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Problem 1. Find all prime numbers $p$ for which there exist positive integers $x, y$ and $z$ such that the number
$$
x^{p}+y^{p}+z^{p}-x-y-z
$$
is a product of exactly three distinct prime numbers.
|
Solution. Let $A=x^{p}+y^{p}+z^{p}-x-y-z$. For $p=2$, we take $x=y=4$ and $z=3$. Then $A=30=2 \cdot 3 \cdot 5$. For $p=3$ we can take $x=3$ and $y=2$ and $z=1$. Then again $A=30=2 \cdot 3 \cdot 5$. For $p=5$ we can take $x=2$ and $y=1$ and $z=1$. Again $A=30=2 \cdot 3 \cdot 5$.
Assume now that $p \geqslant 7$. Workin... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to find all prime numbers p where there exist positive integers x, y, z such that the expression x^p + y^p + z^p - x - y - z is a product of exactly three distinct prime numbers. Hmm, let me break this down step by step.
First, since p is a prime number, possible values for p are 2, 3, 5, 7, et... |
Problem 1. Find all prime numbers $p$ for which there exist positive integers $x, y$ and $z$ such that the number
$$
x^{p}+y^{p}+z^{p}-x-y-z
$$
is a product of exactly three distinct prime numbers.
|
Solution. Let $A=x^{p}+y^{p}+z^{p}-x-y-z$. For $p=2$, we take $x=y=4$ and $z=3$. Then $A=30=2 \cdot 3 \cdot 5$. For $p=3$ we can take $x=3$ and $y=2$ and $z=1$. Then again $A=30=2 \cdot 3 \cdot 5$. For $p=5$ we can take $x=2$ and $y=1$ and $z=1$. Again $A=30=2 \cdot 3 \cdot 5$.
Assume now that $p \geqslant 7$. Workin... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to find all prime numbers p such that there exist positive integers x, y, z where the expression x^p + y^p + z^p - x - y - z is a product of exactly three distinct prime numbers. Hmm, let's break this down step by step.
First, since p is a prime number, possible values are 2, 3, 5, 7, etc. The ... |
Problem 2. Let $a, b$ be two distinct real numbers and let $c$ be a positive real number such that
$$
a^{4}-2019 a=b^{4}-2019 b=c .
$$
Prove that $-\sqrt{c}<a b<0$.
|
Solution. Firstly, we see that
$$
2019(a-b)=a^{4}-b^{4}=(a-b)(a+b)\left(a^{2}+b^{2}\right)
$$
Since $a \neq b$, we get $(a+b)\left(a^{2}+b^{2}\right)=2019$, so $a+b \neq 0$. Thus
$$
\begin{aligned}
2 c & =a^{4}-2019 a+b^{4}-2019 b \\
& =a^{4}+b^{4}-2019(a+b) \\
& =a^{4}+b^{4}-(a+b)^{2}\left(a^{2}+b^{2}\right) \\
& ... | proof | Algebra | proof | Yes | Incomplete | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to prove that if two distinct real numbers a and b, and a positive real number c satisfy the equations a⁴ - 2019a = b⁴ - 2019b = c, then the product ab is between -√c and 0. Hmm, let me start by understanding what the problem is saying.
First, both a and b are distinct real numbers, and c is po... |
Problem 2. Let $a, b$ be two distinct real numbers and let $c$ be a positive real number such that
$$
a^{4}-2019 a=b^{4}-2019 b=c .
$$
Prove that $-\sqrt{c}<a b<0$.
|
Solution. Firstly, we see that
$$
2019(a-b)=a^{4}-b^{4}=(a-b)(a+b)\left(a^{2}+b^{2}\right)
$$
Since $a \neq b$, we get $(a+b)\left(a^{2}+b^{2}\right)=2019$, so $a+b \neq 0$. Thus
$$
\begin{aligned}
2 c & =a^{4}-2019 a+b^{4}-2019 b \\
& =a^{4}+b^{4}-2019(a+b) \\
& =a^{4}+b^{4}-(a+b)^{2}\left(a^{2}+b^{2}\right) \\
& ... | proof | Algebra | proof | Yes | Incomplete | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I have this problem here where there are two distinct real numbers, a and b, and a positive real number c. The equations given are a⁴ - 2019a = c and b⁴ - 2019b = c. I need to prove that -√c is less than ab, which is less than 0. Hmm, interesting. Let me try to unpack this step by step.
First, since a... |
Problem 3. Triangle $A B C$ is such that $A B<A C$. The perpendicular bisector of side $B C$ intersects lines $A B$ and $A C$ at points $P$ and $Q$, respectively. Let $H$ be the orthocentre of triangle $A B C$, and let $M$ and $N$ be the midpoints of segments $B C$ and $P Q$, respectively. Prove that lines $H M$ and $... |
Solution. We have
$$
\angle A P Q=\angle B P M=90^{\circ}-\angle M B P=90^{\circ}-\angle C B A=\angle H C B
$$
and
$$
\angle A Q P=\angle M Q C=90^{\circ}-\angle Q C M=90^{\circ}-\angle A C B=\angle C B H
$$
From these two equalities, we see that the triangles $A P Q$ and $H C B$ are similar. Moreover, since $M$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to prove that lines HM and AN meet on the circumcircle of triangle ABC. Let me start by understanding the problem step by step.
First, we have triangle ABC with AB < AC. The perpendicular bisector of BC intersects AB at P and AC at Q. H is the orthocenter of ABC. M is the midpoint of BC, which ... |
Problem 3. Triangle $A B C$ is such that $A B<A C$. The perpendicular bisector of side $B C$ intersects lines $A B$ and $A C$ at points $P$ and $Q$, respectively. Let $H$ be the orthocentre of triangle $A B C$, and let $M$ and $N$ be the midpoints of segments $B C$ and $P Q$, respectively. Prove that lines $H M$ and $... |
Solution. We have
$$
\angle A P Q=\angle B P M=90^{\circ}-\angle M B P=90^{\circ}-\angle C B A=\angle H C B
$$
and
$$
\angle A Q P=\angle M Q C=90^{\circ}-\angle Q C M=90^{\circ}-\angle A C B=\angle C B H
$$
From these two equalities, we see that the triangles $A P Q$ and $H C B$ are similar. Moreover, since $M$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Alright, so I have this geometry problem here. Let me read it again to make sure I understand all the parts.
We have triangle ABC where AB is less than AC. That's important because it tells me that triangle ABC isn't isoceles at A; AC is longer. Then, the perpendicular bisector of side BC intersects lines AB ... |
Problem 4. A $5 \times 100$ table is divided into 500 unit square cells, where $n$ of them are coloured black and the rest are coloured white. Two unit square cells are called adjacent if they share a common side. Each of the unit square cells has at most two adjacent black unit square cells. Find the largest possible... |
Solution. If we colour all the cells along all edges of the board together with the entire middle row except the second and the last-but-one cell, the condition is satisfied and there are 302 black cells. The figure below exhibits this colouring for the $5 \times 8$ case.
![](https://cdn.mathpix.com/cropped/2024_06_0... | 302 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to figure out the maximum number of black cells, n, in a 5x100 grid where each black cell has at most two adjacent black cells. Hmm, let me start by understanding the problem better.
First, the grid is 5 rows by 100 columns, which makes 500 cells total. Each cell can be black or white. The key ... |
"\nProblem 4. A $5 \\times 100$ table is divided into 500 unit square cells, where $n$ of them are c(...TRUNCATED) | "\nSolution. If we colour all the cells along all edges of the board together with the entire middle(...TRUNCATED) | 302 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | "<think>\nOkay, so I need to find the largest possible number of black cells (n) in a 5x100 grid whe(...TRUNCATED) |
"\nProblem 1. A trapezoid $A B C D(A B \\| C D, A B>C D)$ is circumscribed. The incircle of the tria(...TRUNCATED) | "## Solution.\n\nVersion 1. Let $I$ be the incenter of triangle $A B C$ and $R$ be the common point (...TRUNCATED) | proof | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | "<think>\nOkay, let's see. I need to prove that the incenter of the trapezoid ABCD lies on the line (...TRUNCATED) |
"\nProblem 1. A trapezoid $A B C D(A B \\| C D, A B>C D)$ is circumscribed. The incircle of the tria(...TRUNCATED) | "## Solution.\n\nVersion 1. Let $I$ be the incenter of triangle $A B C$ and $R$ be the common point (...TRUNCATED) | proof | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | "<think>\nAlright, so I have this problem about a trapezoid ABCD where AB is parallel to CD and AB i(...TRUNCATED) |
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