id stringlengths 4 4 | problem_markdown stringlengths 36 3.59k | solutions_markdown listlengths 0 10 | images images listlengths 0 15 | country stringclasses 58
values | competition stringlengths 3 108 ⌀ | topics_flat listlengths 0 12 | language stringclasses 18
values | problem_type stringclasses 4
values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0gjv | Let $n \ge 2$ be an integer. Killer is a game played by a dealer and $n$ players. The game begins with the dealer designating one of the $n$ players a killer and keeping this information a secret. Every player knows that the killer exists among the $n$ players. The dealer can make as many public announcements as he wis... | [] | Thailand | Selected Problems from Thailand Training Camp | [
"Discrete Mathematics > Logic",
"Algebra > Abstract Algebra > Field Theory"
] | English | proof and answer | a) Yes, it can be played with five players. b) Yes, it can also be played with four players. | |
0er1 | If $f(x) = 2^x$, what is the value of $\frac{f(x+1) + f(x)}{f(x)}$? | [
"$$\n\\frac{f(x + 1) + f(x)}{f(x)} = \\frac{f(x + 1)}{f(x)} + 1 = \\frac{2^{x+1}}{2^x} + 1 = 2^1 + 1 = 3.\n$$"
] | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Intermediate Algebra > Exponential functions"
] | English | final answer only | 3 | |
041a | Let $A$ be the closed domain on the plane delimited by three lines $x = 1$, $y = 0$, and $y = t(2x - t)$, where $0 < t < 1$. Prove that the surface of any triangle inside the domain $A$ with $P(t, t^2)$ and $Q(1, 0)$ as two of its vertices cannot exceed $\frac{1}{4}$. | [
"It is easy to observe that the domain is a closed triangle. Its three vertices are $B\\left(\\frac{t}{2}, 0\\right)$, $Q(1, 0)$ and $C(1, t(2-t))$. Pick a point $X$ inside the $\\triangle BQC$, then the area of $\\triangle PQX$ is equal to half of the product of $PQ$ with the distance from $X$ to $PQ$. So the area... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0jx6 | Problem:
Let $ABC$ be a triangle with $AB = 13$, $BC = 14$, and $CA = 15$. Let $\ell$ be a line passing through two sides of triangle $ABC$. Line $\ell$ cuts triangle $ABC$ into two figures, a triangle and a quadrilateral, that have equal perimeter. What is the maximum possible area of the triangle? | [
"Solution:\n\nThere are three cases: $\\ell$ intersects $AB, AC$, $\\ell$ intersects $AB, BC$, and $\\ell$ intersects $AC, BC$. These cases are essentially identical, so let $\\ell$ intersect segment $AB$ at $M$ and segment $AC$ at $N$.\n\nThen the condition is equivalent to\n$$\n\\begin{gathered}\nAM + MN + AN = M... | United States | February 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 1323/26 | |
0hix | Problem:
Which is larger, $A=20^{19^{20}}$ or $B=19^{20^{19}}$? (Here, $a^{b^{c}}$ means $a^{\left(b^{c}
ight)}$ and not $\left(a^{b}
ight)^{c}$.) | [
"Solution:\n\nThe answer is that $A$ is larger. First,\n$$\n\\begin{aligned}\n\\left(\\frac{19}{20}\\right)^{20} & = 0.95^{20} = 0.9025^{10} > 0.9^{10} \\\\\n& = 0.81^{5} = 0.8^{5} = 0.32768 > 0.05 = \\frac{1}{20}\n\\end{aligned}\n$$\nwhich implies that $19^{20} > 20^{19}$. Thus, $A = 20^{19^{20}} > 20^{20^{19}} > ... | United States | Berkeley Math Circle | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | A | |
0iln | Problem:
A positive real number $x$ is such that
$$
\sqrt[3]{1-x^{3}}+\sqrt[3]{1+x^{3}}=1
$$
Find $x^{2}$. | [
"Solution:\nAnswer: $\\frac{\\sqrt[3]{28}}{3}$. Cubing the given equation yields\n$$\n1=\\left(1-x^{3}\\right)+3 \\sqrt[3]{\\left(1-x^{3}\\right)\\left(1+x^{3}\\right)}\\left(\\sqrt[3]{1-x^{3}}+\\sqrt[3]{1+x^{3}}\\right)+\\left(1+x^{3}\\right)=2+3 \\sqrt[3]{1-x^{6}}\n$$\nThen $\\frac{-1}{3}=\\sqrt[3]{1-x^{6}}$, so ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | ∛28 / 3 | |
079t | Determine all integers $x, y$ satisfying the equation
$$
(y^3 + xy - 1)(x^2 + x - y) = (x^3 - xy + 1)(y^2 + x - y).
$$ | [
"We have\n$$\n\\begin{aligned}\n\\frac{x^3 - xy + 1}{x^2 + x - y} &= \\frac{y^3 + xy - 1}{y^2 + x - y} \\Rightarrow x - \\frac{x^2 - 1}{x^2 + x - y} = y + \\frac{y^2 - 1}{y^2 + x - y} \\\\\n\\Rightarrow x - y &= \\frac{x^2 - 1}{x^2 + x - y} + \\frac{y^2 - 1}{y^2 + x - y}. \\quad (*)\n\\end{aligned}\n$$\nNow we have... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (-2,-1), (-1,0), (0,0), (0,1), (-1,-1), (1,1), (1,2) | |
04ck | Determine the set of all real numbers $x$ that satisfy the equation
$$
\left\lfloor \frac{2x-1}{3} \right\rfloor + \left\lfloor \frac{4x+1}{6} \right\rfloor = \frac{5x-4}{3},
$$
where $\lfloor a \rfloor$ denotes the largest integer not greater than $a$. | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | x ∈ { -2/5, 1/5, 4/5, 7/5, 2 } | |
0axq | Problem:
The figure shows a semicircle, where $B$ is a point on its diameter $AC$ so that $AB = 6$ and $BC = 12$. The line perpendicular to this diameter meets the semicircle at $D$. The semicircles with diameters $AB$ and $BC$, drawn as shown, meet $AD$ and $CD$ at $E$ and $F$, respectively. Find the distance between... | [
"Solution:\n\nObserve that $DEFB$ is a rectangle so $EF = DB$. However, $DB^2 = AB \\times BC = 72$. Thus, $EF = 6\\sqrt{2}$."
] | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof and answer | 6√2 | |
0kt6 | Problem:
Alice thinks of four positive integers $a \leq b \leq c \leq d$ satisfying $\{ab + cd, ac + bd, ad + bc\} = \{40, 70, 100\}$. What are all the possible tuples $(a, b, c, d)$ that Alice could be thinking of? | [
"Solution:\n\nSince $ab \\cdot cd = ac \\cdot bd = ad \\cdot bc$, the largest sum among $ab + cd, ac + bd, ad + bc$ will be the one with the largest difference between the two quantities, so $ab + cd = 100$, $ac + bd = 70$, $ad + bc = 40$.\n\nConsider the sum of each pair of equations, which gives $(a + b)(c + d) =... | United States | HMMT November 2022 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1, 4, 6, 16) | |
00f3 | Consider all the triangles $A B C$ which have a fixed base $A B$ and whose altitude from $C$ is a constant $h$. For which of these triangles is the product of its altitudes a maximum? | [
"Let $h_{a}$ and $h_{b}$ be the altitudes from $A$ and $B$, respectively. Then\n$$\n\\begin{aligned}\nA B \\cdot h \\cdot A C \\cdot h_{b} \\cdot B C \\cdot h_{a} &= 8 \\cdot (\\text{area of } \\triangle A B C)^3 \\\\\n&= (A B \\cdot h)^3,\n\\end{aligned}\n$$\nwhich is a constant. So the product $h \\cdot h_{a} \\c... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | If the fixed height is at most half the base length, the maximum occurs for the triangle with a right angle at the third vertex. If the fixed height exceeds half the base length, the maximum occurs for the isosceles triangle with equal sides from the base endpoints (the third vertex on the perpendicular bisector of the... | |
0782 | Suppose an acute scalene triangle $ABC$ has incentre $I$ and incircle touching $BC$ at $D$. Let $Z$ be the antipode of $A$ in the circumcircle of $ABC$. Point $L$ is chosen on the internal angle bisector of $\angle BZC$ such that $AL = LI$. Let $M$ be midpoint of arc $BZC$, and let $V$ be the midpoint of $ID$. Prove th... | [
"Let $N$ be the midpoint of arc $BAC$. Note that $Z, L, N$ are collinear; further, $AM \\parallel LZ$ because they are reflections in the circumcenter $O$ of $\\triangle ABC$. Therefore $\\angle IML = \\angle MLZ$. So it is sufficient to prove that $\\angle IVM = \\angle NLM$; in fact we will prove that $\\triangle... | India | IMO TST Day 4 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"G... | null | proof only | null | |
0990 | Problem:
Suma unor numere naturale impare consecutive este $392$. Aflați aceste numere. | [
"Solution:\n\nNotăm aceste numere: $2m+1, 2m+3, 2m+5, \\ldots, 2m+1+2k$. Numerele $m, k$ le aflăm din relația\n$$\n2m+1 + 2m+3 + 2m+5 + \\cdots + 2m+1+2k = 392\n$$\nsau\n$$\n\\frac{2m+1 + 2m+1+2k}{2} \\cdot (k+1) = 392 \\Leftrightarrow (2m+k+1)(k+1) = 392 = 2^{3} \\cdot 7^{2}\n$$\nDacă $k+1$ ar fi impar, atunci și ... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | [195, 197]; [95, 97, 99, 101]; [15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41] | |
06us | Let $m, n \geqslant 2$ be integers. Let $f\left(x_{1}, \ldots, x_{n}\right)$ be a polynomial with real coefficients such that
$$
f\left(x_{1}, \ldots, x_{n}\right)=\left\lfloor\frac{x_{1}+\ldots+x_{n}}{m}\right\rfloor \text{ for every } x_{1}, \ldots, x_{n} \in\{0,1, \ldots, m-1\} .
$$
Prove that the total degree of $f... | [
"We transform the problem to a single variable question by the following\n\nLemma. Let $a_{1}, \\ldots, a_{n}$ be nonnegative integers and let $G(x)$ be a nonzero polynomial with $\\operatorname{deg} G \\leqslant a_{1}+\\ldots+a_{n}$. Suppose that some polynomial $F\\left(x_{1}, \\ldots, x_{n}\\right)$ satisfies\n$... | IMO | IMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0cbk | Find the smallest positive integer $n$ for which there exists three pairwise distinct positive integers $a, b, c$, such that $n = a + b + c$ and $(a+b)(b+c)(c+a)$ is a perfect cube. | [
"Let $n$ be the required minimum value, and $a, b, c \\in \\mathbb{N}_{\\ge 1}$ pairwise distinct, such that $n = a + b + c$ and $(a+b)(b+c)(c+a) = k^3$, where $k \\in \\mathbb{N}$.\n\nIf $a, b, c$ are all odd, then $n \\ge 1+3+5=9$. If $n=9$, then $\\{a, b, c\\} = \\{1, 3, 5\\}$, which leads to $(a+b)(b+c)(c+a) = ... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - FIRST SELECTION TEST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 10 | |
0l7w | Let $ABCDE$ be a convex pentagon with $AB = 14$, $BC = 7$, $CD = 24$, $DE = 13$, $EA = 26$, and $\angle B = \angle E = 60^\circ$. For each point $X$ in the plane, define $f(X) = AX + BX + CX + DX + EX$. The least possible value of $f(X)$ can be expressed as $m + n\sqrt{p}$, where $m$ and $n$ are positive integers and $... | [] | United States | 2025 AIME I | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 55 | |
081v | Problem:
Determinare tutte le terne di interi positivi $(x, y, z)$ che soddisfano il seguente sistema:
$$
\left\{\begin{array}{l}
45 x y^{2}=8 z^{3} \\
x y z<1000
\end{array}\right.
$$ | [
"Solution:\n\nL'unica soluzione è $(6,10,15)$. Osserviamo anzitutto che $45$ divide $z^{3}$, e quindi $z$ è divisibile per $3 \\cdot 5=15$. Analogamente $8$ divide $x y^{2}$, e quindi $x$ e $y$ non possono essere dispari entrambi. Poniamo $z=15 w$ e distinguiamo 3 casi:\n\n(1) $x$ dispari: allora $8$ divide $y^{2}$... | Italy | Progetto Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (6,10,15) | |
09y6 | Problem:
Gegeven is een koordenvierhoek $A B C D$ met $|A B|=|B C|$. Punt $E$ ligt op de boog $C D$ waar $A$ en $B$ niet op liggen. Het snijpunt van $B E$ en $C D$ noemen we $P$, het snijpunt van $A E$ en $B D$ noemen we $Q$. Bewijs dat $P Q \| A C$.
 | [
"Solution:\nOmdat $|A B|=|B C|$ geldt $\\angle A E B=\\angle B D C$, dus $\\angle Q E P=\\angle A E B=\\angle B D C=\\angle Q D P$, waaruit volgt dat $Q P E D$ een koordenvierhoek is. Dus $\\angle Q P D=\\angle Q E D=\\angle A E D=\\angle A C D$. Dit betekent wegens F-hoeken dat $Q P$ en $A C$ evenwijdig zijn."
] | Netherlands | Selectietoets | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02tl | Problem:
Dez crianças decidem correr ao redor de um lago circular com $200\ \mathrm{~m}$ de perímetro. No início da corrida, as dez crianças estão paradas ocupando posições distintas e cada uma delas correrá no sentido horário ou anti-horário, a depender de sua vontade, com velocidade de $\frac{200}{k}\ \mathrm{~m} / ... | [
"Solution:\n\nSe uma criança tem velocidade $\\frac{200}{k}\\ \\mathrm{~m} / \\mathrm{min}$, então ela demorará $k$ minutos para dar uma volta completa no lago. Tendo isso em mente, se $k_{1}, k_{2}, \\ldots, k_{10}$ denotam os inteiros associados às velocidades das 10 crianças e $M$ é um múltiplo comum de todos el... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof only | null | |
0l61 | Problem:
Point $P$ lies inside square $ABCD$ such that the areas of $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, and $\triangle PDA$ are $1$, $2$, $3$, and $4$, in some order. Compute $PA \cdot PB \cdot PC \cdot PD$. | [
"Solution:\n\nLet $h_{1}$, $h_{2}$, $h_{3}$, and $h_{4}$ be the lengths of the altitudes from $P$ to sides $AB$, $BC$, $CD$, and $DA$, respectively. Then, the problem statement implies that $\\{h_{1}, h_{2}, h_{3}, h_{4}\\} = \\{x, 2x, 3x, 4x\\}$ for some $x$. Furthermore, the area of the square is $1 + 2 + 3 + 4 =... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 8√10 | |
0l69 | Problem:
In the coordinate plane, a closed lattice loop of length $2n$ is a sequence of lattice points $P_{0}, P_{1}, P_{2}, \ldots , P_{2n}$ such that $P_{0}$ and $P_{2n}$ are both the origin and $P_{i}P_{i + 1} = 1$ for each $i$. A closed lattice loop of length $2026$ is chosen uniformly at random from all such loops... | [
"Solution:\nWe claim that if $2026$ is replaced with $2n$, the answer is $\\frac{2n}{n + 1}$.\n\nWrite the path as a sequence of $U$, $D$, $L$, and $R$ moves. The possible sequences that can result are precisely those with an equal number of $U$'s and $D$'s, and an equal number of $R$'s and $L$'s. We first project ... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 1013/507 | |
0ie1 | Problem:
Find the largest positive integer $n$ such that $1+2+3+\cdots+n^{2}$ is divisible by $1+2+3+\cdots+n$. | [
"Solution:\nThe statement is\n$$\n\\frac{n(n+1)}{2} \\mid \\frac{n^{2}(n^{2}+1)}{2} \\Leftrightarrow n+1 \\mid n(n^{2}+1) = n^{3}+n.\n$$\nBut $n+1$ also divides $(n+1)(n^{2}-n+2) = n^{3}+n+2$, so $n+1$ must divide $2$. Hence, $n$ cannot be greater than $1$. And $n=1$ clearly works, so that is the answer."
] | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1 | |
0974 | Problem:
Fie șirul $\left(a_{n}\right)_{n=0}^{\infty}$ astfel încât $a_{1}=1$ și $a_{m+n}+a_{m-n}=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right), \quad \forall m \geq n \geq 0$. Să se determine $a_{2020}$. | [
"Solution:\nConsiderând $m=n$, avem $a_{2 m}+a_{0}=\\frac{1}{2}\\left(a_{2 m}+a_{2 m}\\right)=a_{2 m}, \\quad \\forall m \\geq 0$, ceea ce implică $a_{0}=0$.\n\nSimilar, considerând $n=0$, obținem $a_{m}+a_{m}=\\frac{1}{2}\\left(a_{2 m}+a_{0}\\right)=\\frac{1}{2} a_{2 m}$, adică $a_{2 m}=4 a_{m}, \\forall m \\geq 0... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2020^2 | |
09lp | A four-digit number $\overline{ABBA}$ is defined as a *good number* if it is divisible by the two-digit number $\overline{BB}$. Given that the digits $A$ and $B$ are allowed to be the same, determine the total number of *good numbers*. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 31 | |
08v0 | How many quadruples $(a, b, c, d)$ of positive integers lying in between $1$ and $9$ ($1$ and $9$ inclusive) are there which satisfy $0 < b - a < c - b < d - c$? | [
"Let $x = b - a$, $y = c - b$, $z = d - c$. Then, $x, y, z$ are integers satisfying $0 < x < y < z$. Furthermore, since $x + y + z = d - a \\le 9 - 1 = 8$, we see that the triple $(x, y, z)$ must be one of the following $4$ triples:\n$$\n(1, 2, 3),\\ (1, 2, 4),\\ (1, 2, 5),\\ (1, 3, 4).\n$$\nWhen $(x, y, z) = (1, 2... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 7 | |
0dvp | Problem:
Na voljo imamo 6 različnih barv in veliko kock. Posamezno kocko pobarvamo z vsemi 6 barvami, in sicer vsako mejno ploskev z 1 barvo. Največ koliko kock lahko pobarvamo, če naj bo vsaka izmed njih drugače pobarvana? (Če lahko 1 izmed pobarvanih kock zasučemo tako, da so barve mejnih ploskev enako razporejene k... | [
"Solution:\n\nDenimo, da imamo na voljo rdečo, modro, zeleno, rumeno, oranžno in belo barvo. Kocke postavimo tako, da imajo spodnjo mejno ploskev pobarvano z belo barvo. Za zgornjo ploskev imamo na voljo katero koli izmed preostalih 5 barv.\n\nSedaj se vprašamo, koliko kock ima lahko zgornjo ploskev pobarvano z nek... | Slovenia | 47. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 30 | |
05a3 | Do there exist integers $x$ and $y$ such that:
a. $x^2 + (x + 1)^2 + (x + 2)^2 = y^2$?
b. $x^2 + (x + 1)^2 + (x + 2)^2 + (x + 3)^2 = y^2$? | [
"Answer: (a) No; (b) No.\n\na. The integers $x$, $x+1$, $x+2$ are congruent to $0$, $1$, $2$ modulo $3$ in some order. The squares of these integers are congruent to $0$, $1$, $1$ modulo $3$, respectively. Hence the l.h.s. of the equation is congruent to $2$ while the r.h.s. is congruent to $0$ or $1$ modulo $3$. T... | Estonia | Estonian Math Competitions | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | (a) No; (b) No. | |
0bpc | Problem:
Punctul $P$ este în interiorul triunghiului $ABC$, iar dreptele $AP$, $BP$, $CP$ taie laturile $BC$, $AC$, $AB$ în $A_1$, $B_1$, respectiv $C_1$. Se ştie că
$$
s\left(PBA_1\right) + s\left(PCB_1\right) + s\left(PAC_1\right) = \frac{1}{2} s(ABC)
$$
unde cu $s(XYZ)$ s-a notat aria triunghiului $XYZ$. Arătaţi că... | [
"Solution:\n\nAvem $\\frac{s\\left(PBA_1\\right)}{s\\left(ABA_1\\right)} = \\frac{PA_1}{AA_1} = \\frac{s(BPC)}{s(BAC)}$.\n\nNotând $s(BPC) = s_a$ şi analoagele, rezultă $\\frac{s\\left(PBA_1\\right)}{s_c + s\\left(PBA_1\\right)} = \\frac{s_a}{s}$, unde $s = s(ABC)$, deci $s\\left(PBA_1\\right) = \\frac{s_a s_c}{s -... | Romania | Olimpiada Naţională de Matematică, Etapa Naţională | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
03eo | The inscribed circle in $\triangle ABC$ ($AC \neq BC$) is tangent to its sides $AB$, $BC$, and $CA$ at points $D$, $E$, and $F$, respectively. Let $P$ be the foot of the perpendicular from $D$ to $EF$ ($P \in EF$). If the circles circumscribed about $\triangle ABC$ and $\triangle EFC$ intersect for the second time at p... | [
"From the fact that $C$ lies on the circle circumscribed about $\\triangle FEQ$ and $CE = CF$ it follows that $CQ$ is an exterior bisector of $\\angle FQE$ and it remains to prove that $QP$ is a bisector of $\\angle FQE$, i.e. $QF : QE = FP : PE$. From $\\angle QFC = \\angle QEC$ and $\\angle QAC = \\angle QBC$ it ... | Bulgaria | Bulgarian Winter Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0dya | Problem:
Janez je porabil $\frac{3}{5}$ denarja, ki ga je imel, nato še $\frac{5}{9}$ ostanka ter še $\frac{3}{8}$ novega ostanka. Ostalo mu je 80 evrov. Koliko denarja je imel na začetku? | [
"Solution:\n\nNaj bo $x$ prvotni znesek. Izračunamo delež denarja, ki ga ima Janez po prvem izdatku, to je $\\frac{2}{5} x$. Nato izračunamo nov delež denarja, ki ga ima Janez po drugem izdatku, to je $\\frac{8}{45} x$. Po tretjem izdatku ima Janez še $\\frac{1}{9} x$ denarja. Nastavimo enačbo $\\frac{1}{9} x=80$. ... | Slovenia | Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 720 | |
01ic | $A$ is a 20-element set of residue classes modulo $397$. Prove that for any non-negative integer $n$ there exist $x_1, x_2, x_3, x_4 \in A$, such that $x_1 \neq x_2$ and
$$
(x_1 - x_2)n \equiv x_3 - x_4 \pmod{397}
$$ | [
"Observe that $397$ is a prime. If $n$ is divisible by $397$, the statement is trivial.\n\nFix an arbitrary positive integer $n$ not divisible by $397$. If for $x_1, x_2, x_3, x_4 \\in A$, where $(x_1, x_4)$ and $(x_2, x_3)$ are different ordered pairs, we have\n$$\nx_1 n + x_4 \\equiv x_2 n + x_3 \\pmod{397},\n$$\... | Baltic Way | Baltic Way 2021 Shortlist | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | null | proof only | null | |
0agd | Find all positive integers $n$ such that $n2^{n+1}+1$ is a perfect square. | [
"Clearly, $n2^{n+1}+1$ is odd, so, if this number is a perfect square then\n$n2^{n+1}+1=(2x+1)^2$, $x \\in \\mathbb{N}$, whence $n2^{n-1}=x(x+1)$.\n\nThe integers $x$ and $x+1$ are coprime,\nso one of them must be divisible by $2^{n-1}$,\nwhich means that the other must be at most\n$n$. This shows that $2^{n-1} \\l... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 3 | |
01yk | Two circles $\omega_1$ and $\omega_2$ intersect each other at points $X$ and $Y$. Two lines pass through $Y$: one intersects $\omega_1$ and $\omega_2$ for the second time at points $A$ and $B$ respectively and another intersects $\omega_1$ and $\omega_2$ for the second time at points $C$ and $D$ respectively. The line ... | [
"The quadrilaterals $ACYP$ and $BDQY$ are cyclic, so\n$$\n\\angle ACD = \\angle ACY = \\angle YPQ \\text{ and } \\angle ABD = \\angle YBD = \\angle YQP.\n$$\nSince the triangle $YPQ$ is isosceles, $\\angle YPQ = \\angle YQP$ whence $\\angle ACD = \\angle ABD$. So the quadrilateral $ACBD$ is cyclic. Therefore\n$$\n\... | Belarus | Belarus2022 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01he | We call an ordered pair $(x, y)$ of real numbers *auroral* if the equations $x = y(3 - y)^2$ and $y = x(3 - x)^2$ hold simultaneously.
Find all integers $k$ for which there exists an auroral pair $(x, y)$ of real numbers with $x + y = k$. | [
"The numbers we are looking for are $k \\in \\{0, 3, 4, 5, 8\\}$.\nA pair $(x, x) \\in \\mathbb{R}^2$ is auroral if and only if $x = x(3 - x)^2$, and it is easy to see that this cubic equation has the solution set $\\{0, 2, 4\\}$. These pairs give us 0, 4 and 8 as possible values for $k$.\n\nNext we investigate the... | Baltic Way | Baltic Way 2021 Shortlist | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | null | proof and answer | {0, 3, 4, 5, 8} | |
0hcz | Given a fixed natural number $n > 1$, 2019 natural numbers are placed around the circle in such a way that the product of any two neighboring numbers is a perfect $n$-th power. Is it always the case that the product of any (not necessarily neighboring) two numbers is also a perfect $n$-th power?
(A. Nikolaev, B. Rubly... | [
"For the odd $n = 2l + 1$ one can prove the statement as follows:\n$$\nP^2 = (a_1a_2)(a_2a_3)(a_3a_4)\\dots(a_{2018}a_{2019})(a_{2019}a_1) = m^{2l+1},\n$$\nwhere $P = a_1a_2\\dots a_{2019}$. From here it follows that $P = a_1a_2\\dots a_{2019} = s^{2l+1}$. Therefore,\n$$\na_1 = \\frac{P}{(a_2a_3)(a_4a_5)\\dots(a_{2... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Yes | |
0bmr | Two players play a game on a regular $n$-gon, where $n \ge 4$. Three counters start on consecutive vertices. In turn, each player slides one counter along any number of edges to another vertex of the polygon, without jumping over another counter, to strictly increase the area of the triangle formed by the counters. The... | [
"Since the game is identical for both players, has finitely many possible states and always terminates, we can label the possible states Wins or Losses according as whether a player faced with that position has a winning strategy or not. A state is a Win if and only if there is some legal move taking the state to a... | Romania | THE 2015 Seventh ROMANIAN MASTER OF MATHEMATICS | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | The first player wins if and only if the exponent of 2 in n − 3 is odd (i.e., e2(n−3) is odd). | |
0a0u | Each pupil in the Netherlands is given a finite number of cards. On each card, there is a real number in the interval $[0, 1]$. (The numbers on different cards do not have to be different.) Find the smallest real number $c > 0$ for which the following holds, independent of the numbers on the cards each person has been ... | [
"Suppose one of the pupils has been given $1001$ cards, each containing the number $\\frac{1000}{1001}$. Since the sum of the cards is $1000$, this pupil should be able to distribute the cards among the $100$ boxes. Because of the pigeonhole principle, there is at least one box with $11$ cards. The sum of these $11... | Netherlands | IMO Team Selection Test 3 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 11 - 1/91 | |
0d97 | Fibonacci sequences is defined as
$$
\left\{\begin{array}{l}
f_{1}=1,\ f_{2}=2 \\
f_{n+1}=f_{n}+f_{n-1},\ n \geq 2
\end{array}\right.
$$
1. Prove that every positive integer can be represented as sum of several distinct Fibonacci number.
2. A positive integer is called Fib-unique if the way to represent it as sum of se... | [
"1. We will prove by induction that every positive integer $k$ can be written as sum of several distinct Fibonacci numbers.\nThe case $k=1$ is obvious, we have $1=f_{1}$.\nAssume that every positive integer less than $n$ can be written as sum of several distinct Fibonacci numbers, consider $k=n$.\nWe can find $f_{i... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Fib-unique numbers are f1, f2, and f_k − 1 for k ≥ 3. | |
01tq | Given a prime number $p$ such that the number $2p$ is equal to the sum of the squares of some four consecutive positive integers. Prove that $p-7$ is divisible by 36. | [
"Let $n-1$, $n$, $n+1$, and $n+2$ denote four consecutive positive integers with the sum of the squares equaled $2p$. Then\n$$\n2p = (n-1)^2 + n^2 + (n+1)^2 + (n+2)^2 = 4n^2 + 4n + 6,\n$$\nwhence $p = 2n^2 + 2n + 3 = 2n(n+1) + 3$. Now if one of the numbers $n$ and $n+1$ is divisible by $3$, then $p$ is divisible by... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0g9n | 定義函數 $f : (0,1) \to (0,1)$ 如下
$$
f(x) = \begin{cases} x + \frac{1}{2} & \text{若 } x < \frac{1}{2}, \\ x^2 & \text{若 } x \ge \frac{1}{2}. \end{cases}
$$
令 $a, b$ 為兩實數且 $0 < a < b < 1$. 定義兩數列如下:
$a_0 = a, b_0 = b, \text{且 } a_n = f(a_{n-1}), b_n = f(b_{n-1}), n = 1, 2, \dots$
試證: 存在一正整數 $n$ 使得
$(a_n - a_{n-1})(b_n - b_{n... | [
"設存在 $0 < a < b < 1$ 使不存在正整數 $n$ 讓 $(a_n - a_{n+1})(b_n - b_{n+1}) < 0$.\n$$\nf(x) - x \\begin{cases} > 0 & \\text{若 } x < \\frac{1}{2}, \\\\ < 0 & \\text{若 } x \\ge \\frac{1}{2}. \\end{cases}\n$$\n則對給定的正整數 $n$, 必有 $a_n, b_n \\in (0, \\frac{1}{2})$ 或 $a_n, b_n \\in [\\frac{1}{2}, 1)$.\n設 $d_n = b_n - a_n$, 則若 $a_n,... | Taiwan | 二〇一五數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0a03 | Problem:
Aan het plafond van een kamer hangen 15 lampen, genummerd van 1 tot en met 15. In het begin zijn de lampen allemaal uit. In een andere kamer zijn 15 schakelaars: een schakelaar voor lamp 1 en 2, een schakelaar voor lamp 2 en 3, een schakelaar voor lamp 3 en 4, enzovoorts, tot en met een schakelaar voor lamp 1... | [
"Solution:\n\nMet drie keer lopen kan Raymond het nooit met zekerheid weten. Immers, als je voor elke schakelaar noteert of hij in de drie rondes in de beginstand of juist de omgezette stand staat, dan kun je $2^{3}=8$ verschillende patronen krijgen. Er zijn echter 15 schakelaars, dus er zijn meerdere schakelaars m... | Netherlands | IMO-selectietoets II | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | 4 | |
0k0j | Problem:
Jeffrey writes the numbers $1$ and $100000000=10^{8}$ on the blackboard. Every minute, if $x, y$ are on the board, Jeffrey replaces them with
$$
\frac{x+y}{2} \text{ and } 2\left(\frac{1}{x}+\frac{1}{y}\right)^{-1}
$$
After $2017$ minutes the two numbers are $a$ and $b$. Find $\min (a, b)$ to the nearest int... | [
"Solution:\n\nNote that the product of the integers on the board is a constant. Indeed, we have that\n$$\n\\frac{x+y}{2} \\cdot 2\\left(\\frac{1}{x}+\\frac{1}{y}\\right)^{-1}=x y\n$$\nTherefore, we expect that the answer to the problem is approximately $\\sqrt{1 \\cdot 10^{8}}=10^{4}$.\nTo be more rigorous, we have... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | final answer only | 10000 | |
0a5k | Problem:
Find all integers $a, b$ such that
$$a^{2} + b = b^{2022}.$$ | [
"Solution:\nLet $g = \\gcd (b, b^{2021} - 1)$. Since we cannot have both $b$ and $(b^{2021} - 1)$ being zero, $g$ must be a positive integer. Since $g \\mid b$ we must have $g \\mid b^{2021}$ and therefore\n$$g \\mid (b^{20221}) - (b^{2021} - 1) = 1.$$ \nHence $g = 1$ and thus $b$ and $(b^{2021} - 1)$ are coprime.... | New Zealand | New Zealand Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (a, b) = (0, 0) and (0, 1) | |
0g6z | 令 $ABCD$ 為一凸四邊形, 其中邊 $AD$ 與 $BC$ 不平行。假設以 $AB$ 和 $CD$ 為直徑的兩圓交於四邊形 $ABCD$ 內部的兩點 $E, F$。由 $E$ 分別對直線 $AB, BC$ 和 $CD$ 作垂線得三個垂足 $Q, R$ 與 $S$。令圓 $\omega_E$ 表示通過 $Q, R$ 與 $S$ 三點之圓。同理, 圓 $\omega_F$ 表示通過由 $F$ 分別對直線 $CD, DA$ 和 $AB$ 三邊作垂線所得三個垂足的圓。試證: 線段 $EF$ 之中點落在兩圓 $\omega_E$ 和 $\omega_F$ 的兩個交點之連線上。 | [
"(1) 令 $P, Q, R, S$ 分別為 $E$ 對 $DA, AB, BC, CD$ 的垂足。注意到 $P, Q, A, E$ 四點共圓，故 $\\angle QPE = \\angle QAE$。同理，$\\angle QRE = \\angle QBE$。\n因此，$\\angle QPE + \\angle QRE = \\angle QAE + \\angle QBE = 90^\\circ$ (因為 $E$ 在以 $AB$ 為直徑的圓上。) 同理，$\\angle SPE = \\angle SRE = 90^\\circ$，因此 $\\angle QPS + \\angle QRS = 90^\\circ... | Taiwan | 二〇一二數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0drx | A group of tourists get on 10 buses in the outgoing trip. The same group of tourists get on 8 buses in the return trip. Assuming each bus carries at least 1 tourist, prove that there are at least 3 tourists such that each of them has taken a bus in the return trip that has more people than the bus he has taken in the o... | [
"We prove the general case where there are $p$ outgoing buses and $q$ returning buses with $p > q$. The number of such tourists is $p - q + 1$. Let $T$ be the set of all tourists. For each $t \\in T$, let $p_t$ be the number of tourists in the bus that $t$ takes in the outgoing trip and let $q_t$ be the number of t... | Singapore | Singapur | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | at least 3 | |
06uf | Let $A B C D E$ be a convex pentagon such that $A B = B C = C D$, $\angle E A B = \angle B C D$, and $\angle E D C = \angle C B A$. Prove that the perpendicular line from $E$ to $B C$ and the line segments $A C$ and $B D$ are concurrent. | [
"Throughout the solution, we refer to $\\angle A, \\angle B, \\angle C, \\angle D$, and $\\angle E$ as internal angles of the pentagon $A B C D E$. Let the perpendicular bisectors of $A C$ and $B D$, which pass respectively through $B$ and $C$, meet at point $I$. Then $B D \\perp C I$ and, similarly, $A C \\perp B ... | IMO | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0elv | An equilateral triangle and a regular hexagon have equal perimeters. What is the ratio of the area of the triangle to the area of the hexagon? | [
"Let the perimeter be $6k$. The equilateral triangle will have side length $2k$, and will have an area four times the area of a triangle with side length $k$.\n\nThe hexagon will have side length $k$. By joining the vertices of the hexagon to the centre, it can be divided exactly into six equilateral triangles of s... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2:3 | |
06gk | For any positive integer $n$, let $a_1, a_2, \dots, a_m$ be all the positive divisors of $n$, where $m \ge 1$. If there exist $m$ integers $b_1, b_2, \dots, b_m$ such that $n = \sum_{i=1}^{m} (-1)^{b_i} a_i$, then we say that $n$ is a *good number*. Prove that there exists a good number with exactly 2013 distinct prime... | [
"We say that a positive integer $n$ is *excellent* if it is good, and we can find integers $c_1, c_2, \\dots, c_m$ such that $\\sum_{j=1}^{m} (-1)^{c_j} a_j = 0$. Let $p_1, p_2, \\dots, p_k$ be distinct primes larger than 3. We prove by induction on $k$ that the integer $6p_1p_2\\cdots p_k$ is excellent.\n\nFor the... | Hong Kong | CHKMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
071j | Problem:
The in-circle of triangle $ABC$ touches the sides $BC$, $CA$ and $AB$ in $K$, $L$ and $M$ respectively. The line through $A$ and parallel to $LK$ meets $MK$ in $P$ and the line through $A$ and parallel to $MK$ meets $LK$ in $Q$. Show that the line $PQ$ bisects the sides $AB$ and $AC$ of triangle $ABC$. | [
"Solution:\n\nLet $AP$, $AQ$ produced meet $BC$ in $D$, $E$ respectively.\n\n\n\nSince $MK$ is parallel to $AE$, we have $\\angle AEK = \\angle MKB$. Since $BK = BM$, both being tangents to the circle from $B$, $\\angle MKB = \\angle BMK$. This with the fact that $MK$ is parallel to $AE$ gi... | India | INMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
08c4 | Problem:
Sia $n \geq 2$ un numero intero. Consideriamo le soluzioni $(a, b, c)$ del sistema di equazioni
$$
\left\{\begin{array}{l}
n = a + b - c \\
n = a^{2} + b^{2} - c^{2}
\end{array}\right.
$$
dove $a, b, c$ sono numeri interi. Dimostrare che c'è almeno una soluzione e che ci sono un numero finito di soluzioni. | [
"Solution:\n\nDalla prima equazione ricaviamo $c = a + b - n$; sostituendo nella seconda equazione otteniamo\n$$\nn = a^{2} + b^{2} - (a + b - n)^{2} = -2 a b + 2 a n + 2 b n - n^{2}.\n$$\nRiarrangiando, si ottiene\n$$\n(a - n)(b - n) = a b - n a - n b + n^{2} = \\frac{n^{2} - n}{2} = \\frac{n(n-1)}{2}\n$$\nIl memb... | Italy | XXXIII Olimpiade Italiana di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof only | null | |
0532 | Publisher Soothsayer published a reference book claiming that for each real number $x$ and positive even number $n$ the equality $(1+x)^n \ge 2^n x$ holds. Is this claim true? | [
"*Answer:* No.\n\nThe inequality does not hold for example when $x = \\frac{1}{2}$ and $n = 4$."
] | Estonia | Open Contests | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | No | |
00vw | Let $k$ be a positive integer and let $n_1, n_2, \dots, n_k > 1$ be $k$ distinct integers such that
$$
n_1! + \dots + n_k! = a^{n_1}
$$
for some positive integer $a$. Prove that at least one of $n_1, n_2, \dots, n_k$ is prime. | [
"If $k = 1$, then $n_1! = a^{n_1}$. Since $n_1 > 1$, then $2|n_1!$ and so $2|a^{n_1}$ and therefore $2|a$. Then\n$$\nn_1 \\le v_2(a^{n_1}) = v_2(n_1!) = \\left\\lfloor \\frac{n_1}{2} \\right\\rfloor + \\left\\lfloor \\frac{n_1}{2^2} \\right\\rfloor + \\dots < \\frac{n_1}{2} + \\frac{n_1}{2^2} + \\dots = n_1,\n$$\na... | Balkan Mathematical Olympiad | 42nd Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0kiy | Problem:
Determine the number of convex polygons all of whose sides are the square roots of positive integers which can be inscribed in a unit circle. Polygons that are rotations or reflections of each other are considered the same. | [
"Solution:\n\nAny chord in a unit circle has length at most $2$ (if it is a diameter), so all side lengths must be $1$, $\\sqrt{2}$, $\\sqrt{3}$, or $2$. We can do casework based on the number of sides:\n\n- Triangles: The possible combinations of side lengths are $(\\sqrt{2}, \\sqrt{2}, 2)$, $(1,1, \\sqrt{3})$, $(... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 14 | |
0b9c | Let $\gamma_1$ and $\gamma_2$ be two circles tangent at point $T$, and let $\ell_1$ and $\ell_2$ be two lines through $T$. The lines $\ell_1$ and $\ell_2$ meet again $\gamma_1$ at points $A$ and $B$, respectively, and $\gamma_2$ at points $A_1$ and $B_1$, respectively. Let further $X$ be a point in the complement of $\... | [
"Let the circle $ATX$ and the line $A_2B_1$ meet again at the point $Y$, and let the circle $BTX$ and the line $A_1B_2$ meet again at the point $Z$. Notice that the lines $AB, AY$ and $AZ$ are all parallel to the line $A_1B_1$ to deduce that the points $A, B, Y$ and $Z$ are collinear.\n\nNext, consider the cyclic q... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cy... | English | proof only | null | |
0cq7 | Find all real numbers $a$ such that for every positive integer $n$, the number $a n(n+2)(n+4)$ is also integer. (O. Podlipsky)
Найдите все числа $a$ такие, что для любого натурального $n$ число $a n(n + 2)(n + 4)$ будет целым. (О. Подлипский) | [
"Ответ. $a = \\frac{k}{3}$, где $k$ — любое целое число.\n\nПодставив $n = 1$ и $n = 2$, получаем, что числа $15a$ и $48a$ — целые. Значит, и число $48a - 3 \\cdot 15a = 3a$ — тоже целое. Таким образом, $a = k/3$ для некоторого целого $k$.\n\nОсталось показать, что все числа такого вида подходят. Действительно, одн... | Russia | Russian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English, Russian | proof and answer | a = k/3 for any integer k | |
0c8b | Let $ABC$ be a triangle with a right angle at $C$, let $I$ be its incenter, and let $H$ be the orthogonal projection of $C$ on $AB$. The incircle $\omega$ of the triangle $ABC$ is tangent to the sides $BC$, $CA$, and $AB$ at $A_1$, $B_1$, and $C_1$, respectively. Let $E$ and $F$ be the reflections of $C$ in the lines $... | [
"The line $C_1A_1$ is parallel to the external angle bisector of $\\angle B$, so the reflection in $C_1A_1$ maps the segment $A_1C$ to the segment $A_1E$ parallel to $AB$.\n\nSimilarly, $B_1F \\parallel AB$. Notice also that $A_1E = A_1C = B_1C = B_1F = r$, where $r$ is the inradius of $\\triangle ABC$.\nLet $M$ be... | Romania | Romanian Master of Mathematics | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0egq | Problem:
Kot $\alpha$ je za $33^\circ$ manjši od svojega komplementarnega kota. Koliko je velik komplementarni kot?
(A) $29^\circ 50^\prime$
(B) $61^\circ 50^\prime$
(C) $61,5^\circ$
(D) $29,5^\circ$
(E) $61,30^\circ$ | [
"Solution:\nZapis obeh enačb $\\alpha = \\beta - 33^\\circ$ in $\\alpha + \\beta = 90^\\circ$. Z reševanjem sistema dobimo, da je $\\beta = 61,5^\\circ$."
] | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
0l1f | Problem:
A right rectangular prism of silly powder has dimensions $20 \times 24 \times 25$. Jerry the wizard applies 10 bouts of highdroxylation to the box, each of which increases one dimension of the silly powder by 1 and decreases a different dimension of the silly powder by 1, with every possible choice of dimensio... | [
"Solution:\nConsider the expected change in volume by one bout of highdroxylation. Let the initial dimensions of the silly powder be $a$, $b$, and $c$.\n\nShould $a$ increase to $a+1$ and $b$ decrease to $b-1$, the volume of the silly powder will change from $a b c$ to\n\n$$\n(a+1)(b-1)c = a b c + b c - a c - c\n$$... | United States | HMMT November 2024 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 11770 | |
0670 | Let $c(O,R)$ be a circle, $AB$ a diameter and $\Gamma$ the midpoint of the arch $\widehat{AB}$. We draw the circle $c_I(K,KO)$, where $K$ is a point of the segment $OA$ and we consider the tangents $\Gamma\Delta$, $\Gamma O$ from $\Gamma$ to the circle $c_I(K,KO)$. The line $K\Delta$ intersects the circle $c(O,R)$ at t... | [
"Let $\\hat{E}_i = \\Gamma \\hat{E} Z$. Then from the orthogonal triangle $\\Gamma \\Delta E : \\hat{\\Gamma}_1 = 90^\\circ - \\hat{E}_1$ (1)\nSince $\\Gamma \\hat{O} Z = \\hat{O}_1$ is the corresponding central angle of $\\hat{E}_1 = \\Gamma \\hat{E} Z$, then\n$\\hat{O}_1 = 2\\hat{E}_1$ and from the isosceles tria... | Greece | Selection Examination A | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ftn | Problem:
Sei $S$ die Menge aller $n$-Tupel $(X_{1}, \ldots, X_{n})$, wobei $X_{1}, \ldots, X_{n}$ Teilmengen von $\{1,2, \ldots, 1000\}$ sind, die nicht alle verschieden sein müssen, und die auch leer sein können. Für $a=(X_{1}, \ldots, X_{n}) \in S$ bezeichne
$$
E(a)=\text{Anzahl Elemente von } X_{1} \cup \ldots \cup... | [
"Solution:\n\nWir zählen die Anzahl Paare $(a, x)$ auf zwei Arten, wobei $a=(X_{1}, \\ldots, X_{n}) \\in S$ ein $n$-Tupel ist und $x \\in X_{1} \\cup \\ldots \\cup X_{n}$.\n\na. Für ein festes $a \\in S$ gibt es per Definition genau $E(a)$ solche Elemente $x$. Daher ist die gesuchte Anzahl gerade\n$$\n\\sum_{a \\in... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 1000(2^{n}-1)2^{999 n} | |
0ema | There are 10 fencers taking part in a tournament. Each fencer duels with each other exactly once. Is it possible that such a moment exists when no two fencers have participated in the same number of duels? | [
"Imagine the fencers as nodes in a graph. There is an edge between two nodes if those fencers have duelled. If everyone is to have duelled a different number of opponents, then since there are only 10 fencers, they must have each duelled a distinct number from $\\{0, 1, 2, \\ldots, 8, 9\\}$. In any graph, the sum o... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | No | |
06oi | Determine the smallest number $M$ such that the inequality
$$
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leq M\left(a^{2}+b^{2}+c^{2}\right)^{2}
$$
holds for all real numbers $a, b, c$. | [
"We first consider the cubic polynomial\n$$\nP(t)=t b\\left(t^{2}-b^{2}\\right)+b c\\left(b^{2}-c^{2}\\right)+c t\\left(c^{2}-t^{2}\\right)\n$$\nIt is easy to check that $P(b)=P(c)=P(-b-c)=0$, and therefore\n$$\nP(t)=(b-c)(t-b)(t-c)(t+b+c)\n$$\nsince the cubic coefficient is $b-c$. The left-hand side of the propose... | IMO | IMO 2006 Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 9*sqrt(2)/32 | |
02w8 | Problem:
Os ângulos internos de um quadrilátero não convexo formado por uma linha poligonal que não se auto-intersecta são $45^{\circ}$, $45^{\circ}$, $45^{\circ}$ e $225^{\circ}$. Mostre que os pontos médios de seus lados formam um quadrado.
 | [
"Solution:\n\nSejam $A B C D$ o quadrilátero não convexo, com $\\angle A D C=225^{\\circ}$ e $E, F, G$ e $H$ os pontos médios dos lados $A B, A D, C D$ e $B C$, respectivamente. Além disso, sejam $X$ e $Y$ os pontos de interseção de $C D$ e $B D$ com $A B$ e $A C$, respectivamente.\n\n\n\nC... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
025q | Problem:
Um rei está em um dos cantos de um tabuleiro $m \times n$. Dois jogadores movem o rei alternadamente para qualquer casa ainda não visitada. O primeiro jogador que não puder mais mover o rei perde. Determine, em função das dimensões do tabuleiro, quem possui a estratégia vencedora.
Observação: No xadrez, o re... | [
"Solution:\n\nSejam $A$ o primeiro e $B$ o segundo jogador. Dizemos que um jogador possui uma estatégia vencedora se ele pode sempre garantir a vitória independentemente de como o outro jogador faça seus movimentos. Vamos provar que:\n(i) se $m n$ é par, então $A$ possui a estratégia vencedora;\n(ii) se $m n$ é ímp... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | If the total number of squares is even, the first player has a winning strategy; if it is odd, the second player has a winning strategy. | |
0clr | Let $\mathbb{Z}$ denote the set of integers and let $S \subset \mathbb{Z}$ be the set of integers that are at least $10^{100}$. Fix a positive integer $c$. Determine all functions $f: S \to \mathbb{Z}$ satisfying
$$
f(xy + c) = f(x) + f(y) \quad \text{for all } x, y \in S.
$$ | [
"Observe that if $x_1, y_1, x_2, y_2 \\in S$ with $x_1 y_1 = x_2 y_2$ then\n$$\nf(x_1) + f(y_1) = f(x_2) + f(y_2). \\quad (1)\n$$\n\nThis tells us that for $u, v, w \\in S$,\n$f(uv) + f(w) = f(u) + f(vw)$, so $f(uv) - f(u) - f(v) = f(vw) - f(w) - f(v)$.\nNotice the RHS is independent of $u$ so the same must be true... | Romania | Seventeenth ROMANIAN MASTER OF MATHEMATICS | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | f(x) = 0 for all x in S | |
0e7i | Problem:
Poišči vsa praštevila $p$ in $q$, za katera je $p^{4}-q^{6}$ potenca praštevila. (Števili 7 in 8 sta potenci praštevila, 6 pa ne.) | [
"Solution:\n\nPišimo $p^{4}-q^{6}=r^{n}$ za neko praštevilo $r$ in naravno število $n$. Izraz $p^{4}-q^{6}$ najprej razstavimo kot $p^{4}-q^{6}=(p^{2}-q^{3})(p^{2}+q^{3})$. Ker je $p^{4}-q^{6}>0$, praštevili $p$ in $q$ ne moreta biti enaki, torej sta si tuji. Naj bo $d$ največji skupni delitelj števil $p^{2}-q^{3}$... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | p=3, q=2 | |
071s | Problem:
Given any nine integers show that it is possible to choose, from among them, four integers $a, b, c, d$ such that $a+b-c-d$ is divisible by $20$. Further show that such a selection is not possible if we start with eight integers instead of nine. | [
"Solution:\nSuppose there are four numbers $a, b, c, d$ among the given nine numbers which leave the same remainder modulo $20$. Then $a+b \\equiv c+d \\pmod{20}$ and we are done.\n\nIf not, there are two possibilities:\n\n(1) We may have two disjoint pairs $\\{a, c\\}$ and $\\{b, d\\}$ obtained from the given nine... | India | INMO | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0l0s | Problem:
Two points are chosen independently and uniformly at random from the interior of the X-pentomino shown below. Compute the probability that the line segment between these two points lies entirely within the X-pentomino.
Proposed by: Benjamin Shimabukuro
Answer: $\frac{21}{25}=0.84$ | [
"Solution:\nHere are the cases in which the line segment is guaranteed to lie entirely within the X-pentomino:\n- At least one point lies in center square.\n- The two points lie in the same square.\n- The two points are in opposite edge squares (north and south, or east and west).\n\nBy symmetry, the segment will l... | United States | HMMT November 2024 | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 21/25 | |
008c | Determine if there are triples $(x, y, z)$ of real numbers such that:
$$
x + y + z = 7 \text{ and } xy + yz + zx = 11.
$$
If the answer is yes, find the minimum and maximum value of $z$ in such triples. | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | Yes; z ranges from −1/3 to 5, so min z = −1/3 and max z = 5. | |
0c01 | Let $n$ be a natural number with $n \ge 2$, and let $a_1, a_2, \dots, a_n \in (1, \infty)$ be $n$ real numbers. Prove that the function $f: [0, \infty) \to \mathbb{R}$, defined by
$$
f(x) = (a_1 a_2 \dots a_n)^x - a_1^x - a_2^x - \dots - a_n^x,
$$
for any $x \in [0, \infty)$, is strictly increasing. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Precalculus > Functions"
] | null | proof only | null | |
01n9 | Point $M$ is marked inside the convex quadrilateral $ABCD$ so that the ratio of the areas of the triangles $AMC$ and $BMD$ is equal to the ratio of the tangents of the angles $AMC$ and $BMD$, i.e. $S(AMC) : S(BMD) = \tg \angle AMC : \tg \angle BMD$.
Prove that $AM^2 + MC^2 + BD^2 = AC^2 + BM^2 + MD^2$ if $M$ does not b... | [
"Since\n$$\nS(AMC) = 0.5 AM \\cdot MC \\sin \\angle AMC, \\quad S(BMD) = 0.5 BM \\cdot MD \\sin \\angle BMD,\n$$\nwe have\n$$\nAM \\cdot MC = 2S(AMC)/\\sin \\angle AMC, \\quad BM \\cdot MD = 2S(BMD)/\\sin \\angle BMD. \\quad (1)\n$$\n\n\n\nBy the cosine law,\n$$\nAC^2 = AM^2 + MC^2 - 2AM \\... | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
01sj | Find all positive integers $n$ ($n < 24$) such that the set $M = \{n, n + 1, \dots, 24\}$ can be partitioned into some disjoint subsets so that in each subset one of the numbers is equal to the sum of all other numbers of this subset.
(V. Kaskevich) | [
"A number $a$ from a subset of the desired partition is called major if $a$ is equal to the sum of all other numbers of this subset. All numbers in each subset must be distinct, since one of them is major we see that there exist at least three numbers in each subset. Let $k$ be the number of the subsets of the desi... | Belarus | FINAL ROUND | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | n = 1 and n = 4 | |
007v | Let $a$, $b$, $c$, $d$ be distinct positive integers such that $a^b$ divides $b^c$, $b^c$ divides $c^d$ and $c^d$ divides $d^a$.
a) Is it possible to determine which is the least one of the numbers $a$, $b$, $c$, $d$?
b) Is it possible to determine which is the greatest one of the numbers $a$, $b$, $c$, $d$? | [
"a) The answer is yes, even under the weaker assumptions $a^b \\le b^c$, $b^c \\le c^d$, $c^d \\le d^a$. The least number is $b$.\n\nWe need the inequality $\\sqrt[n]{n} > \\sqrt[n+1]{n+1}$ which holds for all $n \\ge 3$.\nIt is equivalent to $(1+\\frac{1}{n})^n < n$ and can be obtained by standard induction. The b... | Argentina | Mathematical Olympiad Rioplatense | [
"Number Theory > Divisibility / Factorization",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | a) Yes. The least is b. b) No. The greatest cannot be determined. | |
05p1 | Problem:
Soient $a$, $b$ et $c$ trois nombres strictement positifs tels que $a + b + c + abc = 4$. Montrer que
$$
\left(1 + \frac{a}{b} + ca\right)\left(1 + \frac{b}{c} + ab\right)\left(1 + \frac{c}{a} + bc\right) \geqslant 27
$$ | [
"Solution:\n\nEn utilisant l'inégalité arithmético-géométrique, on trouve que\n$$\n4 = a + b + c + abc \\geqslant 4\\left((abc)^2\\right)^{\\frac{1}{4}}\n$$\net donc $abc \\leqslant 1$. Ensuite, en utilisant encore l'inégalité arithmético-géométrique,\n$$\n\\left(1 + \\frac{a}{b} + ca\\right)\\left(1 + \\frac{b}{c}... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0awl | Problem:
Solve the inequality $\log_{7}\left(\frac{1}{x^{2}-3}\right) \geq 0$. | [
"Solution:\nWe need all values of $x$ such that $x^{2}-3>0$ and $\\frac{1}{x^{2}-3} \\geq 1$.\n\nThe first has solution $(-\\infty,-\\sqrt{3}) \\cup (\\sqrt{3}, \\infty)$, while the second has solution $[-2,2]$.\n\nThe intersection of these two sets is $[-2,-\\sqrt{3}) \\cup (\\sqrt{3}, 2]$."
] | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | [-2, -sqrt(3)) ∪ (sqrt(3), 2] | |
0b9q | Let $\alpha$ be an irrational number. For any $n \in \mathbb{N}^*$ let $a_n = \{n\alpha\}$ and define the sequence $(x_n)_{n \ge 1}$ by $x_n = (a_2 - a_1)(a_3 - a_2) \cdots (a_{n+1} - a_n)$. Show that the sequence is convergent and find its limit. | [
"We claim that $\\{x + y\\} - \\{y\\}$ is equal to $\\{x\\}$ or $\\{x\\} - 1$, for any real numbers $x$ and $y$. Write $x + y = [y] + [x] + \\{x\\} + \\{y\\}$ to get $\\{x + y\\} - \\{y\\} = \\{\\{x\\} + \\{y\\}\\} - \\{y\\}$. Hence if $\\{x\\} + \\{y\\} < 1$ then $\\{x + y\\} - \\{y\\} = \\{x\\}$, while if $2 > \\... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 0 | |
051g | For a given positive integer $n$ one has to choose positive integers $a_0, a_1, \dots$ so that the following conditions hold:
(1) $a_i = a_{i+n}$ for any $i$;
(2) $a_i$ is not divisible by $n$ for any $i$;
(3) $a_{i+a_i}$ is divisible by $a_i$ for any $i$.
For which positive integers $n > 1$ is this possible only if t... | [
"Let $n$ be a prime. By condition (1) the sequence $a_0, a_1, \\dots$ contains only finitely many different numbers. If $a_m$ is maximal of them, then by condition (3) $a_{m+a_m}$ must also be maximal. Let us prove that if $a_m$ is maximal of the numbers, then $a_{m+k \\cdot a_m}$ is also maximal for any $k \\ge 0$... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | n is prime | |
016q | Find all polynomials $P(x)$ with real coefficients such that
$$
(x - 2010)P(x + 67) = xP(x)
$$
for every integer $x$. | [
"Taking $x = 0$ in the given equality leads to $-2010P(67) = 0$, implying $P(67) = 0$. Whenever $i$ is an integer such that $1 \\le i < 30$ and $P(i \\cdot 67) = 0$, taking $x = i \\cdot 67$ leads to $(i \\cdot 67 - 2010)P((i+1) \\cdot 67) = 0$; as $i \\cdot 67 < 2010$ for $i < 30$, this implies $P((i+1) \\cdot 67)... | Baltic Way | BALTIC WAY | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | P(x) = c · (x − 67)(x − 2·67)···(x − 30·67), where c is any real constant | |
08u5 | For a positive integer $n$, denote by $S(n)$ the sum of the digits of $n$. If $S(n) = 5$, what is the maximum possible value that $S(n^5)$ can take? | [
"First, let us prove the following lemma.\n\n**Lemma:** For any set of positive integers $m_1, m_2, \\dots, m_k$, we have the inequality\n$$\nS\\left(\\sum_{i=0}^{k} m_i\\right) \\le \\sum_{i=0}^{k} S(m_i).\n$$\nFurthermore, equality holds in the inequality above if the following condition is satisfied:\nFor any no... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 398 | |
0jq1 | Problem:
There are three boxes of stones. Each hour, Sisyphus moves a stone from one box to another. For each transfer of a stone, he receives from Zeus a number of coins equal to the number of stones in the box from which the stone is drawn minus the number of stones in the recipient box, with the stone Sisyphus just... | [
"Solution:\n\nThe myth of Sisyphus suggests the answer: $0$.\nLet $x$ be the amount of money Sisyphus has and $a$, $b$, $c$ the sizes of the boxes. The key observation is that the quantity\n$$\nN = 2x + a^2 + b^2 + c^2\n$$\ndoes not change; for example, after one operation from the first box to the second the quant... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 0 | |
0lbu | Let $M, N$ be two positive integers. We put a light on each unit square of the table $N \times N$ (each light is either on or off). Initially, all the lights are off. In each step, we choose either a column or a row and switch $M$ consecutive lights on the chosen column/row. Find the necessary and sufficient condition ... | [
"We show that the necessary and sufficient condition of the problem is $N$ is divisible by $M$.\n\nThe sufficient condition is easy. If $N$ is divisible by $M$ then $N/M = d$ is a positive integer. In this case, we can divide the $N \\times N$ table into $Nd$ rectangles of size $M \\times 1$. We can easily turn on ... | Vietnam | Vietnamese Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | N is divisible by M | |
073m | Problem:
Find all triples $(p, x, y)$ such that $p^{x} = y^{4} + 4$, where $p$ is a prime and $x, y$ are natural numbers. | [
"Solution:\nWe begin with the standard factorisation\n$$\ny^{4}+4 = \\left(y^{2}-2y+2\\right)\\left(y^{2}+2y+2\\right)\n$$\nThus we have $y^{2}-2y+2 = p^{m}$ and $y^{2}+2y+2 = p^{n}$ for some positive integers $m$ and $n$ such that $m+n = x$. Since $y^{2}-2y+2 < y^{2}+2y+2$, we have $m < n$ so that $p^{m}$ divides ... | India | INMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (5, 1, 1) | |
0k3c | Problem:
A positive integer is called primer if it has a prime number of distinct prime factors. A positive integer is called primest if it has a primer number of distinct primer factors. A positive integer is called prime-minister if it has a primest number of distinct primest factors. Let $N$ be the smallest prime-m... | [
"Solution:\n\nThe answer is $2^{4} \\cdot 3^{3} \\cdot 5^{3} \\cdot 7 = 378000$.\n\nOne heuristic for estimating the answer is that numbers of the form $p^{q} r^{s}$ for primes $p, q, r, s$ with $p \\neq r, q \\neq s$ are primest. Thus, primest numbers are not very rare, so we can expect the answer to be relatively... | United States | HMMT November 2018 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Algorithms"
] | null | final answer only | 378000 | |
09nf | Let $m$ and $n$ be positive integers. Consider an $m \times n$ grid in a standard rectangular coordinate system. A segment is called *good* if it is parallel to a side of the grid. We partition the grid into triangles with vertices at integer coordinates such that each triangle has at least one side that is good, and t... | [
"Without loss of generality, we can assume that $1 \\leq m \\leq n$. We define an \"excellent\" triangle in the partition as one that has exactly two good sides. An excellent triangle thus has two sides of length 1 and forms a right triangle. When $n \\geq 2$, a partition of the $2 \\times n$ grid contains no excel... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Minimum equals 0 if both dimensions are at least two and their product is even; otherwise the minimum equals 2. | |
0gro | Let $a_1, a_2, \dots, a_{1000}$ be a sequence of integers such that $a_1 = 3$, $a_2 = 7$ and for all $n = 2, 3, \dots, 999$
$$
a_{n+1} - a_n = 4(a_1 + a_2)(a_2 + a_3) \cdots (a_{n-1} + a_n).
$$
Find the number of indices $1 \le n \le 1000$ for which $a_n + 2018$ is a perfect square. | [
"For $n \\ge 2$, we have\n$$\n\\frac{a_{n+1} - a_n}{a_n - a_{n-1}} = \\frac{4(a_1 + a_2)(a_2 + a_3) \\cdots (a_{n-1} + a_n)}{4(a_1 + a_2)(a_2 + a_3) \\cdots (a_{n-2} + a_{n-1})} = a_{n-1} + a_n\n$$\nand hence $a_{n+1} = a_n^2 + a_n - a_{n-1}^2$. $a_2 = a_1^2 - 2$ and by induction assuming $a_n = a_{n-1}^2 - 2$ we c... | Turkey | Team Selection Test | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 2 | |
01ao | $$
R(k) = \underbrace{22\dots22}_{k} 1.
$$
Find all positive integers $n$ such that
$$
2013 \mid R(10^n).
$$ | [
"First observe that\n$$\nR(k) = 11 \\cdot \\underbrace{11 \\dots 11}_{k+1}.\n$$\nNext observe that\n$$\n\\underbrace{11 \\dots 11}_{k+1} = \\frac{10^{k+1} - 1}{9}.\n$$\nThis reduces the problem to finding all $n$ such that\n$$\n2013 = 3 \\cdot 11 \\cdot 61 \\mid 11 \\cdot \\frac{10^{10^n+1} - 1}{9}.\n$$\nFor this t... | Baltic Way | Baltic Way 2013 | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | no positive integers n | |
03qr | Find all real numbers $k$, such that the inequality
$$
a^3 + b^3 + c^3 + d^3 + 1 \ge k(a+b+c+d)
$$
holds for any $a, b, c, d \in [-1, +\infty)$. | [
"If $a = b = c = d = -1$, then $-3 \\ge k \\cdot (-4)$. Hence $k \\ge \\frac{3}{4}$.\n\nIf $a = b = c = d = \\frac{1}{2}$, then $4 \\cdot \\frac{1}{8} + 1 \\ge k \\cdot (4 \\cdot \\frac{1}{2})$. Thus $k \\le \\frac{3}{4}$, and so $k = \\frac{3}{4}$.\n\nAt first, we prove that $4x^3 + 1 \\ge 3x$, $x \\in [-1, +\\inf... | China | China Western Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | 3/4 | |
0cp0 | In a canteen, there are $100$ apples of total weight $10$ kg, each apple weighs not less than $25$ g. The barman has to cut them into some pieces and distribute these pieces among $100$ students so that each student receives exactly $100$ g. Prove that the barman can attain this aim in such a way that the weight of eac... | [
"Все веса в решении будут измеряться в граммах. Назовём кусок яблока (или само яблоко) большим, если его вес не меньше $25$.\n\nДокажем индукцией по $n$, что $n$ больших яблок суммарного веса $100n$ можно разрезать на большие куски и раздать $n$ детям поровну.\n\nБаза при $n = 1$ очевидна.\n\nПусть $n > 1$. Рассмот... | Russia | Final round | [
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English; Russian | proof only | null | |
0lft | Problem:
Points $M, N, K$ lie on the sides $B C, C A, A B$ respectively, of a triangle $A B C$, and are different from its vertices. The triangle $M N K$ is called beautiful if the triangles $M N K$ and $A B C$ are similar (with the vertices respectively in this order). Show that if in the triangle $A B C$ there are t... | [
"Solution:\n\nSay $M N_1 K_1$ and $M N_2 K_2$ are such two beautiful triangles. Let $T = N_1 K_1 \\cap N_2 K_2$; $T$ exists, and belongs to the interior of $\\triangle A B C$, since angles at $M$ are equal to $\\angle A$. Then $\\angle N_1 M N_2 = \\angle K_1 M K_2$. We then have $\\angle M N_1 T = \\angle M N_2 T ... | Zhautykov Olympiad | International Zhautykov Olympiad in Sciences | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01xa | The sum of several (not necessarily different) positive integers not exceeding $10$ is equal to $S$.
Find all possible values of $S$ such that these numbers can always be partitioned into two groups with the sums of the numbers in each group not exceeding $80$. | [
"Answer: $S \\le 152$.\nClearly $S \\le 160$. Suppose that $S \\ge 153$ and let $S = 153 + m$, where $0 \\le m \\le 7$. Consider the next collection: $m$ numbers equal $10$ and $17 - m$ numbers equal $9$ (the total sum equals $153 + m$). At least nine of these $17$ numbers will be in the same group. But the sum of ... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | S ≤ 152 | |
06pw | For an integer $m$, denote by $t(m)$ the unique number in $\{1,2,3\}$ such that $m+t(m)$ is a multiple of $3$. A function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfies $f(-1)=0$, $f(0)=1$, $f(1)=-1$ and
$$
f\left(2^{n}+m\right)=f\left(2^{n}-t(m)\right)-f(m) \quad \text{ for all integers } m, n \geq 0 \text{ with } 2... | [
"The given conditions determine $f$ uniquely on the positive integers. The signs of $f(1), f(2), \\ldots$ seem to change quite erratically. However, values of the form $f\\left(2^{n}-t(m)\\right)$ are sufficient to compute directly any functional value. Indeed, let $n>0$ have base $2$ representation $n=2^{a_{0}}+2^... | IMO | 49th International Mathematical Olympiad Spain | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
08cr | Problem:
Quante sono le sequenze di numeri di lunghezza otto composte unicamente da 0 e 1 che contengono il codice 01 ?
(A) 64
(B) 128
(C) 192
(D) 247
(E) 255 | [
"Solution:\n\nLa risposta è (D). Il numero di stringhe di lunghezza 8 formate di cifre 0 e 1 è $2^8$; contiamo quelle che non contengono il codice 01. Ci si rende facilmente conto che se una tale stringa contiene almeno uno 0, allora tutte le cifre a destra di quello 0 sono a loro volta uguali a 0: in caso contrari... | Italy | GARA di FEBBRAIO | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | MCQ | D | |
0jtm | Problem:
Katherine has a piece of string that is $2016$ millimeters long. She cuts the string at a location chosen uniformly at random, and takes the left half. She continues this process until the remaining string is less than one millimeter long. What is the expected number of cuts that she makes? | [
"Solution:\n\nLetting $f(x)$ be the expected number of cuts if the initial length of the string is $x$, we get the integral equation $f(x) = 1 + \\frac{1}{x} \\int_{1}^{x} f(y) d y$. Letting $g(x) = \\int_{1}^{x} f(y) d y$, we get $\\frac{d g}{d x} = 1 + \\frac{1}{x} g(x)$. Using integrating factors, we see that th... | United States | HMMT February | [
"Statistics > Probability > Counting Methods > Other",
"Calculus > Differential Equations > ODEs",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof and answer | 1 + ln(2016) | |
08a0 | Problem:
Siano $x$ e $y$ numeri reali tali che si abbia $x^{2} + 4 y^{2} = 1$; quanto vale come minimo $|x| + 2|y|$?
(A) $\frac{1}{2}$
(B) 1
(C) $\sqrt{2}$
(D) $1 + \frac{\sqrt{2}}{2}$
(E) 2 | [
"Solution:\n\nLa risposta è (B). La quantità $|x| + 2|y|$ è infatti non negativa e il suo quadrato vale $x^{2} + 4 y^{2} + 4|x||y| \\geq x^{2} + 4 y^{2} = 1$, quindi anche $|x| + 2|y| \\geq 1$. Questo valore effettivamente si può ottenere (esattamente quando $|x||y| = 0$, cioè se $x = \\pm 1, y = 0$ o $x = 0, y = \... | Italy | Progetto Olimpiadi della Matematica | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | B | |
0l2w | Problem:
Let $ABCD$ be a convex trapezoid such that $\angle DAB = \angle ABC = 90^{\circ}$, $DA = 2$, $AB = 3$, and $BC = 8$. Let $\omega$ be a circle passing through $A$ and tangent to segment $\overline{CD}$ at point $T$. Suppose that the center of $\omega$ lies on line $BC$. Compute $CT$. | [
"Solution:\n\nLet $A'$ be the reflection of $A$ across $BC$, and let $P = AB \\cap CD$. Then since the center of $\\omega$ lies on $BC$, we have that $\\omega$ passes through $A'$. Thus, by power of a point, $PT^2 = PA \\cdot PA'$. By similar triangles, we have\n$$\n\\frac{PA}{AD} = \\frac{... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4√5 - √7 | |
05is | Problem:
Dans le triangle $ABC$ les bissectrices des angles en $A$ et $C$ rencontrent le cercle circonscrit au triangle $ABC$ respectivement en $A_{0}$ et $C_{0}$. La droite passant par $I$, centre du cercle inscrit dans le triangle $ABC$, et parallèle à la droite $(AC)$ rencontre la droite $(A_{0}C_{0})$ en $P$.
Mon... | [
"Solution:\n\n$\\underline{\\text{Sur la figure on va rajouter } B_{0} \\text{ le point d'intersection de la bissectrice de } \\widehat{ABC} \\text{ avec le cercle :}}$\n\nAvant de commencer, je veux démontrer un petit résultat sur les arcs. Regardez la figure de droite (je vais noter $\\wi... | France | Olympiades Françaises de Mathématiques - Envoi Numéro 1 - Corrigé | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
01no | Ten chess players take part in a chess tournament. Each player plays exactly one game with any other player. For marking results a new system is used at this tournament: a player receives 3 points for a win, 1 point for a draw, and 0 point for a loss (in accordance with the old system a player receives 1 point for a wi... | [
"a) The following table shows the example of the tournament such that grand master $N$ takes the eighth place only. So grand master $A$ can be right.\n\n| | N | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | Points (Σ) | Points (Σ) |\n|---|---|---|---|---|---|---|---|---|---|---|-------------|-------------|\n| | | | ... | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a) Yes. b) No. | |
0bl8 | Prove that $13$ has a multiple whose decimal writing contains only two digits $a$, $b$ and $a$ appears in exactly $3$ places, which are not contiguous. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0682 | Let $c(O, R)$ be a circle and $A, B$ two diametrically opposite points. The bisector of the angle $A\hat{B}O$ intersects the circle $c(O, R)$ at point $C$, the circumcircle of the triangle $AOB$ (say $(c_1)$) at point $K$ and the circumcircle of the triangle $AOC$ (say $(c_2)$) at point $L$. Prove that the point $K$ is... | [
"Segments $OB$, $OC$ are equal, as radii of the circle ($c$), and so the triangle $OBC$ is isosceles. Hence:\n$$\n\\hat{B}_1 = \\hat{C}_1 = \\hat{x} \\qquad (1).\n$$\n\nFigure 3\n$BC$ is the bisector of the angle $OBA$, and hence\n$$\n\\hat{B}_1 = \\hat{B}_2 = \\hat{x} \\qquad (2).\n$$\nThe... | Greece | Selection Examination | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01sa | Find all real $x \ge -1$ such that the inequality
$$
\frac{a_1 + x}{2} \cdot \frac{a_2 + x}{2} \cdot \dots \cdot \frac{a_n + x}{2} \le \frac{a_1 a_2 \dots a_n + x}{2}
$$
holds for all $a_1, \dots, a_n \ge 1$, where $n \ge 2$. | [
"If $x \\ge -1$ satisfies the problem condition, then setting $a_1 = a_2 = \\dots = a_n = 1$ in the given inequality:\n$$\n\\frac{a_1 + x}{2} \\cdot \\frac{a_2 + x}{2} \\cdot \\dots \\cdot \\frac{a_n + x}{2} \\le \\frac{a_1 a_2 \\dots a_n + x}{2} \\quad (1)\n$$\nwe obtain the inequality $\\left(\\frac{1 + x}{2}\\ri... | Belarus | FINAL ROUND | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | x ∈ [-1, 1] |
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