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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
05nz | Problem:
a) Trouver tous les entiers $m \geqslant 1$ et $n \geqslant 1$ tels que $\frac{5^{m}+2^{n+1}}{5^{m}-2^{n+1}}$ soit le carré d'un entier.
b) Plus généralement, trouver tous les entiers $m \geqslant 1$ et $n \geqslant 1$, ainsi que les nombres premiers $p$, tels que $\frac{5^{m}+2^{n} p}{5^{m}-2^{n} p}$ soit le... | [
"Solution:\na) Voir exercice 2 ci-dessus.\n\nb) Supposons que $p=5$. Alors $\\frac{5^{m-1}+2^{n}}{5^{m-1}-2^{n}}$ est le carré d'un entier. D'après la partie a) (qui marchait aussi dans le cas d'entiers $\\geqslant 0$), on a $m=n=2$.\n\nSupposons enfin $p \\neq 2$ et $p \\neq 5$. Soit $d$ le PGCD de $5^{m}+2^{n} p$... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) The only solution is (m, n) = (1, 1).
b) The solutions are exactly (m, n, p) = (1, 1, 2), (2, 2, 5), and (2, 3, 3). | |
06c8 | Let $f$ be a function with the following properties:
(i) $f(n)$ is defined for every positive integer $n$;
(ii) $f(n)$ is a positive integer;
(iii) $f(f(m) + f(n)) = m + n$ for all $m$ and $n$.
Find $f(1997)$. | [
"We have $f(1997) = 1997$.\n\nLabel the equation as follows.\n$$\nf(f(m) + f(n)) = m + n \\quad (1)\n$$\nIf $f(a) = f(b)$ for some $a, b \\in \\mathbb{Z}^+$, then by putting $m = a$ and $m = b$ in (1), we obtain\n$$\na + n = f(f(a) + f(n)) = f(f(b) + f(n)) = b + n.\n$$\nThis implies $a = b$. So $f$ is injective. No... | Hong Kong | HKG TST | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | 1997 | |
0622 | Problem:
Es seien $A$, $B$, $C$, $D$, $E$, $F$ Punkte auf einem Kreis mit $A E \| B D$ und $B C \| D F$. Durch Spiegelung an der Geraden $C E$ gehe der Punkt $D$ in $X$ über. Zeige, dass $X$ so weit von der Geraden $E F$ entfernt ist wie $B$ von $A C$. | [
"Solution:\n\nAlle im folgenden auftretenden Winkel sind als orientiert und modulo $180^{\\circ}$ zu verstehen; dadurch wird eine Betrachtung der Lage der involvierten Punkte zueinander entbehrlich. Die Fußpunkte der Lote von $B$, $X$ auf $A C$, $E F$ seien mit $P$, $Q$ bezeichnet.\n\nStrategie. Zeige die Kongruenz... | Germany | 1. IMO-Auswahlklausur | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0frd | Problem:
Determinar todas las parejas de enteros positivos $(m, n)$ para los cuales es posible colocar algunas piedras en las casillas de un tablero de $m$ filas y $n$ columnas, no más de una piedra por casilla, de manera que todas las columnas tengan la misma cantidad de piedras, y no existan dos filas con la misma c... | [
"Solution:\n\nVeremos que las soluciones son todas las parejas $(m, n)$ con $n \\geq m$ si $m$ es impar, o $n \\geq m-1$ si $m$ es par.\n\nPara $m=1$, es inmediato que cualquier tablero $(1, n)$ es posible, por ejemplo llenándolo completamente de piedras.\n\nNótese la condición necesaria $n \\geq m-1$ : para $m$ fi... | Spain | Spain | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | All pairs where, if the number of rows is odd, the number of columns is at least the number of rows; and if the number of rows is even, the number of columns is at least one less than the number of rows. | |
01e3 | Compute the following product:
$$
\prod_{m=1}^{2018} \frac{(2m-1)^4 + \frac{1}{4}}{(2m)^4 + \frac{1}{4}}
$$ | [
"By applying Sophie-Germain identity we can obtain following equality:\n$$\n\\frac{(2m-1)^4 + \\frac{1}{4}}{(2m)^4 + \\frac{1}{4}} = \\frac{((2m-\\frac{1}{2})^2 + \\frac{1}{4})((2m-\\frac{3}{2})^2 + \\frac{1}{4})}{((2m+\\frac{1}{2})^2 + \\frac{1}{4})((2m-\\frac{1}{2})^2 + \\frac{1}{4})} = \\frac{(2m-\\frac{3}{2})^2... | Baltic Way | Baltic Way shortlist | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | final answer only | 2/(8073^2 + 1) | |
0b6f | Let $f : [0, 1] \to \mathbb{R}$ be an integrable function. If $f(1) = 0$ and $f$ is differentiable in $1$, prove that
$$
\lim_{n \to \infty} n^2 \int_0^1 x^n f(x) dx = -f'(1).
$$ | [
"Let us make the substitution $x = 1 - \\frac{t}{n}$, so that as $x$ goes from $0$ to $1$, $t$ goes from $n(1-0) = n$ to $n(1-1) = 0$ (i.e., $t$ goes from $n$ to $0$). The differential $dx = -\\frac{dt}{n}$.\n\nSo,\n$$\n\\int_0^1 x^n f(x) dx = \\int_{x=0}^{x=1} x^n f(x) dx = \\int_{t=n}^{t=0} \\left(1 - \\frac{t}{n... | Romania | Shortlisted Problems for the Romanian NMO | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Limits"
] | English | proof only | null | |
014e | Problem:
In a triangle $A B C$, points $D, E$ lie on sides $A B, A C$ respectively. The lines $B E$ and $C D$ intersect at $F$. Prove that if
$$
B C^{2}=B D \cdot B A+C E \cdot C A,
$$
then the points $A, D, F, E$ lie on a circle. | [
"Solution:\n\nLet $G$ be a point on the segment $B C$ determined by the condition $B G \\cdot B C = B D \\cdot B A$. (Such a point exists because $B D \\cdot B A < B C^{2}$.) Then the points $A, D, G, C$ lie on a circle. Moreover, we have\n$$\nC E \\cdot C A = B C^{2} - B D \\cdot B A = B C \\cdot (B G + C G) - B C... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0et2 | Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that
$$
f(a^3) + f(b^3) + f(c^3) + 3f(a+b)f(b+c)f(c+a) = (f(a+b+c))^3
$$
for all $a, b, c \in \mathbb{Z}$. | [
"Suppose $f$ satisfies the condition.\nBy taking $(a, b, c) = (0, 0, 0)$, we get $3f(0) + 3f(0)^3 = f(0)^3$, so that either $f(0) = 0$, or $3 = -2f(0)^2$. The latter is not possible in $\\mathbb{Z}$, so we must have $f(0) = 0$.\nBy taking $(a, b, c) = (n, -n, 0)$, we get $f(n^3) + f(-n^3) = 0$, resulting in\n$$\nf(... | South Africa | The South African Mathematical Olympiad, Third Round | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All solutions f: Z -> Z are exactly the following five functions:
1) f(n) = 0 for all integers n.
2) f(n) = n for all integers n.
3) f(n) = -n for all integers n.
4) f(n) = 0 if n is divisible by 3, f(n) = 1 if n ≡ 1 mod 3, and f(n) = -1 if n ≡ 2 mod 3.
5) The negation of (4): f(n) = 0 if n is divisible by 3, f(n) = -1... | |
0a2g | Given are two positive integers $a$ and $b$ with the property that
$$
\frac{b^3}{a^4} \quad \text{and} \quad \frac{a^3}{b^2}
$$
are both integers greater than $1$.
What is the smallest possible value for the sum $a + b$? | [
"The smallest integer greater than $1$ is $2$, so we see that\n$$\n\\frac{b^3}{a^4} \\ge 2 \\quad \\text{and} \\quad \\frac{a^3}{b^2} \\ge 2.\n$$\nIt follows that\n$$\na = \\left(\\frac{b^3}{a^4}\\right)^2 \\cdot \\left(\\frac{a^3}{b^2}\\right)^3 \\ge 2^5 \\quad \\text{and} \\quad b = \\left(\\frac{b^3}{a^4}\\right... | Netherlands | Dutch Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 160 | |
05le | Problem:
On dit qu'une suite $\left(u_{n}\right)_{n \geqslant 1}$ est Sicilienne si $u_{1}$ est un entier strictement positif, et si pour tout $n$,
$$
u_{n+1}= \begin{cases}u_{n} / 2 & \text{ si } u_{n} \text{ est pair, et } \\ u_{n}+\left[\sqrt{u_{n}}\right] & \text{ si } u_{n} \text{ est impair. }\end{cases}$$
Exist... | [
"Solution:\n\nCommençons par une simple remarque. Soit $\\mathrm{f}: \\mathbb{N}^{*} \\rightarrow \\mathbb{N}^{*}$ la fonction qui, à tout entier naturel non nul $n$, associe l'entier $n / 2$ si $n$ est pair, et $n+[\\sqrt{n}]$ si $n$ est impair. Il est clair que l'image $f(n)$ est bien un entier naturel non nul.\n... | France | Olympiades Françaises de Mathématiques - Test de Janvier | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No; every such sequence contains a term equal to one. | |
01e6 | Let $M$ be a subset of a plane sufficing following properties:
1) There is no single line $k$, such that $M \subset k$.
2) For any parallelogram $ABCD$ if $A, B, C \in M$, then $D \in M$.
3) If $A, B \in M$, then $|AB| > 1$.
Prove, that there are two families of parallel lines, such that $M$ is a set consisting of all... | [
"At the beginning we can see that property 3) implies that in any bounded subset of a plane there is only a finite number of points from $M$. (*)\n\nNext, we can see that if for some points $A, B \\in M$ we define by $\\phi$ a translation by vector $\\overrightarrow{AB}$, then for any point $C \\in M$ we also have ... | Baltic Way | Baltic Way shortlist | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
00hx | Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that
$$
f((c+1) x+f(y))=f(x+2 y)+2 c x \quad \text{ for all } x, y \in \mathbb{R}_{>0} .
$$ | [
"We first prove that $f(x) \\geq 2 x$ for all $x>0$. Suppose, for the sake of contradiction, that $f(y)<2 y$ for some positive $y$. Choose $x$ such that $f((c+1) x+f(y))$ and $f(x+2 y)$ cancel out, that is,\n$$\n(c+1) x+f(y)=x+2 y \\Longleftrightarrow x=\\frac{2 y-f(y)}{c}\n$$\nNotice that $x>0$ because $2 y-f(y)>0... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | f(x) = 2x for all positive real x | |
09t0 | Problem:
Gegeven is cirkel $\omega$ met middellijn $A K$. Punt $M$ ligt binnen de cirkel, niet op lijn $A K$. De lijn $A M$ snijdt $\omega$ nogmaals in $Q$. De raaklijn aan $\omega$ in $Q$ snijdt de lijn door $M$ loodrecht op $A K$ in $P$. Punt $L$ ligt op $\omega$ zodat $P L$ een raaklijn is, met $L \neq Q$. Bewijs d... | [
"Solution:\n\nNoem $O$ het middelpunt van $\\omega$ en zij $V$ het snijpunt van $M P$ met $A K$. We bewijzen nu eerst dat $\\angle P V L=\\angle P O L$. Als $V$ en $O$ samenvallen, dan is er niets te bewijzen. Als $V$ en $O$ niet samenvallen, dan geldt $\\angle O V P=90^{\\circ}=\\angle O L P$, dus $O V P L$ of $V ... | Netherlands | IMO-selectietoets III | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fuv | Problem:
Sei $T$ die Menge aller Tripel $(p, q, r)$ von nichtnegativen ganzen Zahlen. Bestimme alle Funktionen $f: T \rightarrow \mathbb{R}$ für die gilt
$$
f(p, q, r)= \begin{cases}0 & \text{ für } p q r=0 \\ 1+\frac{1}{6}\{f(p+1, q-1, r)+f(p-1, q+1, r) \\ +f(p-1, q, r+1)+f(p+1, q, r-1) \\ +f(p, q+1, r-1)+f(p, q-1, r... | [
"Solution:\n\nWir beweisen zuerst, dass höchstens eine solche Funktion existiert. Nehme an, $f$ und $g$ erfüllen die Bedingungen der Aufgabe. Für die Funktion $h=f-g$ gilt dann\n$$\nh(p, q, r)= \\begin{cases}0 & \\text{ für } p q r=0 \\\\ \\frac{1}{6}\\{h(p+1, q-1, r)+h(p-1, q+1, r) \\\\ +h(p-1, q, r+1)+h(p+1, q, r... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | f(p,q,r) = 3pqr/(p+q+r) for p+q+r > 0, and f(0,0,0) = 0 | |
0bfp | The three-element subsets of a seven-element set are colored. If the intersection of two sets is empty then they have different colors. What is the minimum number of colors needed? | [
"Let $A = \\{1, 2, 3, 4, 5, 6, 7\\}$. Two colors are not enough because the sets in the following sequence of three-element subsets of $A$ should have alternating colors: $\\{1, 2, 3\\}$, $\\{4, 5, 6\\}$, $\\{7, 1, 2\\}$, $\\{3, 4, 5\\}$, $\\{6, 7, 1\\}$, $\\{2, 3, 4\\}$, $\\{5, 6, 7\\}$, $\\{1, 2, 3\\}$.\n\nWith t... | Romania | 64th NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 3 | |
04ve | Given two odd positive integers $k$ and $n$. For each two positive integers $i, j$ satisfying $1 \le i \le k$ and $1 \le j \le n$ Martin wrote the fraction $i/j$ on the board. Determine the median of all these fractions, that is a real number $q$ such that if we order all the fractions on the board by their values from... | [
"We show that the median has the value $q = \\frac{k+1}{n+1}$. In the entire solution, $q$ denotes this number.\nSince the given numbers $n$ and $k$ are odd, the fraction with value $q$ is actually written on the board—for example, it is a fraction $\\frac{\\frac{1}{2}(k+1)}{\\frac{1}{2}(n+1)}$.\n\nAccording to the... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | (k+1)/(n+1) | |
08vb | For positive integers $n$, $m$, $k$, let us write $n \equiv m \pmod k$ if $n - m$ is divisible by $k$. Let $A$ be the sum of all positive integers $a$ less than or equal to $2011$ for which $a \equiv 1 \pmod 3$, and $B$ be the sum of all positive integers $b$ less than or equal to $2011$ for which $b \equiv 2 \pmod 3$.... | [
"Note that $2011 \\equiv 1 \\pmod 3$. So, we see that $A$ is the sum of the numbers\n$$\n3 \\cdot 0 + 1,\\ 3 \\cdot 1 + 1,\\ \\dots,\\ 3 \\cdot 669 + 1,\\ 3 \\cdot 670 + 1,\n$$\nwhile $B$ is the sum of the numbers\n$$\n3 \\cdot 0 + 2,\\ 3 \\cdot 1 + 2,\\ \\dots,\\ 3 \\cdot 668 + 2,\\ 3 \\cdot 669 + 2.\n$$\nFor $0 \... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 1341 | |
01lt | Let $g(n)$ be the number of all $n$-digit natural numbers each consisting only of the digits $0, 1, 2, 3$ (but not necessarily all of them) such that the sum of no two neighboring digits equals $2$.
Determine whether $g(2010)$ and $g(2011)$ are divisible by $11$.
(I. Kozlov) | [
"Let $g_0(n)$, $g_1(n)$, $g_2(n)$, $g_3(n)$ be the quantities of the numbers satisfying the given condition which end by the digits $0$, $1$, $2$, $3$ respectively. Then $g(n) = g_0(n) + g_1(n) + g_2(n) + g_3(n)$. By condition,\n$$\n\\begin{align*}\ng_0(n+1) &= g_0(n) + g_1(n) + g_3(n) = g(n) - g_2(n), \\\\\ng_1(n+... | Belarus | Selection and Training Session | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | g(2010) is not divisible by 11; g(2011) is divisible by 11. | |
0eso | Bee, Cee and Vee live in Microphyllia where there are only two types of creatures – those that consistently tell the truth and those that consistently lie. The former creatures are Trudees and the latter Falsees. When I last visited Microphyllia I asked Bee: “Who of you all are Trudees?”
Bee mumbled its answer so I did... | [
"Suppose Cee tells the truth, then Bee said that only one of the three tells the truth, so Bee can't tell the truth or otherwise there will be at least two truth tellers which would be a contradiction. Hence, if Cee tells the truth Bee lies and there must be more than one truth teller, so Vee must be telling the tr... | South Africa | South African Mathematics Olympiad Third Round | [
"Discrete Mathematics > Logic"
] | English | proof and answer | Cee is a Falsee, Vee is a Trudee, and Bee’s status cannot be determined from the given information. | |
0kll | Problem:
Aerith rolls a fair die until she gets a roll that is greater than or equal to her previous roll. Find the expected number of times she will roll the die before stopping. | [
"Solution:\n\nThe number of rolls is always at least $2$ and at most $7$. For there to be at least $k$ rolls, the first $k-1$ rolls have to be distinct and in decreasing order. The number of ways this can happen is $\\binom{6}{k-1}$, and the total number of ways to have $k-1$ rolls is $6^{k-1}$. Thus, the expected ... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | (7/6)^6 | |
0fz7 | Problem:
Sei $ABCD$ ein Sehnenviereck mit Umkreis $k$. Sei $S$ der Schnittpunkt von $AB$ und $CD$ und $T$ der Schnittpunkt der Tangenten an $k$ in $A$ und $C$. Zeige, dass $ADTS$ genau dann ein Sehnenviereck ist, wenn $BD$ die Strecke $AC$ halbiert. | [
"Solution:\n\nWir betrachten zuerst den Fall, dass $T$ auf derselben Seite von $AC$ liegt wie $D$:\n\nNach dem Tangentenwinkelsatz gilt $\\angle TAD = \\angle DCA$. Wir erhalten:\n\n$ADTS$ Sehnenviereck $\\Leftrightarrow \\angle TSD = \\angle TAD \\Leftrightarrow \\angle TSD = \\angle DCA \\Leftrightarrow TS \\para... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06s1 | Let $ABC$ be a triangle with circumcenter $O$ and incenter $I$. The points $D$, $E$ and $F$ on the sides $BC$, $CA$ and $AB$ respectively are such that $BD + BF = CA$ and $CD + CE = AB$. The circumcircles of the triangles $BFD$ and $CDE$ intersect at $P \neq D$. Prove that $OP = OI$. | [
"Let $\\omega_{A}$, $\\omega_{B}$ and $\\omega_{C}$ meet the bisectors $AI$, $BI$ and $CI$ at $A \\neq A'$, $B \\neq B'$ and $C \\neq C'$ respectively. The key observation is that $A'$, $B'$ and $C'$ do not depend on the particular choice of $D$, $E$ and $F$, provided that $BD + BF = CA$, $CD + CE = AB$ and $AE + A... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / C... | null | proof only | null | |
00u8 | For every natural number $x$, let $P(x)$ be the product of the digits of the number $x$. Is there a natural number $n$ such that the numbers $P(n)$ and $P(n^2)$ are non-zero squares of natural numbers, where the number of digits of the number $n$ is equal to
a. $2021$
b. $2022$ | [
"The answers are affirmative in both cases.\n\na.\nTake $n = \\overbrace{33\\cdots3}^{2019}68$. Then $P(n) = (4 \\cdot 3^{1010})^2$. Also,\n$$\n\\begin{aligned}\nn^2 &= \\left( \\frac{10^{2021} - 1}{3} + 35 \\right)^2 = \\frac{(10^{2021} + 104)^2}{9} \\\\\n&= \\frac{10^{4042} + 208 \\cdot 10^{2021} + 10816}{9} \\\\... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | Yes for both 2021 and 2022 | |
04b3 | Determine all pairs of integers $(a, b)$ such that $a (a - b) = b$. | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | (0, 0) and (-2, -4) | |
0jgi | Problem:
Let $H$ be the orthocenter of an acute triangle $ABC$. (The orthocenter is the point at the intersection of the three altitudes. An acute triangle has all angles less than $90^{\circ}$.) Draw three circles: one passing through $A$, $B$ and $H$, another passing through $B$, $C$ and $H$, and finally, one passing... | [
"Solution:\n\n\n\nSee the figure on the left, above. Since $B'$ and $C'$ are each equidistant from $A$ and $H$, the line $B'C'$ is the perpendicular bisector of $AH$. The line $AH$ is also an altitude of $\\triangle ABC$ so $AH$ is also perpendicular to $BC$. Since $BC$ and $B'C'$ are both ... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transfor... | null | proof only | null | |
08zu | How many permutations $p_1, p_2, \dots, p_{2023}$ of $1, 2, \dots, 2023$ satisfy the equation
$$
p_1 + |p_2 - p_1| + |p_3 - p_2| + \dots + |p_{2023} - p_{2022}| + p_{2023} = 4048?
$$ | [
"Let $p_0 = p_{2024} = 0$ and let $t$ be a positive integer such that $p_t = 2023$. The given condition gives\n$$\n\\begin{aligned}\n& 2 = 4048 - (2023 + 2023) \\\\\n&= \\left( p_1 + \\sum_{i=1}^{2022} |p_{i+1} - p_i| + p_{2023} \\right) - \\left( \\sum_{i=0}^{t-1} (p_{i+1} - p_i) + \\sum_{i=t}^{2023} (p_i - p_{i+1... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2021 * 2^2021 | |
0621 | Problem:
Man beweise: Ist $4^{n} \cdot 7 = a^{2} + b^{2} + c^{2} + d^{2}$ mit $n, a, b, c, d \in \mathbb{N} \setminus \{0\}$, dann kann keine der Quadratzahlen die Zahl $4^{n-1}$ unterschreiten. | [
"Solution:\nIst $n=1$, dann ist die Behauptung richtig; die einzigen Lösungen für $a, b, c, d$ sind, abgesehen von der Reihenfolge, die Quadrupel $(1,1,1,5)$, $(1,3,3,3)$ und $(2,2,2,4)$. Für jedes $n \\geq 1$ ist $4^{n} \\cdot 7$ durch $4$ teilbar. Da das Quadrat einer natürlichen Zahl bei der Division durch $4$ n... | Germany | Auswahlwettbewerb zur IMO 2005 | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0i6f | Problem:
How many pairs of integers $(a, b)$, with $1 \leq a \leq b \leq 60$, have the property that $b$ is divisible by $a$ and $b+1$ is divisible by $a+1$? | [
"Solution:\n\nThe divisibility condition is equivalent to $b-a$ being divisible by both $a$ and $a+1$, or, equivalently (since these are relatively prime), by $a(a+1)$. Any $b$ satisfying the condition is automatically $\\geq a$, so it suffices to count the number of values $b-a \\in \\{1-a, 2-a, \\ldots, 60-a\\}$ ... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization"
] | null | final answer only | 106 | |
03ns | Problem:
Let $d(k)$ denote the number of positive integer divisors of $k$. For example, $d(6)=4$ since $6$ has $4$ positive divisors, namely, $1, 2, 3$, and $6$. Prove that for all positive integers $n$,
$$
d(1)+d(3)+d(5)+\cdots+d(2n-1) \leq d(2)+d(4)+d(6)+\cdots+d(2n)
$$ | [
"Solution:\n\nFor any integer $k$ and set of integers $S$, let $f_{S}(k)$ be the number of multiples of $k$ in $S$. We can count the number of pairs $(k, s)$ with $k \\in \\mathbb{N}$ dividing $s \\in S$ in two different ways, as follows:\n- For each $s \\in S$, there are $d(s)$ pairs that include $s$, one for each... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
087i | Problem:
Trovare tutte le terne ordinate di numeri interi positivi $(p, q, n)$ tali che $p, q$ siano primi e $p^{2}+q^{2}=p q n+1$. | [
"Solution:\n\nSupponiamo $p=q$. Sostituendo otteniamo $p^{2}(2-n)=1$ che è impossibile perché $1$ non è diviso da nessun primo. Quindi necessariamente $p$ e $q$ sono diversi; poiché l'equazione è simmetrica in $p$ e $q$ possiamo supporre che $q>p$, cioè $q \\geq p+1$. Scriviamo ora la nostra equazione come:\n$$\np^... | Italy | Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | The ordered triples are (2, 3, 2) and (3, 2, 2). | |
0epk | The sum of three different positive integers is $7$. Their product is
(A) $12$ (B) $10$ (C) $9$ (D) $8$ (E) $5$ | [
"Suppose the numbers are $x$, $y$, $z$, with $0 < x < y < z$ and $x + y + z = 7$. If $x \\ge 2$, then $y \\ge 3$ and $z \\ge 4$, so $x + y + z \\ge 9$, which is too large. Therefore $x = 1$, which leaves $y + z = 6$. Since $2 \\le y < z$, the only possibility is $y = 2$ and $z = 4$. The product $xyz = 1 \\times 2 \... | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | D | |
0he9 | 299 digits $0$ and one digit $1$ are written in a circle. The following moves are allowed:
* from each digit, subtract the sum of the adjacent digits;
* select two digits with exactly two digits between them and increase both by $1$ or decrease both by $1$.
Is it possible to obtain such an arrangement of numbers (afte... | [
"Let us analyse how the recorded moves affect the sum of the digits written in a circle. Let us denote the numbers by $a_1, a_2, \\dots, a_{300}$. The following numbers will be written after the move of the first type: $b_k = a_k - a_{k-1} - a_{k+1}$, $k = 1, \\dots, 300$ ($a_{301} \\equiv a_1$). Let $S = a_1 + a_2... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
00bg | Let $ABC$ be an acute-angled and scalene triangle, $\omega$ its incircle and $\omega'$ the excircle relative to the vertex $A$. The circles $\omega$ and $\omega'$ are tangent to $BC$ at $P$ and $P'$ respectively. Let $\Gamma$ be the circumference passing through $B$ and $C$ that is tangent to $\omega$ in a point $Q$, a... | [
"Let $S$ be the intersection of the lines $QR$ and $BC$.\n\n\nFirst, note that the homothety with center $Q$ which transforms $\\omega$ into $\\Gamma$ maps $P$ to a point in $\\Gamma$ whose tangent is parallel to $BC$, namely, $M$. Then, $Q$, $P$, $M$ are collinear. Then, the equality $BQM ... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"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 | |
0kyl | Problem:
Inside an equilateral triangle of side length $6$, three congruent equilateral triangles of side length $x$ with sides parallel to the original equilateral triangle are arranged so that each has a vertex on a side of the larger triangle, and a vertex on another one of the three equilateral triangles, as shown... | [
"Solution:\n\n\n\nLet $x$ be the side length of the shaded triangles. Note that the centers of the triangles with side lengths $1$ and $6$ coincide; call this common center $O$.\n\nThe distance from $O$ to a side of the equilateral triangle with side length $1$ is $\\frac{\\sqrt{3}}{6}$. Si... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 5/3 | |
01ky | All cells of a $7 \times 7$ table are painted black and white. Per move it is allowed to choose any $n \times n$ square, $1 < n < 7$ (with the sides coinciding with the sides of the cells) and to change the color of all its cells (from black to white and vice versa).
Is it possible to get the table with all white cells... | [
"We separate the table into five parts: the central $3 \\times 3$ square, two $2 \\times 7$ rectangles, and two $3 \\times 2$ rectangles (see Fig. 1).\n\n\nFig. 1\n\n\n\n\n![](attached_image_... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0ay6 | Problem:
Gari is seated in a jeep, and at the moment, has one 10-peso coin, two 5-peso coins, and six 1-peso coins in his pocket. If he picks four coins at random from his pocket, what is the probability that these will be enough to pay for his jeepney fare of 8 pesos? | [
"Solution:\n\nThe only way that Gari will be unable to pay for the fare is if all four coins are 1-peso coins. This has probability $\\frac{\\binom{6}{4}}{\\binom{9}{4}} = \\frac{15}{126} = \\frac{5}{42}$, so there is a $1 - \\frac{5}{42} = \\frac{37}{42}$ chance that it will be enough to pay for his fare."
] | Philippines | Philippine Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 37/42 | |
0fku | Problem:
A cada punto del plano se le asigna un solo color entre siete colores distintos. ¿Existirá un trapecio inscriptible en una circunferencia cuyos vértices tengan todos el mismo color? | [
"Solution:\nLa idea inicial es considerar una circunferencia $C$ de radio $r$ y sobre ella bloques de 8 puntos $A_{1}, A_{2}, \\ldots, A_{8}$ igualmente espaciados; es decir que los arcos $A_{i}A_{i+1}$, $i=1, \\ldots, 7$, tengan igual longitud $\\lambda>0$ (que se elegirá convenientemente) para cada uno de los blo... | Spain | XLIV Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | Yes | |
0bvt | Let $A$, $B \in \mathcal{M}_n(\mathbb{C})$ be matrices such that $A + B$ is invertible and $A^2 = B^2 = O_n$. Show that $n$ is even and the matrices $(AB)^k$, $k = 1, 2, \dots$, have rank $\frac{n}{2}$.
Florin Stănescu | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Linear transformations"
] | English | proof only | null | |
0cgp | Let $p \ge 2$ be an integer number. Prove that the sequence $(x_n)_{n \ge 1}$, defined by $x_1 = a > 0$ and the recurrence relation $x_{n+1} = x_n + \lfloor \frac{p}{x_n} \rfloor$, $n \in \mathbb{N}^*$, is convergent. Determine its limit depending on the values of the parameter $a$. We denote by $[x]$ the integer part ... | [
"The sequence $(x_n)_{n \\ge 1}$ has positive terms (proof by induction). There is $k \\in \\mathbb{N}^*$ such that $x_k > p$. Indeed, if we assume, by reductio ad absurdum, that $x_n \\le p, \\forall n \\in \\mathbb{N}^*$, we obtain $x_{n+1} \\ge x_n+1, \\forall n \\in \\mathbb{N}^*$, so $x_n \\ge a+n-1, \\forall ... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | lim x_n =
- a, if a > p;
- a + floor(p/a), if 0 < a < 1;
- p + {a}, if 1 ≤ a ≤ p and a is not an integer (where {a} is the fractional part of a);
- p + 1, if a ∈ {1,2,…,p}. | |
0da2 | A non-empty subset of $\{1,2, \ldots, n\}$ is called arabic if arithmetic mean of its elements is an integer. Show that the number of arabic subsets of $\{1,2, \ldots, n\}$ has the same parity as $n$. | [
"The solution is based on a simple fact, that if you add the arithmetic mean of the sequence to the sequence, then the arithmetic mean of the sequence will not change, since\n$$\n\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{n}=\\frac{a_{1}+a_{2}+\\cdots+a_{n}+\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{n}}{n+1}\n$$\nDenote by $\\mu(A)$... | Saudi Arabia | Team selection tests for IMO 2018 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
04c4 | A sphere of radius $r$ is inscribed to a tetrahedron. If the altitudes of the tetrahedron are $v_1$, $v_2$, $v_3$ and $v_4$, prove
$$
\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3} + \frac{1}{v_4} = \frac{1}{r}.
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof only | null | |
01xs | A positive integer $n > 10$ has two different positive integer divisors $a$ and $b$ such that $n = a^2 + b$.
Prove that strictly between the numbers $a$ and $b$ there is at least one another divisor of $n$. | [
"From the equality $n = a^2 + b$ it follows that $b$ is divisible by $a$, since $a$ divides both $n$ and $a^2$. Let $b = ma$, where $m > 1$. Then $n = a^2 + ma = a(a + m)$, so $a + m$ divides $n$. Let us show that $a < a + m < b$. Suppose $a + m \\ge b$, then\n$$\na + m \\ge ma \\iff am - a - m + 1 \\le 1 \\iff (a ... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
02js | Problem:
As retas $r$ e $s$ são paralelas, encontre $x$ e $y$:
 | [
"Solution:\n\nTemos $80^{\\circ} + y = 180^{\\circ} \\Rightarrow y = 100^{\\circ}$.\n\nComo as retas $r$ e $s$ são paralelas, segue que, $60^{\\circ} + x + 80^{\\circ} = 180^{\\circ}$, donde $x = 40^{\\circ}$."
] | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | x = 40°, y = 100° | |
00qo | Let $f: N \rightarrow N$ be a function from the positive integers to the positive integers for which $f(1) = 1$, $f(2n) = f(n)$ and $f(2n+1) = f(n)+f(n+1)$ for all $n \in N$. Prove that for any natural number $n$, the number of odd natural numbers $m$ such that $f(m) = n$ is equal to the number of positive integers not... | [
"The crucial observation that solves this problem is that the function $f$ encodes Euclid's algorithm when we view numbers in binary. To make this precise, we will write $g(n)$ for $f(n+1)$, and consider for each integer $n$ the pair $(f(n), g(n))$. If we let $x$ be the binary string representing $n$ then the recur... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
02pq | Problem:
a. Determine $a$, $b$ e $c$ tais que a igualdade
$$
(n+2)^2 = a(n+1)^2 + b n^2 + c(n-1)^2
$$
seja verdadeira qualquer que seja o número $n$.
b. Suponha que $x_1, x_2, \ldots, x_7$ satisfazem o sistema
$$
\left\{\begin{array}{l}
x_1 + 4 x_2 + 9 x_3 + 16 x_4 + 25 x_5 + 36 x_6 + 49 x_7 = 1 \\
4 x_1 + 9 x_2 + 16... | [
"Solution:\n\na. Se um polinômio se anula para infinitos valores, então todos os seus coeficientes são nulos.\nExpandindo a igualdade temos\n$$\nn^2 + 4n + 4 = a(n^2 + 2n + 1) + b n^2 + c(n^2 - 2n + 1)\n$$\nAssim,\n$$\n(a + b + c - 1) n^2 + (2a - 2c - 4) n + (a + c - 4) = 0\n$$\nqualquer que seja o número $n$. Logo... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a=3, b=-3, c=1; desired sum = 334 | |
0cy5 | Let $a>0$. If the system
$$
\left\{\begin{array}{l}
a^{x}+a^{y}+a^{z}=14-a \\
x+y+z=1
\end{array}\right.
$$
has a solution in real numbers, prove that $a \leq 8$. | [
"Assume, by contradiction, that $a>8$.\nThen $14-a<6$. Applying AM-GM inequality, we get\n$$\n6>14-a=a^{x}+a^{y}+a^{z} \\geq 3 a^{\\frac{x+y+z}{3}}>3 \\cdot 8^{\\frac{1}{3}}=3 \\cdot 8^{\\frac{1}{3}}=6,\n$$\na contradiction."
] | Saudi Arabia | SAMC | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof only | null | |
0d6c | Let $ABC$ be a triangle inscribed in the circle $(O)$. The bisector of $\angle BAC$ cuts the circle $(O)$ again at $D$. Let $DE$ be the diameter of $(O)$. Let $G$ be a point on $\operatorname{arc} AB$ which does not contain $C$. The lines $GD$ and $BC$ intersect at $F$. Let $H$ be a point on the line $AG$ such that $FG... | [
"Let $HF$ cut $DE$ at $P$. We have $\\angle HGD = \\angle E = \\angle HPD$ so $HGPD$ is cyclic, we deduce $FH \\cdot FP = FG \\cdot FD = FB \\cdot FC$ so $HBPC$ is cyclic.\n\n\n\nEasily seen $DE$ is perpendicular bisector of $BC$ so $PB = PC$. Hence $HP$ is bisector of $\\angle BHC$. From t... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | English | proof only | null | |
03zi | Prove that for any given positive integers $m$, $n$, there exist infinitely many pairs of coprime positive integers $a$, $b$, such that $a + b \mid am^a + bn^b$. | [
"If $mn = 1$, then the claim is valid. For $mn \\ge 2$, since\n$$\nn^a (am^a + bn^b) = (a+b)n^{a+b} + a((mn)^a - n^{a+b}),\n$$\nit is sufficient to prove the existence of infinitely many coprime number pairs $a$, $b$, such that\n$$\na + b \\mid (mn)^a - n^{a+b}, \\quad (a+b, n) = 1.\n$$\nLet $p = a + b$, we only ne... | China | China Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Dioph... | English | proof only | null | |
03p2 | It is given that complex numbers $z_1$ and $z_2$ satisfy $|z_1| = 2$ and $|z_2| = 3$. If the included angle of their corresponding vectors is $60^\circ$, then $\left|\frac{z_1 + z_2}{z_1 - z_2}\right| = \underline{\hspace{2cm}}$. | [
"By the cosine rule, we obtain\n$$\n|z_1 + z_2| = \\sqrt{|z_1|^2 + |z_2|^2 - 2|z_1||z_2| \\cos 120^\\circ} = \\sqrt{19},\n$$\nand\n$$\n|z_1 - z_2| = \\sqrt{|z_1|^2 + |z_2|^2 - 2|z_1||z_2| \\cos 60^\\circ} = \\sqrt{7}.\n$$\nTherefore,\n$$\n\\left| \\frac{z_1 + z_2}{z_1 - z_2} \\right| = \\frac{\\sqrt{19}}{\\sqrt{7}}... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | final answer only | sqrt(133)/7 | |
0339 | Problem:
a) A set $A$ of positive integers less than $2000000$ is called good if $2000 \in A$ and $a$ divides $b$ for any $a, b \in A$, $a < b$. Find the maximum possible cardinality of a good set.
b) Find the number of the good sets of maximal cardinality. | [
"Solution:\n\na) Let $a_{1} < \\cdots < a_{n-1} < a_{n} = 2000 < a_{n+1} < \\cdots < a_{m}$ be the elements of a good set. Since $a_{i+1} \\geq 2 a_{i}$, then $2000000 > a_{m} \\geq 2^{m-n} 2000$ and hence $m-n \\leq 9$.\n\nOn the other hand, the equality $2000 = 2^{4} 5^{3}$ shows that $a_{i} = 2^{k_{i}} 5^{l_{i}}... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | Maximum cardinality: 17; Number of good sets of maximal cardinality: 350 | |
0faj | Problem:
Show that $x^4 + y^4 + z^2 \geq xyz\sqrt{8}$ for all positive reals $x$, $y$, $z$. | [
"Solution:\nBy AM/GM, $x^4 + y^4 \\geq 2x^2y^2$. Then by AM/GM again, $2x^2y^2 + z^2 \\geq (\\sqrt{8})xyz$."
] | Soviet Union | 1st CIS | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0agb | Let $a$, $b$, $c$ be positive real numbers. Prove that
$$
\frac{a^2 b (b-c)}{a+b} + \frac{b^2 c (c-a)}{b+c} + \frac{c^2 a (a-b)}{c+a} \geq 0.
$$ | [
"Dividing both sides by $a$, $b$, $c$, the inequality is equivalent to\n$$\n\\frac{a(b-c)}{c(a+b)} + \\frac{b(c-a)}{a(b+c)} + \\frac{c(a-b)}{b(c+a)} \\geq 0.\n$$\nBy adding $1+1+1$ to both sides, the inequality becomes\n$$\n\\frac{b(c+a)}{c(a+b)} + \\frac{c(a+b)}{a(b+c)} + \\frac{a(b+c)}{b(c+a)} \\geq 3.\n$$\nBy ap... | North Macedonia | Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
051q | For which natural numbers $n \ge 3$ is it possible to cut a regular $n$-gon into smaller pieces with regular polygonal shape? (The pieces may have different number of sides.) | [
"**Answer:** 3, 4, 6, 12.\n\nA regular triangle can be partitioned into four regular triangles of equal size (Fig. 23), a regular quadrilateral can be partitioned into four regular quadrilaterals with equal size (Fig. 24) and a regular hexagon can be partitioned into six regular triangles of equal size (Fig. 25). B... | Estonia | Final Round of National Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 3, 4, 6, 12 | |
0ell | Problem:
Na 1. nivoju računalniške igre Zakladnica se igralec sprehaja po podzemni zakladnici, ki je sestavljana iz 13 osemkotnih in 12 kvadratnih sob. Edini vhod in izhod iz zakladnice je v osrednji sobi, v preostalih 24 sobah pa je po en zlat cekin (glej sliko). Med vsakima dvema sobama je prehod, ki se zapre takoj,... | [
"Solution:\n\nPreštejmo, koliko vrat ima posamezna soba s cekinom. Opazimo, da je v zakladnici 12 sob s po 3 vrati, 4 sobe s po 4 vrati, 4 sobe s po 5 vrati in 4 sobe s po 8 vrati (srednje sobe ne štejemo, ker nima cekina). Skozi sobo s 3 vrati lahko gre igralec le enkrat, saj bi sicer ostal ujet v njej. S tem pobe... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | 44 | |
03nl | Problem:
Let $S=\{1,4,8,9,16, \ldots\}$ be the set of perfect powers of integers, i.e. numbers of the form $n^{k}$ where $n, k$ are positive integers and $k \geq 2$. Write $S=\left\{a_{1}, a_{2}, a_{3} \ldots\right\}$ with terms in increasing order, so that $a_{1}<a_{2}<a_{3} \cdots$. Prove that there exist infinitely ... | [
"Solution:\nThe idea is that most perfect powers are squares. If $a_{n}=x^{2}$ and $a_{n+1}=(x+1)^{2}$, then $a_{n+1}-a_{n}=2x+1$. Note that $9999 \\mid 2x+1$ is equivalent to $x \\equiv 4999 \\pmod{9999}$. Hence we will be done if we can show that there exist infinitely many $x \\equiv 4999 \\pmod{9999}$ such that... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Other"
] | null | proof only | null | |
0d2l | Solve the following equation where $x$ is a real number:
$$
\left\lfloor x^{2}\right\rfloor - 10 \lfloor x \rfloor + 24 = 0
$$ | [
"Notice first that if $\\lfloor x \\rfloor \\leq 0$ then\n$$\n\\left\\lfloor x^{2} \\right\\rfloor - 10 \\lfloor x \\rfloor + 24 \\geq 24 > 0\n$$\nThis means that if $x$ is a solution to the equation then $\\lfloor x \\rfloor \\geq 1$.\nLet $x$ be a solution to the equation, $m = \\lfloor x \\rfloor \\geq 1$, and $... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | [4, \sqrt{17}) \cup [\sqrt{26}, \sqrt{27}) \cup [6, \sqrt{37}) | |
0di8 | There are $100$ doors labeled with numbers $1, 2, \ldots, 100$. You have $100$ keys labeled with numbers. Each key corresponds to exactly one door. If the key $i$ corresponds to the door $j$, then $|i - j| \le 1$. At each turn, you may pick doors with numbers $i$ and $j$ and check whether the key $i$ corresponds to the... | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | a) yes; b) yes; c) no | |
0gfq | 找出所有整數 $n \ge 1$ 使得存在正整數對 $(a, b)$ 有
$$
\frac{ab + 3b + 8}{a^2 + b + 3} = n
$$
且沒有任何質數的三次方能整除 $a^2 + b + 3$. | [
"The only integer with property is $n = 2$.\nAs $b \\equiv -a^2 - 3 \\pmod{a^2 + b + 3}$, the numerator of the given fraction satisfies\n$$ab + 3b + 8 \\equiv a(-a^2 - 3) + 3(-a^2 - 3) + 8 \\equiv -(a + 1)^3 \\pmod{a^2 + b + 3}$$\nAs $a^2 + b + 3$ is not divisible by $p^3$ for any prime $p$, if $a^2 + b + 3$ divide... | Taiwan | 2022 數學奧林匹亞競賽第一階段選訓營, 國際競賽實作(一) | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Chinese; English | proof and answer | 2 | |
06rh | On a square table of $2011$ by $2011$ cells we place a finite number of napkins that each cover a square of $52$ by $52$ cells. In each cell we write the number of napkins covering it, and we record the maximal number $k$ of cells that all contain the same nonzero number. Considering all possible napkin configurations,... | [
"Let $m=39$, then $2011=52 m-17$. We begin with an example showing that there can exist $3986729$ cells carrying the same positive number.\n\n\n\nTo describe it, we number the columns from the left to the right and the rows from the bottom to the top by $1,2, \\ldots, 2011$. We will denote ... | IMO | 52nd International Mathematical Olympiad 2011 Shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 3986729 | |
0eet | Problem:
Kot $\Varangle BAC$ trikotnika $ABC$ je velik $30^\circ$, $P$ pa je poljubna točka v notranjosti tega trikotnika. Zrcalne slike točke $P$ pri zrcaljenu čez stranice $BC$, $CA$ in $AB$ zaporedoma označimo s $P_A$, $P_B$ in $P_C$. Denimo, da je $P_A P_B P_C$ enakostraničen trikotnik. Dokaži, da je tedaj $\Varan... | [
"Solution:\n\n\n\nNaj bodo $D, E$ in $F$ zaporedoma presečisča daljic $P P_A$, $P P_B$ in $P P_C$ s stranicami $BC$, $CA$ in $AB$. Točke $D, E$ in $F$ so hkrati tudi razpolovišča daljic $P P_A$, $P P_B$ in $P P_C$, zato sta si trikotnika $DEF$ in $P_A P_B P_C$ podobna. Torej je tudi trikotn... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0e29 | Let $ABCDEF$ be a cyclic hexagon, such that $AD$ is the diameter of its circumcircle and the lines $BF$ and $CE$ are parallel. Let $K$ be the intersection of the segments $AC$ and $BD$, let $L$ be the intersection of the segments $AE$ and $DF$ and let $M$ be the midpoint of the segment $KL$. Prove that $|MC| = |ME|$. | [
"The segments $BF$ and $CE$ are parallel, so the quadrilateral $FBCE$ is an isosceles trapezoid and $|BC| = |EF|$. The inscribed angles over the chords $BC$ and $EF$ are equal, $\\angle EDF = \\angle BDC$. Let $\\angle EDF = \\alpha$. The segment $AD$ is the diameter of a circle, so by Thales' theorem we have $\\an... | Slovenia | Selection Examinations for the IMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0cni | 7 cards with numbers $0$, $1$, $2$, $3$, $4$, $5$, $6$ are given. Peter and Basil make moves in turn taking one card by each move; Peter makes the first move. The player who can construct of his cards a decimal number divisible by $17$ earlier than his opponent is declared as a Winner. Determine which of two players ha... | [
"Let us denote the players as $A$ (the first player) and $B$ (his opponent).\n\nWe present a strategy that allows $A$ to guarantee a win. Let him take the digit $3$ on his first move; then $B$ is forced to take $4$ (otherwise $A$ will take it on his second move and win by forming the number $34$). Note that $B$ can... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization"
] | English; Russian | proof and answer | Peter (the first player) has a winning strategy. | |
04gc | A hundred quadratic envelopes, each of different size, are arranged in a way that, for every two different envelopes, the smaller one is either inside of the bigger one, or they are outside of each other. At the same time, in both of the envelopes there can be other envelopes. Two arrangements are considered different ... | [
"Denote the number of arrangements of $n$ envelopes as $K_n$.\nAssume we are given an arrangement of $n$ envelopes. If we remove the smallest envelope, we obtain a possible arrangement of $n-1$ remaining envelopes. On the other hand, if we are given a possible arrangement of the remaining $n-1$ envelopes, we can pu... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 99! | |
04zz | Find all pairs $(a, b)$ of real numbers with $a + b = 1$, which satisfy $(a^2 + b^2)(a^3 + b^3) = a^4 + b^4$. | [
"As $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$, the given equation can be expressed as $(a^2 + b^2)(a^2 - ab + b^2) = a^4 + b^4$. Expanding brackets gives $-a^3b + 2a^2b^2 - ab^3 = 0$, which factorizes to $-ab(a-b)^2 = 0$. Hence $a = 0$ or $b = 0$ or $a-b = 0$. Together with the condition $a+b=1$, we get the following sol... | Estonia | Selected Problems from Open Contests | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (0, 1), (1, 0), (1/2, 1/2) | |
00r4 | Let $n$ be a given even positive integer. First John writes the consecutive square numbers $1^2, 3^2, \dots, (2n-1)^2$ on the blackboard. Then he picks some three of them, say $a_1, a_2, a_3$, erases them and writes the number
$$
1 + \sum_{1 \le i < j \le 3} |a_i - a_j|
$$
on the blackboard. He continues this way repla... | [
"Notice that the initially given numbers are all odd, and that the number that replaces some three of them is again an odd one, so after the replacement we are left again with a set of odd numbers. The same reasoning applies to every replacement, thus all numbers on blackboard are always odd ones. So each one of th... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0gz8 | In a triangle $ABC$ on the side $AC$ the points $F$ and $L$ with $AF = LC < \frac{1}{2}AC$ are chosen. Find an angle $FBL$, if $AB^2 + BC^2 = AL^2 + LC^2$.
**Answer:** $\angle FBL = 90^\circ$. | [
"Let $AB = c$, $BC = a$, $FM = ML = c$ ($M$ is a middle of $AC$), $AF = LC = x$. According to known formula find length of median (Fig.25)\n\n\n\nFig.25\n\n$$\n\\begin{aligned}\nBM^2 = m^2 &= \\frac{1}{4}(2c^2 + 2a^2 - (2b + 2x)^2) = \\frac{1}{4}(2(2b + x)^2 + 2x^2 - (2b + 2x)^2) = \\\\\n&=... | Ukraine | The Problems of Ukrainian Authors | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 90° | |
00e4 | The 400-digit number $N = 20232023\ldots2023$, which is formed by 100 copies of $2023$, is written on a blackboard. Lionel has to erase some of the digits of $N$ in such a way that the resulting number is divisible by $84$, and the largest possible. Determine which digits Lionel should erase. | [
"Notice that $84 = 4 \\times 3 \\times 7$. Let $M$ be the resulting number after Lionel erases some digits. We want $M$ to be even, so we must erase the rightmost $3$. We also want $M$ to be divisible by $3$. For this to happen, the sum of digits of $M$ must be divisible by $3$. Initially, the sum of digits equals ... | Argentina | Rioplatense Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | Erase exactly three digits: the last three of the whole number, the two immediately before it so the number ends with twenty, and the leading two of the third block from the right among the remaining full 2023 blocks. Equivalently, the final number is ninety-six copies of 2023, then 023, then 2023, then 2023, followed ... | |
0ewc | Problem:
Given an $n \times n$ array of numbers. $n$ is odd and each number in the array is $1$ or $-1$. Prove that the number of rows and columns containing an odd number of $-1$s cannot total $n$. | [
"Solution:\nIf we change a $-1$ to $1$, we affect the total number of rows and columns (containing an odd number of $-1$s) by $0$, $2$ or $-2$. After changing all the $-1$s we have total of $0$. Hence the starting total must be even. So it cannot be $n$."
] | Soviet Union | 2nd ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
02ib | Problem:
O símbolo $\odot$ representa uma operação especial com números. Veja alguns exemplos $2 \odot 4=10, 3 \odot 8=27, \quad 4 \odot 27=112, \quad 5 \odot 1=10$. Quanto vale $4 \odot(8 \odot 7)$ ?
(A) 19
(B) 39
(C) 120
(D) 240
(E) 260 | [
"Solution:\n\nTemos que descobrir qual é a regra dessa operação. Note que $2 \\odot 4=10=2 \\times 4+2$, $3 \\odot 8=27=3 \\times 8+3$, $4 \\odot 27=112=4 \\times 27+4$, $5 \\odot 1=10=5 \\times 1+5$.\n\nPodemos concluir que a regra que define a operação $\\odot$ é $a \\odot b = a \\times b + a$.\n\nAssim, temos:\n... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | MCQ | E | |
0a3w | A group of 4050 friends is playing a video game tournament. There are 2025 computers labelled $a_1, \dots, a_{2025}$ in one room and 2025 computers labelled $b_1, \dots, b_{2025}$ in another room at the tournament. The player on computer $a_i$ always plays against the players $b_i, b_{i+2}, b_{i+3}$ and $b_{i+4}$ (in p... | [
"For the opponent computers of $a_i$, we look at the $a_j$ they are playing against, see the following table.\n\n$b_i:$\n| $a_{i-4}$ | $a_{i-3}$ | $a_{i-2}$ | $a_i$ |\n|---|---|---|---|\n\n$b_{i+2}:$\n| $a_{i-2}$ | $a_{i-1}$ | $a_i$ | $a_{i+2}$ |\n|---|---|---|---|\n\n$b_{i+3}:$\n| $a_{i-1}$ | $a_i$ | $a_{i+1}$ | $... | Netherlands | BxMO/EGMO Team Selection Test | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other"
] | null | proof only | null | |
07qp | Aoife and Zoe play a game. On each of Aoife's three turns, Aoife replaces one of the asterisks in the expression $**********$ with one of the digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$ that has not yet been used. At Zoe's turn she can replace two of the asterisks with two different digits not yet selected. Aoif... | [
"Divide the digits into three groups $\\{1, 8, 9\\}$, $\\{2, 3, 4\\}$ and $\\{5, 6, 7\\}$ and note that all the numbers\n\n$189$ $918$ $891$\n$243$ $324$ $432$\n$567$ $756$ $675$\n\nare divisible by $27$. Imagine the $9$ asterisks divided into three groups of three consecutive asterisks. If Aoife plays in one of th... | Ireland | Irish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | Yes | |
0l4h | Problem:
A school needs to elect its president. The school has 121 students, each of whom belongs to one of two tribes: Geometers or Algebraists. Two candidates are running for president: one Geometer and one Algebraist. The Geometers vote only for Geometers and the Algebraists only for Algebraists. There are more Alge... | [
"Solution:\nTo ultimately win the election in step 2, the Geometers must have 6 seats among the 11 representatives who elect the president.\n\nMoving back to step 1, the Geometers need 6 out of 11 votes in a group to elect a representative from that group. Thus, it takes $6 \\times 6=36$ Geometers to elect 6 repres... | United States | 25th Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 36 | |
0e1z | Problem:
Določi vsa cela števila $n$, za katera ima enačba $x^{2}+n x+n+5=0$ le celoštevilske rešitve. | [
"Solution:\n\nKvadratna enačba ima lahko celoštevilske rešitve le, če je diskriminanta popoln kvadrat. Izračunamo lahko $D=n^{2}-4(n+5)=(n-2)^{2}-24$. Torej za neko nenegativno celo število $m$ velja $m^{2}=(n-2)^{2}-24$, oziroma\n$$\n24=(n-2)^{2}-m^{2}=(n+m-2)(n-m-2)\n$$\nKer sta števili $n+m-2$ in $n-m-2$ iste pa... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | -5, -3, 7, 9 | |
0hju | Problem:
Ten fair coins are flipped. Given that there are at least nine heads, what is the probability that all the coins show heads? | [
"Solution:\nThe answer is $\\frac{1}{11}$. Among the $2^{10} = 1024$ sequences of heads and tails, note that\n- There is only $1$ which is all heads, and\n- There are $10$ sequences which have exactly one tails.\nSo there are $11$ possible sequences with at least nine heads, only one of which is all heads. Hence th... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 1/11 | |
0i21 | Problem:
The UC Berkeley math department is about to move into a new, one-story building consisting of a $2001 \times 2001$ square grid of rooms. They would like to install doors between adjacent rooms so that each room has exactly two doors. Prove that this cannot be done. | [
"Solution:\n\nSuppose that it can be done. Color the building in checkerboard fashion, and suppose that we obtain $a$ white rooms and $b$ black rooms. Each door connects a white room with a black room. So, if we consider, for each white room, the number of doors adjoining it, we will count each door exactly once. S... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
07ss | A trapezium $ABCD$, in which $AB$ is parallel to $DC$, is inscribed in a circle of radius $R$ and centre $O$. The non-parallel sides $DA$ and $CB$ are extended to meet at $P$ while diagonals $AC$ and $BD$ intersect at $E$.
Prove that $|OE| \cdot |OP| = R^2$. | [
"Let $F$ be the mid-point if $AB$, then $P$, $E$, $O$, $F$ are collinear. We have $\\angle AOB = 2\\angle ACB$, hence using symmetry, $\\angle FOB = \\angle ACB$. This implies that $CEOB$ is a cyclic quadrilateral, which implies\n$$\n\\angle ECO = \\angle EBO.\n$$\nFrom $\\triangle EOB$ we have $\\angle EBO = \\ang... | Ireland | IRL_ABooklet_2020 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
002j | Determine todas las ternas de números reales $(x, y, z)$ que satisfacen el siguiente sistema de ecuaciones:
$$
\begin{aligned}
xyz &= 8, \\
x^2y + y^2z + z^2x &= 73, \\
x(y - z)^2 + y(z - x)^2 + z(x - y)^2 &= 98.
\end{aligned}
$$ | [] | Argentina | XX Olimpiada Iberoamericana de Matemáticas | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Español | proof and answer | All permutations of (4, 4, 1/2) and (1, 1, 8). | |
009r | Cover a grid square $13 \times 13$ with $2 \times 2$ squares and L-shapes of three unit cells so that the number of L-shapes is least possible.
 | [
"Let a $(2k-1) \\times (2k-1)$ square board be covered as in the statement with $x$ squares $2 \\times 2$ and $y$ shapes L. Denote by $(i, j)$ the cell in row $i$, column $j$, and color black all cells $(i, j)$ with both $i$ and $j$ odd. Thus $k^2$ black cells are obtained. Observe that wherever a $2 \\times 2$ squ... | Argentina | NATIONAL XXX OMA | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 27 | |
0agm | For each integer $n$ ($n \ge 2$) let $f(n)$ denote the sum of all positive integers that are at most $n$ and not relatively prime to $n$.
Prove that $f(n+p) \neq f(n)$ for each such $n$ and for every prime $p$. | [
"Let $m$, $n$ and $k$ be positive integers and $p$ be a prime. Since $(n, k) = 1 \\Leftrightarrow (n, n-k) = 1$, we have\n\n$$\nf(n) = \\frac{n(n+1)}{2} - \\frac{n \\cdot \\phi(n)}{2},\n$$\nwhere $\\phi(n)$ is the Euler function.\nIf $m > n$, $(m, n) = 1$ and $f(m) = f(n)$, then\n$$\nm(m+1-\\phi(m)) = n(n+1-\\phi(n... | North Macedonia | Balkan Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0hks | Problem:
Prove that
$$
1 \cdot 1! + 2 \cdot 2! + \cdots + n \cdot n! = (n+1)! - 1
$$
for all positive integers $n$. | [
"Solution:\n$$\n\\begin{aligned}\n& 1 \\cdot 1! + 2 \\cdot 2! + \\cdots + n \\cdot n! \\\\\n& = (2-1) \\cdot 1! + (3-1) \\cdot 2! + \\cdots + [(n+1)-n] \\cdot n! \\\\\n& = (2 \\cdot 1! - 1!) + (3 \\cdot 2! - 2!) + \\cdots + [(n+1) \\cdot n! - n!] \\\\\n& = (2! - 1!) + (3! - 2!) + \\cdots + [(n+1)! - n!]\n\\end{alig... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
012x | Problem:
Let $ABC$ be an arbitrary triangle and $AMB$, $BNC$, $CKA$ regular triangles outward of $ABC$. Through the midpoint of $MN$ a perpendicular to $AC$ is constructed; similarly through the midpoints of $NK$ resp. $KM$ perpendiculars to $AB$ resp. $BC$ are constructed. Prove that these three perpendiculars inters... | [
"Solution:\n\nLet $O$ be the midpoint of $MN$, and let $E$ and $F$ be the midpoints of $AB$ and $BC$, respectively. As triangle $MBC$ transforms into triangle $ABN$ when rotated $60^{\\circ}$ around $B$ we get $MC = AN$ (it is also a well-known fact). Considering now the quadrangles $AMBN$ and $CMBN$ we get $OE = O... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geome... | null | proof only | null | |
0id5 | Problem:
Eight strangers are preparing to play bridge. How many ways can they be grouped into two bridge games - that is, into unordered pairs of unordered pairs of people? | [
"Solution:\n315\nPutting 8 people into 4 pairs and putting those 4 pairs into 2 pairs of pairs are independent. If the people are numbered from 1 to 8, there are 7 ways to choose the person to pair with person 1. Then there are 5 ways to choose the person to pair with the person who has the lowest remaining number,... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | final answer only | 315 | |
0jvl | Problem:
Let $S=\{1,2, \ldots, 2016\}$, and let $f$ be a randomly chosen bijection from $S$ to itself. Let $n$ be the smallest positive integer such that $f^{(n)}(1)=1$, where $f^{(i)}(x)=f\left(f^{(i-1)}(x)\right)$. What is the expected value of $n$? | [
"Solution:\n\nSay that $n=k$. Then $1, f(1), f^{2}(1), \\ldots, f^{(k-1)}(1)$ are all distinct, which means there are $2015 \\cdot 2014 \\cdots (2016-k+1)$ ways to assign these values. There is $1$ possible value of $f^{k}(1)$, and $(2016-k)!$ ways to assign the image of the $2016-k$ remaining values. Thus the prob... | United States | HMMT November 2016 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 2017/2 | |
0g1l | Problem:
Sei $ABC$ ein gleichschenkliges Dreieck mit Scheitelpunkt $A$ und $AB > BC$. Sei $k$ der Kreis mit Zentrum $A$ durch $B$ und $C$. Sei $H$ der zweite Schnittpunkt von $k$ mit der Höhe des Dreiecks $ABC$ durch $B$. Weiter sei $G$ der zweite Schnittpunkt von $k$ mit der Schwerlinie durch $B$ im Dreieck $ABC$. Se... | [
"\n\nSolution:\n\nSei $M$ der Mittelpunkt der Strecke $AC$. Mit dem Zentriwinkelsatz erhält man $\\angle BGH = \\frac{1}{2} \\angle BAH$. Da das Dreieck $ABH$ gleichschenklig ist und die Strecke $BH$ senkrecht zu der Strecke $AC$ steht, ist $AC$ die Winkelhalbierende von Winkel $BAH$. Daher... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"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 | |
0ehe | Problem:
V trikotniku $ABC$ velja $\angle BAC=\frac{\pi}{3}$ in $|AB|^{2}=|AC|^{2}+|AC| \cdot |BC|$. Določi velikosti ostalih dveh notranjih kotov trikotnika $ABC$. | [
"Solution:\n\nKote trikotnika $ABC$ označimo kot običajno z $\\alpha, \\beta$ in $\\gamma$, torej je $\\alpha=\\frac{\\pi}{3}$. Naj bo $D$ taka točka na premici $AC$, da $C$ leži med $A$ in $D$ in velja $|CD|=|CB|$. Pogoj naloge lahko preoblikujemo v\n$$\n\\frac{|AB|}{|AC|}=\\frac{|AC|+|BC|}{|AB|}=\\frac{|AC|+|CD|}... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | Angle B = 2π/9 and angle C = 4π/9 | |
08z9 | Determine all pairs of positive integers $(x, y)$ which satisfy $3^x - 8^y = 2xy + 1$. | [
"For a prime $p$ and a positive integer $n$, denote by $\\text{ord}_p n$ the largest nonnegative integer $i$ such that $p^i$ divides $n$.\n\nFirst we consider the case $y \\equiv 0 \\pmod 2$. Since $3^z = 8^y + 2xy + 1 \\equiv 1 \\pmod 4$, $x$ is even. Write $x = 2z$ and $y = 2w$ for positive integers $z$ and $w$, ... | Japan | Japan 2022 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (4, 2) | |
02pp | Problem:
(a) Verifique a identidade
$$
(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(a+b)(b+c)(c+a)
$$
(b) Resolva o sistema
$$
\left\{\begin{array}{l}
x+y+z=1 \\
x^{2}+y^{2}+z^{2}=1 \\
x^{3}+y^{3}+z^{3}=1
\end{array}\right.
$$ | [
"Solution:\n\n(a) Vamos expandir $(a+b+c)^{3}$ como $[(a+b)+c]^{3}$.\n$$\n\\begin{aligned}\n{[(a+b)+c]^{3} } & =(a+b)^{3}+3(a+b) c(a+b+c)+c^{3} \\\\\n& =a^{3}+b^{3}+3 a b(a+b)+3(a+b) c(a+b+c)+c^{3} \\\\\n& =a^{3}+b^{3}+c^{3}+3(a+b)[a b+c(a+b+c)] \\\\\n& =a^{3}+b^{3}+c^{3}+3(a+b)\\left(c^{2}+c(a+b)+a b\\right) \\\\\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | (0,0,1), (0,1,0), (1,0,0) | |
0fbt | Problem:
Se colocan $2 n+1$ fichas, blancas y negras, en una fila $(n \geq 1)$. Se dice que una ficha está equilibrada si el número de fichas blancas a su izquierda, más el número de fichas negras a su derecha es $n$. Determina, razonadamente, si el número de fichas que están equilibradas es par o impar. | [
"Solution:\n\nNumeramos las posiciones en la fila desde $1$ hasta $2 n+1$, de izquierda a derecha. Definimos la valoración de una cierta posición $k = 1, 2, \\cdots, 2 n+1$ como el número de fichas blancas a su izquierda más el número de fichas negras a su derecha, con lo que una ficha está equilibrada si y sólo si... | Spain | LIV Olimpiada matemática Española (Concurso Final) | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | odd | |
0ecx | Find all real numbers $x$ which satisfy the equation
$$
2 \sin^2 2x \geq 3 \cos 2x.
$$ | [
"We use the double angle formulas for trigonometric functions to get\n$$\n8 \\sin^2 x \\cos^2 x \\ge 3 \\cos^2 x - 3 \\sin^2 x.\n$$\nUsing the identity $\\cos^2 x = 1 - \\sin^2 x$ we rearrange the inequality to\n$$\n8 \\sin^4 x - 14 \\sin^2 x + 3 \\le 0.\n$$\nIntroducing a new variable $y = \\sin^2 x$ we get $8y^2 ... | Slovenia | National Math Olympiad 2015 – First Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x ∈ [π/6 + kπ, 5π/6 + kπ] for k ∈ ℤ | |
0g0h | Problem:
Soit $ABC$ un triangle non-rectangle avec $M$ le milieu de $BC$. Soit $D$ un point sur la droite $AB$ tel que $CA = CD$ et soit $E$ un point sur la droite $BC$ tel que $EB = ED$. La parallèle à $ED$ passant par $A$ coupe la droite $MD$ au point $I$ et la droite $AM$ coupe la droite $ED$ au point $J$. Montrer ... | [
"Solution:\n\nSoit $P$ le milieu de $AD$. Par propriété des triangles isocèles, on a que le milieu de la base est égal au pied de la hauteur issue du sommet opposé à la base. L'utilisation de cette propriété dans le triangle par construction isocèle en $CACD$, on a que $\\angle CPD = 90^{\\circ}$. Par la réciproque... | Switzerland | IMO-Selektion | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
01j6 | Let $\triangle ABC$ be an acute triangle and $P$ be an inner point of $\triangle ABC$. Let $K$, $L$ and $M$ be the reflections of $P$ across $BC$, $AC$ and $AB$, respectively. Let $D$ and $E$ be the second points of intersection of $\odot(PBC)$ with lines $AB$ and $AC$, respectively. Let lines $MD$ and $LE$ intersect a... | [
"$$\n\\angle BGF = \\angle BGK = \\angle BCK = \\angle PCB = \\angle PDB = \\angle BDM = \\angle BDF.\n$$\nSimilarly, $GFCE$ is cyclic.\n\nLastly, notice that $A$ is the radical centre of circumcircles of $GFBD$, $GFCE$ and $BDCE$. Thus, $F$, $A$ and $K$ are collinear.",
"Construct point $I$ as the second point o... | Baltic Way | Baltic Way 2023 Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
04mx | A trapezium $ABCD$ is given. The bisector of the leg $\overline{BC}$ intersects the leg $\overline{AD}$ at $M$, while the bisector of $\overline{AD}$ intersects $\overline{BC}$ at $N$.
Let $O_1$ and $O_2$ be the circumcentres of triangles $ABN$ and $CDM$, respectively.
Prove that the line $O_1O_2$ bisects the segment $... | [
"Denote by $P$ and $Q$ the midpoints of $\\overline{BC}$ and $\\overline{AD}$, respectively.\n\nWe will show that the quadrilateral $ABNM$ is cyclic.\nFrom $\\angle MQN = \\angle NPM = 90^\\circ$, we get that the quadrilateral $MNPQ$ is cyclic. This implies $\\angle PQM + \\angle MNP = 180^... | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ahe | Find all functions $f: \mathbb{R} \to \mathbb{R}$ which satisfy the conditions:
$$
f(x+y) < f(x) + f(y),
$$
$$
f(f(x)) = [x] + 2.
$$ | [
"Let $f(0) = a$, then $f(a) = f(f(0)) = 2$, $f(2) = f(f(a)) = a + 2$. Continuing this procedure we get that $f(2k) = a + 2k$ and $f(a + 2k) = 2k + 2$. We get $2k + 2 = f(a + 2k) < f(a) + f(2k) = 2 + a + 2k$, from where we get that $a > 0$. If we put $x = y = a$ we get $a + 2a = f(2a) < f(a) + f(a) = 4$ so $3a < 4$.... | North Macedonia | 19-th Macedonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | No such function exists. | |
04s3 | Six teams will take part in a volleyball tournament. Each pair of the teams should play one match. All the matches will be realized in five rounds, each involving three simultaneous matches on the courts numbered 1, 2 and 3. Find the number of all possible draws for such a tournament. By a draw we mean a table $5 \time... | [
"We postpone the question of permutations of the five rounds and the three courts to the end of our solution. Denoting first the teams by numbers $1, 2, 3, 4, 5, 6$ (in a fixed way), we rearrange the five rounds of any satisfactory draw by means of the following numbering: Let $1$ and $2$ be the rounds with matches... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | 5598720 | |
0ctz | From a paper checkered square of size $100 \times 100$, one has cut out along the grid lines 1950 rectangles, each consisting of two cells. Prove that one may cut out from the remaining pieces a $T$-tetramino (□) — perhaps, rotated. (If such figure exists among the remaining pieces, we assume that it can be cut out.) (... | [
"**Hint 1.** Assume that the dominoes are being cut out one by one. At each moment, define the *price* of a remaining cell as the number of its neighbors which are still not cut out, decreased by 2. Initially, the total price of all cells is 19600, and at each step it decreases by at most 10. Thus, at the end there... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof only | null | |
05a7 | For any non-negative integer $i$, denote by $d_i$ the first digit of the number $2^i$. Let $n$ be a positive integer. Prove that there exists a non-zero digit that occurs in the tuple $(d_0, d_1, \dots, d_{n-1})$ less than $\frac{n}{17}$ times. | [
"The claim obviously holds for $n=1$, whence assume in the following that $n \\ge 2$. Let $k$ be the minimal number of occurrences of a non-zero digit in the tuple $(d_0, d_1, \\dots, d_{n-1})$. Then the total number of occurrences of digits $5$, $6$, $7$, $8$, $9$ is at least $5k$. As each digit $5$, $6$, $7$, $8$... | Estonia | Estonian Math Competitions | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0d03 | Show that
$$
\sum_{n=0}^{1006} \frac{2012!}{(n!(1006-n)!)^2}
$$
is a perfect square. | [
"Note that\n$$\n\\sum_{n=0}^{1006} \\frac{2012!}{(n!(1006-n)!)^2} = \\frac{2012!}{(1006!)^2} \\sum_{n=0}^{1006} \\left( \\frac{1006!}{n!(1006-n)!} \\right)^2 = \\left( \\frac{2012!}{1006!} \\right) \\sum_{n=0}^{1006} \\left( \\frac{1006!}{n!} \\right)^2.\n$$\nSince $\\binom{1006}{n} = \\binom{1006}{1006-n}$, we hav... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
07dg | Find all functions $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for all positive real numbers $x$, $y$
$$
\frac{x + f(y)}{x f(y)} = f\left(\frac{1}{y} + f\left(\frac{1}{x}\right)\right)
$$ | [
"The given equation can also be written as\n$$\n\\frac{1}{x} + \\frac{1}{f(y)} = f\\left(\\frac{1}{y} + f\\left(\\frac{1}{x}\\right)\\right).\n$$\nReplacing $x$ with $\\frac{1}{x}$ in the above equality, we obtain\n$$\nP(x, y) : x + \\frac{1}{f(y)} = f\\left(\\frac{1}{y} + f(x)\\right)\n$$\nIf there is some positiv... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = x for all positive real x | |
0krz | Problem:
Compute the remainder when
10002000400080016003200640128025605121024204840968192
is divided by 100020004000800160032. | [
"Solution:\n\nLet $X_{k}$ denote $2^{k}$ except with leading zeroes added to make it four digits long. Let $\\overline{a b c \\cdots}$ denote the number obtained upon concatenating $a, b, c, \\ldots$\n\nWe have\n$$\n2^{6} \\cdot \\overline{X_{0} X_{1} \\ldots X_{5}} = \\overline{X_{6} X_{7} \\ldots X_{11}}\n$$\nThe... | United States | HMMT February | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 40968192 | |
03z5 | Suppose that a line passing the circumcentre $O$ of $\triangle ABC$ intersects $AB$ and $AC$ at points $M$ and $N$, respectively, and $E$ and $F$ are the midpoints of $BN$ and $CM$, respectively. Prove that $\angle EOF = \angle A$. (posed by Tao Pingsheng) | [
"We show that the above conclusion is true for any triangle.\n\nIf $\\triangle ABC$ is right-angled. The conclusion is obvious. In fact, see Fig. 4. 1, where $\\angle ABC = 90^\\circ$. So, the circumcentre $O$ is the midpoint of $AC$, $OA = OB$ and $N = O$. Since $F$ is the midpoint of $CM$, we see that the median ... | China | China Southeastern Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null |
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