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class | generations_count int64 1 8 | correctness_math_verify listlengths 1 8 | correct_count int64 0 8 | generation stringlengths 25 75.7k |
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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.
 in a 5x100 grid where each black cell has at most two adjacent black cells. Let me think through this step by step.
First, the problem states that each cell, whether black or white, has at most two adjacent black cells. Wait, actually, let ... |
Problem 3. Find all the triples of integers $(a, b, c)$ such that the number
$$
N=\frac{(a-b)(b-c)(c-a)}{2}+2
$$
is a power of 2016 .
(A power of 2016 is an integer of the form $2016^{n}$, where $n$ is a non-negative integer.)
|
Solution. Let $a, b, c$ be integers and $n$ be a positive integer such that
$$
(a-b)(b-c)(c-a)+4=2 \cdot 2016^{n}
$$
We set $a-b=-x, b-c=-y$ and we rewrite the equation as
$$
x y(x+y)+4=2 \cdot 2016^{n}
$$
If $n>0$, then the right hand side is divisible by 7 , so we have that
$$
x y(x+y)+4 \equiv 0 \quad(\bmod 7)... | (,b,)=(k+2,k+1,k),k\in\mathbb{Z} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to find all integer triples (a, b, c) such that N, defined as [(a - b)(b - c)(c - a)/2] + 2, is a power of 2016. A power of 2016 means it can be written as 2016^n where n is a non-negative integer. Let me break this down step by step.
First, let me understand the structure of N. It's given by [... |
Problem 3. Find all the triples of integers $(a, b, c)$ such that the number
$$
N=\frac{(a-b)(b-c)(c-a)}{2}+2
$$
is a power of 2016 .
(A power of 2016 is an integer of the form $2016^{n}$, where $n$ is a non-negative integer.)
|
Solution. Let $a, b, c$ be integers and $n$ be a positive integer such that
$$
(a-b)(b-c)(c-a)+4=2 \cdot 2016^{n}
$$
We set $a-b=-x, b-c=-y$ and we rewrite the equation as
$$
x y(x+y)+4=2 \cdot 2016^{n}
$$
If $n>0$, then the right hand side is divisible by 7 , so we have that
$$
x y(x+y)+4 \equiv 0 \quad(\bmod 7)... | (,b,)=(k+2,k+1,k),k\in\mathbb{Z} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, let's see. The problem is asking for all integer triples (a, b, c) such that the number N, defined as [(a - b)(b - c)(c - a)]/2 + 2, is a power of 2016. A power of 2016 means it's like 2016^n where n is a non-negative integer. So, we need to find all triples (a, b, c) where this expression equals 2016^n f... |
Problem 2. Let $n$ three-digit numbers satisfy the following properties:
(1) No number contains the digit 0 .
(2) The sum of the digits of each number is 9 .
(3) The units digits of any two numbers are different.
(4) The tens digits of any two numbers are different.
(5) The hundreds digits of any two numbers are ... |
Solution. Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0 . We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of 6 A's (which means that we add 1 to the current digit) and $2 G$ 's (w... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
true,
true
] | 2 | <think>
Okay, so I need to find the largest possible number of three-digit numbers, let's call this number n, that satisfy all the given conditions. Let me read the problem again and make sure I understand each part.
First, the numbers are three-digit, and they have to meet five conditions. Let me list them again:
1.... |
Problem 2. Let $n$ three-digit numbers satisfy the following properties:
(1) No number contains the digit 0 .
(2) The sum of the digits of each number is 9 .
(3) The units digits of any two numbers are different.
(4) The tens digits of any two numbers are different.
(5) The hundreds digits of any two numbers are ... |
Solution. Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0 . We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of 6 A's (which means that we add 1 to the current digit) and $2 G$ 's (w... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
true,
true
] | 2 | <think>
Okay, let's see. I need to find the largest possible value of n, which is the number of three-digit numbers that satisfy all these conditions. Let's go through the conditions again to make sure I understand them correctly.
First, each number is a three-digit number without any zeros. So each digit—hundreds, te... |
# Problem 1
Let $a, b$ and $c$ be positive real numbers such that $a+b+c=1$. Prove that
$$
\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+6 \geq 2 \sqrt{2}\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)
$$
When does equality hold?
| ## Solution
Replacing $1-a, 1-b, 1-c$ with $b+c, c+a, a+b$ respectively on the right hand side, the given inequality becomes
and equivalently
$$
\left(\frac{b+c}{a}-2 \sqrt{2} \sqrt{\frac{... | =b==\frac{1}{3} | Inequalities | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Alright, let me try to tackle this problem. So, we have three positive real numbers a, b, c such that their sum is 1. We need to prove this inequality involving fractions and square roots. The equality condition is also required. Hmm, okay.
First, let me write down the given condition and the inequality again ... |
# Problem 1
Let $a, b$ and $c$ be positive real numbers such that $a+b+c=1$. Prove that
$$
\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+6 \geq 2 \sqrt{2}\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)
$$
When does equality hold?
| ## Solution
Replacing $1-a, 1-b, 1-c$ with $b+c, c+a, a+b$ respectively on the right hand side, the given inequality becomes
and equivalently
$$
\left(\frac{b+c}{a}-2 \sqrt{2} \sqrt{\frac{... | =b==\frac{1}{3} | Inequalities | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to prove this inequality:
$$
\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+6 \geq 2 \sqrt{2}\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)
$$
given that \( a, b, c \) are positive real numbers with \( a + b + c = 1 \). Also, I have to find... |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$.
Thus we have
$$
180^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, let me try to figure out this geometry problem. So, we have two circles, k₁ and k₂, intersecting at points A and B. There's a common tangent t that touches k₁ at M and k₂ at N. We know that this tangent t is perpendicular to AM, and the length of MN is twice the length of AM. We need to find the measure o... |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$.
Thus we have
$$
180^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to find the measure of angle NMB given the conditions in the problem. Let me start by visualizing the problem and then break it down step by step.
First, there are two circles, k₁ and k₂, intersecting at points A and B. There's a common tangent t that touches k₁ at M and k₂ at N. The tangent li... |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$.
Thus we have
$$
180^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to solve this geometry problem. Let me read it again and make sure I understand all the parts.
We have two circles, k₁ and k₂, intersecting at points A and B. There's a common tangent t that touches k₁ at M and k₂ at N. The tangent t is perpendicular to AM, and the length of MN is twice the len... |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$.
Thus we have
$$
180^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, let's tackle this geometry problem step by step. First, let me understand the problem statement again to make sure I have all the details right.
We have two circles, k₁ and k₂, intersecting at two distinct points A and B. There's a common tangent t to both circles. This tangent touches k₁ at point M and ... |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$.
Thus we have
$$
180^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I have this geometry problem here. Let me read it again and try to visualize. Two circles, k₁ and k₂, intersect at points A and B. There's a common tangent t that touches k₁ at M and k₂ at N. The tangent t is perpendicular to AM, and the length of MN is twice the length of AM. I need to find the measur... |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$.
Thus we have
$$
180^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, let me try to solve this geometry problem step by step. So, we have two intersecting circles, k₁ and k₂, meeting at points A and B. There's a common tangent t that touches k₁ at M and k₂ at N. The tangent t is perpendicular to AM, and the length of MN is twice the length of AM. We need to find the measure... |
## Problem 4
Find all positive integers $x, y, z$ and $t$ such that
$$
2^{x} \cdot 3^{y}+5^{z}=7^{t}
$$
| ## Solution
Reducing modulo 3 we get $5^{z} \equiv 1$, therefore $z$ is even, $z=2 c, c \in \mathbb{N}$.
Next we prove that $t$ is even:
Obviously, $t \geq 2$. Let us suppose that $t$ is odd, say $t=2 d+1, d \in \mathbb{N}$. The equation becomes $2^{x} \cdot 3^{y}+25^{c}=7 \cdot 49^{d}$. If $x \geq 2$, reducing modu... | 3,1,=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3 | [
false,
false,
false
] | 0 | <think>
Okay, let's see. I need to find all positive integers x, y, z, t such that 2^x * 3^y + 5^z = 7^t. Hmm, this looks like an exponential Diophantine equation. These can be tricky, but maybe I can find some patterns or use modular arithmetic to limit the possibilities.
First, since all variables are positive integ... |
## Problem 4
Find all positive integers $x, y, z$ and $t$ such that
$$
2^{x} \cdot 3^{y}+5^{z}=7^{t}
$$
| ## Solution
Reducing modulo 3 we get $5^{z} \equiv 1$, therefore $z$ is even, $z=2 c, c \in \mathbb{N}$.
Next we prove that $t$ is even:
Obviously, $t \geq 2$. Let us suppose that $t$ is odd, say $t=2 d+1, d \in \mathbb{N}$. The equation becomes $2^{x} \cdot 3^{y}+25^{c}=7 \cdot 49^{d}$. If $x \geq 2$, reducing modu... | 3,1,=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3 | [
false,
false,
false
] | 0 | <think>
Okay, let's try to solve this problem where we need to find all positive integers x, y, z, and t such that 2^x * 3^y + 5^z = 7^t. Hmm, so we have an equation involving exponents with different bases. I remember that equations like this often require checking small cases or using modular arithmetic to find contr... |
## Problem 4
Find all positive integers $x, y, z$ and $t$ such that
$$
2^{x} \cdot 3^{y}+5^{z}=7^{t}
$$
| ## Solution
Reducing modulo 3 we get $5^{z} \equiv 1$, therefore $z$ is even, $z=2 c, c \in \mathbb{N}$.
Next we prove that $t$ is even:
Obviously, $t \geq 2$. Let us suppose that $t$ is odd, say $t=2 d+1, d \in \mathbb{N}$. The equation becomes $2^{x} \cdot 3^{y}+25^{c}=7 \cdot 49^{d}$. If $x \geq 2$, reducing modu... | 3,1,=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3 | [
false,
false,
false
] | 0 | <think>
Okay, so I need to find all positive integers x, y, z, and t such that 2^x * 3^y + 5^z = 7^t. Hmm, let's start by understanding what this equation is saying. On the left side, we have the sum of two numbers: one is a product of powers of 2 and 3, and the other is a power of 5. The right side is a power of 7. So... |
Problem 3. Let $A L$ and $B K$ be angle bisectors in the non-isosceles triangle $A B C$ ( $L$ lies on the side $B C, K$ lies on the side $A C$ ). The perpendicular bisector of $B K$ intersects the line $A L$ at point $M$. Point $N$ lies on the line $B K$ such that $L N$ is parallel to $M K$. Prove that $L N=N A$.
|
Solution. The point $M$ lies on the circumcircle of $\triangle A B K$ (since both $A L$ and the perpendicular bisector of $B K$ bisect the arc $B K$ of this circle). Then $\angle C B K=$ $\angle A B K=\angle A M K=\angle N L A$. Thus $A B L N$ is cyclic, whence $\angle N A L=\angle N B L=$ $\angle C B K=\angle N L A$.... | LN=NA | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
true,
true
] | 2 | <think>
Okay, so I need to prove that in triangle ABC, which is not isosceles, the points M and N are constructed in a certain way, and then LN is equal to NA. Let me try to visualize the problem first.
We have triangle ABC with angle bisectors AL and BK. L is on BC, K is on AC. Then, the perpendicular bisector of BK ... |
Problem 3. Let $A L$ and $B K$ be angle bisectors in the non-isosceles triangle $A B C$ ( $L$ lies on the side $B C, K$ lies on the side $A C$ ). The perpendicular bisector of $B K$ intersects the line $A L$ at point $M$. Point $N$ lies on the line $B K$ such that $L N$ is parallel to $M K$. Prove that $L N=N A$.
|
Solution. The point $M$ lies on the circumcircle of $\triangle A B K$ (since both $A L$ and the perpendicular bisector of $B K$ bisect the arc $B K$ of this circle). Then $\angle C B K=$ $\angle A B K=\angle A M K=\angle N L A$. Thus $A B L N$ is cyclic, whence $\angle N A L=\angle N B L=$ $\angle C B K=\angle N L A$.... | LN=NA | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
true,
true
] | 2 | <think>
Okay, so I need to prove that in triangle ABC, which isn't isosceles, with angle bisectors AL and BK, where L is on BC and K is on AC, the perpendicular bisector of BK intersects AL at point M. Then, there's a point N on BK such that LN is parallel to MK, and I have to show that LN equals NA. Hmm, that seems a ... |
Problem 4. A $9 \times 7$ rectangle is tiled with tiles of the two types shown in the picture below (the tiles are composed by three, respectively four unit squares and the L-shaped tiles can be rotated repeatedly with $90^{\circ}$ ).
 that can be used in tiling a 9x7 rectangle. The other tiles are L-shaped, made of three unit squares, and they can be rotated. The problem says n is the number of 2x2 tiles, and ... |
Problem 1. Let $n(n \geq 1)$ be an integer. Consider the equation
$$
2 \cdot\left\lfloor\frac{1}{2 x}\right\rfloor-n+1=(n+1)(1-n x)
$$
where $x$ is the unknown real variable.
(a) Solve the equation for $n=8$.
(b) Prove that there exists an integer $n$ for which the equation has at least 2021 solutions. (For any re... |
Solution. Let $k=\left[\frac{1}{2 x}\right], k \in \mathbb{Z}$.
(a) For $n=8$, the equation becomes
$$
k=\left[\frac{1}{2 x}\right]=8-36 x \Rightarrow x \neq 0 \text { and } x=\frac{8-k}{36}
$$
Since $x \neq 0$, we have $k \neq 8$, and the last relation implies $k=\left[\frac{1}{2 x}\right]=\left[\frac{18}{8-k}\rig... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, let's tackle part (a) first. The problem is to solve the equation for n=8:
2 · floor(1/(2x)) - 8 + 1 = (8 + 1)(1 - 8x)
Simplify both sides first. Left side: 2·floor(1/(2x)) -7. Right side: 9·(1 -8x) = 9 -72x.
So the equation becomes:
2·floor(1/(2x)) -7 = 9 -72x
Let's rearrange terms:
2·floor(1/(2x))... | |
Problem 1. Let $n(n \geq 1)$ be an integer. Consider the equation
$$
2 \cdot\left\lfloor\frac{1}{2 x}\right\rfloor-n+1=(n+1)(1-n x)
$$
where $x$ is the unknown real variable.
(a) Solve the equation for $n=8$.
(b) Prove that there exists an integer $n$ for which the equation has at least 2021 solutions. (For any re... |
Solution. Let $k=\left[\frac{1}{2 x}\right], k \in \mathbb{Z}$.
(a) For $n=8$, the equation becomes
$$
k=\left[\frac{1}{2 x}\right]=8-36 x \Rightarrow x \neq 0 \text { and } x=\frac{8-k}{36}
$$
Since $x \neq 0$, we have $k \neq 8$, and the last relation implies $k=\left[\frac{1}{2 x}\right]=\left[\frac{18}{8-k}\rig... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to solve part (a) where n=8. The equation is given as:
2 · floor(1/(2x)) - 8 + 1 = (8 + 1)(1 - 8x)
First, let me simplify both sides. On the left-hand side (LHS), 2·floor(1/(2x)) minus 8 plus 1. That simplifies to 2·floor(1/(2x)) - 7. On the right-hand side (RHS), (8+1) is 9, so it becomes 9(1... | |
Problem 4. Let $M$ be a subset of the set of 2021 integers $\{1,2,3, \ldots, 2021\}$ such that for any three elements (not necessarily distinct) $a, b, c$ of $M$ we have $|a+b-c|>10$. Determine the largest possible number of elements of $M$.
|
Solution. The set $M=\{1016,1017, \ldots, 2021\}$ has 1006 elements and satisfies the required property, since $a, b, c \in M$ implies that $a+b-c \geqslant 1016+1016-2021=11$. We will show that this is optimal.
Suppose $M$ satisfies the condition in the problem. Let $k$ be the minimal element of $M$. Then $k=|k+k-k|... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
true
] | 5 | <think>
Okay, so I need to find the largest possible subset M of the integers from 1 to 2021 such that for any three elements a, b, c in M, the absolute value |a + b - c| is greater than 10. Hmm, interesting. Let me try to break this down step by step.
First, let's make sure I understand the problem correctly. We have... |
Problem 4. Let $M$ be a subset of the set of 2021 integers $\{1,2,3, \ldots, 2021\}$ such that for any three elements (not necessarily distinct) $a, b, c$ of $M$ we have $|a+b-c|>10$. Determine the largest possible number of elements of $M$.
|
Solution. The set $M=\{1016,1017, \ldots, 2021\}$ has 1006 elements and satisfies the required property, since $a, b, c \in M$ implies that $a+b-c \geqslant 1016+1016-2021=11$. We will show that this is optimal.
Suppose $M$ satisfies the condition in the problem. Let $k$ be the minimal element of $M$. Then $k=|k+k-k|... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
true
] | 5 | <think>
Okay, let me try to figure out this problem. So, we need to find the largest subset M of the integers from 1 to 2021 such that for any three elements a, b, c in M, the absolute value |a + b - c| is greater than 10. Hmm, interesting.
First, let me understand the condition. For any three elements in M, when you... |
Problem 4. Let $M$ be a subset of the set of 2021 integers $\{1,2,3, \ldots, 2021\}$ such that for any three elements (not necessarily distinct) $a, b, c$ of $M$ we have $|a+b-c|>10$. Determine the largest possible number of elements of $M$.
|
Solution. The set $M=\{1016,1017, \ldots, 2021\}$ has 1006 elements and satisfies the required property, since $a, b, c \in M$ implies that $a+b-c \geqslant 1016+1016-2021=11$. We will show that this is optimal.
Suppose $M$ satisfies the condition in the problem. Let $k$ be the minimal element of $M$. Then $k=|k+k-k|... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
true
] | 5 | <think>
Okay, so I need to find the largest subset M of {1,2,3,...,2021} such that for any three elements a, b, c in M, the absolute value |a + b - c| is greater than 10. Hmm, interesting. Let me try to break this down.
First, the problem says that for any three elements in M, the expression |a + b - c| must be greate... |
Problem 4. Let $M$ be a subset of the set of 2021 integers $\{1,2,3, \ldots, 2021\}$ such that for any three elements (not necessarily distinct) $a, b, c$ of $M$ we have $|a+b-c|>10$. Determine the largest possible number of elements of $M$.
|
Solution. The set $M=\{1016,1017, \ldots, 2021\}$ has 1006 elements and satisfies the required property, since $a, b, c \in M$ implies that $a+b-c \geqslant 1016+1016-2021=11$. We will show that this is optimal.
Suppose $M$ satisfies the condition in the problem. Let $k$ be the minimal element of $M$. Then $k=|k+k-k|... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
true
] | 5 | <think>
Okay, so I have this problem here where I need to find the largest possible subset M of the integers from 1 to 2021. The condition is that for any three elements a, b, c in M, the absolute value |a + b - c| must be greater than 10. Hmm, interesting. Let me try to break this down step by step.
First, let me mak... |
Problem 4. Let $M$ be a subset of the set of 2021 integers $\{1,2,3, \ldots, 2021\}$ such that for any three elements (not necessarily distinct) $a, b, c$ of $M$ we have $|a+b-c|>10$. Determine the largest possible number of elements of $M$.
|
Solution. The set $M=\{1016,1017, \ldots, 2021\}$ has 1006 elements and satisfies the required property, since $a, b, c \in M$ implies that $a+b-c \geqslant 1016+1016-2021=11$. We will show that this is optimal.
Suppose $M$ satisfies the condition in the problem. Let $k$ be the minimal element of $M$. Then $k=|k+k-k|... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
true
] | 5 | <think>
Okay, let me try to work through this problem step by step. The problem is about finding the largest possible subset M of the integers from 1 to 2021 such that for any three elements a, b, c in M, the absolute value |a + b - c| is greater than 10. So, I need to figure out how to choose as many numbers as possib... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, so I have this problem where I need to find all distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Hmm, primes... they have to be distinct, so p, q, r can't be the same. Let me start by thinking about possible small primes since exponents are high, which can make numbers big quickly. The p... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, let's see. I need to find distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Hmm, primes are 2, 3, 5, 7, 11, etc. Since they have to be distinct, p, q, r can't be the same. Let me think about how to approach this.
First, maybe I can try plugging in small primes and see if the equation hol... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, so I need to find all distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Alright, primes are numbers like 2, 3, 5, 7, 11, etc. Since they have to be distinct, p, q, and r can't be the same. Let me think about how to approach this.
First, maybe I can try plugging in small prime numbers and... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, let's see. I have to find all distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Hmm, primes are 2, 3, 5, 7, 11, etc. Since they have to be distinct, each of p, q, r has to be different primes.
First, maybe I can start by considering small primes because the equation involves fourth powe... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, let's try to tackle this problem. So, we need to find all distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Hmm, primes are numbers like 2, 3, 5, 7, 11, etc. They have to be distinct, so p, q, and r can't be the same as each other. Let me think step by step.
First, maybe I should try plu... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, let me try to solve this problem. So, we need to find all distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Hmm, primes are numbers like 2, 3, 5, 7, 11, etc. And they have to be distinct, so p, q, r can't be the same. Alright, let's start thinking.
First, maybe I should rearrange the equ... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, let's see. I need to find distinct prime numbers p, q, and r such that 3p^4 - 5q^4 -4r^2 = 26. Hmm, primes are numbers like 2, 3, 5, 7, 11, etc. They have to be distinct, so p, q, r can't be the same. Let's start by thinking about possible primes and see if the equation holds.
First, since primes can be ... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, let's see. I need to find all distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Hmm, primes are 2, 3, 5, 7, 11, etc. Since they have to be distinct, each of p, q, r has to be different primes. The equation involves p to the fourth power, q to the fourth power, and r squared. The coefficie... |
Problem 2. Consider an acute triangle $A B C$ with area S. Let $C D \perp A B \quad(D \in A B)$, $D M \perp A C \quad(M \in A C)$ and $\quad D N \perp B C \quad(N \in B C)$. Denote by $H_{1}$ and $H_{2}$ the orthocentres of the triangles $M N C$ and $M N D$ respectively. Find the area of the quadrilateral $\mathrm{AH}... |
Solution 1. Let $O, P, K, R$ and $T$ be the mid-points of the segments $C D, M N$, $C N, C H_{1}$ and $M H_{1}$, respectively. From $\triangle M N C$ we have that $\overline{P K}=\frac{1}{2} \overline{M C}$ and $P K \| M C$. Analogously, from $\Delta M H_{1} C$ we have that $\overline{T R}=\frac{1}{2} \overline{M C}$ ... | S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, so I need to find the area of quadrilateral AH₁BH₂ in terms of the area S of the original acute triangle ABC. Let me try to break this down step by step.
First, let's visualize the problem. We have triangle ABC, which is acute, so all altitudes are inside the triangle. CD is the altitude from C to AB, me... |
Problem 2. Consider an acute triangle $A B C$ with area S. Let $C D \perp A B \quad(D \in A B)$, $D M \perp A C \quad(M \in A C)$ and $\quad D N \perp B C \quad(N \in B C)$. Denote by $H_{1}$ and $H_{2}$ the orthocentres of the triangles $M N C$ and $M N D$ respectively. Find the area of the quadrilateral $\mathrm{AH}... |
Solution 1. Let $O, P, K, R$ and $T$ be the mid-points of the segments $C D, M N$, $C N, C H_{1}$ and $M H_{1}$, respectively. From $\triangle M N C$ we have that $\overline{P K}=\frac{1}{2} \overline{M C}$ and $P K \| M C$. Analogously, from $\Delta M H_{1} C$ we have that $\overline{T R}=\frac{1}{2} \overline{M C}$ ... | S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, so I have this problem about an acute triangle ABC with area S. The problem involves constructing several perpendiculars and then finding the area of a quadrilateral formed by the orthocenters of two smaller triangles. Let me try to parse this step by step.
First, let me visualize triangle ABC. Since it'... |
Problem 2. Consider an acute triangle $A B C$ with area S. Let $C D \perp A B \quad(D \in A B)$, $D M \perp A C \quad(M \in A C)$ and $\quad D N \perp B C \quad(N \in B C)$. Denote by $H_{1}$ and $H_{2}$ the orthocentres of the triangles $M N C$ and $M N D$ respectively. Find the area of the quadrilateral $\mathrm{AH}... |
Solution 1. Let $O, P, K, R$ and $T$ be the mid-points of the segments $C D, M N$, $C N, C H_{1}$ and $M H_{1}$, respectively. From $\triangle M N C$ we have that $\overline{P K}=\frac{1}{2} \overline{M C}$ and $P K \| M C$. Analogously, from $\Delta M H_{1} C$ we have that $\overline{T R}=\frac{1}{2} \overline{M C}$ ... | S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, so I have this problem here about an acute triangle ABC with area S. There are several perpendiculars drawn: CD is perpendicular to AB, with D on AB. Then from D, they drop two more perpendiculars: DM perpendicular to AC, meeting AC at M, and DN perpendicular to BC, meeting BC at N. Then they mention H1 a... |
Problem 2. Consider an acute triangle $A B C$ with area S. Let $C D \perp A B \quad(D \in A B)$, $D M \perp A C \quad(M \in A C)$ and $\quad D N \perp B C \quad(N \in B C)$. Denote by $H_{1}$ and $H_{2}$ the orthocentres of the triangles $M N C$ and $M N D$ respectively. Find the area of the quadrilateral $\mathrm{AH}... |
Solution 1. Let $O, P, K, R$ and $T$ be the mid-points of the segments $C D, M N$, $C N, C H_{1}$ and $M H_{1}$, respectively. From $\triangle M N C$ we have that $\overline{P K}=\frac{1}{2} \overline{M C}$ and $P K \| M C$. Analogously, from $\Delta M H_{1} C$ we have that $\overline{T R}=\frac{1}{2} \overline{M C}$ ... | S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, so I have this geometry problem here. Let me try to parse it step by step. We have an acute triangle ABC with area S. Then, CD is perpendicular to AB, with D lying on AB. Then from D, we draw DM perpendicular to AC, meeting AC at M, and DN perpendicular to BC, meeting BC at N. Then H1 is the orthocenter o... |
Problem 4. For a positive integer $n$, two players A and B play the following game: Given a pile of $s$ stones, the players take turn alternatively with A going first. On each turn the player is allowed to take either one stone, or a prime number of stones, or a multiple of $n$ stones. The winner is the one who takes ... |
Solution. Denote by $k$ the sought number and let $\left\{s_{1}, \mathrm{~s}_{2}, \ldots, \mathrm{s}_{k}\right\}$ be the corresponding values for $s$. We call each $s_{i}$ a losing number and every other nonnegative integer a winning numbers.
## Clearly every multiple of $n$ is a winning number.
Suppose there are tw... | n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3 | [
true,
true,
true
] | 3 | <think>
Alright, let's try to tackle this problem step by step. So, we have two players, A and B, playing a game with a pile of s stones. They take turns, starting with A. On each turn, a player can take either 1 stone, a prime number of stones, or a multiple of n stones. The person who takes the last stone wins. We ne... |
Problem 4. For a positive integer $n$, two players A and B play the following game: Given a pile of $s$ stones, the players take turn alternatively with A going first. On each turn the player is allowed to take either one stone, or a prime number of stones, or a multiple of $n$ stones. The winner is the one who takes ... |
Solution. Denote by $k$ the sought number and let $\left\{s_{1}, \mathrm{~s}_{2}, \ldots, \mathrm{s}_{k}\right\}$ be the corresponding values for $s$. We call each $s_{i}$ a losing number and every other nonnegative integer a winning numbers.
## Clearly every multiple of $n$ is a winning number.
Suppose there are tw... | n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3 | [
true,
true,
true
] | 3 | <think>
Okay, let me try to figure out this problem. So, we have a game where two players, A and B, take turns picking stones from a pile. The rules are that on each turn, a player can take either 1 stone, a prime number of stones, or a multiple of n stones. The person who takes the last stone wins. We need to determin... |
Problem 4. For a positive integer $n$, two players A and B play the following game: Given a pile of $s$ stones, the players take turn alternatively with A going first. On each turn the player is allowed to take either one stone, or a prime number of stones, or a multiple of $n$ stones. The winner is the one who takes ... |
Solution. Denote by $k$ the sought number and let $\left\{s_{1}, \mathrm{~s}_{2}, \ldots, \mathrm{s}_{k}\right\}$ be the corresponding values for $s$. We call each $s_{i}$ a losing number and every other nonnegative integer a winning numbers.
## Clearly every multiple of $n$ is a winning number.
Suppose there are tw... | n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3 | [
true,
true,
true
] | 3 | <think>
Okay, let's try to figure out this problem. So, we have two players, A and B, who take turns picking stones from a pile. The pile starts with s stones, and they can take either 1 stone, a prime number of stones, or a multiple of n stones on each turn. The person who takes the last stone wins. We need to determi... |
Problem 1. Find all pairs $(a, b)$ of positive integers such that
$$
11 a b \leq a^{3}-b^{3} \leq 12 a b
$$
|
Solution 1. Let $a-b=t$. Due to $a^{3}-b^{3} \geq 11 a b$ we conclude that $a>b$ so $t$ is a positive integer and the condition can be written as
$$
11 b(b+t) \leq t\left[b^{2}+b(b+t)+(b+t)^{2}\right] \leq 12 b(b+t)
$$
Since
$$
t\left[b^{2}+b(b+t)+(b+t)^{2}\right]=t\left(b^{2}+b^{2}+b t+b^{2}+2 b t+t^{2}\right)=3 t... | (5,2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, so I need to find all pairs of positive integers (a, b) that satisfy the inequalities 11ab ≤ a³ - b³ ≤ 12ab. Hmm, let's start by understanding what this means. Both a and b are positive integers, so they must be at least 1. The inequalities involve cubic terms and a product term. Maybe I can manipulate th... |
Problem 1. Find all pairs $(a, b)$ of positive integers such that
$$
11 a b \leq a^{3}-b^{3} \leq 12 a b
$$
|
Solution 1. Let $a-b=t$. Due to $a^{3}-b^{3} \geq 11 a b$ we conclude that $a>b$ so $t$ is a positive integer and the condition can be written as
$$
11 b(b+t) \leq t\left[b^{2}+b(b+t)+(b+t)^{2}\right] \leq 12 b(b+t)
$$
Since
$$
t\left[b^{2}+b(b+t)+(b+t)^{2}\right]=t\left(b^{2}+b^{2}+b t+b^{2}+2 b t+t^{2}\right)=3 t... | (5,2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, let's try to tackle this problem. We need to find all pairs of positive integers (a, b) such that 11ab ≤ a³ - b³ ≤ 12ab. Hmm, so both a and b are positive integers, which means they are at least 1. Let me start by understanding the inequalities given.
First, let's write down the inequalities:
1. a³ - b³... |
Problem 1. Find all pairs $(a, b)$ of positive integers such that
$$
11 a b \leq a^{3}-b^{3} \leq 12 a b
$$
|
Solution 1. Let $a-b=t$. Due to $a^{3}-b^{3} \geq 11 a b$ we conclude that $a>b$ so $t$ is a positive integer and the condition can be written as
$$
11 b(b+t) \leq t\left[b^{2}+b(b+t)+(b+t)^{2}\right] \leq 12 b(b+t)
$$
Since
$$
t\left[b^{2}+b(b+t)+(b+t)^{2}\right]=t\left(b^{2}+b^{2}+b t+b^{2}+2 b t+t^{2}\right)=3 t... | (5,2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, let's see. I have this problem where I need to find all pairs of positive integers (a, b) such that 11ab is less than or equal to a³ - b³, which in turn is less than or equal to 12ab. Hmm, okay. So, the inequalities are 11ab ≤ a³ - b³ ≤ 12ab. Both a and b are positive integers.
First, I need to understa... |
Problem 1. Find all pairs $(a, b)$ of positive integers such that
$$
11 a b \leq a^{3}-b^{3} \leq 12 a b
$$
|
Solution 1. Let $a-b=t$. Due to $a^{3}-b^{3} \geq 11 a b$ we conclude that $a>b$ so $t$ is a positive integer and the condition can be written as
$$
11 b(b+t) \leq t\left[b^{2}+b(b+t)+(b+t)^{2}\right] \leq 12 b(b+t)
$$
Since
$$
t\left[b^{2}+b(b+t)+(b+t)^{2}\right]=t\left(b^{2}+b^{2}+b t+b^{2}+2 b t+t^{2}\right)=3 t... | (5,2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, so I need to find all pairs of positive integers (a, b) such that 11ab is less than or equal to a³ minus b³, which in turn is less than or equal to 12ab. Hmm, let me start by understanding the inequalities given here.
First, the problem states that 11ab ≤ a³ - b³ ≤ 12ab. Both a and b are positive integer... |
Problem 4. We call an even positive integer $n$ nice if the set $\{1,2, \ldots, n\}$ can be partitioned into $\frac{n}{2}$ two-element subsets, such that the sum of the elements in each subset is a power of 3 . For example, 6 is nice, because the set $\{1,2,3,4,5,6\}$ can be partitioned into subsets $\{1,2\},\{3,6\},\... |
Solution. For a nice number $n$ and a given partition of the set $\{1,2, \ldots, n\}$ into twoelement subsets such that the sum of the elements in each subset is a power of 3 , we say that $a, b \in\{1,2, \ldots, n\}$ are paired if both of them belong to the same subset.
Let $x$ be a nice number and $k$ be a (unique)... | 2^{2022}-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1 | [
false
] | 0 | <think>
Okay, so I need to figure out how many nice even positive integers are smaller than \(3^{2022}\). A number \(n\) is nice if the set \(\{1, 2, \ldots, n\}\) can be partitioned into \(\frac{n}{2}\) pairs where each pair sums to a power of 3. The example given is 6, which works because we can pair \(\{1,2\}\), \(\... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, so I have this problem where I need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, that sounds a bit tricky, but let me try to break it down step by step.
First, let's recall what a power of 5 is. A power of 5 is any number of the form 5^k where k is ... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Alright, let's try to tackle this problem. So, we need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, okay. Powers of 5 are numbers like 5, 25, 125, 625, etc., which can be written as 5^k where k is a positive integer. So, both a! + b and b! + a have to be o... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, so I need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, powers of 5 are numbers like 1, 5, 25, 125, 625, etc. So both expressions have to equal some 5^k and 5^m where k and m are non-negative integers. Let me start by considering small values of a and... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, so I need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, powers of 5 are numbers like 5, 25, 125, 625, etc. So both expressions have to equal 5^k for some integers k. Let me start by considering small values for a and b because factorials grow really f... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, so I need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, powers of 5 are numbers like 5, 25, 125, 625, etc. So, each of these expressions has to equal 5^k for some integer k ≥ 1.
First, let's understand what a factorial is. For a positive integer n, ... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, let's try to solve this problem where we need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, powers of 5 are numbers like 5, 25, 125, 625, etc. So, we need a! + b = 5^k and b! + a = 5^m for some non-negative integers k and m. Since a and b are positive... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, let's tackle this problem where we need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, powers of 5 are numbers like 5, 25, 125, 625, etc. So both expressions have to be equal to some 5^k and 5^m where k and m are positive integers.
First, since a and... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, let's try to tackle this problem where we need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, powers of 5, so numbers like 1, 5, 25, 125, 625, etc. Since we're dealing with factorials and additions, maybe starting with small values of a and b will help... |
Problem 1. Find all triples $(a, b, c)$ of real numbers such that the following system holds:
$$
\left\{\begin{array}{l}
a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\
a^{2}+b^{2}+c^{2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
\end{array}\right.
$$
|
Solution. First of all if $(a, b, c)$ is a solution of the system then also $(-a,-b,-c)$ is a solution. Hence we can suppose that $a b c>0$. From the first condition we have
$$
a+b+c=\frac{a b+b c+c a}{a b c}
$$
Now, from the first condition and the second condition we get
$$
(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\rig... | (,b,)=(,\frac{1}{},1),(,\frac{1}{},-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3 | [
false,
false,
false
] | 0 | <think>
Okay, so I need to solve this system of equations where we have triples (a, b, c) of real numbers. The equations are:
First: a + b + c equals the sum of their reciprocals, 1/a + 1/b + 1/c.
Second: The sum of their squares, a² + b² + c², equals the sum of the reciprocals of their squares, 1/a² + 1/b² + 1/c².
... |
Problem 1. Find all triples $(a, b, c)$ of real numbers such that the following system holds:
$$
\left\{\begin{array}{l}
a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\
a^{2}+b^{2}+c^{2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
\end{array}\right.
$$
|
Solution. First of all if $(a, b, c)$ is a solution of the system then also $(-a,-b,-c)$ is a solution. Hence we can suppose that $a b c>0$. From the first condition we have
$$
a+b+c=\frac{a b+b c+c a}{a b c}
$$
Now, from the first condition and the second condition we get
$$
(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\rig... | (,b,)=(,\frac{1}{},1),(,\frac{1}{},-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3 | [
false,
false,
false
] | 0 | <think>
Okay, let's see. I have to find all triples (a, b, c) of real numbers that satisfy both equations:
1. a + b + c = 1/a + 1/b + 1/c
2. a² + b² + c² = 1/a² + 1/b² + 1/c²
Hmm, interesting. So both the sums and the sums of squares of the variables equal the sums of their reciprocals. Let me think about how to appr... |
Problem 1. Find all triples $(a, b, c)$ of real numbers such that the following system holds:
$$
\left\{\begin{array}{l}
a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\
a^{2}+b^{2}+c^{2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
\end{array}\right.
$$
|
Solution. First of all if $(a, b, c)$ is a solution of the system then also $(-a,-b,-c)$ is a solution. Hence we can suppose that $a b c>0$. From the first condition we have
$$
a+b+c=\frac{a b+b c+c a}{a b c}
$$
Now, from the first condition and the second condition we get
$$
(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\rig... | (,b,)=(,\frac{1}{},1),(,\frac{1}{},-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3 | [
false,
false,
false
] | 0 | <think>
Okay, so I need to find all triples (a, b, c) of real numbers that satisfy these two equations:
First equation: a + b + c = 1/a + 1/b + 1/c
Second equation: a² + b² + c² = 1/a² + 1/b² + 1/c²
Hmm, let me think. Maybe I can start by simplifying the first equation. Let me write it out again:
a + b + c = (1/a) ... |
Problem 4. Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number.
|
Solution. It is clear that $p \neq q$. We set
$$
1+\frac{p^{q}-q^{p}}{p+q}=r
$$
and we have that
$$
p^{q}-q^{p}=(r-1)(p+q)
$$
From Fermat's Little Theorem we have
$$
p^{q}-q^{p} \equiv-q \quad(\bmod p)
$$
Since we also have that
$$
(r-1)(p+q) \equiv-r q-q \quad(\bmod p)
$$
from (3) we get that
$$
r q \equiv 0... | (2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to find all pairs of prime numbers (p, q) such that the expression 1 + (p^q - q^p)/(p + q) is a prime number. Hmm, primes, exponents, and fractions... Let me start by understanding the problem step by step.
First, primes are numbers greater than 1 that have no divisors other than 1 and themselv... |
Problem 4. Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number.
|
Solution. It is clear that $p \neq q$. We set
$$
1+\frac{p^{q}-q^{p}}{p+q}=r
$$
and we have that
$$
p^{q}-q^{p}=(r-1)(p+q)
$$
From Fermat's Little Theorem we have
$$
p^{q}-q^{p} \equiv-q \quad(\bmod p)
$$
Since we also have that
$$
(r-1)(p+q) \equiv-r q-q \quad(\bmod p)
$$
from (3) we get that
$$
r q \equiv 0... | (2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I have this problem here: Find all pairs (p, q) of prime numbers such that 1 + (p^q - q^p)/(p + q) is a prime number. Hmm, primes p and q... I need to find all such pairs where this expression results in a prime. Let me start by understanding the problem better.
First, the expression is 1 plus (p^q - ... |
Problem 4. Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number.
|
Solution. It is clear that $p \neq q$. We set
$$
1+\frac{p^{q}-q^{p}}{p+q}=r
$$
and we have that
$$
p^{q}-q^{p}=(r-1)(p+q)
$$
From Fermat's Little Theorem we have
$$
p^{q}-q^{p} \equiv-q \quad(\bmod p)
$$
Since we also have that
$$
(r-1)(p+q) \equiv-r q-q \quad(\bmod p)
$$
from (3) we get that
$$
r q \equiv 0... | (2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to find all pairs of prime numbers (p, q) such that the expression 1 + (p^q - q^p)/(p + q) is a prime number. Hmm, primes p and q, so both are primes. Let me start by understanding the expression step by step.
First, let's write the expression again:
1 + (p^q - q^p)/(p + q)
This needs to be a... |
Problem 4. Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number.
|
Solution. It is clear that $p \neq q$. We set
$$
1+\frac{p^{q}-q^{p}}{p+q}=r
$$
and we have that
$$
p^{q}-q^{p}=(r-1)(p+q)
$$
From Fermat's Little Theorem we have
$$
p^{q}-q^{p} \equiv-q \quad(\bmod p)
$$
Since we also have that
$$
(r-1)(p+q) \equiv-r q-q \quad(\bmod p)
$$
from (3) we get that
$$
r q \equiv 0... | (2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Alright, so I need to find all pairs of prime numbers (p, q) such that the expression 1 + (p^q - q^p)/(p + q) is a prime number. Hmm, okay. Let me start by understanding what this expression actually is. Let's rewrite it for clarity:
The expression is 1 + (p^q - q^p)/(p + q). So, first, compute p^q and q^p, su... |
Problem 4. Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number.
|
Solution. It is clear that $p \neq q$. We set
$$
1+\frac{p^{q}-q^{p}}{p+q}=r
$$
and we have that
$$
p^{q}-q^{p}=(r-1)(p+q)
$$
From Fermat's Little Theorem we have
$$
p^{q}-q^{p} \equiv-q \quad(\bmod p)
$$
Since we also have that
$$
(r-1)(p+q) \equiv-r q-q \quad(\bmod p)
$$
from (3) we get that
$$
r q \equiv 0... | (2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to find all pairs of prime numbers (p, q) such that the expression 1 + (p^q - q^p)/(p + q) is a prime number. Hmm, primes, exponents, and fractions. Let me start by breaking down the problem step by step.
First, since p and q are primes, the possible values they can take are 2, 3, 5, 7, 11, etc... |
Problem 4. Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number.
|
Solution. It is clear that $p \neq q$. We set
$$
1+\frac{p^{q}-q^{p}}{p+q}=r
$$
and we have that
$$
p^{q}-q^{p}=(r-1)(p+q)
$$
From Fermat's Little Theorem we have
$$
p^{q}-q^{p} \equiv-q \quad(\bmod p)
$$
Since we also have that
$$
(r-1)(p+q) \equiv-r q-q \quad(\bmod p)
$$
from (3) we get that
$$
r q \equiv 0... | (2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, let's see. I need to find all pairs of prime numbers (p, q) such that the expression 1 + (p^q - q^p)/(p + q) is a prime number. Hmm, primes... Let me start by understanding the problem step by step.
First, primes are numbers greater than 1 that have no divisors other than 1 and themselves. So p and q can... |
Problem 4. Consider a regular $2 n$-gon $P, A_{1} A_{2} \ldots A_{2 n}$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We ... |
Solution Answer: For $n=2$, the answer is 36 ; for $n=3$, the answer is 30 and for $n \geq 4$, the answer is $6 n$.
Lemma 1. Given a regular $2 n$-gon in the plane and a sequence of $n$ consecutive sides $s_{1}, s_{2}, \ldots, s_{n}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ can... | 6n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, let's try to tackle this problem. So, we have a regular 2n-gon, and we need to color its sides with three different colors. The conditions are that every side is colored, each color is used at least once, and from any external point, you can only see points of at most two different colors. Hmm, that seems... |
Problem 4. Consider a regular $2 n$-gon $P, A_{1} A_{2} \ldots A_{2 n}$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We ... |
Solution Answer: For $n=2$, the answer is 36 ; for $n=3$, the answer is 30 and for $n \geq 4$, the answer is $6 n$.
Lemma 1. Given a regular $2 n$-gon in the plane and a sequence of $n$ consecutive sides $s_{1}, s_{2}, \ldots, s_{n}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ can... | 6n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to solve this problem about coloring the sides of a regular 2n-gon with three colors. The conditions are that each color is used at least once, and from any external point, you can only see points of at most two different colors. Hmm, that's interesting. Let me try to break this down.
First, th... |
Problem 1. Find all prime numbers $a, b, c$ and positive integers $k$ which satisfy the equation
$$
a^{2}+b^{2}+16 \cdot c^{2}=9 \cdot k^{2}+1
$$
| # Solution:
The relation $9 \cdot k^{2}+1 \equiv 1(\bmod 3)$ implies
$$
a^{2}+b^{2}+16 \cdot c^{2} \equiv 1(\bmod 3) \Leftrightarrow a^{2}+b^{2}+c^{2} \equiv 1(\bmod 3)
$$
Since $a^{2} \equiv 0,1(\bmod 3), \quad b^{2} \equiv 0,1(\bmod 3), c^{2} \equiv 0,1(\bmod 3)$, we have:
| $a^{2}$ | 0 | 0 | 0 | 0 | 1 | 1 | 1 |... | (37,3,3,13),(17,3,3,7),(3,37,3,13),(3,17,3,7),(3,3,2,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5 | [
false,
true,
false,
false,
false
] | 1 | <think>
Okay, so I need to find all prime numbers a, b, c and positive integers k that satisfy the equation a² + b² + 16c² = 9k² + 1. Hmm, primes are numbers greater than 1 that have no divisors other than 1 and themselves. The primes start with 2, 3, 5, 7, 11, etc. Since the equation involves squares of a, b, c, and k... |
Problem 1. Find all prime numbers $a, b, c$ and positive integers $k$ which satisfy the equation
$$
a^{2}+b^{2}+16 \cdot c^{2}=9 \cdot k^{2}+1
$$
| # Solution:
The relation $9 \cdot k^{2}+1 \equiv 1(\bmod 3)$ implies
$$
a^{2}+b^{2}+16 \cdot c^{2} \equiv 1(\bmod 3) \Leftrightarrow a^{2}+b^{2}+c^{2} \equiv 1(\bmod 3)
$$
Since $a^{2} \equiv 0,1(\bmod 3), \quad b^{2} \equiv 0,1(\bmod 3), c^{2} \equiv 0,1(\bmod 3)$, we have:
| $a^{2}$ | 0 | 0 | 0 | 0 | 1 | 1 | 1 |... | (37,3,3,13),(17,3,3,7),(3,37,3,13),(3,17,3,7),(3,3,2,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5 | [
false,
true,
false,
false,
false
] | 1 | <think>
Okay, so I need to find all prime numbers a, b, c and positive integers k that satisfy the equation a² + b² + 16c² = 9k² + 1. Let me start by understanding the problem and thinking about possible approaches.
First, all a, b, c are primes. Primes are numbers greater than 1 that have no divisors other than 1 and... |
Problem 1. Find all prime numbers $a, b, c$ and positive integers $k$ which satisfy the equation
$$
a^{2}+b^{2}+16 \cdot c^{2}=9 \cdot k^{2}+1
$$
| # Solution:
The relation $9 \cdot k^{2}+1 \equiv 1(\bmod 3)$ implies
$$
a^{2}+b^{2}+16 \cdot c^{2} \equiv 1(\bmod 3) \Leftrightarrow a^{2}+b^{2}+c^{2} \equiv 1(\bmod 3)
$$
Since $a^{2} \equiv 0,1(\bmod 3), \quad b^{2} \equiv 0,1(\bmod 3), c^{2} \equiv 0,1(\bmod 3)$, we have:
| $a^{2}$ | 0 | 0 | 0 | 0 | 1 | 1 | 1 |... | (37,3,3,13),(17,3,3,7),(3,37,3,13),(3,17,3,7),(3,3,2,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5 | [
false,
true,
false,
false,
false
] | 1 | <think>
Okay, so I need to find all prime numbers a, b, c and positive integers k that satisfy the equation a² + b² + 16c² = 9k² + 1. Hmm, primes and squares... Let me start by breaking this down.
First, primes are numbers greater than 1 that have no divisors other than 1 and themselves. So a, b, c can be 2, 3, 5, 7, ... |
Problem 1. Find all prime numbers $a, b, c$ and positive integers $k$ which satisfy the equation
$$
a^{2}+b^{2}+16 \cdot c^{2}=9 \cdot k^{2}+1
$$
| # Solution:
The relation $9 \cdot k^{2}+1 \equiv 1(\bmod 3)$ implies
$$
a^{2}+b^{2}+16 \cdot c^{2} \equiv 1(\bmod 3) \Leftrightarrow a^{2}+b^{2}+c^{2} \equiv 1(\bmod 3)
$$
Since $a^{2} \equiv 0,1(\bmod 3), \quad b^{2} \equiv 0,1(\bmod 3), c^{2} \equiv 0,1(\bmod 3)$, we have:
| $a^{2}$ | 0 | 0 | 0 | 0 | 1 | 1 | 1 |... | (37,3,3,13),(17,3,3,7),(3,37,3,13),(3,17,3,7),(3,3,2,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5 | [
false,
true,
false,
false,
false
] | 1 | <think>
Okay, let's try to tackle this problem. So, we need to find all prime numbers a, b, c and positive integers k such that the equation a² + b² + 16c² = 9k² + 1 holds. Hmm, primes and squares... Maybe I can start by looking at some small primes and see if they fit.
First, let's list out the primes. The primes are... |
Problem 1. Find all prime numbers $a, b, c$ and positive integers $k$ which satisfy the equation
$$
a^{2}+b^{2}+16 \cdot c^{2}=9 \cdot k^{2}+1
$$
| # Solution:
The relation $9 \cdot k^{2}+1 \equiv 1(\bmod 3)$ implies
$$
a^{2}+b^{2}+16 \cdot c^{2} \equiv 1(\bmod 3) \Leftrightarrow a^{2}+b^{2}+c^{2} \equiv 1(\bmod 3)
$$
Since $a^{2} \equiv 0,1(\bmod 3), \quad b^{2} \equiv 0,1(\bmod 3), c^{2} \equiv 0,1(\bmod 3)$, we have:
| $a^{2}$ | 0 | 0 | 0 | 0 | 1 | 1 | 1 |... | (37,3,3,13),(17,3,3,7),(3,37,3,13),(3,17,3,7),(3,3,2,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5 | [
false,
true,
false,
false,
false
] | 1 | <think>
Alright, let me try to figure out this problem. So, we need to find all prime numbers a, b, c and positive integers k that satisfy the equation a² + b² + 16·c² = 9·k² + 1. Hmm, okay. Let me start by understanding the equation better.
First, all variables a, b, c are primes, and k is a positive integer. The equ... |
Problem 2. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression
$$
A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}
$$
$19^{\text {th ... | ## Solution:
We can rewrite $A$ as follows:
$$
\begin{aligned}
& A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^{2}-b^{2}-c^{2}= \\
& 2\left(\frac{a b+b c+c a}{a b c}\right)-\left(a^{2}+b^{2}+c^{2}\right)=2\left(\frac{a b+b c+c a}{a b c}\right)-\left((a+b+c... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7 | [
true,
true,
true,
true,
true,
true,
true
] | 7 | <think>
Okay, so I have this problem where I need to find the minimum value of the expression A, which is given by (2 - a³)/a + (2 - b³)/b + (2 - c³)/c, and the condition is that a, b, c are positive real numbers adding up to 3. Alright, let me try to think through this step by step.
First, let me rewrite the expressi... |
Problem 2. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression
$$
A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}
$$
$19^{\text {th ... | ## Solution:
We can rewrite $A$ as follows:
$$
\begin{aligned}
& A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^{2}-b^{2}-c^{2}= \\
& 2\left(\frac{a b+b c+c a}{a b c}\right)-\left(a^{2}+b^{2}+c^{2}\right)=2\left(\frac{a b+b c+c a}{a b c}\right)-\left((a+b+c... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7 | [
true,
true,
true,
true,
true,
true,
true
] | 7 | <think>
Okay, let's try to figure out how to find the minimum value of the expression \( A = \frac{2 - a^3}{a} + \frac{2 - b^3}{b} + \frac{2 - c^3}{c} \) given that \( a, b, c \) are positive real numbers and \( a + b + c = 3 \).
First, I need to understand the problem properly. The variables are positive and sum up ... |
Problem 2. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression
$$
A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}
$$
$19^{\text {th ... | ## Solution:
We can rewrite $A$ as follows:
$$
\begin{aligned}
& A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^{2}-b^{2}-c^{2}= \\
& 2\left(\frac{a b+b c+c a}{a b c}\right)-\left(a^{2}+b^{2}+c^{2}\right)=2\left(\frac{a b+b c+c a}{a b c}\right)-\left((a+b+c... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7 | [
true,
true,
true,
true,
true,
true,
true
] | 7 | <think>
Okay, so I have this problem where I need to find the minimum value of the expression A, which is given by (2 - a³)/a + (2 - b³)/b + (2 - c³)/c, and the condition is that a, b, c are positive real numbers adding up to 3. Hmm, let's see.
First, maybe I can simplify the expression A. Let me write each term sepa... |
Problem 2. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression
$$
A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}
$$
$19^{\text {th ... | ## Solution:
We can rewrite $A$ as follows:
$$
\begin{aligned}
& A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^{2}-b^{2}-c^{2}= \\
& 2\left(\frac{a b+b c+c a}{a b c}\right)-\left(a^{2}+b^{2}+c^{2}\right)=2\left(\frac{a b+b c+c a}{a b c}\right)-\left((a+b+c... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7 | [
true,
true,
true,
true,
true,
true,
true
] | 7 | <think>
Okay, so I need to find the minimum value of the expression A = (2 - a³)/a + (2 - b³)/b + (2 - c³)/c, where a, b, c are positive real numbers such that a + b + c = 3. Hmm, let's see.
First, maybe I should try to simplify the expression for A. Let's write each term separately. Each term is (2 - x³)/x, where x ... |
Problem 2. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression
$$
A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}
$$
$19^{\text {th ... | ## Solution:
We can rewrite $A$ as follows:
$$
\begin{aligned}
& A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^{2}-b^{2}-c^{2}= \\
& 2\left(\frac{a b+b c+c a}{a b c}\right)-\left(a^{2}+b^{2}+c^{2}\right)=2\left(\frac{a b+b c+c a}{a b c}\right)-\left((a+b+c... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7 | [
true,
true,
true,
true,
true,
true,
true
] | 7 | <think>
Okay, let's see. I need to find the minimum value of the expression A, which is given by (2 - a³)/a + (2 - b³)/b + (2 - c³)/c, where a, b, c are positive real numbers adding up to 3. Hmm, the constraint is a + b + c = 3.
First, maybe I can simplify the expression A. Let's break it down term by term. For each ... |
Problem 2. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression
$$
A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}
$$
$19^{\text {th ... | ## Solution:
We can rewrite $A$ as follows:
$$
\begin{aligned}
& A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^{2}-b^{2}-c^{2}= \\
& 2\left(\frac{a b+b c+c a}{a b c}\right)-\left(a^{2}+b^{2}+c^{2}\right)=2\left(\frac{a b+b c+c a}{a b c}\right)-\left((a+b+c... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7 | [
true,
true,
true,
true,
true,
true,
true
] | 7 | <think>
Okay, so I need to find the minimum value of the expression \( A = \frac{2 - a^3}{a} + \frac{2 - b^3}{b} + \frac{2 - c^3}{c} \) where \( a, b, c \) are positive real numbers and \( a + b + c = 3 \). Hmm, let's start by understanding the problem and then figuring out a strategy.
First, since \( a, b, c \) are p... |
Problem 2. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression
$$
A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}
$$
$19^{\text {th ... | ## Solution:
We can rewrite $A$ as follows:
$$
\begin{aligned}
& A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^{2}-b^{2}-c^{2}= \\
& 2\left(\frac{a b+b c+c a}{a b c}\right)-\left(a^{2}+b^{2}+c^{2}\right)=2\left(\frac{a b+b c+c a}{a b c}\right)-\left((a+b+c... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7 | [
true,
true,
true,
true,
true,
true,
true
] | 7 | <think>
Okay, so I have this problem here where I need to find the minimum value of the expression A, which is given by (2 - a³)/a + (2 - b³)/b + (2 - c³)/c, and the condition is that a, b, c are positive real numbers adding up to 3. Alright, let's break this down.
First, let me rewrite the expression to see if I can ... |
## Problem 4.
An $L$-figure is one of the following four pieces, each consisting of three unit squares:
A $5 \times 5$ board, consisting of 25 unit squares, a positive integer $k \leq 25$ a... | ## Solution:
We will show that player $\boldsymbol{A}$ wins if $k=1,2,3$, but player $\boldsymbol{B}$ wins if $k=4$. Thus the smallest $k$ for which $\boldsymbol{B}$ has a winning strategy exists and is equal to 4 .
If $k=1$, player $\boldsymbol{A}$ marks the upper left corner of the square and then fills it as follo... | 4 | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Alright, so I need to figure out the minimum number of squares, k, that player B needs to force player A into a position where no matter how the L-figures are placed, there are at least three squares left uncovered. The game is on a 5x5 grid, and players take turns marking squares until k squares are marked. Pl... |
## Problem 4.
An $L$-figure is one of the following four pieces, each consisting of three unit squares:
A $5 \times 5$ board, consisting of 25 unit squares, a positive integer $k \leq 25$ a... | ## Solution:
We will show that player $\boldsymbol{A}$ wins if $k=1,2,3$, but player $\boldsymbol{B}$ wins if $k=4$. Thus the smallest $k$ for which $\boldsymbol{B}$ has a winning strategy exists and is equal to 4 .
If $k=1$, player $\boldsymbol{A}$ marks the upper left corner of the square and then fills it as follo... | 4 | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I have this problem about a game between two players, A and B, on a 5x5 board. The goal is to find the minimum k such that B can ensure that no matter how the game goes (with them marking k squares), any "good" placement of L-figures (which cover exactly three unmarked squares each) will leave at least... |
A2. Let $a$ and $b$ be positive real numbers such that $3 a^{2}+2 b^{2}=3 a+2 b$. Find the minimum value of
$$
A=\sqrt{\frac{a}{b(3 a+2)}}+\sqrt{\frac{b}{a(2 b+3)}}
$$
|
Solution. By the Cauchy-Schwarz inequality we have that
$$
5\left(3 a^{2}+2 b^{2}\right)=5\left(a^{2}+a^{2}+a^{2}+b^{2}+b^{2}\right) \geq(3 a+2 b)^{2}
$$
(or use that the last inequality is equivalent to $(a-b)^{2} \geq 0$ ).
So, with the help of the given condition we get that $3 a+2 b \leq 5$. Now, by the AM-GM i... | \frac{2}{\sqrt{5}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
false
] | 4 | <think>
Okay, so I need to find the minimum value of A, which is given by the expression:
$$
A = \sqrt{\frac{a}{b(3a + 2)}} + \sqrt{\frac{b}{a(2b + 3)}}
$$
subject to the constraint:
$$
3a^2 + 2b^2 = 3a + 2b
$$
where a and b are positive real numbers. Hmm, let's try to tackle this step by step.
First, I should pro... |
A2. Let $a$ and $b$ be positive real numbers such that $3 a^{2}+2 b^{2}=3 a+2 b$. Find the minimum value of
$$
A=\sqrt{\frac{a}{b(3 a+2)}}+\sqrt{\frac{b}{a(2 b+3)}}
$$
|
Solution. By the Cauchy-Schwarz inequality we have that
$$
5\left(3 a^{2}+2 b^{2}\right)=5\left(a^{2}+a^{2}+a^{2}+b^{2}+b^{2}\right) \geq(3 a+2 b)^{2}
$$
(or use that the last inequality is equivalent to $(a-b)^{2} \geq 0$ ).
So, with the help of the given condition we get that $3 a+2 b \leq 5$. Now, by the AM-GM i... | \frac{2}{\sqrt{5}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
false
] | 4 | <think>
Alright, let me try to figure out how to find the minimum value of A. So, the problem says that a and b are positive real numbers satisfying the equation 3a² + 2b² = 3a + 2b. And we need to find the minimum of A, which is this expression with square roots and fractions involving a and b. Hmm, okay.
First, let ... |
A2. Let $a$ and $b$ be positive real numbers such that $3 a^{2}+2 b^{2}=3 a+2 b$. Find the minimum value of
$$
A=\sqrt{\frac{a}{b(3 a+2)}}+\sqrt{\frac{b}{a(2 b+3)}}
$$
|
Solution. By the Cauchy-Schwarz inequality we have that
$$
5\left(3 a^{2}+2 b^{2}\right)=5\left(a^{2}+a^{2}+a^{2}+b^{2}+b^{2}\right) \geq(3 a+2 b)^{2}
$$
(or use that the last inequality is equivalent to $(a-b)^{2} \geq 0$ ).
So, with the help of the given condition we get that $3 a+2 b \leq 5$. Now, by the AM-GM i... | \frac{2}{\sqrt{5}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
false
] | 4 | <think>
Okay, so I need to find the minimum value of A, which is given by the expression:
$$
A=\sqrt{\frac{a}{b(3 a+2)}}+\sqrt{\frac{b}{a(2 b+3)}}
$$
subject to the constraint:
$$
3a^2 + 2b^2 = 3a + 2b
$$
where a and b are positive real numbers. Hmm, let's start by understanding the problem.
First, I notice that b... |
A2. Let $a$ and $b$ be positive real numbers such that $3 a^{2}+2 b^{2}=3 a+2 b$. Find the minimum value of
$$
A=\sqrt{\frac{a}{b(3 a+2)}}+\sqrt{\frac{b}{a(2 b+3)}}
$$
|
Solution. By the Cauchy-Schwarz inequality we have that
$$
5\left(3 a^{2}+2 b^{2}\right)=5\left(a^{2}+a^{2}+a^{2}+b^{2}+b^{2}\right) \geq(3 a+2 b)^{2}
$$
(or use that the last inequality is equivalent to $(a-b)^{2} \geq 0$ ).
So, with the help of the given condition we get that $3 a+2 b \leq 5$. Now, by the AM-GM i... | \frac{2}{\sqrt{5}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
false
] | 4 | <think>
Okay, so I need to find the minimum value of A, which is given by the expression:
$$
A = \sqrt{\frac{a}{b(3a + 2)}} + \sqrt{\frac{b}{a(2b + 3)}}
$$
subject to the constraint:
$$
3a^2 + 2b^2 = 3a + 2b
$$
where a and b are positive real numbers. Hmm, let's start by understanding the problem.
First, the const... |
A2. Let $a$ and $b$ be positive real numbers such that $3 a^{2}+2 b^{2}=3 a+2 b$. Find the minimum value of
$$
A=\sqrt{\frac{a}{b(3 a+2)}}+\sqrt{\frac{b}{a(2 b+3)}}
$$
|
Solution. By the Cauchy-Schwarz inequality we have that
$$
5\left(3 a^{2}+2 b^{2}\right)=5\left(a^{2}+a^{2}+a^{2}+b^{2}+b^{2}\right) \geq(3 a+2 b)^{2}
$$
(or use that the last inequality is equivalent to $(a-b)^{2} \geq 0$ ).
So, with the help of the given condition we get that $3 a+2 b \leq 5$. Now, by the AM-GM i... | \frac{2}{\sqrt{5}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
false
] | 4 | <think>
Okay, so I need to find the minimum value of A, which is given by the expression:
$$
A = \sqrt{\frac{a}{b(3a + 2)}} + \sqrt{\frac{b}{a(2b + 3)}}
$$
subject to the constraint:
$$
3a^2 + 2b^2 = 3a + 2b
$$
where a and b are positive real numbers. Hmm, let me start by understanding the problem better.
First, t... |
C1. Consider a regular $2 n+1$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in... |
Solution. Answer: $n=1$ is clearly a solution, we can just color each side of the equilateral triangle in a different color, and the conditions are satisfied. We prove there is no larger $n$ that fulfills the requirements.
Lemma 1. Given a regular $2 n+1$-gon in the plane, and a sequence of $n+1$ consecutive sides $s... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
true,
true
] | 2 | <think>
Okay, let's try to tackle this problem. So, we have a regular polygon with an odd number of sides, specifically 2n+1 sides. We need to color its sides using three colors with some conditions: each side must be colored, all three colors must be used, and from any external point, you can see at most two different... |
C1. Consider a regular $2 n+1$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in... |
Solution. Answer: $n=1$ is clearly a solution, we can just color each side of the equilateral triangle in a different color, and the conditions are satisfied. We prove there is no larger $n$ that fulfills the requirements.
Lemma 1. Given a regular $2 n+1$-gon in the plane, and a sequence of $n+1$ consecutive sides $s... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
true,
true
] | 2 | <think>
Okay, so I have this problem about coloring the sides of a regular (2n+1)-gon with three colors. The conditions are that each side must be colored, all three colors must be used, and from any external point, you can see at most two different colors. The goal is to find the largest positive integer n for which t... |
C2. Consider a regular $2 n$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in 3... |
Solution. Answer: For $n=2$, the answer is 36 ; for $n=3$, the answer is 30 and for $n \geq 4$, the answer is $6 n$.
Lemma 1. Given a regular $2 n$-gon in the plane and a sequence of $n$ consecutive sides $s_{1}, s_{2}, \ldots, s_{n}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ ca... | 6n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Alright, let's try to tackle this problem step by step. So, we have a regular 2n-gon, and we need to color its sides with 3 colors. The conditions are that each color is used at least once, and from any external point, you can see at most two different colors. Also, two colorings are different if at least one s... |
C2. Consider a regular $2 n$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in 3... |
Solution. Answer: For $n=2$, the answer is 36 ; for $n=3$, the answer is 30 and for $n \geq 4$, the answer is $6 n$.
Lemma 1. Given a regular $2 n$-gon in the plane and a sequence of $n$ consecutive sides $s_{1}, s_{2}, \ldots, s_{n}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ ca... | 6n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to figure out how many valid colorings there are for a regular 2n-gon with three colors, following specific rules. Let me start by understanding the problem thoroughly.
First, the polygon is regular, meaning all sides are equal and all angles are equal. It's a 2n-gon, so there are 2n sides. The... |
Problem G3. Consider triangle $A B C$ such that $A B \leq A C$. Point $D$ on the arc $B C$ of the circumcirle of $A B C$ not containing point $A$ and point $E$ on side $B C$ are such that
$$
\Varangle B A D=\Varangle C A E<\frac{1}{2} \Varangle B A C .
$$
Let $S$ be the midpoint of segment $A D$. If $\Varangle A D E... |
Solution. Let the tangent to the circumcircle of $\triangle A B C$ at point $A$ intersect line $B C$ at $T$. Since $A B \leq A C$ we get that $B$ lies between $T$ and $C$. Since $\Varangle B A T=\Varangle A C B$ and $\Varangle A B T=\Varangle 180^{\circ}-\Varangle A B C$ we get $\Varangle E T A=\Varangle B T A=\Varang... | 2\VarangleBAC=\VarangleBSC | Geometry | proof | Yes | Yes | olympiads | false | 1 | [
false
] | 0 | <think>
Okay, let me try to work through this problem step by step. First, let me understand what's given and what needs to be proven.
We have triangle ABC with AB ≤ AC. So, ABC is a triangle where side AB is not longer than AC. That might mean that angle ACB is less than or equal to angle ABC, but I should be careful... |
NT1. Determine all sets of six consecutive positive integers such that the product of two of them, added to the the product of other two of them is equal to the product of the remaining two numbers.
|
Solution. Exactly two of the six numbers are multiples of 3 and these two need to be multiplied together, otherwise two of the three terms of the equality are multiples of 3 but the third one is not.
Let $n$ and $n+3$ denote these multiples of 3 . Two of the four remaining numbers give remainder 1 when divided by 3 ,... | 1\cdot2+3\cdot6=4\cdot5,\quad2\cdot5+3\cdot6=4\cdot7,\quad7\cdot8+6\cdot9=10\cdot11 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to find all sets of six consecutive positive integers where the product of two of them plus the product of another two equals the product of the last two. Hmm, let's break this down step by step.
First, let's denote the six consecutive positive integers. Since they are consecutive, I can repres... |
NT1. Determine all sets of six consecutive positive integers such that the product of two of them, added to the the product of other two of them is equal to the product of the remaining two numbers.
|
Solution. Exactly two of the six numbers are multiples of 3 and these two need to be multiplied together, otherwise two of the three terms of the equality are multiples of 3 but the third one is not.
Let $n$ and $n+3$ denote these multiples of 3 . Two of the four remaining numbers give remainder 1 when divided by 3 ,... | 1\cdot2+3\cdot6=4\cdot5,\quad2\cdot5+3\cdot6=4\cdot7,\quad7\cdot8+6\cdot9=10\cdot11 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I have this problem: Find all sets of six consecutive positive integers such that the product of two of them added to the product of another two equals the product of the remaining two. Hmm, let me try to break this down.
First, let me parse the question. We need six consecutive positive integers. Let... |
NT3. Find all pairs of positive integers $(x, y)$ such that $2^{x}+3^{y}$ is a perfect square.
|
Solution. In order for the expression $2^{x}+3^{y}$ to be a perfect square, a positive integer $t$ such that $2^{x}+3^{y}=t^{2}$ should exist.
Case 1. If $x$ is even, then there exists a positive integer $z$ such that $x=2 z$. Then
$$
\left(t-2^{z}\right)\left(t+2^{z}\right)=3^{y}
$$
Since $t+2^{z}-\left(t-2^{z}\ri... | (4,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to find all pairs of positive integers (x, y) such that 2^x + 3^y is a perfect square. Hmm, interesting problem. Let me start by understanding what exactly is being asked here. We have two exponents with bases 2 and 3, and their sum has to result in a perfect square. So, for some integer z, 2^x ... |
NT3. Find all pairs of positive integers $(x, y)$ such that $2^{x}+3^{y}$ is a perfect square.
|
Solution. In order for the expression $2^{x}+3^{y}$ to be a perfect square, a positive integer $t$ such that $2^{x}+3^{y}=t^{2}$ should exist.
Case 1. If $x$ is even, then there exists a positive integer $z$ such that $x=2 z$. Then
$$
\left(t-2^{z}\right)\left(t+2^{z}\right)=3^{y}
$$
Since $t+2^{z}-\left(t-2^{z}\ri... | (4,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to find all pairs of positive integers (x, y) such that 2^x + 3^y is a perfect square. Hmm, let's see. I should start by considering small values of x and y to see if I can find any patterns or possible solutions. Then maybe I can generalize from there.
First, let's try x=1. Then 2^1 = 2. So 2 ... |
NT3. Find all pairs of positive integers $(x, y)$ such that $2^{x}+3^{y}$ is a perfect square.
|
Solution. In order for the expression $2^{x}+3^{y}$ to be a perfect square, a positive integer $t$ such that $2^{x}+3^{y}=t^{2}$ should exist.
Case 1. If $x$ is even, then there exists a positive integer $z$ such that $x=2 z$. Then
$$
\left(t-2^{z}\right)\left(t+2^{z}\right)=3^{y}
$$
Since $t+2^{z}-\left(t-2^{z}\ri... | (4,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to find all pairs of positive integers (x, y) such that 2^x + 3^y is a perfect square. Hmm, let me think. First, maybe I should start by checking small values of x and y to see if there are any obvious patterns or solutions.
Let's start with x = 1. Then 2^1 = 2. So we need 2 + 3^y to be a perfe... |
NT3. Find all pairs of positive integers $(x, y)$ such that $2^{x}+3^{y}$ is a perfect square.
|
Solution. In order for the expression $2^{x}+3^{y}$ to be a perfect square, a positive integer $t$ such that $2^{x}+3^{y}=t^{2}$ should exist.
Case 1. If $x$ is even, then there exists a positive integer $z$ such that $x=2 z$. Then
$$
\left(t-2^{z}\right)\left(t+2^{z}\right)=3^{y}
$$
Since $t+2^{z}-\left(t-2^{z}\ri... | (4,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to find all pairs of positive integers (x, y) such that 2^x + 3^y is a perfect square. Hmm, let's start by understanding the problem. We have two exponentials with different bases, 2 and 3, added together, and their sum should be a square number. Both x and y are positive integers, so starting f... |
NT3. Find all pairs of positive integers $(x, y)$ such that $2^{x}+3^{y}$ is a perfect square.
|
Solution. In order for the expression $2^{x}+3^{y}$ to be a perfect square, a positive integer $t$ such that $2^{x}+3^{y}=t^{2}$ should exist.
Case 1. If $x$ is even, then there exists a positive integer $z$ such that $x=2 z$. Then
$$
\left(t-2^{z}\right)\left(t+2^{z}\right)=3^{y}
$$
Since $t+2^{z}-\left(t-2^{z}\ri... | (4,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, let's see. I need to find all pairs of positive integers (x, y) such that 2^x + 3^y is a perfect square. Hmm, interesting problem. Alright, where do I start? Maybe by trying small values of x and y and see if I can spot a pattern or find some solutions. Let's begin.
First, let's consider x=1. Then 2^1 = ... |
NT3. Find all pairs of positive integers $(x, y)$ such that $2^{x}+3^{y}$ is a perfect square.
|
Solution. In order for the expression $2^{x}+3^{y}$ to be a perfect square, a positive integer $t$ such that $2^{x}+3^{y}=t^{2}$ should exist.
Case 1. If $x$ is even, then there exists a positive integer $z$ such that $x=2 z$. Then
$$
\left(t-2^{z}\right)\left(t+2^{z}\right)=3^{y}
$$
Since $t+2^{z}-\left(t-2^{z}\ri... | (4,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to find all pairs of positive integers (x, y) such that 2^x + 3^y is a perfect square. Hmm, let's start by understanding what the problem is asking. We have two exponents here, 2 raised to some power x and 3 raised to some power y, and when we add them together, the result should be a perfect sq... |
NT4. Solve in nonnegative integers the equation $5^{t}+3^{x} 4^{y}=z^{2}$.
|
Solution. If $x=0$ we have
$$
z^{2}-2^{2 y}=5^{t} \Longleftrightarrow\left(z+2^{y}\right)\left(z-2^{y}\right)=5^{t}
$$
Putting $z+2^{y}=5^{a}$ and $z-2^{y}=5^{b}$ with $a+b=t$ we get $5^{a}-5^{b}=2^{y+1}$. This gives us $b=0$ and now we have $5^{t}-1=2^{y+1}$. If $y \geq 2$ then consideration by modulo 8 gives $2 \m... | (,x,y,z)=(1,0,1,3),(0,1,0,2),(2,2,2,13),(0,1,2,7) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Alright, so I need to solve the equation \(5^{t} + 3^{x} 4^{y} = z^{2}\) in nonnegative integers. Hmm, okay. Let me break this down step by step. First, let's understand each term and what constraints they might have.
The equation is made up of three terms: \(5^t\), \(3^x 4^y\), and \(z^2\). All variables \(t,... |
NT4. Solve in nonnegative integers the equation $5^{t}+3^{x} 4^{y}=z^{2}$.
|
Solution. If $x=0$ we have
$$
z^{2}-2^{2 y}=5^{t} \Longleftrightarrow\left(z+2^{y}\right)\left(z-2^{y}\right)=5^{t}
$$
Putting $z+2^{y}=5^{a}$ and $z-2^{y}=5^{b}$ with $a+b=t$ we get $5^{a}-5^{b}=2^{y+1}$. This gives us $b=0$ and now we have $5^{t}-1=2^{y+1}$. If $y \geq 2$ then consideration by modulo 8 gives $2 \m... | (,x,y,z)=(1,0,1,3),(0,1,0,2),(2,2,2,13),(0,1,2,7) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to solve the equation \(5^{t} + 3^{x} 4^{y} = z^{2}\) in nonnegative integers. Hmm, let's break this down step by step.
First, I should consider the different variables involved: t, x, y, and z. All of them are nonnegative integers, which means they can be 0, 1, 2, and so on. The equation comb... |
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