problem stringlengths 1 7.47k | solution stringlengths 0 13.5k | answer stringlengths 1 272 | problem_type stringclasses 8
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G2 In a right trapezoid $A B C D(A B \| C D)$ the angle at vertex $B$ measures $75^{\circ}$. Point $H$ is the foot of the perpendicular from point $A$ to the line $B C$. If $B H=D C$ and $A D+A H=8$, find the area of $A B C D$.
|
Solution: Produce the legs of the trapezoid until they intersect at point $E$. The triangles $A B H$ and $E C D$ are congruent (ASA). The area of $A B C D$ is equal to area of triangle $E A H$ of hypotenuse
$$
A E=A D+D E=A D+A H=8
$$
Let $M$ be the midpoint of $A E$. Then
$$
M E=M A=M H=4
$$
and $\angle A M H=30^... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
NT1 Determine all positive integer numbers $k$ for which the numbers $k+9$ are perfect squares and the only prime factors of $k$ are 2 and 3 .
|
Solution: We have an integer $x$ such that
$$
x^{2}=k+9
$$
$k=2^{a} 3^{b}, a, b \geq 0, a, b \in \mathbb{N}$.
Therefore,
$$
(x-3)(x+3)=k \text {. }
$$
If $b=0$ then we have $k=16$.
If $b>0$ then we have $3 \mid k+9$. Hence, $3 \mid x^{2}$ and $9 \mid k$.
Therefore, we have $b \geq 2$. Let $x=3 y$.
$$
(y-1)(y+1... | {16,27,72,216,432,2592} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT2 A group of $n>1$ pirates of different age owned total of 2009 coins. Initially each pirate (except for the youngest one) had one coin more than the next younger.
a) Find all possible values of $n$.
b) Every day a pirate was chosen. The chosen pirate gave a coin to each of the other pirates. If $n=7$, find the la... | ## Solution:
a) If $n$ is odd, then it is a divisor of $2009=7 \times 7 \times 41$. If $n>49$, then $n$ is at least $7 \times 41$, while the average pirate has 7 coins, so the initial division is impossible. So, we can have $n=7, n=41$ or $n=49$. Each of these cases is possible (e.g. if $n=49$, the average pirate has ... | 1994 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT3 Find all pairs $(x, y)$ of integers which satisfy the equation
$$
(x+y)^{2}\left(x^{2}+y^{2}\right)=2009^{2}
$$
|
Solution: Let $x+y=s, x y=p$ with $s \in \mathbb{Z}^{*}$ and $p \in \mathbb{Z}$. The given equation can be written in the form
$$
s^{2}\left(s^{2}-2 p\right)=2009^{2}
$$
or
$$
s^{2}-2 p=\left(\frac{2009}{s}\right)^{2}
$$
So, $s$ divides $2009=7^{2} \times 41$ and it follows that $p \neq 0$.
If $p>0$, then $2009^{... | (40,9),(9,40),(-40,-9),(-9,-40) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT4 Determine all prime numbers $p_{1}, p_{2}, \ldots, p_{12}, p_{13}, p_{1} \leq p_{2} \leq \ldots \leq p_{12} \leq p_{13}$, such that
$$
p_{1}^{2}+p_{2}^{2}+\ldots+p_{12}^{2}=p_{13}^{2}
$$
and one of them is equal to $2 p_{1}+p_{9}$.
|
Solution: Obviously, $p_{13} \neq 2$, because sum of squares of 12 prime numbers is greater or equal to $12 \times 2^{2}=48$. Thus, $p_{13}$ is odd number and $p_{13} \geq 7$.
We have that $n^{2} \equiv 1(\bmod 8)$, when $n$ is odd. Let $k$ be the number of prime numbers equal to 2 . Looking at equation modulo 8 we g... | (2,2,2,2,2,2,2,3,3,5,7,7,13),(2,2,2,2,2,2,2,3,3,5,7,13,17),(2,2,2,2,2,2,2,3,3,5,7,29,31) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
C2 Can we divide an equilateral triangle $\triangle A B C$ into 2011 small triangles using 122 straight lines? (there should be 2011 triangles that are not themselves divided into smaller parts and there should be no polygons which are not triangles)
| ## Solution
Firstly, for each side of the triangle, we draw 37 equidistant, parallel lines to it. In this way we get $38^{2}=1444$ triangles. Then we erase 11 lines which are closest to the vertex $A$ and parallel to the side $B C$ and we draw 21 lines perpendicular to $B C$, the first starting from the vertex $A$ and... | 2011 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
C4 In a group of $n$ people, each one had a different ball. They performed a sequence of swaps; in each swap, two people swapped the ball they had at that moment. Each pair of people performed at least one swap. In the end each person had the ball he/she had at the start. Find the least possible number of swaps, if: $... | ## Solution
We will denote the people by $A, B, C, \ldots$ and their initial balls by the corresponding small letters. Thus the initial state is $A a, B b, C c, D d, E e(, F f)$. A swap is denoted by the (capital) letters of the people involved.
a) Five people form 10 pairs, so at least 10 swaps are necessary.
In fa... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
C5 A set $S$ of natural numbers is called good, if for each element $x \in S, x$ does not divide the sum of the remaining numbers in $S$. Find the maximal possible number of elements of a good set which is a subset of the set $A=\{1,2,3, \ldots, 63\}$.
|
Solution
Let set $B$ be the good subset of $A$ which have the maximum number of elements. We can easily see that the number 1 does not belong to $B$ since 1 divides all natural numbers. Based on the property of divisibility, we know that $x$ divides the sum of the remaining numbers if and only if $x$ divides the sum ... | 61 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
C7 Consider a rectangle whose lengths of sides are natural numbers. If someone places as many squares as possible, each with area 3 , inside of the given rectangle, such that
the sides of the squares are parallel to the rectangle sides, then the maximal number of these squares fill exactly half of the area of the rect... | ## Solution
Let $A B C D$ be a rectangle with $A B=m$ and $A D=n$ where $m, n$ are natural numbers such that $m \geq n \geq 2$. Suppose that inside of the rectangle $A B C D$ is placed a rectangular lattice consisting of some identical squares whose areas are equals to 3 , where $k$ of them are placed along the side $... | (2,3);(3,4);(3,6);(3,8);(3,10);(3,12) | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A2. Find all positive integers $x, y$ satisfying the equation
$$
9\left(x^{2}+y^{2}+1\right)+2(3 x y+2)=2005
$$
| ## Solution
The given equation can be written into the form
$$
2(x+y)^{2}+(x-y)^{2}=664
$$
Therefore, both numbers $x+y$ and $x-y$ are even.
Let $x+y=2 m$ and $x-y=2 t, t \in \mathbb{Z}$.
Now from (1) we have that $t$ and $t^{2}$ are even and $m$ is odd.
So, if $t=2 k, k \in \mathbb{Z}$ and $m=2 n+1, n \in \mathb... | (x,y)=(11,7)or(x,y)=(7,11) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A3. Find the maximum value of the area of a triangle having side lengths $a, b, c$ with
$$
a^{2}+b^{2}+c^{2}=a^{3}+b^{3}+c^{3}
$$
| ## Solution
Without any loss of generality, we may assume that $a \leq b \leq c$.
On the one hand, Tchebyshev's inequality gives
$$
(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \leq 3\left(a^{3}+b^{3}+c^{3}\right)
$$
Therefore using the given equation we get
$$
a+b+c \leq 3 \text { or } p \leq \frac{3}{2}
$$
where $p$ d... | S\leq\frac{\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A4. Find all the integer solutions of the equation
$$
9 x^{2} y^{2}+9 x y^{2}+6 x^{2} y+18 x y+x^{2}+2 y^{2}+5 x+7 y+6=0
$$
| ## Solution
The equation is equivalent to the following one
$$
\begin{aligned}
& \left(9 y^{2}+6 y+1\right) x^{2}+\left(9 y^{2}+18 y+5\right) x+2 y^{2}+7 y++6=0 \\
& \Leftrightarrow(3 y+1)^{2}\left(x^{2}+x\right)+4(3 y+1) x+2 y^{2}+7 y+6=0
\end{aligned}
$$
Therefore $3 y+1$ must divide $2 y^{2}+7 y+6$ and so it must... | (-2,0),(-3,0),(0,-2),(-1,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A5. Solve the equation
$$
8 x^{3}+8 x^{2} y+8 x y^{2}+8 y^{3}=15\left(x^{2}+y^{2}+x y+1\right)
$$
in the set of integers.
| ## Solution
We transform the equation to the following one
$$
\left(x^{2}+y^{2}\right)(8 x+8 y-15)=15(x y+1)
$$
Since the right side is divisible by 3 , then $3 /\left(x^{2}+y^{2}\right)(8 x+8 y-15)$. But if $3 /\left(x^{2}+y^{2}\right)$, then $3 / x$ and $3 / y, 009$ will wive $15(x y+1)$ and $3 /(x y+1)$, which is... | (x,y)=(1,2)(x,y)=(2,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
G3. Let $A B C D E F$ be a regular hexagon. The points $\mathrm{M}$ and $\mathrm{N}$ are internal points of the sides $\mathrm{DE}$ and $\mathrm{DC}$ respectively, such that $\angle A M N=90^{\circ}$ and $A N=\sqrt{2} \cdot C M$. Find the measure of the angle $\angle B A M$.
| ## Solution
Since $A C \perp C D$ and $A M \perp M N$ the quadrilateral $A M N C$ is inscribed. So, we have
$$
\angle M A N=\angle M C N
$$
Let $P$ be the projection of the point $M$ on the line $C D$. The triangles $A M N$ and $C P M$ are similar implying
$$
\frac{A M}{C P}=\frac{M N}{P M}=\frac{A N}{C M}=\sqrt{2}... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
G6. A point $O$ and the circles $k_{1}$ with center $O$ and radius $3, k_{2}$ with center $O$ and radius 5, are given. Let $A$ be a point on $k_{1}$ and $B$ be a point on $k_{2}$. If $A B C$ is equilateral triangle, find the maximum value of the distance $O C$.
| ## Solution
It is easy to see that the points $O$ and $C$ must be in different semi-planes with respect to the line $A B$.
Let $O P B$ be an equilateral triangle ( $P$ and $C$ on the same side of $O B$ ). Since $\angle P B C$ $=60^{\circ}-\angle A B P$ and $\angle O B A=60^{\circ}-\angle A B P$, then $\angle P B C=\a... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
NT1. Find all the natural numbers $m$ and $n$, such that the square of $m$ minus the product of $n$ with $k$, is 2 , where the number $k$ is obtained from $n$ by writing 1 on the left of the decimal notation of $n$.
| ## Solution
Let $t$ be the number of digits of $n$. Then $k=10^{t}+n$. So
$$
\mathrm{m}^{2}=n\left(10^{t}+\mathrm{n}\right)+2, \text { i.e. } \mathrm{m}^{2}-\mathrm{n}^{2}=10^{t} n+2
$$
This implies that $\mathrm{m}, \mathrm{n}$ are even and both $\mathrm{m}, \mathrm{n}$ are odd.
If $t=1$, then, 4 is divisor of $10... | =11,n=7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT2. Find all natural numbers $n$ such that $5^{n}+12^{n}$ is perfect square.
| ## Solution
By checking the cases $n=1,2,3$ we get the solution $n=2$ and $13^{2}=5^{2}+12^{2}$.
If $n=2 k+1$ is odd, we consider the equation modulo 5 and we obtain
$$
\begin{aligned}
x^{2} & \equiv 5^{2 k+1}+12^{2 k+1}(\bmod 5) \equiv 2^{2 k} \cdot 2(\bmod 5) \\
& \equiv(-1)^{k} \cdot 2(\bmod 5) \equiv \pm 2(\bmod... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT4. Find all the three digit numbers $\overline{a b c}$ such that
$$
\overline{a b c}=a b c(a+b+c)
$$
| ## Solution
We will show that the only solutions are 135 and 144 .
We have $a>0, b>0, c>0$ and
$$
9(11 a+b)=(a+b+c)(a b c-1)
$$
- If $a+b+c \equiv 0(\bmod 3)$ and $a b c-1 \equiv 0(\bmod 3)$, then $a \equiv b \equiv c \equiv 1(\bmod 3)$ and $11 a+b \equiv 0(\bmod 3)$. It follows now that
$$
a+b+c \equiv 0(\bmod 9)... | 135144 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
C4. Let $p_{1}, p_{2}, \ldots, p_{2005}$ be different prime numbers. Let $\mathrm{S}$ be a set of natural numbers which elements have the property that their simple divisors are some of the numbers $p_{1}, p_{2}, \ldots, p_{2005}$ and product of any two elements from $\mathrm{S}$ is not perfect square.
What is the ma... | ## Solution
Let $a, b$ be two arbitrary numbers from $\mathrm{S}$. They can be written as
$$
a=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{2005}^{a_{2005}} \text { and } b=p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{2005}^{\beta_{2005}}
$$
In order for the product of the elements $a$ and $b$ to be a square all the sums ... | 2^{2005} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A4 Let $x, y$ be positive real numbers such that $x^{3}+y^{3} \leq x^{2}+y^{2}$. Find the greatest possible value of the product $x y$.
| ## Solution 1
We have $(x+y)\left(x^{2}+y^{2}\right) \geq(x+y)\left(x^{3}+y^{3}\right) \geq\left(x^{2}+y^{2}\right)^{2}$, hence $x+y \geq x^{2}+y^{2}$. Now $2(x+y) \geq(1+1)\left(x^{2}+y^{2}\right) \geq(x+y)^{2}$, thus $2 \geq x+y$. Because $x+y \geq 2 \sqrt{x y}$, we will obtain $1 \geq x y$. Equality holds when $x=y... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
A5 Determine the positive integers $a, b$ such that $a^{2} b^{2}+208=4\{l c m[a ; b]+g c d(a ; b)\}^{2}$.
| ## Solution
Let $d=\operatorname{gcd}(a, b)$ and $x, y \in \mathbb{Z}_{+}$such that $a=d x, b=d y$. Obviously, $(x, y)=1$. The equation is equivalent to $d^{4} x^{2} y^{2}+208=4 d^{2}(x y+1)^{2}$. Hence $d^{2} \mid 208$ or $d^{2} \mid 13 \cdot 4^{2}$, so $d \in\{1,2,4\}$. Take $t=x y$ with $t \in \mathbb{Z}_{+}$.
Cas... | (,b)\in{(2,12);(4,6);(6,4);(12;2)} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A1 If for the real numbers $x, y, z, k$ the following conditions are valid, $x \neq y \neq z \neq x$ and $x^{3}+y^{3}+k\left(x^{2}+y^{2}\right)=y^{3}+z^{3}+k\left(y^{2}+z^{2}\right)=z^{3}+x^{3}+k\left(z^{2}+x^{2}\right)=2008$, find the product $x y z$.
|
Solution
$x^{3}+y^{3}+k\left(x^{2}+y^{2}\right)=y^{3}+z^{3}+k\left(y^{2}+z^{2}\right) \Rightarrow x^{2}+x z+z^{2}=-k(x+z):(1)$ and $y^{3}+z^{3}+k\left(y^{2}+z^{2}\right)=z^{3}+x^{3}+k\left(z^{2}+x^{2}\right) \Rightarrow y^{2}+y x+x^{2}=-k(y+x):(2)$
- From (1) $-(2) \Rightarrow x+y+z=-k:(*)$
- If $x+z=0$, then from $... | xy1004 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A2 Find all real numbers $a, b, c, d$ such that $a+b+c+d=20$ and $a b+a c+a d+b c+b d+c d=$ 150 .
| ## Solution
$400=(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2 \cdot 150$, so $a^{2}+b^{2}+c^{2}+d^{2}=100$. Now $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}+(c-d)^{2}=3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)-2(a b+$ $a c+a d+b c+b d+c d)=300-300=0$. Thus $a=b=c=d=5$.
| =b===5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A3 Let the real parameter $p$ be such that the system
$$
\left\{\begin{array}{l}
p\left(x^{2}-y^{2}\right)=\left(p^{2}-1\right) x y \\
|x-1|+|y|=1
\end{array}\right.
$$
has at least three different real solutions. Find $p$ and solve the system for that $p$.
| ## Solution
The second equation is invariant when $y$ is replaced by $-y$, so let us assume $y \geq 0$. It is also invariant when $x-1$ is replaced by $-(x-1)$, so let us assume $x \geq 1$. Under these conditions the equation becomes $x+y=2$, which defines a line on the coordinate plane. The set of points on it that s... | p=1orp=-1,0\leqx\leq1,withy=\x | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A4 Find all triples $(x, y, z)$ of real numbers that satisfy the system
$$
\left\{\begin{array}{l}
x+y+z=2008 \\
x^{2}+y^{2}+z^{2}=6024^{2} \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2008}
\end{array}\right.
$$
| ## Solution
The last equation implies $x y z=2008(x y+y z+z x)$, therefore $x y z-2008(x y+y z+z x)+$ $2008^{2}(x+y+z)-2008^{3}=0$.
$(x-2008)(y-2008)(z-2008)=0$.
Thus one of the variable equals 2008. Let this be $x$. Then the first equation implies $y=-z$. From the second one it now follows that $2 y^{2}=6024^{2}-20... | (2008,4016,-4016) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A5 Find all triples $(x, y, z)$ of real positive numbers, which satisfy the system
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{4}{y}+\frac{9}{z}=3 \\
x+y+z \leq 12
\end{array}\right.
$$
| ## Solution
If we multiply the given equation and inequality $(x>0, y>0, z>0)$, we have
$$
\left(\frac{4 x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9 x}{z}\right)+\left(\frac{4 z}{y}+\frac{9 y}{z}\right) \leq 22
$$
From AM-GM we have
$$
\frac{4 x}{y}+\frac{y}{x} \geq 4, \quad \frac{z}{x}+\frac{9 x}{z} \geq 6,... | (2,4,6) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A9 Consider an integer $n \geq 4$ and a sequence of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$. An operation consists in eliminating all numbers not having the rank of the form $4 k+3$, thus leaving only the numbers $x_{3}, x_{7}, x_{11}, \ldots$ (for example, the sequence $4,5,9,3,6,6,1,8$ produces the sequenc... | ## Solution
After the first operation 256 number remain; after the second one, 64 are left, then 16, next 4 and ultimately only one number.
Notice that the 256 numbers left after the first operation are $3,7, \ldots, 1023$, hence they are in arithmetical progression of common difference 4. Successively, the 64 number... | 683 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
G3 The vertices $A$ and $B$ of an equilateral $\triangle A B C$ lie on a circle $k$ of radius 1 , and the vertex $C$ is inside $k$. The point $D \neq B$ lies on $k, A D=A B$ and the line $D C$ intersects $k$ for the second time in point $E$. Find the length of the segment $C E$.
| ## Solution
As $A D=A C, \triangle C D A$ is isosceles. If $\varangle A D C=\varangle A C D=\alpha$ and $\varangle B C E=\beta$, then $\beta=120^{\circ}-\alpha$. The quadrilateral $A B E D$ is cyclic, so $\varangle A B E=180^{\circ}-\alpha$. Then $\varangle C B E=$ $120^{\circ}-\alpha$ so $\varangle C B E=\beta$. Thus... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
G7 Let $A B C$ be an isosceles triangle with $A C=B C$. The point $D$ lies on the side $A B$ such that the semicircle with diameter $[B D]$ and center $O$ is tangent to the side $A C$ in the point $P$ and intersects the side $B C$ at the point $Q$. The radius $O P$ intersects the chord $D Q$ at the point $E$ such that... | ## Solution
We denote $O P=O D=O B=R, A C=B C=b$ and $A B=2 a$. Because $O P \perp A C$ and $D Q \perp B C$, then the right triangles $A P O$ and $B Q D$ are similar and $\varangle B D Q=\varangle A O P$. So, the triangle $D E O$ is isosceles with $D E=O E$. It follows that
$$
\frac{P E}{D E}=\frac{P E}{O E}=\frac{3}... | \frac{6}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
NT1 Find all the positive integers $x$ and $y$ that satisfy the equation
$$
x(x-y)=8 y-7
$$
| ## Solution 1:
The given equation can be written as:
$$
\begin{aligned}
& x(x-y)=8 y-7 \\
& x^{2}+7=y(x+8)
\end{aligned}
$$
Let $x+8=m, m \in \mathbb{N}$. Then we have: $x^{2}+7 \equiv 0(\bmod m)$, and $x^{2}+8 x \equiv 0(\bmod m)$. So we obtain that $8 x-7 \equiv 0(\bmod m) \quad(1)$.
Also we obtain $8 x+8^{2}=8(x... | (x,y)=(63,56) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT3 Let $s(a)$ denote the sum of digits of a given positive integer $a$. The sequence $a_{1}, a_{2}, \ldots a_{n}, \ldots$ of positive integers is such that $a_{n+1}=a_{n}+s\left(a_{n}\right)$ for each positive integer $n$. Find the greatest possible $n$ for which it is possible to have $a_{n}=2008$.
| ## Solution
Since $a_{n-1} \equiv s\left(a_{n-1}\right)$ (all congruences are modulo 9 ), we have $2 a_{n-1} \equiv a_{n} \equiv 2008 \equiv 10$, so $a_{n-1} \equiv 5$. But $a_{n-1}<2008$, so $s\left(a_{n-1}\right) \leq 28$ and thus $s\left(a_{n-1}\right)$ can equal 5,14 or 23 . We check $s(2008-5)=s(2003)=5, s(2008-1... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT4 Find all integers $n$ such that $n^{4}+8 n+11$ is a product of two or more consecutive integers.
|
Solution
We will prove that $n^{4}+8 n+11$ is never a multiple of 3 . This is clear if $n$ is a multiple of 3 . If
$n$ is not a multiple of 3 , then $n^{4}+8 n+11=\left(n^{4}-1\right)+12+8 n=(n-1)(n+1)\left(n^{2}+1\right)+12+8 n$, where $8 n$ is the only term not divisible by 3 . Thus $n^{4}+8 n+11$ is never the prod... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT6 Let $f: \mathbb{N} \rightarrow \mathbb{R}$ be a function, satisfying the following condition:
for every integer $n>1$, there exists a prime divisor $p$ of $n$ such that $f(n)=f\left(\frac{n}{p}\right)-f(p)$. If
$$
f\left(2^{2007}\right)+f\left(3^{2008}\right)+f\left(5^{2009}\right)=2006
$$
determine the value o... | ## Solution
If $n=p$ is prime number, we have
$$
f(p)=f\left(\frac{p}{p}\right)-f(p)=f(1)-f(p)
$$
i.e.
$$
f(p)=\frac{f(1)}{2}
$$
If $n=p q$, where $p$ and $q$ are prime numbers, then
$$
f(n)=f\left(\frac{n}{p}\right)-f(p)=f(q)-f(p)=\frac{f(1)}{2}-\frac{f(1)}{2}=0
$$
If $n$ is a product of three prime numbers, we... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT7 Determine the minimal prime number $p>3$ for which no natural number $n$ satisfies
$$
2^{n}+3^{n} \equiv 0(\bmod p)
$$
| ## Solution
We put $A(n)=2^{n}+3^{n}$. From Fermat's little theorem, we have $2^{p-1} \equiv 1(\bmod p)$ and $3^{p-1} \equiv 1(\bmod p)$ from which we conclude $A(n) \equiv 2(\bmod p)$. Therefore, after $p-1$ steps
at most, we will have repetition of the power. It means that in order to determine the minimal prime num... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT8 Let $a, b, c, d, e, f$ are nonzero digits such that the natural numbers $\overline{a b c}, \overline{d e f}$ and $\overline{a b c d e f}$ are squares.
a) Prove that $\overline{a b c d e f}$ can be represented in two different ways as a sum of three squares of natural numbers.
b) Give an example of such a number.... |
Solution
a) Let $\overline{a b c}=m^{2}, \overline{d e f}=n^{2}$ and $\overline{a b c d e f}=p^{2}$, where $11 \leq m \leq 31,11 \leq n \leq 31$ are natural numbers. So, $p^{2}=1000 \cdot m^{2}+n^{2}$. But $1000=30^{2}+10^{2}=18^{2}+26^{2}$. We obtain the following relations
$$
\begin{gathered}
p^{2}=\left(30^{2}+10... | 225625=475^{2}=450^{2}+150^{2}+25^{2}=270^{2}+390^{2}+25^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT9 Let $p$ be a prime number. Find all positive integers $a$ and $b$ such that:
$$
\frac{4 a+p}{b}+\frac{4 b+p}{a}
$$
and
$$
\frac{a^{2}}{b}+\frac{b^{2}}{a}
$$
are integers.
| ## Solution
Since $a$ and $b$ are symmetric we can assume that $a \leq b$. Let $d=(a, b), a=d u, b=d v$ and $(u, v)=1$. Then we have:
$$
\frac{a^{2}}{b}+\frac{b^{2}}{a}=\frac{d\left(u^{3}+v^{3}\right)}{u v}
$$
Since,
$$
\left(u^{3}+v^{3}, u\right)=\left(u^{3}+v^{3}, v\right)=1
$$
we deduce that $u \mid d$ and $v \... | (,b)={(1,1),(2,2),(p,p),(2p,2p),(5,25),(6,18),(18,6),(25,5),(30,150),(150,30)} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT11 Determine the greatest number with $n$ digits in the decimal representation which is divisible by 429 and has the sum of all digits less than or equal to 11 .
| ## Solution
Let $A=\overline{a_{n} a_{n-1} \ldots a_{1}}$ and notice that $429=3 \cdot 11 \cdot 13$.
Since the sum of the digits $\sum a_{i} \leq 11$ and $\sum a_{i}$ is divisible by 3 , we get $\sum a_{i}=3,6$ or 9. As 11 divides $A$, we have
$$
11 \mid a_{n}-a_{n-1}+a_{n-2}-a_{n-3}+\ldots
$$
in other words $11 \m... | 30030000\ldots | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT12 Solve the equation $\frac{p}{q}-\frac{4}{r+1}=1$ in prime numbers.
| ## Solution
We can rewrite the equation in the form
$$
\begin{gathered}
\frac{p r+p-4 q}{q(r+1)}=1 \Rightarrow p r+p-4 q=q r+q \\
p r-q r=5 q-p \Rightarrow r(p-q)=5 q-p
\end{gathered}
$$
It follows that $p \neq q$ and
$$
\begin{gathered}
r=\frac{5 q-p}{p-q}=\frac{4 q+q-p}{p-q} \\
r=\frac{4 q}{p-q}-1
\end{gathered}
... | (p,q,r)=(3,2,7), | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
COM 2 Natural numbers 1,2,3, .., 2003 are written in an arbitrary sequence $a_{1}, a_{2}, a_{3}, \ldots a_{2003}$. Let $b_{1}=1 a_{1}, b_{2}=2 a_{2}, b_{3}=3 a_{3}, \ldots, b_{2003}=2003 a_{2003}$, and $B$ be the maximum of the numbers $b_{1}, b_{2}, b_{3}, \ldots, b_{2003}$.
a) If $a_{1}=2003, a_{2}=2002, a_{3}=2001... |
Solution: a) Using the inequality between the arithmetical and geometrical mean, we obtain that $b_{n}=n(2004-n) \leq\left(\frac{n+(2004-n)}{2}\right)^{2}=1002^{2}$ for $n=1,2,3, \ldots, 2003$. The equality holds if and only if $n=2004-n$, i.e. $n=1002$. Therefore, $B=b_{1002}=1002 \times(2004-1002)=1002^{2}$. b) Let ... | 1002^2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
87.3. Let $f$ be a strictly increasing function defined in the set of natural numbers satisfying the conditions $f(2)=a>2$ and $f(m n)=f(m) f(n)$ for all natural numbers $m$ and $n$. Determine the smallest possible value of $a$.
|
Solution. Since $f(n)=n^{2}$ is a function satisfying the conditions of the problem, the smallest posiible $a$ is at most 4. Assume $a=3$. It is easy to prove by induction that $f\left(n^{k}\right)=f(n)^{k}$ for all $k \geq 1$. So, taking into account that $f$ is strictly increasing, we get
$$
\begin{gathered}
f(3)^{... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
88.1. The positive integer $n$ has the following property: if the three last digits of $n$ are removed, the number $\sqrt[3]{n}$ remains. Find $n$.
|
Solution. If $x=\sqrt[3]{n}$, and $y, 0 \leq y1000$, and $x>31$. On the other hand, $x^{3}<1000 x+1000$, or $x\left(x^{2}-1000\right)<1000$. The left hand side of this inequality is an increasing function of $x$, and $x=33$ does not satisfy the inequality. So $x<33$. Since $x$ is an integer, $x=32$ and $n=32^{3}=32768... | 32768 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
89.2. Three sides of a tetrahedron are right-angled triangles having the right angle at their common vertex. The areas of these sides are $A, B$, and $C$. Find the total surface area of the tetrahedron.
|
Solution 1. Let $P Q R S$ be the tetrahedron of the problem and let $S$ be the vertex common to the three sides which are right-angled triangles. Let the areas of $P Q S, Q R S$, and $R P S$ be $A, B$, and $C$, respectively. Denote the area of $Q R S$ by $X$. If $S S^{\prime}$ is the altitude from $S$ (onto $P Q R$ ) ... | A+B+C+\sqrt{A^{2}+B^{2}+C^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
91.1. Determine the last two digits of the number
$$
2^{5}+2^{5^{2}}+2^{5^{3}}+\cdots+2^{5^{1991}}
$$
written in decimal notation.
|
Solution. We first show that all numbers $2^{5^{k}}$ are of the form $100 p+32$. This can be shown by induction. The case $k=1$ is clear $\left(2^{5}=32\right)$. Assume $2^{5^{k}}=100 p+32$. Then, by the binomial formula,
$$
2^{5^{k+1}}=\left(2^{5^{k}}\right)^{5}=(100 p+32)^{5}=100 q+32^{5}
$$
and
$$
\begin{gathere... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
92.1. Determine all real numbers $x>1, y>1$, and $z>1$, satisfying the equation
$$
\begin{aligned}
x+y+z+\frac{3}{x-1} & +\frac{3}{y-1}+\frac{3}{z-1} \\
& =2(\sqrt{x+2}+\sqrt{y+2}+\sqrt{z+2})
\end{aligned}
$$
|
Solution. Consider the function $f$,
$$
f(t)=t+\frac{3}{t-1}-2 \sqrt{t+2}
$$
defined for $t>1$. The equation of the problem can be written as
$$
f(x)+f(y)+f(z)=0
$$
We reformulate the formula for $f$ :
$$
\begin{aligned}
f(t) & =\frac{1}{t-1}\left(t^{2}-t+3-2(t-1) \sqrt{t+2}\right) \\
& =\frac{1}{t-1}\left(t^{2}-... | \frac{3+\sqrt{13}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
92.4. Peter has many squares of equal side. Some of the squares are black, some are white. Peter wants to assemble a big square, with side equal to $n$ sides of the small squares, so that the big square has no rectangle formed by the small squares such that all the squares in the vertices of the rectangle are of equal... |
Solution. We show that Peter only can make a $4 \times 4$ square. The construction is possible, if $n=4$ :
Now consider the case $n=5$. We may assume that at least 13 of the 25 squares are ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
93.1. Let $F$ be an increasing real function defined for all $x, 0 \leq x \leq 1$, satisfying the conditions
$$
\begin{aligned}
& F\left(\frac{x}{3}\right)=\frac{F(x)}{2} \\
& F(1-x)=1-F(x)
\end{aligned}
$$
Determine $F\left(\frac{173}{1993}\right)$ and $F\left(\frac{1}{13}\right)$.
|
Solution. Condition (i) implies $F(0)=\frac{1}{2} F(0)$, so $F(0)=0$. Because of condition (ii), $F(1)=1-F(0)=1$. Also $F\left(\frac{1}{3}\right)=\frac{1}{2}$ and $F\left(\frac{2}{3}\right)=1-F\left(\frac{1}{3}\right)=\frac{1}{2}$. Since $F$ is an increasing function, this is possible only if $F(x)=\frac{1}{2}$ for al... | \frac{3}{16},\frac{1}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
93.3. Find all solutions of the system of equations
$$
\left\{\begin{aligned}
s(x)+s(y) & =x \\
x+y+s(z) & =z \\
s(x)+s(y)+s(z) & =y-4
\end{aligned}\right.
$$
where $x, y$, and $z$ are positive integers, and $s(x), s(y)$, and $s(z)$ are the numbers of digits in the decimal representations of $x, y$, and $z$, respect... |
Solution. The first equation implies $x \geq 2$ and the first and third equation together imply
$$
s(z)=y-x-4
$$
So $y \geq x+5 \geq 7$. From (1) and the second equation we obtain $z=2 y-4$. Translated to the values of $s$, these equation imply $s(x) \leq s(2 y) \leq s(y)+1$ and $s(x) \leq s(y)$. We insert these ine... | (2,8,12) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
94.4. Determine all positive integers $n<200$, such that $n^{2}+(n+1)^{2}$ is the square of an integer.
|
Solution. We determine the integral solutions of
$$
n^{2}+(n+1)^{2}=(n+p)^{2}, \quad p \geq 2
$$
The root formula for quadratic equations yields
$$
n=p-1+\sqrt{2 p(p-1)} \geq 2(p-1)
$$
Because $n<200$, we have $p \leq 100$. Moreover, the number $2 p(p-1)$ has to be the square of an integer. If $p$ is odd, $p$ and ... | 20,3,119 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
95.2. Messages are coded using sequences consisting of zeroes and ones only. Only sequences with at most two consecutive ones or zeroes are allowed. (For instance the sequence 011001 is allowed, but 011101 is not.) Determine the number of sequences consisting of exactly 12 numbers.
|
Solution 1. Let $S_{n}$ be the set of acceptable sequences consisting of $2 n$ digits. We partition $S_{n}$ in subsets $A_{n}, B_{n}, C_{n}$, and $D_{n}$, on the basis of the two last digits of the sequence. Sequences ending in 00 are in $A_{n}$, those ending in 01 are in $B_{n}$, those ending in 10 are in $C_{n}$, an... | 466 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
96.2. Determine all real numbers $x$, such that
$$
x^{n}+x^{-n}
$$
is an integer for all integers $n$.
|
Solution. Set $f_{n}(x)=x^{n}+x^{-n}$. $f_{n}(0)$ is not defined for any $n$, so we must have $x \neq 0$. Since $f_{0}(x)=2$ for all $x \neq 0$, we have to find out those $x \neq 0$ for which $f_{n}(x)$ is an integer foe every $n>0$. We note that
$$
x^{n}+x^{-n}=\left(x+x^{-1}\right)\left(x^{n-1}+x^{1-n}\right)-\left... | \frac{}{2}\ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
97.1. Let A be a set of seven positive numbers. Determine the maximal number of triples $(x, y, z)$ of elements of A satisfying $x<y$ and $x+y=z$.
|
Solution. Let $0<a_{1}<a_{2}<\ldots<a_{7}$ be the elements of the set $A$. If $\left(a_{i}, a_{j}, a_{k}\right)$ is a triple of the kind required in the problem, then $a_{i}<a_{j}<a_{i}+a_{j}=a_{k}$. There are at most $k-1$ pairs $\left(a_{i}, a_{j}\right)$ such that $a_{i}+a_{j}=a_{k}$. The number of pairs satisfying... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
97.3. Let $A, B, C$, and $D$ be four different points in the plane. Three of the line segments $A B, A C, A D, B C, B D$, and $C D$ have length a. The other three have length $b$, where $b>a$. Determine all possible values of the quotient $\frac{b}{a}$.
|
Solution. If the three segments of length $a$ share a common endpoint, say $A$, then the other three points are on a circle of radius $a$, centered at $A$, and they are the vertices of an equilateral triangle of side length $b$. But this means that $A$ is the center of the triangle $B C D$, and
$$
\frac{b}{a}=\frac{b... | \frac{\sqrt{5}+1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
98.1. Determine all functions $f$ defined in the set of rational numbers and taking their values in the same set such that the equation $f(x+y)+f(x-y)=2 f(x)+2 f(y)$ holds for all rational numbers $x$ and $y$.
|
Solution. Insert $x=y=0$ in the equation to obtain $2 f(0)=4 f(0)$, which implies $f(0)=0$. Setting $x=0$, one obtains $f(y)+f(-y)=2 f(y)$ of $f(-y)=f(y)$. Then assume $y=n x$, where $n$ is a positive integer. We obtain
$$
f((n+1) x)=2 f(x)+2 f(n x)-f((n-1) x)
$$
In particular, $f(2 x)=2 f(x)+2 f(x)-f(0)=4 f(x)$ and... | f(x)=^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
99.1. The function $f$ is defined for non-negative integers and satisfies the condition
$$
f(n)= \begin{cases}f(f(n+11)), & \text { if } n \leq 1999 \\ n-5, & \text { if } n>1999\end{cases}
$$
Find all solutions of the equation $f(n)=1999$.
|
Solution. If $n \geq 2005$, then $f(n)=n-5 \geq 2000$, and the equation $f(n)=1999$ has no solutions. Let $1 \leq k \leq 4$. Then
$$
\begin{gathered}
2000-k=f(2005-k)=f(f(2010-k)) \\
=f(1999-k)=f(f(2004-k))=f(1993-k)
\end{gathered}
$$
Let $k=1$. We obtain three solutions $1999=f(2004)=f(1998)=f(1992)$. Moreover, $19... | 1999=f(6n),ifonlyifn=1,2,\ldots,334 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
99.2. Consider 7-gons inscribed in a circle such that all sides of the 7-gon are of different length. Determine the maximal number of $120^{\circ}$ angles in this kind of a 7-gon.
|
Solution. It is easy to give examples of heptagons $A B C D E F G$ inscribed in a circle with all sides unequal and two angles equal to $120^{\circ}$. These angles cannot lie on adjacent vertices of the heptagon. In fact, if $\angle A B C=\angle B C D=120^{\circ}$, and arc $B C$ equals $b^{\circ}$, then arcs $A B$ and... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
00.1. In how many ways can the number 2000 be written as a sum of three positive, not necessarily different integers? (Sums like $1+2+3$ and $3+1+2$ etc. are the same.)
|
Solution. Since 3 is not a factor of 2000 , there has to be at least two different numbers among any three summing up to 2000 . Denote by $x$ the number of such sums with three different summands and by $y$ the number of sums with two different summands. Consider 3999 boxes consequtively numbered fron 1 to 3999 such t... | 333333 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
00.2. The persons $P_{1}, P_{1}, \ldots, P_{n-1}, P_{n}$ sit around a table, in this order, and each one of them has a number of coins. In the start, $P_{1}$ has one coin more than $P_{2}, P_{2}$ has one coin more than $P_{3}$, etc., up to $P_{n-1}$ who has one coin more than $P_{n}$. Now $P_{1}$ gives one coin to $P_... |
Solution. Assume that $P_{n}$ has $m$ coins in the start. Then $P_{n-1}$ has $m+1$ coins, ... and $P_{1}$ has $m+n-1$ coins. In every move a player receives $k$ coins and gives $k+1$ coins away, so her net loss is one coin. After the first round, when $P_{n}$ has given $n$ coins to $P_{1}$, $P_{n}$ has $m-1$ coins, $P... | 6or63 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
01.3. Determine the number of real roots of the equation
$$
x^{8}-x^{7}+2 x^{6}-2 x^{5}+3 x^{4}-3 x^{3}+4 x^{2}-4 x+\frac{5}{2}=0
$$
|
Solution. Write
$$
\begin{gathered}
x^{8}-x^{7}+2 x^{6}-2 x^{5}+3 x^{4}-3 x^{3}+4 x^{2}-4 x+\frac{5}{2} \\
=x(x-1)\left(x^{6}+2 x^{4}+3 x^{2}+4\right)+\frac{5}{2}
\end{gathered}
$$
If $x(x-1) \geq 0$, i.e. $x \leq 0$ or $x \geq 1$, the equation has no roots. If $0x(x-1)=\left(x-\frac{1}{2}\right)^{2}-\frac{1}{4} \ge... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
02.2. In two bowls there are in total $N$ balls, numbered from 1 to $N$. One ball is moved from one of the bowls to the other. The average of the numbers in the bowls is increased in both of the bowls by the same amount, $x$. Determine the largest possible value of $x$.
|
Solution. Consider the situation before the ball is moved from urn one to urn two. Let the number of balls in urn one be $n$, and let the sum of numbers in the balls in that urn be $a$. The number of balls in urn two is $m$ and the sum of numbers $b$. If $q$ is the number written in the ball which was moved, the condi... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
02.4. Eva, Per and Anna play with their pocket calculators. They choose different integers and check, whether or not they are divisible by 11. They only look at nine-digit numbers consisting of all the digits 1, 2, .., 9. Anna claims that the probability of such a number to be a multiple of 11 is exactly 1/11. Eva has... |
Solution. We write the numbers in consideration, $n=a_{0}+10 a_{1}+10^{2} a_{2}+\cdots+10^{8} a_{8}$, in the form
$$
\begin{gathered}
a_{0}+(11-1) a_{1}+(99+1) a_{2}+(1001-1) a_{3} \\
+(9999+1) a_{4}+(100001-1) a_{5}+(999999+1) a_{6} \\
\quad+(10000001-1) a_{7}+(99999999+1) a_{8} \\
=\left(a_{0}-a_{1}+a_{2}-a_{3}+a_{... | \frac{11}{126}<\frac{1}{11} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
03.2. Find all triples of integers $(x, y, z)$ satisfying
$$
x^{3}+y^{3}+z^{3}-3 x y z=2003
$$
|
Solution. It is a well-known fact (which can be rediscovered e.g. by noticing that the left hand side is a polynomial in $x$ having $-(y+z)$ as a zero) that
$$
\begin{aligned}
& x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) \\
&=(x+y+z) \frac{(x-y)^{2}+(y-z)^{2}+(z-x)^{2}}{2}
\end{aligne... | (668,668,667) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
03.4. Let $\mathbb{R}^{*}=\mathbb{R} \backslash\{0\}$ be the set of non-zero real numbers. Find all functions $f: \mathbb{R}^{*} \rightarrow \mathbb{R}^{*}$ satisfying
$$
f(x)+f(y)=f(x y f(x+y))
$$
for $x, y \in \mathbb{R}^{*}$ and $x+y \neq 0$.
|
Solution. If $x \neq y$, then
$$
f(y)+f(x-y)=f(y(x-y) f(x))
$$
Because $f(y) \neq 0$, we cannot have $f(x-y)=f(y(x-y) f(x))$ or $x-y=y(x-y) f(x)$. So for all $x \neq y, y f(x) \neq 1$. The only remaining possibility is $f(x)=\frac{1}{x}$. - One easily checks that $f, f(x)=\frac{1}{x}$, indeed satisfies the original ... | f(x)=\frac{1}{x} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
04.1. 27 balls, labelled by numbers from 1 to 27, are in a red, blue or yellow bowl. Find the possible numbers of balls in the red bowl, if the averages of the labels in the red, blue, and yellow bowl are 15, 3 ja 18, respectively.
|
Solution. Let $R, B$, and $Y$, respectively, be the numbers of balls in the red, blue, and yellow bowl. The mean value condition implies $B \leq 5$ (there are at most two balls with a number $3$ ). $R, B$ and $Y$ satisfy the equations
$$
\begin{aligned}
R+B+Y & =27 \\
15 R+3 S+18 Y & =\sum_{j=1}^{27} j=14 \cdot 27=37... | 21,16,11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
05.1. Find all positive integers $k$ such that the product of the digits of $k$, in the decimal system, equals
$$
\frac{25}{8} k-211
$$
|
Solution. Let
$$
a=\sum_{k=0}^{n} a_{k} 10^{k}, \quad 0 \leq a_{k} \leq 9, \text { for } 0 \leq k \leq n-1,1 \leq a_{n} \leq 9
$$
Set
$$
f(a)=\prod_{k=0}^{n} a_{k}
$$
Since
$$
f(a)=\frac{25}{8} a-211 \geq 0
$$
$a \geq \frac{8}{25} \cdot 211=\frac{1688}{25}>66$. Also, $f(a)$ is an integer, and $\operatorname{gcf}... | 7288 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
06.2. The real numbers $x, y$ and $z$ are not all equal and satisfy
$$
x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}=k
$$
Determine all possible values of $k$.
|
Solution. Let $(x, y, z)$ be a solution of the system of equations Since
$$
x=k-\frac{1}{y}=\frac{k y-1}{y} \quad \text { and } \quad z=\frac{1}{k-y}
$$
the equation
$$
\frac{1}{k-y}+\frac{y}{k y-1}=k
$$
to be simplified into
$$
\left(1-k^{2}\right)\left(y^{2}-k y+1\right)=0
$$
is true. So either $|k|=1$ or
$$
... | \1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
06.3. A sequence of positive integers $\left\{a_{n}\right\}$ is given by
$$
a_{0}=m \quad \text { and } \quad a_{n+1}=a_{n}^{5}+487
$$
for all $n \geq 0$. Determine all values of $m$ for which the sequence contains as many square numbers as possible.
|
Solution. Consider the expression $x^{5}+487$ modulo 4. Clearly $x \equiv 0 \Rightarrow x^{5}+487 \equiv 3$, $x \equiv 1 \Rightarrow x^{5}+487 \equiv 0 ; x \equiv 2 \Rightarrow x^{5}+487 \equiv 3$, and $x \equiv 3 \Rightarrow x^{5}+487 \equiv 2$. Square numbers are always $\equiv 0$ or $\equiv 1 \bmod 4$. If there is ... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
07.1. Find one solution in positive integers to the equation
$$
x^{2}-2 x-2007 y^{2}=0
$$
|
Solution. The equation can be written in the form
$$
x(x 2)=223 \cdot(3 y)^{2}
$$
Here the prime number 223 must divide $x$ or $x 2$. In fact, for $x=225$ we get $x(x 2)=$ $15^{2} \cdot 223$, which is equivalent to $223 \cdot(3 y)^{2}$ for $y=5$. Thus, $(x, y)=(225,5)$ is one solution.
| (225,5) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
08.1. Determine all real numbers $A, B$ and $C$ such that there exists a real function $f$ that satisfies
$$
f(x+f(y))=A x+B y+C
$$
for all real $x$ and $y$.
|
Solution. Let $A, B$ and $C$ be real numbers and $f$ a function such that $f(x+f(y))=$ $A x+B y+C$ for all $x$ and $y$. Let $z$ be a real number and set $x=z-f(0)$ and $y=0$. Then
$$
f(z)=f(z-f(0)+f(0))=A(z-f(0))+B \cdot 0+C=A z-A f(0)+C
$$
so there are numbers $a$ and $b$ such that $f(z)=a z+b$ for all $z$. Now $f(... | (A,B,C)=(,^2,)where\neq-1arbitrary,or(A,B,C)=(-1,1,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
08.2. Assume that $n \geq 3$ people with different names sit around a round table. We call any unordered pair of them, say $M$ and $N$, dominating, if
(i) $M$ and $N$ do not sit on adjacent seats, and
(ii) on one (or both) of the arcs connecting $M$ and $N$ along the table edge, all people have names that come alpha... |
Solution. We will show by induction that the number of dominating pairs (hence also the minimal number of dominating pairs) is $n-3$ for $n \geq 3$. If $n=3$, all pairs of people sit on adjacent seats, so there are no dominating pairs. Assume that the number of dominating pairs is $n-3$ for some $n>3$. If there are $n... | n-3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
09.2. On a faded piece of paper it is possible, with some effort, to discern the following:
$$
\left(x^{2}+x+a\right)\left(x^{15}-\ldots\right)=x^{17}+x^{13}+x^{5}-90 x^{4}+x-90
$$
Some parts have got lost, partly the constant term of the first factor of the left side, partly the main part of the other factor. It wo... |
Solution. We denote the polynomial $x^{2}+x+a$ by $P_{a}(x)$, the polynomial forming the other factor of the left side by $Q(x)$ and the polynomial on the right side by $R(x)$. The polynomials are integer valued for every integer $x$. For $x=0$ we get $P_{a}(0)=a$ and $R(0)=-90$, so $a$ is a divisor of $90=2 \cdot 3 \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. Laura has 2010 lamps connected with 2010 buttons in front of her. For each button, she wants to know the corresponding lamp. In order to do this, she observes which lamps are lit when Richard presses a selection of buttons. (Not pressing anything is also a possible selection.) Richard always presses the buttons ... |
Solution. a) Let us say that two lamps are separated, if one of the lamps is turned on while the other lamp remains off. Laura can find out which lamps belong to the buttons if every two lamps are separated. Let Richard choose two arbitrary lamps. To begin with, he turns both lamps on and then varies all the other lam... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.4. A positive integer is called simple if its ordinary decimal representation consists entirely of zeroes and ones. Find the least positive integer $k$ such that each positive integer $n$ can be written as $n=a_{1} \pm a_{2} \pm a_{3} \pm \cdots \pm a_{k}$, where $a_{1}, \ldots, a_{k}$ are simple.
|
Solution. We can always write $n=a_{l}+a_{2}+\cdots+a_{9}$ where $a_{j}$ has 1 's in the places where $n$ has digits greater or equal to $j$ and 0 's in the other places. So $k \leq 9$. To show that $k \geq 9$, consider $n=10203040506070809$. Suppose $n=a_{l}+a_{2}+\cdots+a_{j}-a_{j+l}-a_{j+2}-\cdots-a_{k}$, where $a_... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Problem 2
Let $A B C D$ be a cyclic quadrilateral satisfying $A B=A D$ and $A B+B C=C D$.
Determine $\angle C D A$.
|
Solution 2 Answer: $\angle C D A=60^{\circ}$.
Choose the point $E$ on the segment $C D$ such that $D E=A D$. Then $C E=C D-A D=$ $C D-A B=B C$, and hence the triangle $C E B$ is isosceles.
... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem 3
Find all $a \in \mathbb{R}$ for which there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$, such that
(i) $f(f(x))=f(x)+x$, for all $x \in \mathbb{R}$,
(ii) $f(f(x)-x)=f(x)+$ ax, for all $x \in \mathbb{R}$.
|
Solution 3 Answer: $a=\frac{1 \pm \sqrt{5}}{2}$.
From (i) we get $f(f(f(x))-f(x))=f(x)$. On the other hand (ii) gives
$$
f(f(f(x))-f(x))=f(f(x))+a f(x)
$$
Thus we have $(1-a) f(x)=f(f(x))$. Now it follows by (i) that $(1-a) f(x)=f(x)+x$, and hence $f(x)=-\frac{1}{a} x$, since $a=0$ obviously does not give a solutio... | \frac{1\\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem 4
King George has decided to connect the 1680 islands in his kingdom by bridges. Unfortunately the rebel movement will destroy two bridges after all the bridges have been built, but not two bridges from the same island.
What is the minimal number of bridges the King has to build in order to make sure that ... |
Solution 4 Answer: 2016
An island cannot be connected with just one bridge, since this bridge could be destroyed. Consider the case of two islands, each with only two bridges, connected by a bridge. (It is not possible that they are connected with two bridges, since then they would be isolated from the other islands ... | 2016 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Problem 1
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(f(x) f(1-x))=f(x) \quad \text { and } \quad f(f(x))=1-f(x)
$$
for all real $x$.
|
Solution 1. Notice that $f(f(f(x)))=^{2} 1-f(f(x))={ }^{2} f(x)$. This is equation 3 . By substituting $f(x)$ for $x$ in the first equation we get:
$$
f(\underline{f(x)})={ }^{1} f(f(\underline{f(x)}) f(1-\underline{f(x)}))={ }^{2} f(f(f(x)) f(f(f(x))))={ }^{3} f(f(f(x)) f(x))
$$
Again we substitute $f(x)$ for $x$ a... | f(x)=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Find the smallest positive integer $n$, such that there exist $n$ integers $x_{1}, x_{2}, \ldots, x_{n}$ (not necessarily different), with $1 \leq x_{k} \leq n, 1 \leq k \leq n$, and such that
$$
x_{1}+x_{2}+\cdots+x_{n}=\frac{n(n+1)}{2}, \quad \text { and } \quad x_{1} x_{2} \cdots x_{n}=n!
$$
but $\left... |
Solution. If it is possible to find a set of numbers as required for some $n=k$, then it will also be possible for $n=k+1$ (choose $x_{1}, \ldots, x_{k}$ as for $n=k$, and
let $x_{k+1}=k+1$ ). Thus we have to find a positive integer $n$ such that a set as required exists, and prove that such a set does not exist for $... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. The number 1 is written on the blackboard. After that a sequence of numbers is created as follows: at each step each number $a$ on the blackboard is replaced by the numbers $a-1$ and $a+1$; if the number 0 occurs, it is erased immediately; if a number occurs more than once, all its occurrences are left on t... |
Solution I. Let $S$ be a set of different numbers, all of them less than $2^{n-1}$, and create two new sets as follows: $S_{1}$, consisting of all the numbers in $S$ except
the smallest one, and $S_{2}$, with elements the smallest element of $S$ and all the numbers we get by adding $2^{n-1}$ to each number in $S$. Not... | \binom{n}{\lfloor\frac{n}{2}\rfloor} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Problem 2.
Find the primes $p, q, r$, given that one of the numbers $p q r$ and $p+q+r$ is 101 times the other.
|
Solution. We may assume $r=\max \{p, q, r\}$. Then $p+q+r \leq 3 r$ and $p q r \geq 4 r$. So the sum of the three primes is always less than their product. The only relevant requirement thus is $p q r=101(p+q+r)$. We observe that 101 is a prime. So one of $p, q, r$ must be 101. Assume $r=101$. Then $p q=p+q+101$. This... | {2,101,103} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Problem 4.
An encyclopedia consists of 2000 numbered volumes. The volumes are stacked in order with number 1 on top and 2000 in the bottom. One may perform two operations with the stack:
(i) For $n$ even, one may take the top $n$ volumes and put them in the bottom of the stack without changing the order.
(ii) For... |
Solution 1. (By the proposer.) Let the positions of the books in the stack be $1,2,3, \ldots, 2000$ from the top (and consider them modulo 2000). Notice that both operations fix the parity of the number of the book at a any given position. Operation (i) subtracts an even integer from the number of the book at each pos... | (1000!)^2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a football tournament there are $n$ teams, with $n \geq 4$, and each pair of teams meets exactly once. Suppose that, at the end of the tournament, the final scores form an arithmetic sequence where each team scores 1 more point than the following team on the scoreboard. Determine the maximum possible sco... |
Solution. Note that the total number of games equals the number of different pairings, that is, $n(n-1) / 2$. Suppose the lowest scoring team ends with $k$ points. Then the total score for all teams is
$$
k+(k+1)+\cdots+(k+n-1)=n k+\frac{(n-1) n}{2}
$$
Some games must end in a tie, for otherwise, all team scores wou... | n-2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LIV OM - II - Task 3
Given is the polynomial $ W(x) = x^4 - 3x^3 + 5x^2 - 9x $. Determine all pairs of different integers $ a $, $ b $ satisfying the equation | Notice that $ W(x) = (x - 1)(x - 2)(x^2 + 3) - 6 $. For $ n > 3 $, the following inequalities therefore hold
and moreover
From this, we obtain
Thus, the values of the polynomial $ W $ at the points $ -2, \pm 3, \pm 4, \ldots $ are pairwise distinct. We directly calculate that
and from inequality... | (-1,0),(0,-1),(1,2),(2,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XXIV OM - I - Problem 8
Find a polynomial with integer coefficients of the lowest possible degree, for which $ \sqrt{2} + \sqrt{3} $ is a root. | We will find a polynomial with integer coefficients, of which the root is the number $ a = \sqrt{2} + \sqrt{3} $. It is, of course, a root of the polynomial
This polynomial, however, does not have integer coefficients. Let's multiply the polynomial (1) by $ x - \sqrt{2} + \sqrt{3} $. We get $ (x - \sqrt{2})^2 - 3... | x^4-10x^2+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XLVII OM - I - Problem 2
A palindromic number is defined as a natural number whose decimal representation read from left to right is the same as when read from right to left. Let $ (x_n) $ be the increasing sequence of all palindromic numbers. Determine all prime numbers that are divisors of at least one of the differ... | The decimal representation of the palindromic number $ x_n $ (respectively: ($ 2m $)-digit or ($ 2m-1 $)-digit) looks like this:
or
($ c_i \in \{ 0,1,2,3,4,5,6,7,8,9\} $ for $ i = 1, \ldots ,m;\ c_1 \ne 0 $). There are three possible situations:
Case I. $ c_m \ne 9 $. If $ x_n $ has the form (1), then $ x_{n+1} $ (th... | 2,5,11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
VI OM - I - Problem 9
Present the polynomial $ x^4 + x^3 + x^2 + x + 1 $ as the difference of squares of two polynomials of different degrees with real coefficients. | If the polynomial $ W (x) = x^4 + x^3 + x^2 + x + 1 $ is equal to the difference $ U(x)^2 - V(x)^2 $, where $ U(x) $ and $ V(x) $ are polynomials of different degrees, then the polynomial $ U(x) $ must be of the second degree and the polynomial $ V(x) $ must be of the first or zero degree. In this case, the polynomial ... | x^4+x^3+x^2+x+1=(x^2+\frac{1}{2}x+1)^2-(\frac{1}{2}x)^2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XLVI OM - I - Problem 11
Given are natural numbers $ n > m > 1 $. From the set $ \{1,2, \ldots ,n\} $, we draw $ m $ numbers without replacement. Calculate the expected value of the difference between the largest and the smallest drawn number. | An elementary event is the selection of an $m$-element subset from an $n$-element set. These elementary events are equally probable; there are $ {n}\choose{m} $ of them. Let the largest and smallest selected number be denoted by $a$ and $b$, respectively. The random variable $X$ under consideration, which is the differ... | \frac{(n-1)}{n-+1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Given an integer $ c \geq 1 $. To each subset $ A $ of the set $ \{1,2, \ldots ,n\} $, we assign a number $ w(A) $ from the set $ \{1,2, \ldots ,c\} $ such that the following condition is satisfied:
Let $ a(n) $ be the number of such assignments. Calculate $ \lim_{n\to \infty}\sqrt[n]{a(n)} $.
Note: $ \min(x,y) $ is t... | Let $ M $ be the set $ \{1,2,\ldots,n\} $, and $ M_i $ be the $(n-1)$-element set obtained from $ M $ by removing the number $ i $:
Given the assignment as mentioned in the problem, let's denote the value $ w(M) $ by $ m $, and the value $ w(M_i) $ by $ f(i) $ (for $ i = 1,2, \ldots ,n $). According to the conditions ... | c | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LV OM - III - Task 3
In a certain tournament, $ n $ players participated $ (n \geq 3) $. Each played against each other exactly once, and there were no draws. A three-element set of players is called a draw triplet if the players can be numbered in such a way that the first won against the second, the second against t... | Instead of determining the maximum number of draw triples, we will determine the minimum number of non-draw triples.
Let's number the players from 1 to $ n $ and assume that player number $ i $ won $ x_i $ times. Then the total number of matches is
Each non-draw triple is uniquely determined by the player who won... | \frac{1}{24}n(n^2-1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXIV OM - I - Problem 10
Find the smallest natural number $ n > 1 $ with the following property: there exists a set $ Z $ consisting of $ n $ points in the plane such that every line $ AB $ ($ A, B \in Z $) is parallel to some other line $ CD $ ($ C, D \in Z $). | We will first prove that the set $ Z $ of vertices of a regular pentagon has the property given in the problem, that is, $ n \leq 5 $. We will show that each side of the regular pentagon is parallel to a certain diagonal and vice versa, each diagonal is parallel to a corresponding side.
It suffices to prove that $ AB \... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXV - I - Task 1
During World War I, a battle took place near a certain castle. One of the shells destroyed a statue of a knight with a spear standing at the entrance to the castle. This happened on the last day of the month. The product of the day of the month, the month number, the length of the spear expressed in f... | The last day of the month can only be $28$, $29$, $30$, or $31$. Of these numbers, only $29$ is a divisor of the number $451,066 = 2 \cdot 7 \cdot 11 \cdot 29 \cdot 101$. Therefore, the battle took place on February $29$ in a leap year. During World War I, only the year $1916$ was a leap year. From the problem statemen... | 1714 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
XXI OM - III - Task 1
Diameter $ \overline{AB} $ divides the circle into two semicircles. On one semicircle, n points $ P_1 P_2, \ldots, P_n $ are chosen such that $ P_1 $ lies between $ A $ and $ P_2 $, $ P_2 $ lies between $ P_1 $ and $ P_3 $, $ \ldots $, $ P_n $ lies between $ P_{n-1} $ and $ B $. How should point ... | Notice that the sum of the areas of triangles $ CP_1P_2, CP_2P_3, CP_3P_4, \ldots, CP_{n-1}P_n $ is equal to the sum of the areas of the polygon $ P_1P_2 \ldots P_n $ and triangle $ CP_1P_n $ (Fig. 13). The first area does not depend on the choice of point $ C $. The second area will be maximal when the distance from p... | C | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LVI OM - I - Problem 10
Among all subsets of a fixed $ n $-element set $ X $, we sequentially draw with replacement three sets $ A $, $ B $, $ C $. Each time, the probability of drawing any of the $ 2^n $ subsets of set $ X $ is equally likely. Determine the most probable number of elements in the set $ A\cap B\cap C ... | The set of elementary events $\Omega$ consists of all triples $(A,B,C)$, where $A$, $B$, and $C$ are subsets of a given $n$-element set $S$. The selection of each of the $(2^n)^3 = 8^n$ triples $(A,B,C)$ is equally probable. Let $X_k$ $(k = 0,1,2,\ldots,n)$ denote the event that the triple of sets $(A,B,C)$ satisfies t... | [\frac{1}{8}(n+1)] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXII OM - I - Problem 10
Given is a table with $ n $ rows and $ n $ columns. The number located in the $ m $-th column and $ k $-th row is equal to $ n(k - 1) + m $. How should $ n $ numbers be chosen, one from each row and each column, so that the product of these numbers is the largest? | Let's denote $ a_{k,m} = n (k - 1) + m $, where $ 1 \leq k, m \leq n $. Suppose we select the number in the column with number $ m_i $ from the $ i $-th row, where $ i = 1, 2, \ldots, n $. From the conditions of the problem, it follows that the sequence $ m_1, m_2, \ldots, m_n $ has distinct terms. Each such sequence c... | a_{1n}a_{2n-1}\ldotsa_{n1}=n(2n-1)(3n-2)(4n-3)\ldots(n^2-(n-1)) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
V OM - I - Problem 7
In the plane, a line $ p $ and points $ A $ and $ B $ are given. Find a point $ M $ on the line $ p $ such that the sum of the squares $ AM^2 + BM^2 $ is minimized. | The geometric locus of points $ X $ in the plane for which $ AX^2 + BX^2 $ has a given value $ 2k^2 $ is a certain circle whose center lies at the midpoint $ S $ of segment $ AB $; the radius of this circle is larger the larger $ k $ is.
The smallest circle centered at $ S $ that intersects a line $ p $ is the circle t... | M | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXIX OM - I - Problem 12
Determine the least upper bound of such numbers $ \alpha \leq \frac{\pi}{2} $, that every acute angle $ MON $ of measure $ \alpha $ and every triangle $ T $ on the plane have the following property. There exists a triangle $ ABC $ isometric to $ T $ such that the side $ \overline{AB} $ is para... | First, note that if $Q$ is the midpoint of side $\widehat{AC}$, then $\measuredangle ABQ \leq \frac{\pi}{6}$. Indeed, denoting by $D$ the projection of point $A$ onto the line $BQ$, we get $AD \leq AQ = \frac{1}{2} AC \leq \frac{1}{2} AB$ (Fig. 9).
Therefore, $\sin \measuredangle ABQ = \frac{AD}{AB} \leq \frac{1}{2}$. ... | \frac{\pi}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LVII OM - I - Problem 4
Participants in a mathematics competition solved six problems, each graded with one of the scores 6, 5, 2, 0. It turned out that
for every pair of participants $ A, B $, there are two problems such that in each of them $ A $ received a different score than $ B $.
Determine the maximum number ... | We will show that the largest number of participants for which such a situation is possible is 1024. We will continue to assume that the permissible ratings are the numbers 0, 1, 2, 3 (instead of 5 points, we give 4, and then divide each rating by 2).
Let $ P = \{0,1,2,3\} $ and consider the set
Set $ X $ obviou... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
II OM - III - Task 1
A beam of length $ a $ has been suspended horizontally by its ends on two parallel ropes of equal length $ b $. We rotate the beam by an angle $ \varphi $ around a vertical axis passing through the center of the beam. By how much will the beam be raised? | When solving geometric problems, a properly executed drawing is an important aid to our imagination. We represent spatial figures through mappings onto the drawing plane. There are various ways of such mapping. In elementary geometry, we most often draw an oblique parallel projection of the figure; in many cases, it is... | \sqrt{b^2-(\frac{} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LVII OM - III - Problem 2
Determine all positive integers $ k $ for which the number $ 3^k+5^k $ is a power of an integer with an exponent greater than 1. | If $ k $ is an even number, then the numbers $ 3^k $ and $ 5^k $ are squares of odd numbers, giving a remainder of 1 when divided by 4. Hence, the number $ 3^k + 5^k $ gives a remainder of 2 when divided by 4, and thus is divisible by 2 but not by $ 2^2 $. Such a number cannot be a power of an integer with an exponent ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXIV OM - I - Problem 12
In a class of n students, a Secret Santa event was organized. Each student draws the name of the person for whom they are to buy a gift, so student $ A_1 $ buys a gift for student $ A_2 $, $ A_2 $ buys a gift for $ A_3 $, ..., $ A_k $ buys a gift for $ A_1 $, where $ 1 \leq k \leq n $. Assumin... | In the result of any drawing, each student draws a certain student from the same class (possibly themselves), and different students draw different students. Therefore, the result of each drawing defines a certain one-to-one mapping (permutation) of the set of all students in the class onto itself. Conversely, each per... | \frac{1}{n} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
VIII OM - I - Task 6
Find a four-digit number, whose first two digits are the same, the last two digits are the same, and which is a square of an integer. | If $ x $ is the number sought, then
where $ a $ and $ b $ are integers satisfying the inequalities $ 0 < a \leq 9 $, $ 0 \leq b \leq 9 $. The number $ x $ is divisible by $ 11 $, since
Since $ x $ is a perfect square, being divisible by $ 11 $ it must be divisible by $ 11^2 $, so the number
is divisib... | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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