y\right| \geq x + y - 1. \quad (1)\]
The inequality is strict if \(x\) and \(y\) are integers, and it holds wi... |
IMOSL-2021-A4 | Show that for all real numbers \(x_{1},\ldots ,x_{n}\) the following inequality holds:
\[\sum_{i = 1}^{n}\sum_{j = 1}^{n}\sqrt{|x_{i} - x_{j}|}\leqslant \sum_{i = 1}^{n}\sum_{j = 1}^{n}\sqrt{|x_{i} + x_{j}|}.\] | Solution 1. If we add \(t\) to all the variables then the left- hand side remains constant and the right- hand side becomes
\[H(t):= \sum_{i = 1}^{n}\sum_{j = 1}^{n}\sqrt{|x_{i} + x_{j} + 2t|}.\]
Let \(T\) be large enough such that both \(H(- T)\) and \(H(T)\) are larger than the value \(L\) of the left- hand sid... |
IMOSL-2021-A5 | Let \(n \geq 2\) be an integer, and let \(a_1, a_2, \ldots , a_n\) be positive real numbers such that \(a_1 + a_2 + \cdots + a_n = 1\) . Prove that
\[\sum_{k = 1}^{n} \frac{a_k}{1 - a_k} (a_1 + a_2 + \cdots + a_{k - 1})^2 < \frac{1}{3}.\] | Solution 1. For all \(k \leq n\) , let
\[s_k = a_1 + a_2 + \dots +a_k\qquad \mathrm{and}\qquad b_k = \frac{a_ks_{k - 1}^2}{1 - a_k},\]
with the convention that \(s_0 = 0\) . Note that \(b_k\) is exactly a summand in the sum we need to estimate. We shall prove the inequality
\[b_k < \frac{s_k^3 - s_{k - 1}^3}{3... |
IMOSL-2021-A6 | Let \(A\) be a finite set of (not necessarily positive) integers, and let \(m \geq 2\) be an integer. Assume that there exist non- empty subsets \(B_{1}, B_{2}, B_{3}, \ldots , B_{m}\) of \(A\) whose elements add up to the sums \(m^{1}, m^{2}, m^{3}, \ldots , m^{m}\) , respectively. Prove that \(A\) contains at least \... | Solution. Let \(A = \{a_{1}, \ldots , a_{k}\}\) . Assume that, on the contrary, \(k = |A| < m / 2\) . Let
\[s_{i} := \sum_{j: a_{j} \in B_{i}} a_{j}\]
be the sum of elements of \(B_{i}\) . We are given that \(s_{i} = m^{i}\) for \(i = 1, \ldots , m\) .
Now consider all \(m^{m}\) expressions of the form
\[f(... |
IMOSL-2021-A7 | Let \(n \geq 1\) be an integer, and let \(x_0, x_1, \ldots , x_{n+1}\) be \(n + 2\) non- negative real numbers that satisfy \(x_i x_{i+1} - x_{i-1}^2 \geq 1\) for all \(i = 1, 2, \ldots , n\) . Show that
\[x_0 + x_1 + \dots +x_n + x_{n + 1} > \left(\frac{2n}{3}\right)^{3 / 2}.\] | Solution 1.
Lemma 1.1. If \(a, b, c\) are non- negative numbers such that \(ab - c^2 \geq 1\) , then
\[(a + 2b)^2 \geq (b + 2c)^2 + 6.\]
Proof. \((a + 2b)^2 - (b + 2c)^2 = (a - b)^2 + 2(b - c)^2 + 6(ab - c^2) \geq 6\) .
Lemma 1.2. \(\sqrt{1} + \dots + \sqrt{n} > \frac{2}{3} n^{3 / 2}\) .
Proof. Bernoulli... |
IMOSL-2021-A8 | Determine all functions \(f:\mathbb{R}\to \mathbb{R}\) that satisfy
\[(f(a) - f(b))\left(f(b) - f(c)\right)\left(f(c) - f(a)\right) = f(ab^{2} + bc^{2} + ca^{2}) - f(a^{2}b + b^{2}c + c^{2}a)\]
for all real numbers \ | Answer: \(f(x) = \alpha x + \beta\) or \(f(x) = \alpha x^{3} + \beta\) where \(\alpha \in \{- 1,0,1\}\) and \(\beta \in \mathbb{R}\)
Solution. It is straightforward to check that above functions satisfy the equation. Now let \(f(x)\) satisfy the equation, which we denote \(E(a,b,c)\) . Then clearly \(f(x) + C\) also... |
IMOSL-2021-C1 | Let \(S\) be an infinite set of positive integers, such that there exist four pairwise distinct \(a,b,c,d\in S\) with \(\gcd (a,b)\neq \gcd (c,d)\) . Prove that there exist three pairwise distinct \(x,y,z\in S\) such that \(\gcd (x,y) = \gcd (y,z)\neq \gcd (z,x)\) . | Solution. There exists \(\alpha \in S\) so that \(\{\gcd (\alpha ,s)\mid s\in S\) \(s\neq \alpha \}\) contains at least two elements. Since \(\alpha\) has only finitely many divisors, there is a \(d\mid \alpha\) such that the set \(B = \{\beta \in\) \(S\mid \gcd (\alpha ,\beta) = d\}\) is infinite. Pick \(\gamma \in S\... |
IMOSL-2021-C2 | Let \(n \geq 3\) be an integer. An integer \(m \geq n + 1\) is called \(n\) - colourful if, given infinitely many marbles in each of \(n\) colours \(C_1, C_2, \ldots , C_n\) , it is possible to place \(m\) of them around a circle so that in any group of \(n + 1\) consecutive marbles there is at least one marble of colo... | Answer: \(m_{max} = n^2 - n - 1\) .
Solution. First suppose that there are \(n(n - 1) - 1\) marbles. Then for one of the colours, say blue, there are at most \(n - 2\) marbles, which partition the non- blue marbles into at most \(n - 2\) groups with at least \((n - 1)^2 > n(n - 2)\) marbles in total. Thus one of the... |
IMOSL-2021-C3 | A thimblergger has 2021 thimbles numbered from 1 through 2021. The thimbles are arranged in a circle in arbitrary order. The thimblergger performs a sequence of 2021 moves; in the \(k^{\mathrm{th}}\) move, he swaps the positions of the two thimbles adjacent to thimble \(k\) .
Prove that there exists a value of \(k\)... | Solution. Assume the contrary. Say that the \(k^{\mathrm{th}}\) thimble is the central thimble of the \(k^{\mathrm{th}}\) move, and its position on that move is the central position of the move.
Step 1: Black and white colouring.
Before the moves start, let us paint all thimbles in white. Then, after each move, w... |
IMOSL-2021-C4 | The kingdom of Anisotropy consists of \(n\) cities. For every two cities there exists exactly one direct one- way road between them. We say that a path from \(X\) to \(Y\) is a sequence of roads such that one can move from \(X\) to \(Y\) along this sequence without returning to an already visited city. A collection of ... | Solution 1. We write \(X \to Y\) (or \(Y \leftarrow X\) ) if the road between \(X\) and \(Y\) goes from \(X\) to \(Y\) . Notice that, if there is any route moving from \(X\) to \(Y\) (possibly passing through some cities more than once), then there is a path from \(X\) to \(Y\) consisting of some roads in the route. In... |
IMOSL-2021-C5 | Let \(n\) and \(k\) be two integers with \(n > k \geq 1\) . There are \(2n + 1\) students standing in a circle. Each student \(S\) has \(2k\) neighbours—namely, the \(k\) students closest to \(S\) on the right, and the \(k\) students closest to \(S\) on the left.
Suppose that \(n + 1\) of the students are girls, and... | Solution. We replace the girls by 1's, and the boys by 0's, getting the numbers \(a_{1}, a_{2}, \ldots , a_{2n + 1}\) arranged in a circle. We extend this sequence periodically by letting \(a_{2n + 1 + k} = a_{k}\) for all \(k \in \mathbb{Z}\) . We get an infinite periodic sequence
\[\ldots ,a_{1},a_{2},\ldots ,a_{2... |
IMOSL-2021-C6 | A hunter and an invisible rabbit play a game on an infinite square grid. First the hunter fixes a colouring of the cells with finitely many colours. The rabbit then secretly chooses a cell to start in. Every minute, the rabbit reports the colour of its current cell to the hunter, and then secretly moves to an adjacent ... | Answer: Yes, there exists a colouring that yields a winning strategy for the hunter.
Solution. A central idea is that several colourings \(C_{1}, C_{2}, \ldots , C_{k}\) can be merged together into a single product colouring \(C_{1} \times C_{2} \times \dots \times C_{k}\) as follows: the colours in the product colo... |
IMOSL-2021-C7 | Consider a checkered \(3m \times 3m\) square, where \(m\) is an integer greater than 1. A frog sits on the lower left corner cell \(S\) and wants to get to the upper right corner cell \(F\) . The frog can hop from any cell to either the next cell to the right or the next cell upwards.
Some cells can be sticky, and t... | Solution for part (a). In the following example the square is divided into \(m\) stripes of size \(3 \times 3m\) . It is easy to see that \(X\) is a minimal blocking set. The first and the last stripe each contains \(3m - 1\) cells from the set \(X\) ; every other stripe contains \(3m - 2\) cells, see Figure 1. The tot... |
IMOSL-2021-G1 | Let \(ABCD\) be a parallelogram such that \(AC = BC\) . A point \(P\) is chosen on the extension of the segment \(AB\) beyond \(B\) . The circumcircle of the triangle \(ACD\) meets the segment \(PD\) again at \(Q\) , and the circumcircle of the triangle \(APQ\) meets the segment \(PC\) again at \(R\) . Prove that the l... | Common remarks. The introductory steps presented here are used in all solutions below.
Since \(AC = BC = AD\) , we have \(\angle ABC = \angle BAC = \angle ACD = \angle ADC\) . Since the quadrilaterals \(APRQ\) and \(AQCD\) are cyclic, we obtain
\[\angle CRA = 180^{\circ} - \angle ARP = 180^{\circ} - \angle AQP = ... |
IMOSL-2021-G2 | Let \(ABCD\) be a convex quadrilateral circumscribed around a circle with centre \(I\) . Let \(\omega\) be the circumcircle of the triangle \(ACI\) . The extensions of \(BA\) and \(BC\) beyond \(A\) and \(C\) meet \(\omega\) at \(X\) and \(Z\) , respectively. The extensions of \(AD\) and \(CD\) beyond \(D\) meet \(\ome... | Solution. The point \(I\) is the intersection of the external bisector of the angle \(TCZ\) with the circumcircle \(\omega\) of the triangle \(TCZ\) , so \(I\) is the midpoint of the arc \(TCZ\) and \(IT = IZ\) . Similarly, \(I\) is the midpoint of the arc \(YAX\) and \(IX = IY\) . Let \(O\) be the centre of \(\omega\)... |
IMOSL-2021-G3 | Let \(n\) be a fixed positive integer, and let S be the set of points \((x,y)\) on the Cartesian plane such that both coordinates \(x\) and \(y\) are nonnegative integers smaller than \(2n\) (thus \(|S| = 4n^{2}\) ). Assume that \(\mathcal{F}\) is a set consisting of \(n^{2}\) quadrilaterals such that all their vertice... | Answer: The largest possible sum of areas is \(\Sigma (n):= \frac{1}{3} n^{2}(2n + 1)(2n - 1)\) .
Common remarks. Throughout all solutions, the area of a polygon \(P\) will be denoted by \([P]\) . We say that a polygon is legal if all its vertices belong to S. Let \(O = \left(n - \frac{1}{2},n - \frac{1}{2}\right)\)... |
IMOSL-2021-G3.5 | Let \(n\) be a fixed positive integer, and let S be the set of points \((x,y)\) on the Cartesian plane such that both coordinates \(x\) and \(y\) are nonnegative integers smaller than \(2n\) (thus \(|S| = 4n^{2}\) ). Assume that \(\mathcal{F}\) is a set of polygons such that all vertices of polygons in \(\mathcal{F}\) ... | Answer: The largest possible sum of areas is \(\Sigma (n):= \frac{1}{3} n^{2}(2n + 1)(2n - 1)\) .
Common remarks. Throughout all solutions, the area of a polygon \(P\) will be denoted by \([P]\) . We say that a polygon is legal if all its vertices belong to S. Let \(O = \left(n - \frac{1}{2},n - \frac{1}{2}\right)\)... |
IMOSL-2021-G4 | Let \(A B C D\) be a quadrilateral inscribed in a circle \(\Omega\) . Let the tangent to \(\Omega\) at \(D\) intersect the rays \(B A\) and \(B C\) at points \(E\) and \(F\) , respectively. A point \(T\) is chosen inside the triangle \(A B C\) so that \(T E\parallel C D\) and \(T F\parallel A D\) . Let \(K\neq D\) be a... | Solution 1. Let the segments \(T E\) and \(T F\) cross \(A C\) at \(P\) and \(Q\) , respectively. Since \(P E\parallel C D\) and \(E D\) is tangent to the circumcircle of \(A B C D\) , we have
\[\angle E P A = \angle D C A = \angle E D A,\]
and so the points \(A\) , \(P\) , \(D\) , and \(E\) lie on some circle \(... |
IMOSL-2021-G5 | Let \(A B C D\) be a cyclic quadrilateral whose sides have pairwise different lengths. Let \(O\) be the circumcentre of \(A B C D\) . The internal angle bisectors of \(\angle A B C\) and \(\angle A D C\) meet \(A C\) at \(B_{1}\) and \(D_{1}\) , respectively. Let \(O_{B}\) be the centre of the circle which passes throu... | Common remarks. We introduce some objects and establish some preliminary facts common for all solutions below.
Let \(\Omega\) denote the circle \((A B C D)\) , and let \(\gamma_{B}\) and \(\gamma_{D}\) denote the two circles from the problem statement (their centres are \(O_{B}\) and \(O_{D}\) , respectively). Clear... |
IMOSL-2021-G6 | Determine all integers \(n \geq 3\) satisfying the following property: every convex \(n\) - gon whose sides all have length 1 contains an equilateral triangle of side length 1. | Answer: All odd \(n \geq 3\) .
Solution. First we show that for every even \(n \geq 4\) there exists a polygon violating the required statement. Consider a regular \(k\) - gon \(A_{0}A_{1},\ldots A_{k - 1}\) with side length 1. Let \(B_{1},B_{2},\ldots ,B_{n / 2 - 1}\) be the points symmetric to \(A_{1},A_{2},\ldots... |
IMOSL-2021-G7 | A point \(D\) is chosen inside an acute- angled triangle \(ABC\) with \(AB > AC\) so that \(\angle BAD = \angle DAC\) . A point \(E\) is constructed on the segment \(AC\) so that \(\angle ADE = \angle DCB\) . Similarly, a point \(F\) is constructed on the segment \(AB\) so that \(\angle ADF = \angle DBC\) . A point \(X... | Common remarks. Let \(Q\) be the isogonal conjugate of \(D\) with respect to the triangle \(ABC\) . Since \(\angle BAD = \angle DAC\) , the point \(Q\) lies on \(AD\) . Then \(\angle QBA = \angle DBC = \angle FDA\) , so the points \(Q\) , \(D\) , \(F\) , and \(B\) are concyclic. Analogously, the points \(Q\) , \(D\) , ... |
IMOSL-2021-G8 | Let \(\omega\) be the circumcircle of a triangle \(ABC\) , and let \(\Omega_{A}\) be its excircle which is tangent to the segment \(BC\) . Let \(X\) and \(Y\) be the intersection points of \(\omega\) and \(\Omega_{A}\) . Let \(P\) and \(Q\) be the projections of \(A\) onto the tangent lines to \(\Omega_{A}\) at \(X\) a... | Solution 1. Let \(D\) be the point of tangency of \(BC\) and \(\Omega_{A}\) . Let \(D^{\prime}\) be the point such that \(DD^{\prime}\) is a diameter of \(\Omega_{A}\) . Let \(R^{\prime}\) be (the unique) point such that \(AR^{\prime} \perp BC\) and \(R^{\prime}D^{\prime} \parallel BC\) . We shall prove that \(R^{\prim... |
IMOSL-2021-N2 | Let \(n \geqslant 100\) be an integer. The numbers \(n, n + 1, \ldots , 2n\) are written on \(n + 1\) cards, one number per card. The cards are shuffled and divided into two piles. Prove that one of the piles contains two cards such that the sum of their numbers is a perfect square. | Solution. To solve the problem it suffices to find three squares and three cards with numbers \(a, b, c\) on them such that pairwise sums \(a + b, b + c, a + c\) are equal to the chosen squares. By choosing the three consecutive squares \((2k - 1)^{2}, (2k)^{2}, (2k + 1)^{2}\) we arrive at the triple
\[(a,b,c) = \le... |
IMOSL-2021-N3 | Find all positive integers \(n\) with the following property: the \(k\) positive divisors of \(n\) have a permutation \((d_{1},d_{2},\ldots ,d_{k})\) such that for every \(i = 1,2,\ldots ,k\) , the number \(d_{1} + \dots +d_{i}\) is a perfect square. | Answer: \(n = 1\) and \(n = 3\)
Solution. For \(i = 1,2,\ldots ,k\) let \(d_{1} + \ldots +d_{i} = s_{i}^{2}\) , and define \(s_{0} = 0\) as well. Obviously \(0 = s_{0}< s_{1}< s_{2}< \ldots < s_{k}\) , so
\[s_{i}\geqslant i\quad \mathrm{and}\quad d_{i} = s_{i}^{2} - s_{i - 1}^{2} = (s_{i} + s_{i - 1})(s_{i} - s_{... |
IMOSL-2021-N4 | Alice is given a rational number \(r > 1\) and a line with two points \(B \neq R\) , where point \(R\) contains a red bead and point \(B\) contains a blue bead. Alice plays a solitaire game by performing a sequence of moves. In every move, she chooses a (not necessarily positive) integer \(k\) , and a bead to move. If ... | Answer: All \(r = (b + 1) / b\) with \(b = 1, \ldots , 1010\) .
Solution. Denote the red and blue beads by \(\mathcal{R}\) and \(B\) , respectively. Introduce coordinates on the line and identify the points with their coordinates so that \(R = 0\) and \(B = 1\) . Then, during the game, the coordinate of \(\mathcal{R... |
IMOSL-2021-N5 | Prove that there are only finitely many quadruples \((a,b,c,n)\) of positive integers such that
\[n! = a^{n - 1} + b^{n - 1} + c^{n - 1}.\] | Solution. For fixed \(n\) there are clearly finitely many solutions; we will show that there is no solution with \(n > 100\) . So, assume \(n > 100\) . By the AM- GM inequality,
\[n! = 2n(n - 1)(n - 2)(n - 3)\cdot (3\cdot 4\dots (n - 4))\] \[\qquad \leqslant 2(n - 1)^{4}\left(\frac{3 + \dots +(n - 4)}{n - 6}\right)^... |
IMOSL-2021-N6 | Determine all integers \(n \geq 2\) with the following property: every \(n\) pairwise distinct integers whose sum is not divisible by \(n\) can be arranged in some order \(a_{1}, a_{2}, \ldots , a_{n}\) so that \(n\) divides \(1 \cdot a_{1} + 2 \cdot a_{2} + \dots + n \cdot a_{n}\) . | Answer: All odd integers and all powers of 2.
Solution. If \(n = 2^{k}a\) , where \(a \geq 3\) is odd and \(k\) is a positive integer, we can consider a set containing the number \(2^{k} + 1\) and \(n - 1\) numbers congruent to 1 modulo \(n\) . The sum of these numbers is congruent to \(2^{k}\) modulo \(n\) and ther... |
IMOSL-2021-N7 | Let \(a_{1},a_{2},a_{3},\ldots\) be an infinite sequence of positive integers such that \(a_{n + 2m}\) divides \(a_{n} + a_{n + m}\) for all positive integers \(n\) and \(m\) . Prove that this sequence is eventually periodic, i.e. there exist positive integers \(N\) and \(d\) such that \(a_{n} = a_{n + d}\) for all \(n... | Solution. We will make repeated use of the following simple observation:
Lemma 1. If a positive integer \(d\) divides \(a_{n}\) and \(a_{n - m}\) for some \(m\) and \(n > 2m\) , it also divides \(a_{n - 2m}\) . If \(d\) divides \(a_{n}\) and \(a_{n - 2m}\) , it also divides \(a_{n - m}\) .
Proof. Both parts are o... |
IMOSL-2021-N8 | For a polynomial \(P(x)\) with integer coefficients let \(P^{1}(x) = P(x)\) and \(P^{k + 1}(x) = P(P^{k}(x))\) for \(k \geq 1\) . Find all positive integers \(n\) for which there exists a polynomial \(P(x)\) with integer coefficients such that for every integer \(m \geq 1\) , the numbers \(P^{m}(1), \ldots , P^{m}(n)\)... | Answer: All powers of 2 and all primes.
Solution. Denote the set of residues modulo \(\ell\) by \(\mathbb{Z}_{\ell}\) . Observe that \(P\) can be regarded as a function \(\mathbb{Z}_{\ell} \to \mathbb{Z}_{\ell}\) for any positive integer \(\ell\) . Denote the cardinality of the set \(P^{m}(\mathbb{Z}_{\ell})\) by \(... |
IMOSL-2022-A1 | Let \((a_{n})_{n\geqslant 1}\) be a sequence of positive real numbers with the property that
\[(a_{n + 1})^{2} + a_{n}a_{n + 2}\leqslant a_{n} + a_{n + 2}\]
for all positive integers \(n\) . Show that \(a_{2022}\leqslant 1\) | Solution. We begin by observing that \((a_{n + 1})^{2} - 1\leqslant a_{n} + a_{n + 2} - a_{n}a_{n + 2} - 1\) , which is equivalent to
\[(a_{n + 1})^{2} - 1\leqslant (1 - a_{n})(a_{n + 2} - 1).\]
Suppose now that there exists a positive integer \(n\) such that \(a_{n + 1} > 1\) and \(a_{n + 2} > 1\) . Since \((a_{... |
IMOSL-2022-A3 | Let \(\mathbb{R}_{>0}\) be the set of positive real numbers. Find all functions \(f\colon \mathbb{R}_{>0}\to \mathbb{R}_{>0}\) such that, for every \(x\in \mathbb{R}_{>0}\) , there exists a unique \(y\in \mathbb{R}_{>0}\) satisfying
\[x f(y) + y f(x)\leqslant 2.\] | Answer: The function \(f(x) = 1 / x\) is the only solution.
Solution 1. First we prove that the function \(f(x) = 1 / x\) satisfies the condition of the problem statement. The AM- GM inequality gives
\[\frac{x}{y} +\frac{y}{x}\geqslant 2\]
for every \(x,y > 0\) , with equality if and only if \(x = y\) . This m... |
IMOSL-2022-A4 | Let \(n \geqslant 3\) be an integer, and let \(x_{1}, x_{2}, \ldots , x_{n}\) be real numbers in the interval \([0, 1]\) . Let \(s = x_{1} + x_{2} + \ldots + x_{n}\) , and assume that \(s \geqslant 3\) . Prove that there exist integers \(i\) and \(j\) with \(1 \leqslant i < j \leqslant n\) such that
\[2^{j - i}x_{i}... | Solution.
Let \(1 \leqslant a < b \leqslant n\) be such that \(2^{b - a}x_{a}x_{b}\) is maximal. This choice of \(a\) and \(b\) implies that \(x_{a + t} \leqslant 2^{t}x_{a}\) for all \(1 - a \leqslant t \leqslant b - a - 1\) , and similarly \(x_{b - t} \leqslant 2^{t}x_{b}\) for all \(b - n \leqslant t \leqslant b ... |
IMOSL-2022-A6 | Let \(\mathbb{R}\) be the set of real numbers. We denote by \(\mathcal{F}\) the set of all functions \(f\colon \mathbb{R}\to \mathbb{R}\) such that
\[f(x + f(y)) = f(x) + f(y)\]
for every \(x,y\in \mathbb{R}\) . Find all rational numbers \(q\) such that for every function \(f\in \mathcal{F}\) , there exists some ... | Answer: The desired set of rational numbers is \(\left\{{\frac{n + 1}{n}}:n\in \mathbb{Z},n\neq 0\right\}\) .
Solution. Let \(Z\) be the set of all rational numbers \(q\) such that for every function \(f\in \mathcal{F}\) , there exists some \(z\in \mathbb{R}\) satisfying \(f(z) = qz\) . Let further
\[S = \left\{\... |
IMOSL-2022-A7 | For a positive integer \(n\) we denote by \(s(n)\) the sum of the digits of \(n\) . Let \(P(x) = x^{n} + a_{n - 1}x^{n - 1} + \dots + a_{1}x + a_{0}\) be a polynomial, where \(n \geqslant 2\) and \(a_{i}\) is a positive integer for all \(0 \leqslant i \leqslant n - 1\) . Could it be the case that, for all positive inte... | Answer: No. For any such polynomial there exists a positive integer \(k\) such that \(s(k)\) and \(s(P(k))\) have different parities.
Solution. With the notation above, we begin by choosing a positive integer \(t\) such that
\[10^{t} > \max \left\{\frac{100^{n - 1}a_{n - 1}}{(10^{\frac{1}{n - 1}} - 9^{\frac{1}{n ... |
IMOSL-2022-A8 | For a positive integer \(n\) , an \(n\) - sequence is a sequence \((a_{0},\ldots ,a_{n})\) of non-negative integers satisfying the following condition: if \(i\) and \(j\) are non-negative integers with \(i + j\leqslant n\) , then \(a_{i} + a_{j}\leqslant n\) and \(a_{a_{i} + a_{j}} = a_{i + j}\) .
Let \(f(n)\) be th... | Answer: Such constants exist with \(\lambda = 3^{1 / 6}\) ; we will discuss appropriate values of \(c_{1}\) and \(c_{2}\) in the solution below.
Solution. In order to solve this, we will give a complete classification of \(n\) - sequences.
Let \(k = \lfloor n / 2\rfloor\) . We will say that an \(n\) - sequence is... |
IMOSL-2022-C2 | The Bank of Oslo issues coins made out of two types of metal: aluminium (denoted \(A\) ) and copper (denoted \(C\) ). Morgane has \(n\) aluminium coins, and \(n\) copper coins, and arranges her \(2n\) coins in a row in some arbitrary initial order. Given a fixed positive integer \(k \leq 2n\) , she repeatedly performs ... | Answer: All pairs \((n,k)\) such that \(n\leq k\leq \frac{3n + 1}{2}\)
Solution. Define a block to be a maximal subsequence of consecutive coins made out of the same metal, and let \(M^{b}\) denote a block of \(b\) coins of metal \(M\) . The property that there is at most one aluminium coin adjacent to a copper coin... |
IMOSL-2022-C4 | Let \(n > 3\) be a positive integer. Suppose that \(n\) children are arranged in a circle, and \(n\) coins are distributed between them (some children may have no coins). At every step, a child with at least 2 coins may give 1 coin to each of their immediate neighbours on the right and left. Determine all initial distr... | Answer: All distributions where \(\sum_{i = 1}^{n}i c_{i} = \frac{n(n + 1)}{2} \pmod {n}\) , where \(c_{i}\) denotes the number of coins the \(i\) - th child starts with.
Solution 1.
Number the children \(1,\ldots ,n\) , and denote the number of coins the \(i\) - th child has by \(c_{i}\) . A step of this process... |
IMOSL-2022-C5 | Let \(m, n \geq 2\) be integers, let \(X\) be a set with \(n\) elements, and let \(X_{1}, X_{2}, \ldots , X_{m}\) be pairwise distinct non- empty, not necessary disjoint subsets of \(X\) . A function \(f: X \to \{1, 2, \ldots , n + 1\}\) is called nice if there exists an index \(k\) such that
\[\sum_{x\in X_{k}}f(x)... | Solution. For a subset \(Y\subseteq X\) , we write \(f(Y)\) for \(\textstyle \sum_{y\in Y}f(y)\) . Note that a function \(f:X\to\) \(\{1,\ldots ,n + 1\}\) is nice, if and only if \(f\left(X_{i}\right)\) is maximized by a unique index \(i\in \{1,\ldots ,m\}\)
We will first investigate the set \(\mathcal{F}\) of funct... |
IMOSL-2022-C6 | Let \(n\) be a positive integer. We start with \(n\) piles of pebbles, each initially containing a single pebble. One can perform moves of the following form: choose two piles, take an equal number of pebbles from each pile and form a new pile out of these pebbles. For each positive integer \(n\) , find the smallest nu... | Answer: 1 if \(n\) is a power of two, and 2 otherwise.
Solution 1. The solution we describe is simple, but not the most effective one.
We can combine two piles of \(2^{k - 1}\) pebbles to make one pile of \(2^{k}\) pebbles. In particular, given \(2^{k}\) piles of one pebble, we may combine them as follows:
\[2... |
IMOSL-2022-C8 | Alice fills the fields of an \(n\times n\) board with numbers from 1 to \(n^{2}\) , each number being used exactly once. She then counts the total number of good paths on the board. A good path is a sequence of fields of arbitrary length (including 1) such that:
(i) The first field in the sequence is one that is onl... | Answer: \(2n^{2} - 2n + 1\) .
Solution. We will call any field that is only adjacent to fields with larger numbers a well. Other fields will be called non- wells. Let us make a second \(n\times n\) board \(B\) where in each field we will write the number of good sequences which end on the corresponding field in the ... |
IMOSL-2022-C9 | Let \(\mathbb{Z}_{\geq 0}\) be the set of non-negative integers, and let \(f:\mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}\to \mathbb{Z}_{\geq 0}\) be a bijection such that whenever \(f(x_{1},y_{1}) > f(x_{2},y_{2})\) , we have \(f(x_{1} + 1,y_{1}) > f(x_{2} + 1,y_{2})\) and \(f(x_{1},y_{1} + 1) > f(x_{2},y_{2} + 1)\) ... | Answer: The optimal bounds are \(2500\leqslant N\leqslant 7500\) .
Solution. We defer the constructions to the end of the solution. Instead, we begin by characterizing all such functions \(f\) , prove a formula and key property for such functions, and then solve the problem, providing constructions.
Characterizat... |
IMOSL-2022-G1 | Let \(A B C D E\) be a convex pentagon such that \(B C = D E\) . Assume there is a point \(T\) inside \(A B C D E\) with \(T B = T D\) , \(T C = T E\) and \(\angle T B A = \angle A E T\) . Let lines \(C D\) and \(C T\) intersect line \(A B\) at points \(P\) and \(Q\) , respectively, and let lines \(C D\) and \(D T\) in... | Solution 1. By the conditions we have \(B C = D E\) , \(C T = E T\) and \(T B = T D\) , so the triangles \(T B C\) and \(T D E\) are congruent, in particular \(\angle B T C = \angle D T E\) .
In triangles \(T B Q\) and \(T E S\) we have \(\angle T B Q = \angle S E T\) and \(\angle Q T B = 180^{\circ} - \angle B T C ... |
IMOSL-2022-G2 | In the acute- angled triangle \(A B C\) , the point \(F\) is the foot of the altitude from \(A\) , and \(P\) is a point on the segment \(A F\) . The lines through \(P\) parallel to \(A C\) and \(A B\) meet \(B C\) at \(D\) and \(E\) , respectively. Points \(X \neq A\) and \(Y \neq A\) lie on the circles \(A B D\) and \... | Solution 1. Let \(A^{\prime}\) be the intersection of lines \(B X\) and \(C Y\) . By power of a point, it suffices to prove that \(A^{\prime}B\cdot A^{\prime}X = A^{\prime}C\cdot A^{\prime}Y\) , or, equivalently, that \(A^{\prime}\) lies on the radical axis of the circles \(A B D X\) and \(A C E Y\) .
From \(D A = D... |
IMOSL-2022-G3 | Let \(ABCD\) be a cyclic quadrilateral. Assume that the points \(Q\) , \(A\) , \(B\) , \(P\) are collinear in this order, in such a way that the line \(AC\) is tangent to the circle \(ADQ\) , and the line \(BD\) is tangent to the circle \(BCP\) . Let \(M\) and \(N\) be the midpoints of \(BC\) and \(AD\) , respectively.... | Solution 1. We first prove that triangles \(ADQ\) and \(CDB\) are similar. Since \(ABCD\) is cyclic, we have \(\angle DAQ = \angle DCB\) . By the tangency of \(AC\) to the circle \(AQD\) we also have \(\angle CBD = \angle CAD = \angle AQD\) . The claimed similarity is proven.
Let \(R\) be the midpoint of \(CD\) . Po... |
IMOSL-2022-G4 | Let \(A B C\) be an acute- angled triangle with \(A C > A B\) , let \(O\) be its circumcentre, and let \(D\) be a point on the segment \(B C\) . The line through \(D\) perpendicular to \(B C\) intersects the lines \(A O\) , \(A C\) and \(A B\) at \(W\) , \(X\) and \(Y\) , respectively. The circumcircles of triangles \(... | Solution 1. Let \(A O\) intersect \(B C\) at \(E\) . As \(E D W\) is a right- angled triangle and \(O\) is on \(W E\) , the condition \(O W = O D\) means \(O\) is the circumcentre of this triangle. So \(O D = O E\) which establishes that \(D\) , \(E\) are reflections in the perpendicular bisector of \(B C\) .
Now ob... |
IMOSL-2022-G5 | Let \(ABC\) be a triangle, and let \(\ell_{1}\) and \(\ell_{2}\) be two parallel lines. For \(i = 1,2\) , let \(\ell_{i}\) meet the lines \(BC\) , \(CA\) , and \(AB\) at \(X_{i}\) , \(Y_{i}\) , and \(Z_{i}\) , respectively. Suppose that the line through \(X_{i}\) perpendicular to \(BC\) , the line through \(Y_{i}\) per... | Solution 1. Throughout the solutions, \(\hat{\times} (p,q)\) will denote the directed angle between lines \(p\) and \(q\) , taken modulo \(180^{\circ}\) .
Let the vertices of \(\Delta_{i}\) be \(D_{i},E_{i},F_{i}\) , such that lines \(E_{i}F_{i}\) , \(F_{i}D_{i}\) and \(D_{i}E_{i}\) are the perpendiculars through \(... |
IMOSL-2022-G6 | In an acute- angled triangle \(A B C\) , point \(H\) is the foot of the altitude from \(A\) . Let \(P\) be a moving point such that the bisectors \(k\) and \(\ell\) of angles \(P B C\) and \(P C B\) , respectively, intersect each other on the line segment \(A H\) . Let \(k\) and \(A C\) meet at \(E\) , let \(\ell\) and... | Solution 1. Let the reflections of the line \(B C\) with respect to the lines \(A B\) and \(A C\) intersect at point \(K\) . We will prove that \(P\) , \(Q\) and \(K\) are collinear, so \(K\) is the common point of the varying line \(P Q\) .
Let lines \(B E\) and \(C F\) intersect at \(I\) . For every point \(O\) an... |
IMOSL-2022-G7 | Let \(A B C\) and \(A^{\prime}B^{\prime}C^{\prime}\) be two triangles having the same circumcircle \(\omega\) , and the same orthocentre \(H\) . Let \(\Omega\) be the circumcircle of the triangle determined by the lines \(A A^{\prime}\) , \(B B^{\prime}\) and \(C C^{\prime}\) . Prove that \(H\) , the centre of \(\omega... | Solution. In what follows, \(\mathcal{x}(p,q)\) will denote the directed angle between lines \(p\) and \(q\) , taken modulo \(180^{\circ}\) . Denote by \(O\) the centre of \(\omega\) . In any triangle, the homothety with ratio \(- \frac{1}{2}\) centred at the centroid of the triangle takes the vertices to the midpoints... |
IMOSL-2022-G8 | Let \(AA'BCC'B'\) be a convex cyclic hexagon such that \(AC\) is tangent to the incircle of the triangle \(A'B'C'\) , and \(A'C'\) is tangent to the incircle of the triangle \(ABC\) . Let the lines \(AB\) and \(A'B'\) meet at \(X\) and let the lines \(BC\) and \(B'C'\) meet at \(Y\) .
Prove that if \(XBYB'\) is a co... | Solution. Denote by \(\omega\) and \(\omega '\) the incircles of \(\triangle ABC\) and \(\triangle A'B'C'\) and let \(I\) and \(I'\) be the centres of these circles. Let \(N\) and \(N'\) be the second intersections of \(BI\) and \(B'I'\) with \(\Omega\) , the circumcircle of \(A'BCC'B'A\) , and let \(O\) be the centre ... |
IMOSL-2022-N3 | Let \(a > 1\) be a positive integer, and let \(d > 1\) be a positive integer coprime to \(a\) . Let \(x_{1} = 1\) and, for \(k \geqslant 1\) , define
\[x_{k + 1} = \left\{ \begin{array}{l l}{x_{k} + d} & {\mathrm{if~}a\mathrm{~doesn't~divide~}x_{k},}\\ {x_{k} / a} & {\mathrm{if~}a\mathrm{~divides~}x_{k}.} \end{array... | Answer: \(n\) is the exponent with \(d < a^{n} < ad\) .
Solution 1. By trivial induction, \(x_{k}\) is coprime to \(d\) .
By induction and the fact that there can be at most \(a - 1\) consecutive increasing terms in the sequence, it also holds that \(x_{k} < da\) if \(x_{k} = x_{k - 1} + d\) and that \(x_{k} < d\... |
IMOSL-2022-N4 | Find all triples of positive integers \((a,b,p)\) with \(p\) prime and
\[a^{p} = b! + p.\] | Answer: \((2,2,2)\) and \((3,4,3)\) .
Solution 1. Clearly, \(a > 1\) . We consider three cases.
Case 1: We have \(a< p\) . Then we either have \(a\leqslant b\) which implies \(a\mid a^{p} - b! = p\) leading to a contradiction, or \(a > b\) which is also impossible since in this case we have \(b!\leqslant a!< a^{p... |
IMOSL-2022-N5 | For each \(1 \leqslant i \leqslant 9\) and \(T \in \mathbb{N}\) , define \(d_{i}(T)\) to be the total number of times the digit \(i\) appears when all the multiples of 1829 between 1 and \(T\) inclusive are written out in base 10.
Show that there are infinitely many \(T \in \mathbb{N}\) such that there are precisely... | Solution. Let \(n := 1829\) . First, we choose some \(k\) such that \(n \mid 10^{k} - 1\) . For instance, any multiple of \(\phi (n)\) would work since \(n\) is coprime to 10. We will show that either \(T = 10^{k} - 1\) or \(T = 10^{k} - 2\) has the desired property, which completes the proof since \(k\) can be taken t... |
IMOSL-2022-N6 | Let \(Q\) be a set of prime numbers, not necessarily finite. For a positive integer \(n\) consider its prime factorisation; define \(p(n)\) to be the sum of all the exponents and \(q(n)\) to be the sum of the exponents corresponding only to primes in \(Q\) . A positive integer \(n\) is called special if \(p(n) + p(n + ... | Solution. Let us call two positive integers \(m, n\) friends if \(p(m) + p(n)\) and \(q(m) + q(n)\) are both even integers. We start by noting that the pairs \((p(k), q(k))\) modulo 2 can take at most 4 different values; thus, among any five different positive integers there are two which are friends.
In addition, b... |
IMOSL-2022-N7 | Let \(k\) be a positive integer and let \(S\) be a finite set of odd prime numbers. Prove that there is at most one way (modulo rotation and reflection) to place the elements of \(S\) around a circle such that the product of any two neighbors is of the form \(x^{2} + x + k\) for some positive integer \(x\) . | Solution. Let us allow the value \(x = 0\) as well; we prove the same statement under this more general constraint. Obviously that implies the statement with the original conditions.
Call a pair \(\{p,q\}\) of primes with \(p\neq q\) special if \(p q = x^{2} + x + k\) for some nonnegative integer \(x\) . The followi... |
IMOSL-2022-N8 | Prove that \(5^{n} - 3^{n}\) is not divisible by \(2^{n} + 65\) for any positive integer \(n\) . | Solution 1. Let \(n\) be a positive integer, and let \(m = 2^{n} + 65\) . For the sake of contradiction, suppose that \(m \mid 5^{n} - 3^{n}\) , so \(5^{n} \equiv 3^{n} \pmod {m}\) .
Notice that if \(n\) is even, then \(3 \mid m\) , but \(3 \nmid 5^{n} - 3^{n}\) , contradiction. So, from now on we assume that \(n\) ... |
IMOSL-2023-A2 | Let \(\mathbb{R}\) be the set of real numbers. Let \(f\colon \mathbb{R}\to \mathbb{R}\) be a function such that
\[f(x + y)f(x - y)\geqslant f(x)^{2} - f(y)^{2}\]
for every \(x,y\in \mathbb{R}\) . Assume that the inequality is strict for some \(x_{0},y_{0}\in \mathbb{R}\)
Prove that \(f(x)\geqslant 0\) for ever... | Common remarks. We will say that \(f\) has constant sign, if \(f\) satisfies the conclusion of the problem.
Solution 1. We introduce the new variables \(s:= x + y\) and \(t:= x - y\) . Equivalently, \(x = \frac{s + t}{2}\) and \(y = \frac{s - t}{2}\) . The inequality becomes
\[f(s)f(t)\geqslant f\left(\frac{s + t... |
IMOSL-2023-A3 | Let \(x_{1}, x_{2}, \ldots , x_{2023}\) be distinct real positive numbers such that
\[a_{n} = \sqrt{(x_{1} + x_{2} + \cdots + x_{n})\left(\frac{1}{x_{1}} + \frac{1}{x_{2}} + \cdots + \frac{1}{x_{n}}\right)}\]
is an integer for every \(n = 1, 2, \ldots , 2023\) . Prove that \(a_{2023} \geqslant 3034\) . | Solution 1. We start with some basic observations. First note that the sequence \(a_{1}, a_{2}, \ldots , a_{2023}\) is increasing and thus, since all elements are integers, \(a_{n + 1} - a_{n} \geqslant 1\) . We also observe that \(a_{1} = 1\) and
\[a_{2} = \sqrt{(x_{1} + x_{2})\left(\frac{1}{x_{1}} + \frac{1}{x_{2}... |
IMOSL-2023-A4 | Let \(\mathbb{R}_{>0}\) be the set of positive real numbers. Determine all functions \(f\colon \mathbb{R}_{>0}\to \mathbb{R}_{>0}\) such that
\[x\big(f(x) + f(y)\big)\geqslant \big(f(f(x)) + y\big)f(y)\]
for every \(x,y\in \mathbb{R}_{>0}\) | Answer: All functions \(f(x) = \frac{c}{x}\) for some \(c > 0\)
Solution 1. Let \(f\colon \mathbb{R}_{>0}\to \mathbb{R}_{>0}\) be a function that satisfies the inequality of the problem statement. We will write \(f^{k}(x) = f(f(\cdot \cdot \cdot f(x)\cdot \cdot \cdot))\) for the composition of \(f\) with itself \(k\... |
IMOSL-2023-A5 | Let \(a_{1},a_{2},\ldots ,a_{2023}\) be positive integers such that
\(\cdot a_{1},a_{2},\ldots ,a_{2023}\) is a permutation of \(1,2,\ldots ,2023\) , and
\(\cdot |a_{1} - a_{2}|,|a_{2} - a_{3}|,\ldots ,|a_{2022} - a_{2023}|\) is a permutation of \(1,2,\ldots ,2022\)
Prove that \(\max \left(a_{1},a_{2023}\right... | Solution. For the sake of clarity, we consider and prove the following generalisation of the original problem (which is the case \(N = 1012\) ):
Let \(N\) be a positive integer and \(a_{1},a_{2},\ldots ,a_{2N - 1}\) be positive integers such that
\(\cdot a_{1},a_{2},\ldots ,a_{2N - 1}\) is a permutation of \(1,2,... |
IMOSL-2023-A6 | Let \(k \geqslant 2\) be an integer. Determine all sequences of positive integers \(a_{1}, a_{2}, \ldots\) for which there exists a monic polynomial \(P\) of degree \(k\) with non- negative integer coefficients such that
\[P(a_{n}) = a_{n + 1}a_{n + 2}\cdot \cdot \cdot a_{n + k}\]
for every integer \(n \geqslant ... | Answer: The sequence \((a_{n})\) must be an arithmetic progression consisting of positive integers with common difference \(d \geqslant 0\) , and \(P(x) = (x + d) \dots (x + kd)\) .
Common remarks. The following arguments and observations are implicit in the solutions given below.
Suppose the sequence \((a_{n})\)... |
IMOSL-2023-A7 | Let \(N\) be a positive integer. Prove that there exist three permutations \(a_{1}, a_{2}, \ldots , a_{N}\) ; \(b_{1}, b_{2}, \ldots , b_{N}\) ; and \(c_{1}, c_{2}, \ldots , c_{N}\) of \(1, 2, \ldots , N\) such that
\[\left|\sqrt{a_{k}} + \sqrt{b_{k}} + \sqrt{c_{k}} - 2\sqrt{N}\right| < 2023\]
for every \(k = 1, ... | Solution 1. The idea is to approximate the numbers \(\sqrt{1}, \sqrt{2}, \ldots , \sqrt{N}\) by the nearest integer with errors \(< 0.5\) . This gives the following sequence
\[1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 4, \ldots .\]
More precisely, for each \(k \geqslant 1\) , we round \(\sqrt{k^{2} - k + 1}, \ldots , \sqrt{k... |
IMOSL-2023-C1 | Let \(m\) and \(n\) be positive integers greater than 1. In each unit square of an \(m \times n\) grid lies a coin with its tail- side up. A move consists of the following steps:
1. select a \(2 \times 2\) square in the grid;
2. flip the coins in the top-left and bottom-right unit squares;
3. flip the coin in ... | Answer: The answer is all pairs \((m, n)\) satisfying \(3 \mid mn\) .
Solution 1. Let us denote by \((i, j)\) - square the unit square in the \(i^{\mathrm{th}}\) row and the \(j^{\mathrm{th}}\) column. We first prove that when \(3 \mid mn\) , it is possible to make all the coins show head- side up. For integers \(1 ... |
IMOSL-2023-C2 | Determine the maximal length \(L\) of a sequence \(a_{1},\ldots ,a_{L}\) of positive integers satisfying both the following properties:
- every term in the sequence is less than or equal to \(2^{2023}\) , and
- there does not exist a consecutive subsequence \(a_{i},a_{i + 1},\ldots ,a_{j}\) (where \(1\leqslant i\... | Answer: The answer is \(L = 2^{2024} - 1\) .
Solution. We prove more generally that the answer is \(2^{k + 1} - 1\) when \(2^{2023}\) is replaced by \(2^{k}\) for an arbitrary positive integer \(k\) . Write \(n = 2^{k}\) .
We first show that there exists a sequence of length \(L = 2n - 1\) satisfying the properti... |
IMOSL-2023-C5 | Elisa has 2023 treasure chests, all of which are unlocked and empty at first. Each day, Elisa adds a new gem to one of the unlocked chests of her choice, and afterwards, a fairy acts according to the following rules:
- if more than one chests are unlocked, it locks one of them, or
- if there is only one unlocked ... | Solution 1. We will prove that such a constant \(C\) exists when there are \(n\) chests for \(n\) an odd positive integer. In fact we can take \(C = n - 1\) . Elisa's strategy is simple: place a gem in the chest with the fewest gems (in case there are more than one such chests, pick one arbitrarily).
For each intege... |
IMOSL-2023-C6 | Let \(N\) be a positive integer, and consider an \(N \times N\) grid. A right- down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right- up path is a sequence of grid cells such that each cell is either one cell to the righ... | Solution 1. We define a good parallelogram to be a parallelogram composed of two isosceles right- angled triangles glued together as shown below.
Given any partition into \(k\) right- down or right- up paths, we can find a corresponding packing of good parallelograms that leaves an area of \... |
IMOSL-2023-C7 | The Imomi archipelago consists of \(n \geqslant 2\) islands. Between each pair of distinct islands is a unique ferry line that runs in both directions, and each ferry line is operated by one of \(k\) companies. It is known that if any one of the \(k\) companies closes all its ferry lines, then it becomes impossible for... | Answer: The largest \(k\) is \(k = \lfloor \log_{2}n\rfloor\) .
Solution. We reformulate the problem using graph theory. We have a complete graph \(K_{n}\) on \(n\) nodes (corresponding to islands), and we want to colour the edges (corresponding to ferry lines) with \(k\) colours (corresponding to companies), so tha... |
IMOSL-2023-G1 | Let \(A B C D E\) be a convex pentagon such that \(\angle A B C = \angle A E D = 90^{\circ}\) . Suppose that the midpoint of \(C D\) is the circumcentre of triangle \(A B E\) . Let \(O\) be the circumcentre of triangle \(A C D\) .
Prove that line \(A O\) passes through the midpoint of segment \(B E\) . | Solution 1 (Area Ratio).
Let \(M\) be the midpoint of \(C D\) and \(X = B C\cap E D\) . Since \(\angle A B X = \angle A E X = 90^{\circ}\) , \(A X\) is a diameter of the circumcircle of \(\triangle A B E\) so the midpoint of \(A X\) is the circumcentre of \(\triangle A B E\) . Therefore, the... |
IMOSL-2023-G2 | Let \(ABC\) be a triangle with \(AC > BC\) . Let \(\omega\) be the circumcircle of triangle \(ABC\) and let \(r\) be the radius of \(\omega\) . Point \(P\) lies on segment \(AC\) such that \(BC = CP\) and point \(S\) is the foot of the perpendicular from \(P\) to line \(AB\) . Let ray \(BP\) intersect \(\omega\) again ... | Solution 1 (Similar Triangles).
First observe that
\[\angle DPA = \angle BPC^{CP\equiv CB}\angle CBP = \angle CBD = \angle CAD = \angle PAD\]
so \(DP = DA\) . Thus there is a symmetry in the problem statement swapping \((A, D) \leftrightarrow (B, C)\) .
Let \(O\) be the centre of \(... |
IMOSL-2023-G3 | Let \(ABCD\) be a cyclic quadrilateral with \(\angle BAD < \angle ADC\) . Let \(M\) be the midpoint of the arc \(CD\) not containing \(A\) . Suppose there is a point \(P\) inside \(ABCD\) such that \(\angle ADB = \angle CPD\) and \(\angle ADP = \angle PCB\) .
Prove that lines \(AD, PM, BC\) are concurrent. | Solution 1. Let \(X\) and \(Y\) be the intersection points of \(AM\) and \(BM\) with \(PD\) and \(PC\) respectively. Since \(ABCDM\) is cyclic and \(CM = MD\) , we have
\[\angle XAD = \angle MAD = \angle CBM = \angle CBY.\]
Combining this with \(\angle ADX = \angle YCB\) , we get \(\angle DXA = \angle BYC\) , and... |
IMOSL-2023-G4 | Let \(A B C\) be an acute- angled triangle with \(A B< A C\) . Denote its circumcircle by \(\Omega\) and denote the midpoint of arc \(C A B\) by \(S\) . Let the perpendicular from \(A\) to \(B C\) meet \(B S\) and \(\Omega\) at \(D\) and \(E\neq A\) respectively. Let the line through \(D\) parallel to \(B C\) meet line... | Solution 1 (Triangles in Perspective). Let \(S^{\prime}\) be the midpoint of arc \(B C\) of \(\Omega\) , diametrically opposite to \(S\) so \(S S^{\prime}\) is a diameter in \(\Omega\) and \(A S^{\prime}\) is the angle bisector of \(\angle B A C\) . Let the tangent of \(\omega\) at \(P\) meet \(\Omega\) again at \(Q\ne... |
IMOSL-2023-G5 | Let \(ABC\) be an acute- angled triangle with circumcircle \(\omega\) and circumcentre \(O\) . Points \(D \neq B\) and \(E \neq C\) lie on \(\omega\) such that \(BD \perp AC\) and \(CE \perp AB\) . Let \(CO\) meet \(AB\) at \(X\) , and \(BO\) meet \(AC\) at \(Y\) .
Prove that the circumcircles of triangles \(BXD\) a... | Solution 1 (Reflections).
Note that \(AO = OC\) implies the lines \(AO\) , \(XO\) are reflections of each other about the line parallel to \(AC\) through \(O\) , which is the perpendicular bisector of \(BD\) . Call this line \(\ell\) .
Let \(P \neq X\) be the second intersection of circle \(\odot BXD\) with line ... |
IMOSL-2023-G6 | Let \(ABC\) be an acute- angled triangle with circumcircle \(\omega\) . A circle \(\Gamma\) is internally tangent to \(\omega\) at \(A\) and also tangent to \(BC\) at \(D\) . Let \(AB\) and \(AC\) intersect \(\Gamma\) at \(P\) and \(Q\) respectively. Let \(M\) and \(N\) be points on line \(BC\) such that \(B\) is the m... | Solution 1 (Similar Triangles).
Let \(MP\) and \(NQ\) intersect \(AD\) at \(K_{1}\) and \(K_{2}\) respectively. By applying Menelaus' theorem to triangle \(ABD\) and line \(MPK_{1}\) , we have
\[\frac{A K_{1}}{K_{1}D} = \frac{A P}{P B}\cdot \frac{B M}{M D} = \frac{A P}{2P B}\]
and simi... |
IMOSL-2023-G7 | Let \(ABC\) be an acute, scalene triangle with orthocentre \(H\) . Let \(\ell_{a}\) be the line through the reflection of \(B\) with respect to \(CH\) and the reflection of \(C\) with respect to \(BH\) . Lines \(\ell_{b}\) and \(\ell_{c}\) are defined similarly. Suppose lines \(\ell_{a}\) , \(\ell_{b}\) , and \(\ell_{c... | Solution 1.
We write \(\triangle P_{1}P_{2}P_{3}\stackrel {\perp}{\sim}\triangle Q_{1}Q_{2}Q_{3}\) (resp. \(\triangle P_{1}P_{2}P_{3}\sim \triangle Q_{1}Q_{2}Q_{3}\) ) to indicate that two triangles are directly (resp. oppositely) similar. We use directed angles throughout denoted with \(\an... |
IMOSL-2023-G8 | Let \(ABC\) be an equilateral triangle. Points \(A_{1}\) , \(B_{1}\) , \(C_{1}\) lie inside triangle \(ABC\) such that triangle \(A_{1}B_{1}C_{1}\) is scalene, \(BA_{1} = A_{1}C\) , \(CB_{1} = B_{1}A\) , \(AC_{1} = C_{1}B\) and
\[\angle B A_{1}C + \angle C B_{1}A + \angle A C_{1}B = 480^{\circ}.\]
Lines \(BC_{1}\... | Solution. Let \(\delta_{A},\delta_{B},\delta_{C}\) be the circumcircles of \(\triangle AA_{1}A_{2}\) , \(\triangle BB_{1}B_{2}\) , \(\triangle CC_{1}C_{2}\) . The general strategy of the solution is to find two different points having equal power with respect to \(\delta_{A},\delta_{B},\delta_{C}\) .
Claim. \(A_{1}\... |
IMOSL-2023-N1 | Determine all positive, composite integers \(n\) that satisfy the following property: if the positive divisors of \(n\) are \(1 = d_{1}< d_{2}< \dots < d_{k} = n\) , then \(d_{i}\) divides \(d_{i + 1} + d_{i + 2}\) for every \(1\leqslant i\leqslant k - 2\) . | Answer: \(n = p^{r}\) is a prime power for some \(r\geqslant 2\)
Solution 1. It is easy to see that such an \(n = p^{r}\) with \(r\geqslant 2\) satisfies the condition as \(d_{i} = p^{i - 1}\) with \(1\geqslant i\geqslant k = r + 1\) and clearly
\[p^{i - 1}\mid p^{i} + p^{i + 1}.\]
Now, let us suppose that the... |
IMOSL-2023-N3 | For positive integers \( n \) and \( k \ge 2 \) define \( E_k(n) \) as the greatest exponent \( r \) such that \( k^r \) divides \( n! \). Prove that there are infinitely many \( n \) such that \( E_{10}(n) > E_9(n) \) and infinitely many \( m \) such that \( E_{10}(m) < E_9(m) \). | Solution 1. We let \( v_p(m) \) denote the \( p \)-adic valuation of \( m \). By Legendre’s Formula we know, for \( p \) prime, that
\[
v_p(n!) = \lfloor n/p \rfloor + \lfloor n/p^2 \rfloor + \cdots.
\]
We can see that \( E_9(n) = \left\lfloor \frac{v_3(n!)}{2} \right\rfloor. \)
Since \( v_5(n!) \le v_2(n!) \) and \( E... |
IMOSL-2023-N5 | Let \(a_{1}< a_{2}< a_{3}< \dots\) be positive integers such that \(a_{k + 1}\) divides \(2(a_{1} + a_{2} + \dots +a_{k})\) for every \(k\geq 1\) . Suppose that for infinitely many primes \(p\) , there exists \(k\) such that \(p\) divides \(a_{k}\) . Prove that for every positive integer \(n\) , there exists \(k\) such... | Solution. For every \(k\geq 2\) define the quotient \(b_{k} = 2(a_{1} + \dots +a_{k - 1}) / a_{k}\) , which must be a positive integer. We first prove the following properties of the sequence \((b_{k})\) :
Claim 1. We have \(b_{k + 1}\leqslant b_{k} + 1\) for all \(k\geq 2\)
Proof. By subtracting \(b_{k}a_{k} = 2... |
IMOSL-2023-N6 | A sequence of integers \(a_{0},a_{1},a_{2},\ldots\) is called kawaii, if \(a_{0} = 0,a_{1} = 1\) , and, for any positive integer \(n\) , we have
\[(a_{n + 1} - 3a_{n} + 2a_{n - 1})(a_{n + 1} - 4a_{n} + 3a_{n - 1}) = 0.\]
An integer is called kawaii if it belongs to a kawaii sequence.
Suppose that two consecuti... | Solution 1. We start by rewriting the condition in the problem as:
\[a_{n + 1} = 3a_{n} - 2a_{n - 1},\mathrm{~or~}a_{n + 1} = 4a_{n} - 3a_{n - 1}.\]
We have \(a_{n + 1}\equiv a_{n}\) or \(a_{n - 1}\) (mod 2) and \(a_{n + 1}\equiv a_{n - 1}\) or \(a_{n}\) (mod 3) for all \(n\geqslant 1\) . Now, since \(a_{0} = 0\)... |
IMOSL-2023-N7 | Let \(a, b, c, d\) be positive integers satisfying
\[\frac{ab}{a + b} +\frac{cd}{c + d} = \frac{(a + b)(c + d)}{a + b + c + d}.\]
Determine all possible values of \(a + b + c + d\) . | Answer: The possible values are the positive integers that are not square- free.
Solution.
First, note that if we take \(a = \ell\) , \(b = k\ell\) , \(c = k\ell\) , \(d = k^{2}\ell\) for some positive integers \(k\) and \(\ell\) , then we have
\[\frac{ab}{a + b} +\frac{cd}{c + d} = \frac{k\ell^{2}}{\ell + k\e... |
IMOSL-2023-N8 | Let \(\mathbb{Z}_{>0}\) be the set of positive integers. Determine all functions \(f\colon \mathbb{Z}_{>0}\to \mathbb{Z}_{>0}\) such that
\[f^{b f(a)}(a + 1) = (a + 1)f(b)\]
holds for all \(a,b\in \mathbb{Z}_{>0}\) , where \(f^{k}(n) = f(f(\cdot \cdot f(n)\cdot \cdot \cdot))\) denotes the composition of \(f\) wit... | Answer: The only function satisfying the condition is \(f(n) = n + 1\) for all \(n\in \mathbb{Z}_{>0}\)
Let \(P(a,b)\) be the equality in the statement.
Solution 1. We divide the solution into 5 steps.
Step 1. ( \(f\) is injective)
Claim 1. For any \(a\geq 2\) , the set \(\{f^{n}(a)\mid n\in \mathbb{Z}_{>0}... |
IMOSL-2024-A1 | Determine all real numbers \(\alpha\) such that the number
\[|\alpha | + |2\alpha | + \dots +|n\alpha |\]
is a multiple of \(n\) for every positive integer \(n\) . (Here \(|z|\) denotes the greatest integer less than or equal to \(z\) .) | Answer: All even integers satisfy the condition of the problem and no other real number \(\alpha\) does so.
Solution 1. First we will show that even integers satisfy the condition. If \(\alpha = 2m\) where \(m\) is an integer then
\[|\alpha | + |2\alpha | + \dots +|n\alpha | = 2m + 4m + \dots +2mn = mn(n + 1)\] ... |
IMOSL-2024-A2 | Let \(n\) be a positive integer. Find the minimum possible value of
\[S = 2^{0}x_{0}^{2} + 2^{1}x_{1}^{2} + \cdot \cdot \cdot +2^{n}x_{n}^{2},\]
where \(x_{0}, x_{1}, \ldots , x_{n}\) are nonnegative integers such that \(x_{0} + x_{1} + \dots + x_{n} = n\) . | Answer: The minimum value is \(\frac{n(n + 1)}{2}\) .
Solution 1. For a fixed \(n\) , let \(f(n)\) denote the minimum possible value of \(S\) . Consider the following variant: among all infinite sequences of nonnegative integers \(x_{0}, x_{1}, \ldots\) , only finitely many of which are nonzero, satisfying \(x_{0} +... |
IMOSL-2024-A3 | Decide whether for every sequence \((a_{n})\) of positive real numbers,
\[\frac{3^{a_{1}} + 3^{a_{2}} + \cdot \cdot \cdot + 3^{a_{n}}}{(2^{a_{1}} + 2^{a_{2}} + \cdot \cdot \cdot + 2^{a_{n}})^{2}} < \frac{1}{2024} \quad (1)\]
is true for at least one positive integer \(n\) . | Comment. The question can be asked in several forms, as follows:
(i) students could be asked, as above, to show the existence of such an \(n\) ;
(ii) students could be asked to show that this happens for all sufficiently large \(n\) ;
(iii) students could be given a concrete positive integer \(N\) and asked to... |
IMOSL-2024-A4 | Let \(\mathbb{Z}_{>0}\) be the set of all positive integers. Determine all subsets \(\mathcal{S}\) of \(\{2^{0},2^{1},2^{2},\ldots \}\) for which there exists a function \(f\colon \mathbb{Z}_{>0}\to \mathbb{Z}_{>0}\) such that
\[\mathcal{S} = \{f(a + b) - f(a) - f(b) \mid a,b\in \mathbb{Z}_{>0}\} .\] | Answer: \(\mathcal{S}\) can be any subset of size 1 or 2.
Common remarks. For this problem, it is convenient to use notation such as \(\{a,b,c\}\) for multisets rather than sets, and the subset relation is likewise that for multisets. Both solutions use the following property of powers of 2: if \(2^{a} + 2^{b} = 2^{... |
IMOSL-2024-A5 | Find all periodic sequences \(a_{1}\) , \(a_{2}\) , ... of real numbers such that the following conditions hold for all \(n \geqslant 1\) :
\[a_{n + 2} + a_{n}^{2} = a_{n} + a_{n + 1}^{2}\qquad \mathrm{and}\qquad |a_{n + 1} - a_{n}|\leqslant 1.\] | Answer: The sequences satisfying the conditions of the problem are:
\[c, - c, c, - c, \dots ,\] \[d, d, d, d, \dots ,\]
where \(c \in [- \frac{1}{2}, \frac{1}{2}]\) and \(d\) is any real number.
Solution 1. We rewrite the first condition as
\[a_{n + 2} + a_{n + 1} = (a_{n + 1} + a_{n})(a_{n + 1} - a_{n} + 1... |
IMOSL-2024-A6 | Let \(a_{0}, a_{1}, a_{2}, \ldots\) be an infinite strictly increasing sequence of positive integers such that for each \(n \geqslant 1\) we have
\[a_{n} \in \left\{\frac{a_{n - 1} + a_{n + 1}}{2}, \sqrt{a_{n - 1} \cdot a_{n + 1}}\right\} .\]
Let \(b_{1}, b_{2}, \ldots\) be an infinite sequence of letters defined... | Common remarks. In fact, all known proofs proceed by showing that the eventual period of the sequence \((b_{n})\) always consists of some number of occurrences of \(G\) (possibly zero) followed by an \(A\) .
Such sequences of any period \(p \geqslant 1\) exist. Indeed, consider the sequence
\[\ldots , k^{p}, k^{p... |
IMOSL-2024-A7 | Let \(\mathbb{Q}\) be the set of rational numbers. Let \(f\colon \mathbb{Q}\to \mathbb{Q}\) be a function such that the following property holds: for all \(x\) , \(y\in \mathbb{Q}\) ,
\[f(x + f(y)) = f(x) + y\qquad \mathrm{or}\qquad f(f(x) + y) = x + f(y).\]
Determine the maximum possible number of elements of \(... | Answer: 2 is the maximum number of elements.
Common remarks. Suppose that \(f\) is a function satisfying the condition of the problem. We will use the following throughout all solutions.
- \(a \sim b\) if either \(f(a) = b\) or \(f(b) = a\) ,
- \(a \to b\) if \(f(a) = b\) ,
- \(P(x, y)\) to denote the propo... |
IMOSL-2024-A8 | Let \(p \neq q\) be coprime positive integers. Determine all infinite sequences \(a_{1}, a_{2}, \ldots\) of positive integers such that the following conditions hold for all \(n \geq 1\) :
\[\max (a_{n},a_{n + 1},\ldots ,a_{n + p}) - \min (a_{n},a_{n + 1},\ldots ,a_{n + p})\] \[\max (a_{n},a_{n + 1},\ldots ,a_{n + q... | Answer: The only such sequences are \(a_{n} = n + C\) , where \(C\) is a nonnegative integer.
Common remarks.
- Denote by \(a_{[i,j]}\) the subsequence \(a_{i}, a_{i + 1}, \ldots , a_{j}\) .
- Without loss of generality, in each solution we suppose \(p < q\) . It can be convenient to treat the case where \(p =... |
IMOSL-2024-C1 | Let \(n\) be a positive integer. A class of \(n\) students run \(n\) races, in each of which they are ranked with no draws. A student is eligible for a rating \((a,b)\) for positive integers \(a\) and \(b\) if they come in the top \(b\) places in at least \(a\) of the races. Their final score is the maximum possible va... | Answer: The maximum possible sum is \(\frac{n(n - 1)}{2}\) .
Solution 1. The answer can be achieved by the students finishing in the same order in every race. To show that this is the maximum, we will apply a series of modifications to the results of the races, each of which does not decrease the total score, such t... |
IMOSL-2024-C2 | Let \(n\) be a positive integer. The integers 1, 2, 3, ..., \(n^{2}\) are to be written in the cells of an \(n \times n\) board such that each integer is written in exactly one cell and each cell contains exactly one integer. For every integer \(d\) with \(d \mid n\) , the \(d\) - division of the board is the division ... | Answer: The even cool numbers are \(n = 2^{k}\) where \(k\) is a positive integer.
Solution. We first show by induction that \(n = 2^{k}\) is a cool number. The base case of \(n = 2\) is trivial as there is no such \(d\) .
For induction, assume that \(2^{k}\) is a cool number. We construct a numbering of a \(2^{k... |
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