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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
05ij | Problem:
Trouver cent entiers positifs distincts $n_{1}, \ldots, n_{100}$ tels que
$$
1=\frac{1}{n_{1}}+\frac{1}{n_{2}}+\cdots+\frac{1}{n_{100}}
$$ | [
"Solution:\nNous montrerons par récurrence la propriété suivante :\nPour tout entier $k \\geqslant 3$ on peut trouver $k$ entiers positifs $n_{1}<n_{2}<\\cdots<n_{k}$ tels que\n$$\n1=\\frac{1}{n_{1}}+\\frac{1}{n_{2}}+\\cdots+\\frac{1}{n_{k}}\n$$\net que $n_{k}$ soit pair.\n\nPour $k=3$, il suffit de prendre $n_{1}=... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0728 | Problem:
If $\alpha$ is a real root of the equation $x^{5}-x^{3}+x-2=0$, prove that $\left[\alpha^{6}\right]=3$. (For any real number $a$, we denote by $[a]$ the greatest integer not exceeding $a$.) | [
"Solution:\n\nSuppose $\\alpha$ is a real root of the given equation. Then\n$$\n\\alpha^{5}-\\alpha^{3}+\\alpha-2=0\n$$\nThis gives $\\alpha^{5}-\\alpha^{3}+\\alpha-1=1$ and hence $(\\alpha-1)\\left(\\alpha^{4}+\\alpha^{3}+1\\right)=1$.\n\nObserve that $\\alpha^{4}+\\alpha^{3}+1 \\geq 2 \\alpha^{2}+\\alpha^{3}=\\al... | India | INMO 2004 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3 | |
0hud | Problem:
Aerith has an unlimited number of coins of values $5$, $m-1$, and $m$, and $71$ is the greatest value that cannot be made with a combination of these coins. Find all possible values of $m$. | [
"Solution:\n\nBy the Chicken McNugget Theorem, the largest coin value that can't be achieved with $5$ and $m-1$ is\n$$\n5(m-1)-5-(m-1)=4m-9 \\geq 71,\n$$\nso $4m \\geq 80$ and thus $m \\geq 20$.\n\nSince $71$ is not possible, there cannot be a number less than $71$ which is $1 \\pmod{5}$ and divisible by $m$ or $m-... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 20 and 38 | |
0cid | Let $ABC$ be an isosceles triangle with $AB = AC$ and $\angle BAC = 20^\circ$. Points $X$ and $Y$ are taken on side $AC$ such that $AX = YC = BC$. Find $\angle XBY$.
Mihaela Berindeanu | [] | Romania | 75th NMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 20° | |
05a8 | Determine all triples $(a,b,c)$ of integers which satisfy the equation
$$ (a-b)^3(a+b)^2 = c^2 + 2(a-b) + 1. $$ | [
"Answer: $(0,1,0), (-1,0,0)$.\n\nSubstituting $x = a - b$, $y = a + b$ we obtain the equation\n$$\nx^3 y^2 = c^2 + 2x + 1,\n$$\nwhich suffices to be solved in integers such that $x$ and $y$ have equal parity.\nWe shall consider the equivalent equation\n$$\nx(x^2 y^2 - 2) = c^2 + 1.\n$$\nIf both $x$ and $y$ are even... | Estonia | Estonian Math Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | (0,1,0), (-1,0,0) | |
08jh | Problem:
The real numbers $x$ and $y$ satisfy the equalities
$$
\sqrt{3x}\left(1+\frac{1}{x+y}\right)=2, \quad \sqrt{7y}\left(1-\frac{1}{x+y}\right)=4\sqrt{2}
$$
Find the numerical value of the ratio $y / x$. | [] | JBMO | The second selection test for JMBO 2003 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 6 | |
0l52 | Let $S$ be the set of vertices of a regular $24$-gon. Find the number of ways to draw $12$ segments of equal lengths so that each vertex in $S$ is an endpoint of exactly one of the $12$ segments. | [] | United States | AIME II | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 113 | |
0jx3 | Problem:
Let $ABCD$ be a quadrilateral with side lengths $AB=2$, $BC=3$, $CD=5$, and $DA=4$. What is the maximum possible radius of a circle inscribed in quadrilateral $ABCD$? | [
"Solution:\nLet the tangent lengths be $a, b, c, d$ so that\n$$\n\\begin{aligned}\n& a+b=2 \\\\\n& b+c=3 \\\\\n& c+d=5 \\\\\n& d+a=4\n\\end{aligned}\n$$\nThen $b=2-a$ and $c=1+a$ and $d=4-a$. The radius of the inscribed circle of quadrilateral $ABCD$ is given by\n$$\n\\sqrt{\\frac{abc+abd+acd+bcd}{a+b+c+d}} = \\sqr... | United States | February 2017 | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 2*sqrt(30)/7 | |
09ts | Problem:
Vind alle functies $f: \mathbb{R} \rightarrow \mathbb{R}$ waarvoor
$$
f\left(x^{2}\right)-f\left(y^{2}\right) \leq (f(x)+y)(x-f(y))
$$
voor alle $x, y \in \mathbb{R}$. | [
"Solution:\n\nOplossing I. Invullen van $x=y=0$ geeft $0 \\leq f(0) \\cdot -f(0)$. Maar kwadraten zijn nietnegatief, dus hieruit volgt $f(0)^{2}=0$ en daarmee $f(0)=0$. Nu geeft $x=0$ en $y=t$ dat $-f\\left(t^{2}\\right) \\leq t \\cdot -f(t)$, terwijl $x=t$ en $y=0$ geeft dat $f\\left(t^{2}\\right) \\leq f(t) \\cdo... | Netherlands | IMO-selectietoets III | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | f(x) = x for all real x; f(x) = -x for all real x | |
0jz7 | Problem:
Ashwin the frog is traveling on the $xy$-plane in a series of $2^{2017}-1$ steps, starting at the origin. At the $n^{\text{th}}$ step, if $n$ is odd, then Ashwin jumps one unit to the right. If $n$ is even, then Ashwin jumps $m$ units up, where $m$ is the greatest integer such that $2^{m}$ divides $n$. If Ash... | [
"Solution:\n\nNotice that since $v_{2}(x)=v_{2}\\left(2^{2017}-x\\right)$, the path divides the rectangle bonded by the coordinate axes and the two lines passing through Ashwin's final location parallel to the axes. The answer is therefore half of the product of the coordinates of Ashwin's final coordinates. The $x... | United States | HMMT November 2017 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2^{2015}*(2^{2017}-2018) | |
0jch | Problem:
Amy and Ben need to eat 1000 total carrots and 1000 total muffins. The muffins can not be eaten until all the carrots are eaten. Furthermore, Amy can not eat a muffin within 5 minutes of eating a carrot and neither can Ben. If Amy eats 40 carrots per minute and 70 muffins per minute and Ben eats 60 carrots pe... | [
"Solution:\n\nAnswer: 23.5 or $47 / 2$\n\nAmy and Ben will continuously eat carrots, then stop (not necessarily at the same time), and continuously eat muffins until no food is left. Suppose that Amy and Ben finish eating the carrots in $T_{1}$ minutes and the muffins $T_{2}$ minutes later; we wish to find the mini... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | final answer only | 47/2 | |
0f1v | Problem:
1000 numbers are written in line 1, then further lines are constructed as follows. If the number $m$ occurs in line $n$, then we write under it in line $n + 1$, each time it occurs, the number of times that $m$ occurs in line $n$. Show that lines 11 and 12 are identical. Show that we can choose numbers in lin... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
05vu | Problem:
Soit $\mathcal{S}$ un ensemble infini d'entiers naturels non nuls contenant quatre entiers $a, b, c, d$ deux à deux distincts tels que $\operatorname{pgcd}(a, b) \neq \operatorname{pgcd}(c, d)$. Démontrer que $\mathcal{S}$ contient trois entiers $x, y, z$ deux à deux distincts tels que $\operatorname{pgcd}(x,... | [
"Solution:\n\nCi-dessous, on note $m \\wedge n$ le plus grand diviseur commun à deux entiers $m$ et $n$. On considère alors le graphe colorié infini dont les sommets sont les éléments de $\\mathcal{S}$, et où chaque paire de sommets $(u, v)$ distincts l'un de l'autre est reliée par une arête de couleur $u \\wedge v... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0cbe | Let $ABCD$ be a cyclic quadrangle and let $M$ and $N$ be the midpoints of the sides $AD$ and $BC$, respectively. The circle through $A$ and $D$ tangent to $AC$ crosses the line $AB$ again at $P$ and the circle through $B$ and $C$ tangent to $BD$ crosses the line $AB$ again at $Q$. Let $t_A$ be the tangent of the circle... | [
"We first prove that the triangles $ADP$ and $CDB$ are similar. Since $ABCD$ is cyclic, $\\angle DAP = \\angle DCB$; and since $AC$ is tangent to the circle $ADP$, $\\angle CBD = \\angle CAD = \\angle APD$. Consequently, the triangles $ADP$ and $CDB$ are similar.\n\nLet $R$ be the midpoint of $CD$. The points $M$ a... | Romania | THE 73rd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS - SECOND SELECTION TEST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
0iy5 | Problem:
Let $s(n)$ denote the number of 1's in the binary representation of $n$. Compute
$$
\frac{1}{255} \sum_{0 \leq n<16} 2^{n}(-1)^{s(n)}
$$ | [
"Solution:\nNotice that if $n<8$, $(-1)^{s(n)} = (-1) \\cdot (-1)^{s(n+8)}$ so the sum becomes\n$$\n\\frac{1}{255}\\left(1-2^{8}\\right) \\sum_{0 \\leq n<8} 2^{n}(-1)^{s(n)} = 45.\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | final answer only | 45 | |
03po | Let a line with the inclination angle of $60^\circ$ be drawn through the focus $F$ of the parabola $y^2 = 8(x + 2)$. If the two intersection points of the line and the parabola are $A$ and $B$, and the perpendicular bisector of the chord $AB$ intersects the $x$-axis at the point $P$, then the length of the segment $PF$... | [
"It follows from the property of the focus of a parabola that $F = (0, 0)$. Then the equation of the straight line through points $A$ and $B$ will be $y = \\sqrt{3}x$. Substitute it into the parabola equation, and then obtain\n$$\n3x^2 - 8x - 16 = 0.\n$$\nLet $E$ be the midpoint of the chord $AB$, then the $x$-coor... | China | China Mathematical Competition (Shaanxi) | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | MCQ | A | |
0as7 | Problem:
There is a triple $k, m, n$ of positive integers without common factors such that $k \log_{400} 5 + m \log_{400} 2 = n$. Find the sum $k + m + n$. | [
"Solution:\n\n(ans. $(k, m, n) = (2, 4, 1),\\ k + m + n = 7$.\n\n$\\log_{400} 5^{k} 2^{m} = \\log_{400} 400^{n} \\Rightarrow 5^{k} 2^{m} = 400^{n} = (2^{4} 5^{2})^{n} \\Rightarrow m = 4n,\\ k = 2n \\Rightarrow (k, m, n) = (2, 4, 1)$."
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 7 | |
02em | Show that a convex polyhedron with an odd number of faces has at least one face with an even number of edges. | [
"Count the number $N$ of pairs $(F, e)$, where $F$ is a face and $e$ an edge belonging to $F$. Each edge belongs to two faces, so $N$ is even. Hence the number of faces with an odd number of edges must be even. The total number of faces is odd, so the number of faces with an even number of edges must be odd. In par... | Brazil | XII OBM | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof only | null | |
0171 | The circles $C_1$ and $C_2$ intersect at $A$ and $B$. The points $P$ and $Q$ are on $C_2$, $P$ in the interior and $Q$ in the exterior of $C_1$. The lines $AP$ and $BP$ meet $C_1$ also at $X$ and $Y$, respectively, while the lines $QA$ and $QB$ meet $C_1$ also at $Z$ and $T$. Show that $XY = ZT$. | [
"Let $\\angle APB = \\alpha$, $\\angle AYB = \\beta$. Then $\\angle YPA = 180^\\circ - \\alpha$ and $\\angle XAY = 180^\\circ - (180^\\circ - \\alpha) - \\beta = \\alpha - \\beta$. As $AQBP$ is cyclic, $\\angle AQB = 180^\\circ - \\alpha$. As $\\angle ATB = \\beta$, $\\angle ZAT = \\beta + 180^\\circ - \\alpha = 18... | Baltic Way | BALTIC WAY | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0eir | Problem:
Dano je naravno število $m$. Zaporedje $a_{n}$ je definirano s pogojema $a_{1}=m$ in $a_{n+1} = \left[\frac{a_{n}^{2} + 10 a_{n}}{2 a_{n} + 4}\right]$ za vse $n \in \mathbb{N}$. Dokaži, da je zaporedje $a_{n}$ od nekega člena dalje konstantno.
Opomba. Za poljubno realno število $x$ oznaka $[x]$ označuje najv... | [
"Solution:\n\nČe je $a_{n} \\geq 0$ za nek $n$, tedaj je $\\frac{a_{n}^{2} + 10 a_{n}}{2 a_{n} + 4} \\geq 0$ in zato je tudi $a_{n+1} \\geq 0$. Ker je $a_{1} = m > 0$, so torej vsi členi zaporedja $a_{n}$ nenegativna cela števila.\n\nZa majhna naravna števila $m$ izračunamo nekaj členov zaporedja $a_{n}$:\n- pri $m... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"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"
] | null | proof only | null | |
05mk | Problem:
On dit qu'un entier strictement positif $n$ est amusant si pour tout diviseur strictement positif $d$ de $n$, l'entier $d+2$ est premier. Déterminer tous les entiers amusants dont le nombre de diviseurs est maximum. | [
"Solution:\n\nSoit $n$ un entier amusant et $p$ un diviseur premier de $n$. Alors $p$ est impair, puisque $p+2$ est premier.\n\nSupposons que $p \\geqslant 5$. Alors $p+2$ est premier, et $p+2>3$, donc $p+2$ n'est pas divisible par $3$. On en déduit que $p$ n'est pas congru à $1$ modulo $3$. Il est clair que $p$ n'... | France | Olympiades FRANçaises DE Mathématiques | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | 135 | |
0d8t | Find all pairs of primes $(p, q)$ such that
$$
p^{3} - q^{5} = (p + q)^{2}.
$$ | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | (p, q) = (7, 3) | |
032q | Problem:
Find all values of $x \in (-\pi, \pi)$ such that the numbers $2^{\sin x}$, $2-2^{\sin x+\cos x}$ and $2^{\cos x}$ are consecutive terms of a geometric progression. | [
"Solution:\nThe numbers form a geometric progression if and only if\n$$\n2^{\\sin x} \\cdot 2^{\\cos x} = \\left(2 - 2^{\\sin x+\\cos x}\\right)^2\n$$\nThis is equivalent to $4^{\\sin x+\\cos x} - 5 \\cdot 2^{\\sin x+\\cos x} + 4 = 0$. Setting $y = 2^{\\sin x+\\cos x}$ gives $y^2 - 5y + 4 = 0$, i.e., $y_1 = 4$ or $... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | x = -π/4 or x = 3π/4 | |
016s | Let $ABC$ be a scalene and non-right triangle. Let $A'$ be the second intersection point of the median drawn from $A$ with the circumcircle of the triangle. Let the tangents to the circumcircle of $ABC$ at points $A$ and $A'$ intersect at $A''$. Similarly define points $B''$ and $C''$. Prove that $A''$, $B''$, $C''$ ar... | [
"Let $M$ be the centroid and $O$ the circumcentre of the triangle $ABC$. As $ABC$ is scalene, $M \\neq O$. We will show that the points $A''$, $B''$ and $C''$ lie on the radical axis of the circumcircle of $ABC$ and the circle with diameter $MO$.\nLet $N$ be the intersection of the line $A''O$ and line $AA'$. Now $... | Baltic Way | BALTIC WAY | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
... | null | proof only | null | |
0gkr | Let $m, n$ be positive integers. There are $n$ piles of gold coins, where the $i^{th}$ pile contains $a_i (> 0)$ coins ($i = 1, \dots, n$). Consider the following game:
**Step 1** Bob selects sets $B_1, B_2, \dots, B_n$ such that each set $B_i$ is a nonempty subset of $\{1, 2, \dots, m\}$.
**Step 2** Alice selects a se... | [
"At the end of the game, the number of gold coins received by Bob less that of Alice is equal to $\\sum_{i=1}^{n}(-1)^{|B_i \\cap S|} \\cdot a_i$. The number of Alice coins is greater than that of Bob if and only if $\\sum_{i=1}^{n}(-1)^{|B_i \\cap S|} \\cdot a_i < 0$.\n\nSuppose (for contradiction purposes) that f... | Thailand | The 10th Thailand Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
022p | Problem:
O macaco diz para a raposa:
- Você vê as 3 pessoas que estão correndo lá longe? Eu sei que o produto de suas idades é $2450$; e que a soma das idades é o dôbro da sua idade. Você pode me dizer suas idades?
- Não, responde a raposa.
- E se eu te disser que o mais jovem dos três é o único louro, você pode agora... | [
"Solution:\n\n$2450$ é o produto dos números primos $1, 2, 5, 5, 7, 7$. As 3 idades correspondem a uma combinação particular desses números ou de seus produtos.\n\nA raposa não pode descobrir as idades no início porque pelo menos duas dessas combinações têm por soma o dobro de sua idade. De todas as combinações pos... | Brazil | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Logic"
] | null | proof and answer | The fox is 32 years old; the three ages are 5, 10, and 49. | |
0a73 | Problem:
Let $m_{n}$ be the smallest value of the function
$$
f_{n}(x)=\sum_{k=0}^{2 n} x^{k}
$$
Show that $m_{n} \rightarrow \frac{1}{2}$, as $n \rightarrow \infty$. | [
"Solution:\nFor $n>1$,\n$$\n\\begin{gathered}\nf_{n}(x)=1+x+x^{2}+\\cdots \\\\\n=1+x\\left(1+x^{2}+x^{4}+\\cdots\\right)+x^{2}\\left(1+x^{2}+x^{4} \\cdots\\right) \\\\\n=1+x(1+x) \\sum_{k=0}^{n-1} x^{2 k}\n\\end{gathered}\n$$\nFrom this we see that $f_{n}(x) \\geq 1$, for $x \\leq -1$ and $x \\geq 0$. Consequently,... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1/2 | |
02t2 | Problem:
Encontre dois inteiros positivos $x$ e $y$ tais que:
$$
\frac{1-\sqrt{2}+\sqrt{3}}{1+\sqrt{2}-\sqrt{3}}=\frac{\sqrt{x}+\sqrt{y}}{2}
$$ | [
"Solution:\n$$\n\\begin{aligned}\n\\frac{1-\\sqrt{2}+\\sqrt{3}}{1+\\sqrt{2}-\\sqrt{3}} \\cdot \\frac{\\sqrt{2}}{\\sqrt{2}} & =\\frac{-2+\\sqrt{2}+\\sqrt{6}}{2+\\sqrt{2}-\\sqrt{6}} \\\\\n& =\\frac{-2+\\sqrt{2}+\\sqrt{6}}{2+\\sqrt{2}-\\sqrt{6}} \\cdot \\frac{2 \\sqrt{2}+2+2 \\sqrt{3}}{2 \\sqrt{2}+2+2 \\sqrt{3}} \\\\\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | x = 2, y = 6 | |
0e75 | Let $ABC$ be an acute triangle. The bisector of the side $AB$ intersects the lines $BC$ and $CA$ at the points $X$ and $Y$, respectively, and the bisector of the side $AC$ intersects the lines $BC$ and $AB$ at the points $Z$ and $W$, respectively. Prove that the points $X, Y, Z$ and $W$ are concyclic. | [
"The solution uses directed angles. Let $E$ be the midpoint of the side $AB$ and let $F$ be the midpoint of the side $AC$. Since $EF$ is a midline of the triangle $ABC$, $EF$ is parallel to $BC$. Because the lines $XY$ and $ZW$ are bisectors of the sides $AB$ and $AC$, respectively, the points $E, X$ and $Y$ as wel... | Slovenia | Selection Examinations for the IMO 2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0frh | Problem:
Demostrar que todos los números racionales pueden expresarse como suma de algunas fracciones de la forma $\frac{n-1}{n+2}$, con $n \geq 0$ entero, admitiendo repetir sumandos. | [
"Solution:\n\nLlamamos $f(n)=\\frac{n-1}{n+2}$, y denotamos por $A$ al conjunto de números que pueden expresarse como suma de elementos $f(n)$, para ciertos $n \\geq 0$. Probaremos que $A=\\mathbb{Q}$.\n\nTodo número racional puede escribirse como un cociente $\\frac{k}{n}$, con $k$ entero (positivo, negativo o nul... | Spain | Spain | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other"
] | null | proof only | null | |
0cil | Determine all functions $f: \mathbb{N}^* \to \mathbb{N}$ that satisfy the following properties:
a) the expression $n^3 - n f(n) - 1$ is a cube of a natural number for all $n \in \mathbb{N}^*$;
b) the function $g: \mathbb{N}^* \to \mathbb{N}$ defined by $g(n) = 4n - f(n)$ is increasing. | [] | Romania | 75th NMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | f(n) = 3n − 3 for all n ∈ ℕ* | |
013a | Problem:
Given a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of non-negative real numbers satisfying the conditions
(1) $a_{n} + a_{2n} \geq 3n$
(2) $a_{n+1} + n \leq 2 \sqrt{a_{n} \cdot (n+1)}$
for all indices $n = 1, 2, \ldots$
a. Prove that the inequality $a_{n} \geq n$ holds for every $n \in \mathbb{N}$.
b. Give a... | [
"Solution:\n\na.\nNote that the inequality\n$$\n\\frac{a_{n+1} + n}{2} \\geq \\sqrt{a_{n+1} \\cdot n}\n$$\nholds, which together with the second condition of the problem gives\n$$\n\\sqrt{a_{n+1} \\cdot n} \\leq \\sqrt{a_{n} \\cdot (n+1)}\n$$\nThis inequality simplifies to\n$$\n\\frac{a_{n+1}}{a_{n}} \\leq \\frac{n... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | For all indices, the sequence satisfies a_n at least n; one example is a_n equal to n plus one. | |
02mk | Problem:
Coloque os números $663, 664, 665, 666, 667, 668, 669, 670$ e $671$, sem repetir, numa tabela $3 \times 3$, de tal maneira que a soma em cada linha, em cada coluna e em cada diagonal seja $2001$. Caso isso não seja possível, justifique sua resposta. | [
"Solution:\n\nVamos analisar se é possível dispor os números $663, 664, 665, 666, 667, 668, 669, 670$ e $671$ em uma tabela $3 \\times 3$ de modo que a soma de cada linha, coluna e diagonal seja $2001$.\n\nA soma de todos os números é:\n$$\n663 + 664 + 665 + 666 + 667 + 668 + 669 + 670 + 671 = (663 + 671) + (664 + ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Yes. One valid arrangement is: [[664, 669, 668], [671, 667, 663], [666, 665, 670]]. | |
033x | Problem:
Find all values of $a$ such that the equation
$$
\sqrt{\left(4 a^{2}-4 a-1\right) x^{2}-2 a x+1}=1-a x-x^{2}
$$
has exactly two solutions. | [
"Solution:\nSquaring the equation\n$$\n\\sqrt{\\left(4 a^{2}-4 a-1\\right) x^{2}-2 a x+1}=1-a x-x^{2}\n$$\ngives the equation\n$$\nx^{2}\\left(x^{2}+2 a x-3 a^{2}+4 a-1\\right)=0\n$$\nwith roots $x_{1}=0$, $x_{2}=1-3 a$ and $x_{3}=a-1$. It is clear that $x_{1}=0$ is a root of (1) for any $a$. On the other hand, $x_... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a = 1/3, a = 1/2, and a in (5/6, 3/2] excluding 1 | |
03ho | Problem:
i. $15$ chairs are equally placed around a circular table on which are name cards for $15$ guests. The guests fail to notice these cards until after they have sat down, and it turns out that no one is sitting in the correct seat. Prove that the table can be rotated so that at least two of the guests are simul... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0kwn | Problem:
The cells of a $5 \times 5$ grid are each colored red, white, or blue. Sam starts at the bottom-left cell of the grid and walks to the top-right cell by taking steps one cell either up or to the right. Thus, he passes through 9 cells on his path, including the start and end cells. Compute the number of colori... | [
"Solution:\n\nLet $c_{i, j}$ denote the cell in the $i$-th row from the bottom and the $j$-th column from the left, so Sam starts at $c_{1,1}$ and is traveling to $c_{5,5}$. The key observation (from, say, trying small cases) is that\n\nClaim. For $1 \\leq i, j < 5$, the cells $c_{i+1, j}$ and $c_{i, j+1}$ must be ... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 1680 | |
0isj | Problem:
Let $f(x) = \sin^6\left(\frac{x}{4}\right) + \cos^6\left(\frac{x}{4}\right)$ for all real numbers $x$. Determine $f^{(2008)}(0)$ (i.e., $f$ differentiated 2008 times and then evaluated at $x=0$). | [
"Solution:\n\nWe have\n$$\n\\begin{aligned}\n\\sin^6 x + \\cos^6 x &= \\left(\\sin^2 x + \\cos^2 x\\right)^3 - 3 \\sin^2 x \\cos^2 x \\left(\\sin^2 x + \\cos^2 x\\right) \\\\\n&= 1 - 3 \\sin^2 x \\cos^2 x = 1 - \\frac{3}{4} \\sin^2 2x = 1 - \\frac{3}{4}\\left(\\frac{1 - \\cos 4x}{2}\\right) \\\\\n&= \\frac{5}{8} + ... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Trigonometric functions"
] | null | proof and answer | 3/8 | |
034s | Problem:
On the sides of an acute $\triangle ABC$ of area $1$ points $A_1 \in BC$, $B_1 \in CA$ and $C_1 \in AB$ are chosen so that
$$
\Varangle CC_1B = \Varangle AA_1C = \Varangle BB_1A = \varphi,
$$
where the angle $\varphi$ is acute. The segments $AA_1$, $BB_1$ and $CC_1$ meet at points $M$, $N$ and $P$.
a) Prove ... | [
"Solution:\n\nSet $AA_1 \\cap BB_1 = M$, $BB_1 \\cap CC_1 = N$ and $CC_1 \\cap AA_1 = P$. Using the standard notation for the angles of $\\triangle ABC$, we have\n$$\n\\Varangle PMN = \\varphi - \\Varangle B_1BC = \\varphi - (\\varphi - \\gamma) = \\gamma\n$$\nAnalogously, we get $\\Varangle MNP = \\alpha$ and $\\V... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > ... | null | proof and answer | 75 degrees | |
0jtn | Problem:
We want to design a new chess piece, the American, with the property that (i) the American can never attack itself, and (ii) if an American $A_{1}$ attacks another American $A_{2}$, then $A_{2}$ also attacks $A_{1}$. Let $m$ be the number of squares that an American attacks when placed in the top left corner o... | [
"Solution:\nAnswer: $1024$\nSince one of the Americans must be in the top left corner, that eliminates $m$ squares from consideration for placing additional Americans. So $m+n$ is at most $64$, which implies $m n$ can be at most $1024$. To achieve $1024$, we can color a chessboard the normal way, and say that an Am... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 1024 | |
0c0r | Determine how many possible ways can one put two distinct natural numbers, which are less or equal to $9$ and have the sum equal to $10$, in a $10 \times 10$ table, such that the two numbers belong to distinct lines and to distinct columns. | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 32400 | |
06g4 | For any positive real numbers $a$, $b$, $c$, prove that
$$
(ab + bc + ca) \left( \frac{1}{(a+b)^2} + \frac{1}{(b+c)^2} + \frac{1}{(c+a)^2} \right) \ge \frac{9}{4}.
$$ | [
"(Iran TST 1996) Since the expressions are homogeneous, WLOG we may assume $ab + bc + ca = 1$. We need to minimize\n$$\nf(a, b, c) = \\frac{1}{(a+b)^2} + \\frac{1}{(b+c)^2} + \\frac{1}{(c+a)^2}.\n$$\nWLOG assume $a \\ge b \\ge c$. Let $t$ be the positive root to $t^2 + 2ct = 1$, so that $(t)(t) + (t)(c) + (c)(t) = ... | Hong Kong | CHKMO | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
00e0 | Ana and Beto play the following game on a $2022 \times 2022$ grid. First Ana colors some unit segments on the grid red, so that no cell has two red sides who share a vertex. Next, Beto must draw a blue path that joins two of the four corners of the grid, following the sides of the grid and without using any red segment... | [
"We will show that Beto has a winning strategy. We will prove that he can connect the bottom left corner with the bottom right or the top left corner.\n\nWe say that a row is *full red* if every vertical line in it is red.\n\n| R | R | R | | | | R | R | R |\n|---|---|---|---|---|---|---|---|---|\n\nAnalogousl... | Argentina | XXXIII Cono Sur Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Beto | |
0g7i | 設有一個 $2013 \times 2013$ 的棋盤, 其格線的方向為南北/東西向。在每一個方格內填入一個朝東或朝南的箭頭。每一分鐘內, 對所有的箭頭 $A$, 如果 $A$ 指向相鄰的某方格, 且該方格中的箭頭 $B$ 的方向與 $A$ 不同, 則下一分鐘將 $A$ 換成另一個方向的箭頭 (如果 $A$ 原本朝南則換成朝東, 如果 $A$ 原本朝東則換成朝南); 否則箭頭 $A$ 不動。所有這樣的變換在每一分鐘同時執行。
試證：在有限時間之後，所有的箭頭都不能再變換方向。並求出還能改變至少一個箭頭方向的最長可能時間。 | [
"本題的答案是最多在前 $4024$ 分鐘內可以有箭頭方向變換產生。\n\n我們首先證明：自第 $4025$ 分鐘起, 所有箭頭的方向都固定下來, 不再產生變換。\n\n在棋盤上建立坐標系：每一個方格的坐標爲 $(i, j)$，其中 $0 \\le i, j \\le 2012$，且 $(0, 0)$ 方格在棋盤的東南角。我們宣稱：若 $x + y < m$，則方格 $(x, y)$ 內的箭頭自第 $m$ 分鐘起都不再改變方向。\n\n我們對 $m$ 作數學歸納法。$m = 1$ 的情形自動成立，因為 $(0, 0)$ 方格內的箭頭並未指向其他方格。\n\n現在我們假設命題對正整數 $m$ 成立, 亦即只要 $x + y <... | Taiwan | 二〇一三數學奧林匹亞競賽第三階段選訓營, 獨立研究(一) | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 4024 | |
0ee4 | Which quadratic equation has roots $1 - \sqrt{2}$ and $\frac{1}{\sqrt{2}-1}$?
(A) $x^2 + 2x - 1 = 0$ (B) $x^2 + \sqrt{2}x - 1 = 0$ (C) $x^2 + x - 1 = 0$
(D) $x^2 - 2x - 1 = 0$ (E) $x^2 - x + 2 = 0$ | [
"Rationalize the second root to get $\\frac{1}{\\sqrt{2}-1} = \\frac{\\sqrt{2}+1}{2-1} = \\sqrt{2} + 1$. These roots belong to the equation $(x - (1 - \\sqrt{2})) (x - (\\sqrt{2} + 1)) = 0$. Expanding the left side we get $x^2 - 2x - 1 = 0$. The correct answer is (D)."
] | Slovenia | Slovenija 2016 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | D | |
06ci | Determine all positive rational numbers $r \neq 1$ such that $r^{\frac{1}{r-1}}$ is rational. | [
"$r$ can be any number of the form $1 + \\frac{1}{k}$, where $k$ is an integer different from $0$, $-1$.\n\nIf $r > 1$, let $r = 1 + \\frac{m}{n}$ where $m, n \\in \\mathbb{Z}^+$ and $(m, n) = 1$. Since\n$$\nr^{\\frac{1}{r-1}} = \\left(\\frac{n+m}{n}\\right)^{\\frac{n}{m}}\n$$\nand $(n+m, n) = (m, n) = 1$, we must ... | Hong Kong | CHKMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | r = 1 + 1/n for integers n ≥ 1, or r = 1 − 1/n for integers n > 1 | |
01oz | Three of six segments (three sides and three medians of a triangle) are painted red, and three others are painted blue.
Can one construct a triangle using the segments of the same color as its sides? | [
"Answer: yes, one can.\n\nLet $G$ be a gravicenter of the triangle $ABC$, and $A_1$, $B_1$, $C_1$ be the midpoints of the sides $BC$, $AC$, $AB$ respectively. Denote the sides and the medians of the triangle $ABC$ in the following way: $AB = c$, $BC = a$, $CA = b$, $AA_1 = d$, $BB_1 = e$, $CC_1 = f$.\n\nSuppose tha... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"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 | Yes | |
0cpd | A triangle $ABC$ has perimeter $4$. Points $X$ and $Y$ are chosen on the rays $AB$ and $AC$ respectively so that $AX = AY = 1$. Segments $BC$ and $XY$ intersect at point $M$. Prove that the perimeter of one of the triangles $ABM$ and $ACM$ equals $2$. (V. Shmarov)
Периметр треугольника $ABC$ равен $4$. На лучах $AB$ и... | [
"Поскольку отрезки $BC$ и $XY$ пересекаются, без ограничения общности можно считать, что $AB > AX$ и $AC < AY$.\nПусть вневписанная окружность $\\omega$ исходного треугольника касается стороны $BC$ в точке $R$, а продолжений сторон $AB$ и $AC$ в точках $P$ и $Q$ соответственно. Пусть $J$ — центр $\\omega$, а $r$ — ... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English, Russian | proof only | null | |
0erk | All my Maths tests are out of the same total, and the average of my marks in the four tests so far is $75\%$. What percentage do I need in a fifth test so that my overall average will be $80\%$?
(A) $100$
(B) $90$
(C) $85$
(D) $80$
(E) $75$ | [
"Let the five marks be $a$, $b$, $c$, $d$ and $e$.\n\n$$\n\\frac{a+b+c+d+e}{5} = 80 \\therefore a+b+c+d+e = 400\n$$\nBut $\\frac{a+b+c+d}{4} = 75 \\therefore a+b+c+d = 300$, and so $e = 100$"
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | MCQ | A | |
07bk | Square $ABCD$ is given. Points $N$ and $P$ are selected on sides $AB$ and $AD$, respectively, such that $PN = NC$, and point $Q$ is selected on segment $AN$ such that $\angle NCB = \angle QPN$. Prove that $\angle BCQ = \frac{1}{2} \angle PQA$. | [
"Let $E$ be the intersection point of $PQ$ and $BC$. According to the problem assumption, $PN = NC$ and so $\\angle NPC = \\angle PCN$. On the other hand, we know $\\angle QPN = \\angle NCB$. From these we conclude that $EPC$ is an isosceles triangle. Therefore, its altitudes $PS$ and $CK$ have equal length. So $CK... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
084t | Problem:
Un numero si dice "moderno" se, in base 10, può essere espresso concatenando "un po'" di scritture decimali di $2006$: ad esempio $200620062006$ è moderno, mentre $20200606$ e $2006200$ non lo sono. Quante cifre ha il più piccolo quadrato perfetto moderno positivo?
(A) 32
(B) 64
(C) 100
(D) 1000
(E) non esis... | [
"Solution:\n\nLa risposta è (E). Un numero moderno è sempre divisibile per $2006$, e il quoziente è della forma $1000100010001 \\ldots 10001$, quindi dispari. Un numero moderno è quindi sempre pari, ma non è mai divisibile per $4$, e non può quindi essere un quadrato perfetto."
] | Italy | Progetto Olimpiadi di Matematica 2006 GARA di SECONDO LIVELLO | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | MCQ | E | |
00f1 | Determine all functions $f$ from the reals to the reals for which
(1) $f(x)$ is strictly increasing,
(2) $f(x)+g(x)=2x$ for all real $x$, where $g(x)$ is the composition inverse function to $f(x)$.
(Note: $f$ and $g$ are said to be composition inverses if $f(g(x))=x$ and $g(f(x))=x$ for all real $x$.) | [
"Denote by $f_n$ the $n$th iterate of $f$, that is, $f_n(x)=\\underbrace{f(f(\\ldots f}_{n \\text{ times }}(x)))$.\nPlug $x \\rightarrow f_{n+1}(x)$ in (2): since $g\\left(f_{n+1}(x)\\right)=g\\left(f\\left(f_n(x)\\right)\\right)=f_n(x)$,\n$$\nf_{n+2}(x)+f_n(x)=2 f_{n+1}(x),\n$$\nthat is,\n$$\nf_{n+2}(x)-f_{n+1}(x)... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1989 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x + c for any real constant c | |
095u | Problem:
Numerele de forma $\frac{k(k+1)}{2}$, unde $k \in \mathbb{N}$, se numesc triunghiulare. Să se stabilească, dacă 2018 este suma a două numere triunghiulare. | [
"Solution:\n\nFie $\\frac{m(m+1)}{2}+\\frac{n(n+1)}{2}=2018$, cu $m, n \\in \\mathbb{N}$. De aici se obţine consecutiv:\n$$\n\\begin{gathered}\nm^{2}+m+n^{2}+n=4036 ;\\left(4 m^{2}+4 m+1\\right)+\\left(4 n^{2}+4 n+1\\right)=4 \\cdot 4036+2 \\\\\n(2 m+1)^{2}+(2 n+1)^{2}=16146\n\\end{gathered}\n$$\nPartea dreaptă a e... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2018 is not a sum of two triangular numbers. | |
0379 | Problem:
Let $n$ be a positive integer. Find the number of all finite strictly increasing sequences $a_{0}=1, a_{1}, \ldots, a_{k}=2.3^{n}$ of positive integers with the following property: $\prod_{i=1}^{k}\left[\frac{a_{i}+a_{i-1}-1}{a_{i-1}}\right]=2.3^{n}$, where $[x]$ is the integral part of $x$. | [
"Solution:\nLet $m$ and $n$ be positive integers and $n=m q+r, 0 \\leq r<m$. Then $\\left[\\frac{n+m-1}{m}\\right]=q+1+\\left[\\frac{r-1}{m}\\right]$ and $\\frac{n}{m}=q+\\frac{r}{m}$. Hence $\\left[\\frac{n+m-1}{m}\\right] \\geq \\frac{n}{m}$ with equality if and only if $m$ divides $n$.\n\nApplying the above ineq... | Bulgaria | Spring Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2^{n-1}(n+2) | |
0fj5 | Problem:
Sean $a$, $b$, y $c$ números reales. Prueba que si $x^{3} + a x^{2} + b x + c$ tiene tres raíces reales, entonces $3b \leq a^{2}$. | [
"Solution:\nSean $\\alpha$, $\\beta$, $\\gamma$ las raíces y supongamos $\\alpha \\geq \\beta \\geq \\gamma$.\n$$\nx^{3} + a x^{2} + b x + c = (x - \\alpha)(x - \\beta)(x - \\gamma) = x^{3} + (-\\alpha - \\beta - \\gamma) x^{2} + (\\alpha \\beta + \\alpha \\gamma + \\beta \\gamma) x - \\alpha \\beta \\gamma\n$$\nAs... | Spain | Viernes 19 de enero de 2001 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0id3 | Let $A = (0,0,0)$ be the origin in the three dimensional coordinate space. The *weight* of a point is the sum of the absolute values of its coordinates. A point is a *primitive lattice point* if all its coordinates are integers with their greatest common divisor equal to 1. A square $ABCD$ is called a *unbalanced primi... | [
"**First Solution:** Let $(a, b, c)$ be a **Pythagorean Triple**, that is, $a, b, c$ are positive integers with $a^2 + b^2 = c^2$. The key facts are\n(i) for any positive integers $m$ and $n$ such that $m > n$, $(m^2 - n^2, 2mn, m^2 + n^2)$ is a Pythagorean triple; and\n(ii) the vector $[c-a, c-b, a+b-c]$ has integ... | United States | USA IMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Divisibility / Factoriz... | null | proof only | null | |
02v3 | Problem:
Seja $n>1$ um inteiro positivo, chamamos de $d(n)$ a quantidade de divisores positivos de $n$. Para calcular $d(n)$, basta escrever a fatoração de $n$ em potências de primos distintos e multiplicar os sucessores dos expoentes. Por exemplo, para $2016$ temos a fatoração $2016=2^{5} \cdot 3^{2} \cdot 7^{1}$ e $... | [
"Solution:\n\na. Se $n$ é um quadrado perfeito, então existe um inteiro positivo $m$ tal que $n=m^{2}$. Na fatoração em primos de $n$, todos os expoentes são pares, pois cada fator primo de $m$ aparece o dobro de vezes na fatoração de $n$. Por exemplo, se $m=2^{2} \\cdot 3$, então $n=m^{2}=\\left(2^{2} \\cdot 3\\ri... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | [36, 100, 196, 225, 256] | |
0f2a | Problem:
A polynomial is monic if its leading coefficient is $1$. Two polynomials $p(x)$ and $q(x)$ commute if $p(q(x)) = q(p(x))$.
(1) Find all monic polynomials of degree $3$ or less which commute with $x^2 - k$.
(2) Given a monic polynomial $p(x)$, show that there is at most one monic polynomial of degree $n$ whic... | [] | Soviet Union | ASU | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | null | |
0f3h | Problem:
Define the sequence $\{a\}_{n}$ of positive integers as follows. $a_{1} = m$. $a_{n + 1} = a_{n}$ plus the product of the digits of $a_{n}$. For example, if $m = 5$, we have $5, 10, 10, \ldots$. Is there an $m$ for which the sequence is unbounded? | [
"Solution:\n\nPut $p(n)$ for the product of the digits of $n$. We show that, for sufficiently large $n$, a sequence starting below it cannot get past the \"gap\" from $10^{n}$ to $10^{n} + 10^{n - 1}$. For suppose $N$ is the last member of the sequence below the gap. Then $N$ has at most $n$ digits, so $p(N) \\leq ... | Soviet Union | ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | null | proof and answer | No; for every starting value the sequence is bounded. | |
0i4f | Problem:
How many real solutions are there to the equation
$$
||||x|-2|-2|-2|=||||x|-3|-3|-3| ?
$$ | [
"Solution:\n6. The graphs of the two sides of the equation can be graphed on the same plot to reveal six intersection points."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 6 | |
0hwe | Problem:
Suppose $a, b, c, d$, and $e$ are objects that we can multiply together, but the multiplication doesn't necessarily satisfy the associative law, i.e. $(x y) z$ does not necessarily equal $x(y z)$. How many different ways are there to interpret the product $a b c d e$? | [
"Solution:\n$C$ (number of letters $-1$) $= C(4) = 14$."
] | United States | null | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | final answer only | 14 | |
0ero | A target consists of two circles with the same centre. The larger circle has radius $12$ cm. When shots hit the target randomly, $403$ out of $900$ land in the inner circle. The radius of the inner circle is closest to
(A) $6$ cm
(B) $7$ cm
(C) $8$ cm
(D) $9$ cm
(E) $10$ cm | [
"If the radius is $r$ cm then\n$$\n\\frac{\\text{area smaller circle}}{\\text{area larger circle}} = \\frac{\\pi r^2}{\\pi \\cdot 12^2} = \\frac{r^2}{12^2} = \\frac{403}{900} \\approx \\frac{4}{9}\n$$\nSo $r^2$ is close to $\\frac{2^2 \\cdot 12^2}{3^2}$ which means $r$ is $\\frac{2 \\cdot 12}{3} = 8$"
] | South Africa | South African Mathematics Olympiad First Round | [
"Geometry > Plane Geometry > Circles"
] | English | MCQ | C | |
0ks0 | Problem:
Let $S$ be a finite set of positive real numbers. If $S$'s average is at most $1$ but its product is at least $0.9$, show that any three elements of $S$ can form the sides of a triangle. | [
"Solution:\n\nAssume otherwise for the sake of contradiction, i.e. that there exist $x, y, z \\in S$ so that $x + y \\leq z$. For fixed $z$, the product $x y$ is then maximized when $x = y$. Then, if $x, y, z$ have average $a$, their product is at most that when $x + y = z$, which happens at $x = 3a/4$, $y = 3a/4$,... | United States | Berkeley Math Circle | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0bl6 | Problem:
O "formă L" este oricare din următoarele patru piese, fiecare constând din trei pătrăţele unitate:
Se dau: o tablă $5 \times 5$, constând din $25$ de pătrăţele unitate, un număr natural nenul $k \leq 25$ şi o colecţie nelimitată de "forme L".
Doi jucători, $A$ şi $B$, joacă următo... | [] | Romania | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 6 | |
08aj | Problem:
Davide fa il seguente gioco: parte da un numero intero compreso tra 1 e 99 e ad ogni mossa sostituisce il numero $n$ che ha al momento con il numero formato dalle ultime due cifre di $51 n+50$ (o solo dall'ultima cifra, se la penultima è 0). Quanti numeri diversi può ottenere al massimo nel corso delle prime ... | [
"Solution:\n\nLa risposta è $\\mathbf{( A )}$. Supponiamo di partire da un numero $n$ dispari: allora lo possiamo scrivere come $n=2k+1$, e moltiplicando per 51 ed aggiungendo 50 troviamo $102k+101=100k+100+(2k+1)$. Ora $100k+100$ è un numero che termina con due zeri, quindi le ultime due cifre di $51n+50$ sono le ... | Italy | Progetto Olimpiadi della Matematica - Gara di Febbraio | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | A | |
0es8 | For every positive integer $n$, determine the greatest possible value of the quotient
$$
\frac{1 - x^n - (1-x)^n}{x(1-x)^n + (1-x)x^n}
$$
where $0 < x < 1$. | [
"It is convenient to write $y = 1 - x$, so that $x + y = 1$. We show that the maximum of the resulting expression\n$$\n\\frac{1 - x^n - y^n}{xy^n + yx^n}\n$$\nis attained when $x = y = \\frac{1}{2}$. The value in this case is $2^n - 2$. We first rewrite the expression as follows:\n$$\n\\begin{aligned} \\frac{1 - x^... | South Africa | South African Mathematics Olympiad Third Round | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | 2^n - 2 | |
0ksz | Problem:
All positive integers whose binary representations (excluding leading zeroes) have at least as many $1$'s as $0$'s are put in increasing order. Compute the number of digits in the binary representation of the $200$th number. | [
"Solution:\n\nWe do a rough estimation. There are $255$ positive integers with at most $8$ digits and a majority of them, but not more than $200$, satisfy the property. Meanwhile, there are $511$ positive integers with at most $9$ digits, and a majority of them satisfy this property. Thus, the answer must be greate... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 9 | |
0c5q | Around point $O$ one considers the angles $\widehat{A_0OA_1} = 1^\circ$, $\widehat{A_1OA_2} = 2^\circ$, $\widehat{A_2OA_3} = 3^\circ, \dots, \widehat{A_{25}OA_{26}} = 26^\circ$ and $\widehat{A_{26}OA_0}$.
a) Determine the measure of $\widehat{A_{26}OA_0}$.
b) For what integers $n$, such that $1 \le n \le 25$, one has... | [
"a.\n\nThe sum of all angles around point $O$ is $360^\\circ$.\n\nWe have:\n$$\n\\widehat{A_0OA_1} + \\widehat{A_1OA_2} + \\widehat{A_2OA_3} + \\cdots + \\widehat{A_{25}OA_{26}} + \\widehat{A_{26}OA_0} = 360^\\circ\n$$\n\nThe sum $\\widehat{A_0OA_1} + \\widehat{A_1OA_2} + \\cdots + \\widehat{A_{25}OA_{26}}$ is $1^\... | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a) 9 degrees; b) n = 18, 19, 20, 21, 22, 23, 24, 25 | |
02p5 | Problem:
Joãozinho escreveu os números de $1$ até $100000$ no quadro, depois foi trocando cada número pela soma de seus algarismos e repetiu este processo até obter uma lista de $100000$ números de um algarismo. Por exemplo, começando pelo número $7234$ obtemos $7+2+3+4=16$ e $1+6=7$.
a. Que número ficou no lugar do n... | [
"Solution:\n\na. $98765 \\longrightarrow 9+8+7+6+5=35 \\longrightarrow 3+5=8$.\n\nb. Observemos que um número e a soma de seus algarismos deixam o mesmo resto quando divididos por $9$. Assim, depois de terminar todo o processo vamos obter uma lista da forma\n$$\n1,2,3,4,5,6,7,8,9,1,2,3,4, \\ldots, 8,9,1\n$$\nAssim ... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a: 8; b: 11111; c: 1 | |
0kh5 | Which of the following conditions is sufficient to guarantee that integers $x$, $y$, and $z$ satisfy the equation
$$
x(x - y) + y(y - z) + z(z - x) = 1?
$$
(A) $x > y$ and $y = z$
(B) $x = y - 1$ and $y = z - 1$
(C) $x = z + 1$ and $y = x + 1$
(D) $x = z$ and $y - 1 = x$
(E) $x + y + z = 1$ | [
"**Answer (D):** The given equation is equivalent to $2x(x - y) + 2y(y - z) + 2z(z - x) = 2$, which can be rewritten as $(x - y)^2 + (y - z)^2 + (z - x)^2 = 2$. This equation has an integer solution if and only if two of the squares are 1 and one is 0. This in turn means that two of the variables must be equal and ... | United States | Fall 2021 AMC 10 B | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | D | |
0927 | Problem:
The expression
$$
\pm \square \pm \square \pm \square \pm \square \pm \square \pm \square
$$
is written on the blackboard. Two players, $A$ and $B$, play a game, taking turns. Player $A$ takes the first turn. In each turn, the player on turn replaces a symbol $\square$ by a positive integer. After all the sym... | [
"Solution:\n\nWe will call a number good if it has a divisor in the set $\\{11,12, \\ldots, 18\\}$.\n\nWe will show that player $B$ has a winning strategy. Namely, $B$ plays $18!$ in both his first and second move. In his last move he plays a number $x$ (which we will specify later), which ensures that each possibl... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | Player B | |
09dv | Let $a_n$ be an arithmetic progression and $a_1^2, a_2^2, a_3^2$ are terms of the progression. Prove that all terms of the progression are integers. | [
"Suppose that arithmetic progression $a_n$ satisfies given condition and let $d$ be common difference of the progression.\n\nIf $d = 0$ then $a_n = 0, \\forall n \\in \\mathbb{N}$ or $a_n = 1, \\forall n \\in \\mathbb{N}$.\n\nLet $d \\neq 0$. $a_1^2, a_2^2, a_3^2$ are terms of the given progression $\\Leftrightarro... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Sequences and Series"
] | English | proof only | null | |
0jwa | Problem:
Two ordered pairs $(a, b)$ and $(c, d)$, where $a, b, c, d$ are real numbers, form a basis of the coordinate plane if $a d \neq b c$. Determine the number of ordered quadruples $(a, b, c, d)$ of integers between $1$ and $3$ inclusive for which $(a, b)$ and $(c, d)$ form a basis for the coordinate plane. | [
"Solution:\n\nAny pair of distinct points will form a basis except when $(a, b)$ and $(c, d)$ are both from $\\{(1,1),(2,2),(3,3)\\}$, so the answer is $9 \\cdot 8 - 3 \\cdot 2 = 66$."
] | United States | HMMT November 2017 | [
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Vectors",
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 66 | |
02qb | Problem:
Estefânia tem cinco cartas marcadas com as letras $A, B, C, D$ e $E$, empilhadas nessa ordem de cima para baixo. Ela embaralha as cartas pegando as duas de cima e colocando-as, com a ordem trocada, embaixo da pilha. A figura mostra o que acontece nas duas primeiras vezes em que ela embaralha as cartas.
![](a... | [
"Solution:\n\nO leitor pode verificar que, se Estefânia embaralhar as cartas 6 vezes, elas voltarão à posição inicial. Como $74=12 \\times 6+2$, embaralhar as cartas 74 vezes tem o mesmo efeito que fazê-lo duas vezes, o que deixa a carta $E$ no topo da pilha."
] | Brazil | Nível 2 | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Other"
] | null | MCQ | E | |
0cv7 | Let $AM$ be a median of an acute triangle $ABC$, and let $BH$ be the altitude. The line through $M$ perpendicular to $AM$ meets the ray $HB$ at $K$. Given that $\angle MAC = 30^\circ$, prove that $AK = BC$. | [
"The points $A$, $H$, $M$, and $K$ lie on a circle $\\omega$ with diameter $AK$ (see Fig. 4). Since $\\angle MAH = 30^\\circ$, we have $HM = AK/2$. On the other hand, $BC = 2HM$.",
"First solution. Since $\\angle ANK = \\angle AMK = 90^\\circ$, the points $A$, $H$, $M$, and $K$ lie on a circle $\\omega$ with diam... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distan... | English; Russian | proof only | null | |
04mk | We say that a circular arrangement of positive integers is alternating if every number is either smaller, or larger than both of its neighbours. We call a pair of adjacent numbers good if, upon its removal, the remaining numbers form an alternating arrangement.
The numbers $1$ to $300$ are placed on a circle in an alt... | [
"Let $a$, $p$, $q$ and $b$ be four consecutive numbers in an alternating arrangement. Assume $p > q$. Then $a < p$ and $q < b$.\n\nThe pair $(p, q)$ is not good if and only if $a > b$. Therefore, we conclude that $(p, q)$ is not good if and only if $p$ is the largest, and $q$ the smallest number of the quadruple $(... | Croatia | Croatia_2018 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 151 | |
031w | Problem:
Let $ABCD$ be a circumscribed quadrilateral and let $P$ be the orthogonal projection of its incenter on the diagonal $AC$. Prove that $\Varangle APB = \Varangle APD$. | [
"Solution:\nWe shall use the following fact: if $\\alpha, \\beta, \\gamma$ and $\\delta$ are angles such that $\\sin \\alpha \\sin \\delta = \\sin \\beta \\sin \\gamma$ and $\\alpha + \\beta = \\gamma + \\delta < 180^{\\circ}$, then $\\alpha = \\gamma$ and $\\beta = \\delta$.\n\nDenote by $M, N, R$ and $S$ the tang... | Bulgaria | Team selection test for 44. IMO | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
080x | Problem:
Fissato un intero $n>1$, Alberto e Barbara giocano il seguente gioco:
1. Alberto sceglie un intero positivo;
2. Barbara sceglie un intero maggiore di 1 che sia multiplo o sottomultiplo del numero di Alberto (compreso il numero stesso);
3. Alberto restituisce a Barbara il numero da lei detto, eventualmente ag... | [
"Solution:\n\nBarbara riesce a pronunciare il numero $n$ entro 50 mosse (in realtà ne sono necessarie al più 8), indipendentemente dalla strategia di Alberto, se e solo se $n$ è un multiplo di 6.\n\nSupponiamo infatti che $n$ sia un multiplo di 6. Allora si ha che:\n- se ad un certo passo Alberto sceglie un numero ... | Italy | Cesenatico | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | Barbara can force a win if and only if the target number is a multiple of 6. | |
0e6s | Which two points lie on the graph of a linear function $y = bx + 1$ where $b$ is some non-zero real number?
(A) $(0,1)$ and $(\frac{1}{b},0)$
(B) $(0,b)$ and $(-\frac{1}{b},0)$
(C) $(0,1)$ and $(b,0)$
(D) $(0,1)$ and $(-\frac{1}{b},0)$
(E) $(0,-\frac{1}{b})$ and $(1,0)$ | [
"Only two among the given points always fulfill the equation. These are $(0, 1)$ and $(-\\frac{1}{b}, 0)$."
] | Slovenia | National Math Olympiad 2012 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
07vq | Suppose that $a, b, c > 0$ and $a + b + c = 3$. Prove that $a^2 + b^2 + c^2 + 6abc \le 9$. | [
"We may homogenise in two steps the inequality we wish to show. First, we use $(a+b+c)^2 = 9$ to replace the RHS and obtain the equivalent inequality\n$$\na^2 + b^2 + c^2 + 6abc \\le (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) + 3abc.\n$$\nThis is equivalent to $3abc \\le ab + bc + ca$. We now multiply by $3 ... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0ewp | Problem:
The natural numbers $m$ and $n$ are relatively prime. Prove that the greatest common divisor of $m + n$ and $m^2 + n^2$ is either $1$ or $2$. | [
"Solution:\n\nIf $d$ divides $m + n$ and $m^2 + n^2$, then it also divides $(m + n)^2 - (m^2 + n^2) = 2mn$ and hence also $2m(m + n) - 2mn = 2m^2$ and $2n(m + n) - 2mn = 2n^2$. But $m$ and $n$ are relatively prime, so $m^2$ and $n^2$ are also. Hence $d$ must divide $2$."
] | Soviet Union | 3rd ASU | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0hs6 | Problem:
For a positive integer $n$, let $f(n)$ be the number of divisors of $n$ which are perfect squares, and let $g(n)$ be the number of divisors of $n$ which are perfect cubes. Determine whether there exists an integer $n$ such that
$$
\frac{f(n)}{g(n)}=2012
$$ | [
"Solution:\n\nIf $n$ has the prime factorization\n$$\nn=p_{1}^{e_{1}} p_{2}^{e_{2}} \\cdots p_{k}^{e_{k}}\n$$\nthen the divisors of $n$ which are perfect squares are the numbers of the form\n$$\np_{1}^{2 a_{1}} p_{2}^{2 a_{2}} \\cdots p_{k}^{2 a_{k}}\n$$\nwhere each $a_{i}$ is an integer such that $0 \\leq a_{i} \\... | United States | Berkeley Math Circle | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | Yes; for example n = 2^{4022} * 3^{2680} * 5^{1786} * 7^{1190} * 11^{792} * 13^{528} * 17^{352} * 19^{234} * 23^{156} * 29^{104} * 31^{68} * 37^{44} * 41^{28} * 43^{18} * 47^{12} * 53^{8} * 59^{4} * 61^{2}. | |
0g2w | Problem:
Zeige, dass keine Funktion $f: \mathbb{Z} \rightarrow \mathbb{Z}$ existiert, sodass für alle ganzen Zahlen $m, n$ gilt:
$$
f(m+f(n))=f(m)-n
$$ | [
"Solution:\n\nWir nehmen an, dass die Funktionalgleichung eine Lösung hat und führen dies zu einem Widerspruch. Setzen wir $m=0$ ein, dann erhalten wir die Gleichung\n$$\nf(f(n))=f(0)-n\n$$\nDaraus folgt sofort, dass $f$ eine surjektive Funktion ist ($f$ ist sogar bijektiv, das brauchen wir aber nicht für die Lösun... | Switzerland | SMO 2019 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0h3w | Is it possible to paint each cell of an $8 \times 8$ table with one of $16$ colors so that for each two colors there are two cells painted with these colors and having a common side? | [
"У таблиці є $112$ одиничних відрізків, що розділяють її сусідні клітинки. Але з $16$ даних кольорів можна утворити $120$ пар кольорів, і для кожної з цих пар повинні знайтись дві сусідні клітинки, що неможливо.\n\n*Відповідь:* Не можна."
] | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Not possible | |
0gfe | 找出所有合成數 $n$, 使得 $n$ 的比 $1$ 大的正因數擺在一個圓上, 相鄰的兩個都不會互質。 | [
"當 $n = p^{\\alpha}, \\alpha \\ge 2$ 時, 依 $p, p^2, \\dots, p^{\\alpha}$ 排列, 顯然滿足條件。\n\n當 $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_k^{\\alpha_k}, k \\ge 3$ 時, 先把 $p_1 p_2 p_3 \\cdots p_k p_1$ 依次排列在圓上, 然後把 $p_j$ 的倍數排列在 $p_{j-1} p_j$ 與 $p_j p_{j+1}$ 之間。得到的排列滿足條件。\n\n最後當 $n = p^a q^b, a > 1$，先把 $pq$ 與 $p^2 q$ 排在圓... | Taiwan | 國際奧林匹亞競賽第三次訓練營 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Chinese; English | proof and answer | All composite numbers except those equal to the product of two distinct primes. | |
08qm | Problem:
Find all prime numbers $p$ such that
$$
(x+y)^{19}-x^{19}-y^{19}
$$
is a multiple of $p$ for any positive integers $x, y$. | [
"Solution:\nIf $x = y = 1$ then $p$ divides\n$$\n2^{19} - 2 = 2\\left(2^{18} - 1\\right) = 2\\left(2^{9} - 1\\right)\\left(2^{9} + 1\\right) = 2 \\cdot 511 \\cdot 513 = 2 \\cdot 3^{3} \\cdot 7 \\cdot 19 \\cdot 73\n$$\nIf $x = 2$, $y = 1$ then\n$$\np \\mid 3^{19} - 2^{19} - 1\n$$\nWe will show that $3^{19} - 2^{19} ... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2, 3, 7, 19 | |
039u | For any positive integer $m > 1$ denote by $f(m)$ the sum of all positive integers which are less than $m$ and co-prime to $m$. Find all positive integers $n$ such that there exist positive integers $k$ and $l$ such that $f(n^k) = n^l$. | [
"If $m > 2$ then the positive integers which are less than $m$ and co-prime to $m$ can be partitioned into pairs $(k, m-k)$, $m-k \\neq k$. The number of these pairs is $\\frac{\\varphi(m)}{2}$ and the sum of the numbers in every pair is $m$. Therefore $f(m) = \\frac{m\\varphi(m)}{2}$, which is true for $m = 2$ as ... | Bulgaria | Fall Mathematical Competition | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 2, 3, 4, 6 | |
0dmr | Problem:
Нека је $A$ бесконачан подскуп скупа природних бројева. Одредити све природне бројеве $n$ такве да за свако $a \in A$ важи
$$
a^{n}+a^{n-1}+\ldots+a^{1}+1 \mid a^{n!}+a^{(n-1)!}+\ldots+a^{1!}+1
$$ | [
"Solution:\n\nОзначимо $P(x)=x^{n}+x^{n-1}+\\cdots+1$ и $Q(x)=x^{n!}+\\cdots+x^{1!}+1$; нека је $Q(x)=C(x) P(x)+R(x)$, где су $C$ и $R$ полиноми са целим коефицијентима и $\\operatorname{deg} R<\\operatorname{deg} P$. По услову задатка $P(a) \\mid Q(a)$, и самим тим $P(a) \\mid R(a)$, за бесконачно много целих број... | Serbia | Serbian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | [1, 2] | |
03ek | Let $ABC$ be a triangle, satisfying $2AC = AB + BC$. If $O$ and $I$ are its circumcenter and incenter, show that $\angle OIB = 90^\circ$.
(Konstantin Delchev) | [
"Let $D = BI \\cap (ABC)$. We apply Ptolemy's theorem\n$$\nAB \\cdot DC + BC \\cdot AD = AC \\cdot BD\n$$\nwhich implies $BD = 2DA$. From $AD = DI = DC$ it yields $I$ is midpoint of $BD$ so $\\angle OIB = 90^\\circ$.\n$\\square$"
] | Bulgaria | 2 Bulgarian Winter Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof only | null | |
0lcc | 1. Prove that there exists infinitely many positive integer $t$ such that $t$ both $2012t+1$ and $2013t+1$ are perfect square.
2. Let $m, n$ be positive integers such that both $mn+1$ and $(m+1)n+1$ are perfect square. Prove that $n$ is divisible by $8(2m+1)$. | [
"1. Let $d = (2012t+1, 2013t+1)$ then it is easy to see that $d=1$. Thus, both $2012t+1$ and $2013t+1$ are perfect square if and only if $(2012t+1)(2013t+1) = y^2$ for some positive integer $y$. We have\n$$\n\\begin{aligned}\n(2012t+1)(2013t+1) &= y^2 \\\\\n&\\Leftrightarrow 4 \\cdot 2012^2 \\cdot 2013^2 t^2 + 4 \\... | Vietnam | Vietnamese Mathematical Competitions | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Recu... | null | proof only | null | |
02yt | Problem:
O segmento $AB$ de comprimento $16~\mathrm{cm}$ é diâmetro de um círculo de centro $O$. Uma reta secante corta o círculo em $C$ e $D$ e a reta $AB$ em $P$, como indica a figura a seguir. Se $OD = DP$ e $\angle APC = 18^\circ$, qual o valor do ângulo $\angle AOC$?
 | [
"Solution:\n\nComo $OD = DP$, segue que $\\angle DOP = \\angle OPD = 18^\\circ$. Pelo Teorema do Ângulo Externo, segue que $\\angle ODC = 2 \\cdot 18^\\circ = 36^\\circ$. O triângulo $COD$ é isósceles, pois $CO$ e $OD$ são raios do círculo. Assim, $\\angle OCD = \\angle CDO = 36^\\circ$. Novamente, pelo Teorema do ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 54° | |
0156 | Problem:
Find all finite sets of positive integers with at least two elements such that for any two numbers $a, b$ ($a > b$) belonging to the set, the number $\frac{b^{2}}{a-b}$ belongs to the set, too. | [
"Solution:\nLet $X$ be a set we seek for, and $a$ be its minimal element. For each other element $b$ we have $\\frac{a^{2}}{b-a} \\geq a$, hence $b \\leq 2a$. Therefore all the elements of $X$ belong to the interval $[a, 2a]$. So the quotient of any two elements of $X$ is at most $2$.\n\nNow consider two biggest el... | Baltic Way | Baltic Way 2008 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other"
] | null | proof and answer | {n, 2n} for any positive integer n | |
00d7 | Determinar el mayor valor posible de
$$
S = a_1 a_2 a_3 + a_4 a_5 a_6 + \dots + a_{2017} a_{2018} a_{2019} + a_{2020},
$$
donde $(a_1, a_2, a_3, \dots, a_{2020})$ es una permutación de $(1, 2, 3, \dots, 2020)$. | [
"De inspeccionar casos pequeños conjeturamos que el máximo es $S = 2020 \\cdot 2019 \\cdot 2018 + 2017 \\cdot 2016 \\cdot 2015 + \\dots + 4 \\cdot 3 \\cdot 2 + 1$.\n\nVamos a probarlo.\nPrimero observamos que si tenemos $a > b > c > d$ entonces el mayor valor para el producto de tres de ellos más el cuarto es $abc ... | Argentina | Nacional OMA | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | 2020*2019*2018 + 2017*2016*2015 + ... + 4*3*2 + 1 | |
021g | Problem:
Let $a_{0}, a_{1}, \ldots, a_{10}$ be integers such that, for each $i \in \{0,1,\ldots,2047\}$, there exists a subset $S \subseteq \{0,1,\ldots,10\}$ with
$$
\sum_{j \in S} a_{j} \equiv i \pmod{2048}.
$$
Show that for each $i \in \{0,1,\ldots,10\}$, there is exactly one $j \in \{0,1,\ldots,10\}$ such that $a_{... | [
"Solution:\nWe denote by $\\nu_{2}(a)$ the 2-adic valuation of the integer $a$. Let us prove by induction the more general statement that, for $n \\in \\mathbb{N}_{>0}$, if $a_{0}, a_{1}, \\ldots, a_{n-1}$ are integers such that, for each $i \\in \\{0,1,\\ldots,2^{n}-1\\}$, there exists a subset $S \\subseteq \\{0,... | Benelux Mathematical Olympiad | 17th Benelux Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0c2o | What is the maximum number of rooks one can place on a chessboard such that any rook attacks exactly two other rooks? (We say that two rooks attack each other if they are on the same line or on the same column and between them there are no other rooks.) | [
"We say that there can be two types of rooks on the chessboard: a rook of type **T1** is a rook that is attacked from two perpendicular lines; a rook is of type **T2** if it is attacked by two rooks situated on the same line, but opposite directions.\nSuppose one can place $m$ rooks of type **T1** and $n$ of type *... | Romania | 69th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 16 | |
0d50 | Let $ABC$ be a triangle and $D$ a point on the side $BC$. The tangent line to the circumcircle of the triangle $ABD$ at the point $D$ intersects the side $AC$ at $E$. The tangent line to the circumcircle of the triangle $ACD$ at the point $D$ intersects the side $AB$ at $F$. Prove that the point $A$ and the circumcente... | [
"Let $\\alpha = \\angle BAC$, $\\beta = \\angle CBA$, $\\gamma = \\angle ACB$ and $\\theta = BAD$. Because line $DE$ is tangent to the circumcircle of triangle $ABD$, we have $\\angle EDA = \\angle DBA = \\beta$. Because line $DF$ is tangent to the circumcircle of triangle $ADC$, we have $\\angle ADF = \\angle ACD ... | Saudi Arabia | SAMC 2015 | [
"Geometry > Plane Geometry > Circles > Tangents",
"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 c... | English, Arabic | proof only | null | |
02cj | Problem:
Uma mesa circular tem 60 cadeiras em sua volta. Existem $N$ pessoas sentadas nessas cadeiras de tal modo que a próxima pessoa a se sentar vai ter que se sentar ao lado de alguém. Qual é o menor valor possível para $N$? | [
"Solution:\n\nSe a próxima pessoa a se sentar vai ter que se sentar ao lado de uma cadeira ocupada, isso significa que existem no máximo 2 cadeiras desocupadas consecutivas. Veja na figura: as cadeiras ocupadas estão representadas por quadradinhos brancos e as desocupadas por quadradinhos pretos. Podemos então pens... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 20 | |
03fb | Given is a triangle $ABC$ and two points $D \in AC$, $E \in BD$ such that $\angle DAE = \angle AED = \angle ABC$. Show that $BE = 2CD$ iff $\angle ACB = 90^\circ$.
(Nikolay Nikolov) | [
"Let $\\alpha = \\angle A$, $\\beta = \\angle B$ and $\\gamma = \\angle C$. Then\n$$\n\\frac{BE}{\\sin(\\alpha - \\beta)} = \\frac{AB}{\\sin(\\pi - \\beta)}, \\quad \\frac{CD}{\\sin(\\alpha - \\beta)} = \\frac{BC}{\\sin 2\\beta}, \\quad \\frac{AB}{\\sin \\gamma} = \\frac{BC}{\\sin \\alpha}\n$$\nand so\n$$\n\\frac{B... | Bulgaria | Bulgarian Spring Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
04i8 | Five teams are playing at a football tournament, each playing with the other exactly once. Each victory is worth 3 points, defeat 0 points and a draw 1 point. Is it possible that at the end of the tournament, in the final ranking, each team except the last has exactly two points more than the next one? | [] | Croatia | Croatia Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0b5e | Let $ABCDE$ be a convex pentagon with $AB + CD = BC + DE$, and $k$ a circle centered on side $AE$, tangent to sides $AB$, $BC$, $CD$ and $DE$ at points $P$, $Q$, $R$ and $S$ respectively. Prove that lines $PS$ and $AE$ are parallel. | [
"$$\nAB + CD = (AP + PB) + (CR + RD) = AP + (BP + CR + DR)\n$$\n$$\nBC + DE = (BQ + QC) + (DS + SE) = ES + (BQ + CQ + DS).\n$$\nUsing the fact that tangents from a point to a circle are of equal length, one gets $AP = ES$. Denoting by $O$ and $r$ the center, respectively radius of $k$, one gets that right-angled tr... | Romania | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0h5m | 2015 candies are placed along a circle and numbered $1$ to $2015$ clockwise. Andriy and Olesia play the following game. In each turn, a player can take either $2$ or $3$ candies with consecutive numbers ($1$ and $2015$ are also considered "consecutive"). The player who can't make a move loses. Who has a winning strateg... | [
"Andriy takes $2$ (or $3$) consecutive candies. Then Olesia takes $3$ (or $2$) diametrically opposite candies, so that there are $1005$ candies on both sides between the groups of taken candies. Then she just copies Andriy's moves on the other part. If Andriy can make a move, she can as well, so she will not lose a... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Olesia |
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