problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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|---|---|---|---|---|---|---|---|---|
Let's say a positive integer is simple if it has only the digits 1 or 2 (or both). How many numbers less than 1 million are simple? | With 1 digit, we have the simple numbers 1 and 2; with 2 digits, we have the $2^{2}=4$ simple numbers 11, 12, 21, and 22; with 3 digits, we have the $2^{3}=8$ simple numbers 111, 112, 121, 122, 211, 212, 221, and 222. With 4 digits, we have $2^{4}=16$ simple numbers, with 5 digits, we have $2^{5}=32$ simple numbers, an... | 126 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Henrique bought chocolate bars for $\mathrm{R} \$ 1.35$ each. He paid with a $\mathrm{R} \$ 10.00$ bill and received change less than $\mathrm{R} \$ 1.00$. How many bars did he buy? | As $8 \times 1.35 = 10.8$ is greater than 10, Henrique bought 7 chocolate bars and received $10 - 7 \times 1.35 = 0.55$ dollars, or 55 cents, in change. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A certain concrete mixture is made of cement, sand, and soil, in the ratio of $1: 3: 5$ per kilogram. How many kilograms of this mixture can be made with 5 kilograms of cement?
(a) $13 \frac{1}{3}$
(b) 15
(c) 25
(d) 40
(e) 45 | The correct option is (e).
According to the data in the problem, we mix $1 \mathrm{~kg}$ of cement with $3 \mathrm{~kg}$ of sand and $5 \mathrm{~kg}$ of soil. This is equivalent to mixing $5 \mathrm{~kg}$ of cement with $15 \mathrm{~kg}$ of sand and $25 \mathrm{~kg}$ of soil, and this mixture weighs $5+15+25=45 \mathr... | 45 | Algebra | MCQ | Yes | Yes | olympiads | false |
The ants Maricota and Nandinha are strolling on a balcony whose floor is made up of rectangular tiles measuring $4 \mathrm{~cm}$ in width by $6 \mathrm{~cm}$ in length, as indicated in the figure. Maricota starts from point $M$, Nandinha starts from $N$, and both walk only along the edges of the tiles, following the pa... | (a) The journey from $M$ to $N$ comprises 14 lengths and 12 widths of the tiles. Therefore, its length is $14 \times 6 + 12 \times 4 = 84 + 48 = 132 \mathrm{~cm}$.
Since the two ants travel the same distance, each one must walk $132 \div 2 = 66 \mathrm{~cm}$.
(b) Let's follow, from the beginning, the path taken by Ma... | 66 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The sum of three numbers is 100, two are prime and one is the sum of the other two.
(a) What is the largest of the three numbers?
(b) Give an example of such three numbers.
(c) How many solutions exist for this problem? | (a) Initially observe that, since the sum of the three numbers is 100 and the largest of them is equal to the sum of the other two, then twice the largest number is 100, that is, the largest number is 50.
(b) Since 50 is not a prime number, the other two numbers are primes and their sum is 50. For example, 3 and 47 ar... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a school, one quarter of the students play only volleyball, one third play only football, 300 practice both sports, and 1/12 play neither of these two sports.
(a) How many students are there in the school?
(b) How many students play only football?
(c) How many students play football?
(d) How many students practi... | The total number of students in the school is given by the fraction $12 / 12$, which we can graphically represent by a rectangle divided into 12 equal parts.
Let V, F, and NE denote the nu... | 900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Teacher Ana went to buy cheese bread to honor the students who won in the OBMEP, given that
- every 100 grams of cheese bread cost $R \$ 3.20$ and correspond to 10 cheese breads; and
- each person eats, on average, 5 cheese breads.
In addition to the teacher, 16 students, one monitor, and 5 parents will be present at... | (a) The teacher + 16 students + 1 supervisor + 5 parents $=23$ people will eat the cheese bread. To ensure that each person can eat at least 5 cheese breads, it is necessary to buy, at a minimum, $5 \times 23=115$ cheese breads. On average, each cheese bread weighs $\frac{100}{10}=10$ grams, so it will be necessary to ... | 1200 | Other | math-word-problem | Yes | Yes | olympiads | false |
Among the numbers 712, 1262, and 1680, which is the only one that can be written as a product of four consecutive natural numbers? | First, note that if three numbers are consecutive, then one of them is divisible by 3. Therefore, any number that is the product of three or more consecutive numbers must be divisible by 3. However, among the given numbers, only 1680 is divisible by 3, and furthermore,
$$
1680=2^{4} \times 3 \times 5 \times 7=5 \times... | 1680 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The year 2002 is a palindrome because it does not change when read from right to left.
373 and 1221
were palindromic years.
(a) What will be the next palindromic year after 2002?
(b) The last palindromic year, 1991, was odd. When will the next odd palindromic year be?
(c) The last prime palindromic year occurred m... | (a) The next one is 2112.
(b) The next odd palindrome is 3003.
(c) To be prime, the palindrome cannot have four digits, as every four-digit palindrome is of the form $a b b a$, which is divisible by 11, since
$$
\begin{aligned}
a b b a & =a 00 a+b b 0=a \times 1001+b \times 110=a \times 11 \times 91+b \times 11 \tim... | 10301 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In how many ways can we distribute 14 identical candies among three children so that each child receives at least three candies? | First, we need to determine the number of ways to obtain 14 as the sum of three integer parts, each of which is greater than or equal to 3, that is,
![](https://cdn.mathpix.com/cropped/202... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A Math test starts at 12:35 and lasts for $4 \frac{5}{6}$ hours. At what time does the test end? | First, let's observe that
$$
\frac{5}{6} \text{ h} = \frac{5}{6} \times 60 \text{ min} = 50 \text{ min}
$$
so the exam lasted $4 \text{ h} 50 \text{ min}$. Adding the hours and minutes, we get
$$
12 \text{ h} 35 \text{ min} + 4 \text{ h} 50 \text{ min} = 16 \text{ h} 85 \text{ min}
$$
But, $85 \text{ min} = 1 \text... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
What is the smallest five-digit number divisible by 4 that can be formed with the digits 1, 2, 3, 4, and 9? | A number is only divisible by 4 if the number formed by its last two digits is divisible by 4. Thus, using only the digits 1, 2, 3, 4, and 9, the only possibilities are 12, 24, 32, or 92. Since 9 is the largest digit, we should place it "as far to the right as possible," meaning that 9 must be the digit in the tens pla... | 13492 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Antônio has a parrot that performs fantastic calculations with integers. When Antônio whispers certain numbers into its ear, the parrot multiplies that number by 5, then adds 14, divides the result by 6, and finally subtracts 1, shouting the result afterward. However, the parrot knows nothing about decimals, so sometim... | (a) We have $8 \xrightarrow{\times 5} 40 \xrightarrow{+14} 54 \xrightarrow{\dot{\circ} 6} 9 \xrightarrow{-1} 8$. Therefore, the parrot shouts 8.
(b) We should perform the inverse operation of what the parrot did, starting from the last operation, that is, add 1 to the number, multiply the number by 6, then subtract 14... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Roberto wants to write the number 111111 as a product of two numbers, neither of which ends in 1. Is this possible? Why? | Factoring 111111, we obtain $111111=3 \times 7 \times 11 \times 13 \times 37$. It follows that it is indeed possible to write the number 111111 as a product of two factors, neither of which ends in 1. For example, $111111=3 \times 37037$. But there are other possibilities, such as, for example, $111111=7 \times 15873$.... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A number is said to be balanced if one of its digits is the arithmetic mean of the others. For example, 132, 246, and 777 are balanced. How many balanced three-digit numbers exist? | Note that if a balanced number has three distinct non-zero digits, then with the same digits, we obtain six balanced numbers. For this, it is enough to swap the digits' positions. For example, 123, 132, 213, 231, 312, and 321.
If one of the three digits of the balanced number is 0, then with these digits, we obtain on... | 121 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A class has 22 male students and 18 female students. During the holidays, $60 \%$ of the students in this class went to do community work. At a minimum, how many female students participated in this work?
(a) 1
(b) 2
(c) 4
(d) 6
(e) 8 | The correct answer is (b).
The total number of students in this class is $22+18=40$, of which $60\%$ went to do community work, that is, $0.6 \times 40=24$. The minimum number of female students who participated in this work is obtained when the number of male students who participated is maximum, that is, when all 22... | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
By joining four identical trapezoids, which have non-parallel sides of equal length and bases measuring 50 and $30 \mathrm{~cm}$, as shown in the given figure, we can form a square of $2500 \mathrm{~cm}^{2}$ in area, which has a square "hole" in the middle. What is the
.
By joining the four trapezoids, we form a square with a side length of $50 \mathrm{~cm}$ and, therefore, an area of $2500 \mathrm{~cm}^{2}$. Since the "hole" square has a side length of $30 \mathrm{~cm}$, its area is $30 \times 30=900 \mathrm{~cm}^{2}$. Thus, the area of each of the four tra... | 400 | Geometry | MCQ | Yes | Yes | olympiads | false |
Discover the rule used for the filled cells and complete the table. What is the value of A?
| 0 | 1 | 2 | 3 | 4 |
| :---: | :---: | :---: | :---: | :---: |
| 1 | 2 | 5 | 10 | |
| 2 | | | | |
| 3 | | | | |
| 4 | | | | $\mathbf{A}$ | | Observe that in each square formed by four smaller squares, the number at the bottom right is the sum of the other three numbers. Thus, we fill in the table.
| 0 | 1 | 2 | 3 | 4 |
| :---: | :---: | :---: | :---: | :---: |
| 1 | 2 | 5 | 10 | $3+4+10=17$ |
| 2 | $1+2+2=5$ | $2+5+5=12$ | $5+10+12=27$ | $10+17+27=54$ |
| ... | 360 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A baker went to the market to buy eggs to make 43 cakes, all with the same recipe, which requires fewer than nine eggs. The seller notices that if he tries to wrap the eggs the baker bought in groups of two, or three, four, five, or six eggs, there is always one egg left over. How many eggs does she use in each cake? W... | Since the 43 cakes have the same recipe, the number of eggs the baker needs is a multiple of 43. On the other hand, this number is also a multiple of 2, 3, 4, 5, and 6, plus 1. The LCM of $2,3,4,5$ and 6 is 60, but $60+1=61$ is not a multiple of 43. We need to find a number with these two properties:
- it is a multipl... | 301 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The sum on the side is incorrect. To correct it, it is enough to replace a certain digit in all the places where it appears in the equation with another digit. Which is the incorrect digit and what is its correct substitute? | At first inspection, we can admit that the three digits to the right of the numbers are correct, that is, the digits $0,1,3,4,5$, 6 and 8 are correct. Therefore, among the digits 2, 7, and 9, one of them is wrong. The digit 9 is correct, because if we change it, the sum with 2 will not be correct. Thus, only 2 and 7 re... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The digits 1, 2, 3, 4, and 5 were used, each one only once, to write a certain five-digit number $a b c d e$ such that $a b c$ is divisible by $4, b c d$ is divisible by 5, and $c d$ is divisible by 3. Find this number. | For $a b c$ to be divisible by 4, its last two digits must form a number divisible by 4. Since the digits are 1, 2, 3, 4, and 5, the only possibilities are $b c=12, b c=24, b c=32$, and $b c=52$. On the other hand, numbers divisible by 5 end in 0 or 5. Since 0 is not included, it follows that $d=5$, because $b c d$ is ... | 12453 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a certain city with almost thirty thousand inhabitants, exactly two ninths of the inhabitants are men who only practice sports on weekends and two fifteenths are women who only practice sports on weekends. The number of inhabitants who do not practice sports is five times those who practice sports regularly. With th... | The total number of inhabitants in this city is almost $30000 \mathrm{e}$ and is divisible by 9 and 15. Therefore, it must end in 0 or 5, and the sum of its digits must be a multiple of 9. Since 29970 is the largest number less than 30000 that has factors of 9 and 15, we can assume that this is the total population of ... | 29970 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the luminous mechanism of the figure, each of the eight buttons can light up in green or blue. The mechanism works as follows: when turned on, all buttons light up blue, and if we press a button, that button and its two neighbors change color. If we turn on the mechanism and successively press buttons 1, 3, and 5, h... | The correct answer is (c).
The table shows the color of each button at each step.
| | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ | $\mathbf{5}$ | $\mathbf{6}$ | $\mathbf{7}$ | $\mathbf{8}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| start | blue | blue | blue | blue |... | 5 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
A six-digit number starts with 1. If we move this digit 1 from the first to the last position on the right, we get a new six-digit number, which is three times the original number. What is this number? | The problem is to determine the digits $b, c, d, e$ and $f$ such that the number $b c d$ and $f 1$ is three times $1 b c d e f$.
At the beginning, we see that $f=7$, and from there, we can discover each of the digits, as follows.
 $30^{... | The correct answer is (b).
We have $A \widehat{O} C+C \widehat{O} E=90^{\circ}$ and $C \widehat{O} E=D \widehat{O} Y$. Therefore, $A \widehat{O} C=90^{\circ}-D \widehat{O} Y$. Since $D \widehat{O} Y$ is between $40^{\circ}$ and $50^{\circ}$, it follows that $A \widehat{O} C$ is between $90^{\circ}-50^{\circ}=40^{\circ... | 40 | Geometry | MCQ | Yes | Yes | olympiads | false |
Find the measure of angle $B \widehat{A} D$, knowing that $D \widehat{A} C=39^{\circ}, A B=A C$ and $A D=B D$.
 | Given $A B=A C$, the triangle $\triangle A B C$ is isosceles, so $A \widehat{B} C=A \widehat{C} B$. Since $A D=B D$, the triangle $\triangle A B D$ is also isosceles, so $A \widehat{B} D=B \widehat{A} D$. Therefore,
$$
A \widehat{C} B=A \widehat{B} C=A \widehat{B} D=B \widehat{A} D
$$
.
 | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
Elisa has 24 science books and others of mathematics and literature. If Elisa had one more mathematics book, then one ninth of her books would be mathematics and one quarter literature. If Elisa has fewer than 100 books, how many mathematics books does she have? | Let $N$ be the total number of books Elisa has. Since $N+1$ is a multiple of 9 and 4, it follows that $N+1$ is a multiple of 36. Therefore, $N+1$ is 36 or 72, as Elisa has fewer than 100 books. If $N=35$, then the number of math books is $36 \div 9-1=3$ and the number of literature books is $36 \div 4=9$. However, this... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Laura and her grandmother Ana have just discovered that last year their ages were divisible by 8 and that next year they will be divisible by 7. Grandma Ana is not yet a centenarian. What is Laura's age? | Next year, Laura and her grandmother will be two years older than they were last year. Therefore, their ages last year are multiples of 8 that, when added to 2, give multiples of 7. Let's find these numbers.
$$
\begin{array}{rllllllllllll}
\text { multiples of } 7: & 7 & 14 & 21 & 28 & 35 & 42 & 49 & 56 & 63 & \ldots ... | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
O dobro de um número dividido por 5 deixa resto 1. Qual é o resto da divisão desse número por 5 ?
The double of a number divided by 5 leaves a remainder of 1. What is the remainder of the division of this number by 5? | Solution 1: The double of the number sought is a multiple of 5 increased by 1. Since multiples of 5 end in 0 or 5, the double ends in 1 or 6. But the double is an even number, so it ends in 6. Thus, the number ends in 3 or 8 and, therefore, when divided by 5, leaves a remainder of 3.
Solution 2: We know that the integ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Real numbers $a, b, c, d$ satisfy the equation $a b + b c + c d + d a = 16$.
a) Prove that among the numbers $a, b, c, d$, there are two with a sum of at most 4.
b) What is the minimum value that the sum $a^{2} + b^{2} + c^{2} + d^{2}$ can have? | SOLUTION. a) From the equality $16=a b+b c+c d+d a=(a+c)(b+d)$, it follows that both sums $a+c$ and $b+d$ cannot be greater than 4 at the same time. Therefore, at least one of the sums $a+c$ or $b+d$ must have the required property.
b) Let us use the general equality
$$
a^{2}+b^{2}+c^{2}+d^{2}=\frac{1}{2}(a-b)^{2}+\f... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. $V$ each of the four rooms there are several items. Let $n \geqq 2$ be a natural number. We move one $n$-th of the items from the first room to the second. Then, one $n$-th of the (new) number of items is moved from the second room to the third. Similarly, from the third room to the fourth, and from the fourth back ... | SOLUTION. When analyzing the number of items after individual steps, we will proceed "backwards." First, let's show how the numbers of items in two rooms before the transfer can be determined from the numbers of items in the rooms after the transfer. Let's say that before the transfer from room $A$ to room $B$, there a... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Nela and Jana choose a natural number $k$ and then play a game with a $9 \times 9$ table. Starting with Nela, each player, on their turn, selects an empty cell and writes a zero in it. Jana, on her turn, writes a one in some empty cell. Additionally, after each of Nela's moves, Jana makes $k$ moves. If at any point ... | SOLUTION. Let's first show that in the case $k=3$, Jana wins. We will work with squares $A_{1}, A_{2}$, and $A_{3}$ of size $3 \times 3$ (Fig. 3). A $3 \times 3$ square is considered covered if there is exactly one one in each of its rows and columns. If Jana covers the squares $A_{1}, A_{2}$, and $A_{3}$ without playi... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. For positive real numbers $a, b, c$ it holds that
$$
a b+b c+c a=16, \quad a \geqq 3 .
$$
Find the smallest possible value of the expression $2 a+b+c$. | 3. Let's modify the square of the expression $V=2a+b+c$, which is clearly positive. We will conveniently use the given relationship $ab+bc+ca=16$:
$$
\begin{aligned}
V^{2} & =(2a+b+c)^{2}=4a^{2}+b^{2}+c^{2}+4ab+4ac+2bc= \\
& =4a^{2}+b^{2}-2bc+c^{2}+4(ab+bc+ca)=4a^{2}+(b-c)^{2}+4 \cdot 16 .
\end{aligned}
$$
According ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let's say a subset $\mathrm{P}$ of the set $\mathrm{M}=\{1,2,3, \ldots, 42\}$ is halfish if it contains 21 elements and each of the 42 numbers in the sets $\mathrm{P}$ and $\mathrm{Q}=\{7 x ; x \in \mathrm{P}\}$ gives a different remainder when divided by 43. Determine the number of halfish subsets of the set M.
(J... | SOLUTION. We will treat all numbers as residue classes modulo the prime number 43.
First, we note that the sets $\mathrm{P}$ and $\mathrm{Q}$ form a disjoint partition of the set $\mathrm{M}$. Let $\mathrm{P}$ be the half-set and let $x \in \mathrm{P}$, for which then $7 x \in \mathbf{Q}$.
We will show that $7^{2} x ... | 128 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Determine the number of all four-digit natural numbers that are divisible by six and in whose notation exactly two ones appear. | 1. For a number to be divisible by six, it must be even and have a digit sum divisible by three. Let's denote $b$ as the digit in the units place (which must be even, $b \in\{0,2,4,6,8\}$) and $a$ as the digit that, together with the digits $1,1 (a \neq 1)$, is on the first three places of a four-digit number that meet... | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Determine in how many ways the vertices of a regular nonagon $A B C D E F G H I$ can be assigned numbers from the set $\{17,27,37,47,57,67,77,87,97\}$ such that each number is assigned to a different vertex and the sum of the numbers assigned to each set of three consecutive vertices is divisible by three. | 2. First, let us realize that the numbers 27, 57, and 87 are divisible by three, the numbers 37, 67, and 97 give a remainder of 1 when divided by three, and finally, the numbers 17, 47, and 77 give a remainder of 2 when divided by three. Let us denote $\overline{0}=\{27,57,87\}, \overline{1}=\{37,67,97\}$, and $\overli... | 1296 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the smallest four-digit number $\overline{a b c d}$ such that the difference $(\overline{a b})^{2}-(\overline{c d})^{2}$ is a three-digit number written with three identical digits. | SOLUTION. Let us first look for a solution to the equation for the smallest three-digit number with the same digits, which we will immediately factorize into prime factors:
$$
\overline{a b}^{2}-\overline{c d}^{2}=(\overline{a b}+\overline{c d})(\overline{a b}-\overline{c d})=111=37 \cdot 3
$$
The first parenthesis i... | 2017 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Determine the largest possible number of non-empty, pairwise disjoint sets with the same sum of elements, into which the set can be divided:
a) $\{1,2, \ldots, 2017\}$,
b) $\{1,2, \ldots, 2018\}$.
If a set consists of a single number, we consider it as the sum of its elements. | SOLUTION. Let's first look at the sets with the smallest possible number of elements; we can probably have at most one one-element set - otherwise, those two one-element sets would not have the same sum of elements. The other sets are therefore at least two-element.
In part a), let's take the one-element set to be the... | 1009 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Determine the largest integer $n$, for which it is possible to fill a square table $n \times n$ with natural numbers from 1 to $n^{2}$ in such a way that in every $3 \times 3$ square part of it, at least one square of an integer is written. | SOLUTION. It will be easy for us to fill the $11 \times 11$ table in the required way; it is enough to choose nine out of the 11 squares of integers $1^{2}, 2^{2}, \ldots, 11^{2}$ and place them on nine cells of the table with coordinates
$$
(3,3),(3,6),(3,9), \quad(6,3),(6,6),(6,9), \quad(9,3),(9,6),(9,9)
$$
and fil... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. The arithmetic mean of several distinct prime numbers is 27. Determine the largest prime number among them. | 1. Let $P$ be the set of prime numbers under investigation, and first show that $2 \notin P$. The number 2 is the only prime number that is not odd. If $2 \in P$, the sum of all primes in $P$ (an odd number of them) would be even, and the sum of an even number of them would be odd, so the arithmetic mean could not be t... | 139 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. A natural number will be called magical if and only if it can be expressed as the sum of two three-digit numbers written with the same digits but in reverse order. For example, the number 1413 is magical because $1413=756+657$; the smallest magical number is 202.
a) Determine the number of all magical numbers.
b) ... | SOLUTION. Taking the number 1413 as an example, we see that it is not always easy to tell whether a given three- or four-digit number is magical or not. Therefore, let's first look at how a magical number $x$ can be expressed using the digits of those three-digit numbers $\overline{a b c}$ and $\overline{c b a}$, which... | 187000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. On the board, there are several (not necessarily distinct) prime numbers such that their product is $2020$ times greater than their sum. Determine their smallest possible number. (Patrik Bak) | SOLUTION. Let the prime numbers on the board be denoted as $p_{1}, p_{2}, \ldots, p_{n}$. According to the problem statement, we have
$$
2020 \cdot\left(p_{1}+p_{2}+\ldots+p_{n}\right)=p_{1} p_{2} \ldots p_{n}
$$
The left side of equation (1) is divisible by the number $2020=2 \cdot 2 \cdot 5 \cdot 101$. Therefore, t... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. If a, b, c are distinct non-negative real numbers, what is the smallest possible number of distinct numbers among the numbers $a+b, b+c, c+a, a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+a^{2}$? (Patrik Bak) | SOLUTION. The problem is symmetric in the variables $a, b, c$: changing their order only changes the order of the six numbers being examined. In the first part of the solution, we will assume that $a < b < c$ and $a > 1$. Then it will hold that
$$
11$ will satisfy the condition $a^{2}=a+1$. A simple calculation reveal... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Along a circle, 16 real numbers are arranged with a sum of 7.
a) Prove that there exists a segment of five consecutive numbers with a sum of at least 2.
b) Determine the smallest $k$ such that in the described situation, one can always find a segment of $k$ consecutive numbers with a sum of at least 3. | 2. a) Among 16 numbers written along a circle, there are exactly 16 segments of five consecutive numbers (if we select any of the written numbers and mark the numbers along the circle sequentially as the first, second, ..., sixteenth, the first segment will consist of the first to fifth numbers, the second segment will... | 7 | Combinatorics | proof | Yes | Yes | olympiads | false |
4. The sum of 74 (not necessarily distinct) real numbers from the interval $\langle 4,10\rangle$ is 356. Determine the greatest possible value of the sum of their squares.
(Zdeněk Pezlar) | SOLUTION. Let $x_{1}, x_{2}, \ldots, x_{74}$ be the numbers from the problem. For each $i$, from the assumption $x_{i} \in \langle 4,10\rangle$, it clearly follows that $\left(x_{i}-4\right)\left(10-x_{i}\right) \geqq 0$, which after expanding gives $x_{i}^{2} \leqq 14 x_{i}-40$. By summing these inequalities for all $... | 2024 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. On the table lie 54 piles of stones with $1,2,3, \ldots, 54$ stones. In each step, we select any pile, say with $k$ stones, and remove it from the table along with $k$ stones from each pile that has at least $k$ stones. For example, after the first step, if we select the pile with 52 stones, the piles remaining on t... | 1. If in each step we choose the pile with the most stones, we will gradually remove piles with $54, 53, 52, \ldots$ stones, and after 53 steps, only one pile with one stone will remain on the table.
We can prove that regardless of the procedure, the last pile will always contain a single stone. We will show that afte... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A square table is divided into $16 \times 16$ cells. A knight moves on it in two directions: to the right or down, alternating jumps of two and three cells (that is, no two consecutive jumps are of the same length). It starts with a jump of length two from the top-left cell. In how many different ways can the knight... | SOLUTION. In the course of its journey, the grasshopper must move a total of 15 squares to the right and 15 squares down. In total, it will move 30 squares, so it will repeat the pair of jumps of length $2+3=5$ six times. More precisely, its individual jumps will have lengths in sequence
$$
2,3,2,3,2,3,2,3,2,3,2,3 \te... | 412 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Given a rectangle $A B C D$ with perimeter $o$. In its plane, find the set of all points whose sum of distances from the lines $A B, B C, C D, D A$ is equal to $\frac{2}{3} o$. | SOLUTION. The required value of the sum of four distances can be written in the form
$$
\frac{2}{3} o=\frac{1}{6} o+\frac{1}{2} o=\frac{1}{6} o+|A B|+|B C| \text {. }
$$
For any point in the strip determined by the lines $A B$ and $C D$, it holds that the sum of its distances from these two parallel lines is equal to... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. What is the maximum number of numbers that can be selected from the set $\{1,2, \ldots, 2019\}$ such that the product of any three of the selected numbers is not divisible by nine? Provide an example of a suitable subset and explain why it cannot have more elements.
(Aleš Kobza) | 1. Among the selected numbers, there must not be any multiple of nine, and at most one of them can be divisible by three, but not by nine. Therefore, we can select all numbers that are not divisible by three and add $\mathrm{k}$ to them any number that is divisible by three, but not by nine.
Conclusion. The largest po... | 1347 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Determine the smallest natural number $n$, for which the following holds: If some natural number has at least $n$ three-digit multiples, then 840 is one of them.
(Michal Rolínek) | Solution. From the factorization $840=2^{3} \cdot 3 \cdot 5 \cdot 7$ we see that the smallest natural number that is not a divisor of 840 is the number $9=3^{2}$. The three-digit multiples of the number 9 are
$$
108=9 \cdot 12, 117=9 \cdot 13, \ldots, 999=9 \cdot 111 \text {, }
$$
in total there are $111-12+1=100$ (a... | 101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. An unconventional piece, which we will call the "sick lady," threatens any square in the row and column where it stands, while on the diagonal, it only threatens the adjacent squares. How many "sick ladies" do we need to place on an $8 \times 8$ chessboard so that they threaten all unoccupied squares?
 are called directly threatened if they lie in the column or row where the queen stands; threatened fields that are not directly threatened are called indirectly threatened. Each queen thus threatens 15 fields directly (including the... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. We understand a palindrome as a natural number that reads the same forwards and backwards, for example, 16 261. Find the largest four-digit palindrome whose square is also a palindrome. | SOLUTION. Each four-digit palindrome $p=\overline{a b b a}$ can be written in the form
$$
p=a \cdot 1001+b \cdot 110
$$
where $a \in\{1,2, \ldots, 9\}$ and $b \in\{0,1,2, \ldots, 9\}$. Then the square of the number $\overline{a b b a}$ has the form
$$
\begin{aligned}
p^{2}= & a^{2} \cdot 1002001+2 a b \cdot 110110+b... | 2002 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The average of Pažout's grades is exactly 3. If we did not include three of Pažout's fives in the average, the average of his grades would be exactly 2. Determine the maximum number of ones that Pažout could have received. (Possible grades are 1, 2, 3, 4, 5.) | SOLUTION. Let $s$ be the sum of all Pážout's grades and $p$ their count. We do not know either of these numbers, but we know that
$$
\frac{s}{p}=3 \quad \text { or } \quad s=3 p .
$$
According to the second sentence of the problem, we set up another equation for the unknowns $s, p$ and immediately simplify it:
$$
\f... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Consider any $3 \times 3$ table filled with positive integers such that the sum of the numbers $v$ in each row and each column is 10. How many numbers $v$ in such a table can be: a) the same, b) different?
(Ján Mazák) | SOLUTION. a) Six identical numbers can be found, for example, in the following table that meets the conditions of the problem:
| 8 | 1 | 1 |
| :--- | :--- | :--- |
| 1 | 8 | 1 |
| 1 | 1 | 8 |
If the table contained at least seven identical numbers, one of the rows would contain three of these seven identical numbers.... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find all real numbers $p$ for which the system of inequalities
$$
\begin{aligned}
& 25+2 x^{2} \leqq 13 y+10 z-p \\
& 25+3 y^{2} \leqq 6 z+10 x \\
& 25+4 z^{2} \leqq 6 x+5 y+p
\end{aligned}
$$
with unknowns $x, y, z$ has a solution in the set of real numbers. | 1. By adding all three inequalities, we obtain the inequality
$$
75+2 x^{2}+3 y^{2}+4 z^{2} \leqq 16 x+18 y+16 z
$$
from which, after "completing the square," we get
$$
2(x-4)^{2}+3(y-3)^{2}+4(z-2)^{2} \leqq 0 .
$$
This inequality, which is a consequence of the given system of inequalities, clearly holds only when ... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Find all four-digit numbers $\overline{a b c d}$ that are divisible by each of the two-digit numbers $\overline{a b}, \overline{b c}, \overline{c d}$, where the digits $a, b, c, d$ are odd and not all the same.
The school - written part of the first round in category A takes place
on Tuesday, December 5, 2000
so ... | 3. From the expression $\overline{a b c d}=100 \cdot \overline{a b}+\overline{c d}$, it follows that the divisibility conditions for the numbers $\overline{a b}$ and $\overline{c d}$ are satisfied if and only if $\overline{c d} \mid 100 \cdot \overline{a b}$ and $\overline{a b} \mid \overline{c d}$, i.e., if and only i... | 1155 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In rectangle $A B C D$ with sides $|A B|=9,|B C|=8$, there are mutually touching circles $k_{1}\left(S_{1}, r_{1}\right)$ and $k_{2}\left(S_{2}, r_{2}\right)$ such that $k_{1}$ touches sides $A D$ and $C D$, and $k_{2}$ touches sides $A B$ and $B C$.
a) Prove that $r_{1}+r_{2}=5$.
b) Determine the smallest and lar... | 1. a) Through point $S_{1}$, draw a line parallel to side $AD$ and denote its intersections with sides $AB$ and $CD$ as $M$ and $N$. Similarly, through point $S_{2}$, draw a line parallel to $AB$ and denote its intersections with sides $AD$ and $BC$ as $K$ and $L$; the intersection of lines $KL$ and $MN$ is denoted as ... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Find the greatest natural number $d$ that has the property that for any natural number $n$, the value of the expression
$$
V(n)=n^{4}+11 n^{2}-12
$$
is divisible by the number $d$. | SOLUTION. First, let's calculate the values of $V(n)$ for the smallest natural numbers $n$ and write their prime factorizations in a table:
| $n$ | 1 | 2 | 3 | 4 |
| :---: | :---: | :---: | :---: | :---: |
| $V(n)$ | 0 | $48=2^{4} \cdot 3$ | $168=2^{3} \cdot 3 \cdot 7$ | $420=2^{2} \cdot 3 \cdot 5 \cdot 7$ |
From thi... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. On the board, five different positive numbers are written. Determine the maximum number of ways in which pairs can be formed from them, the sum of which equals one of the five numbers written on the board. | 1. Let $a_{1}<a_{2}<a_{3}<a_{4}<a_{5}$ be positive numbers written on the board. The smallest numbers $a_{1}$ and $a_{2}$ clearly cannot be the sum of any two numbers written on the board. The number $a_{3}$ can be obtained as the sum of some pair in at most one way, namely $a_{3}=a_{1}+a_{2}$. The number $a_{4}$ can t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Consider the expression
$$
V(x)=\frac{5 x^{4}-4 x^{2}+5}{x^{4}+1} .
$$
a) Prove that for every real number $x$, $V(x) \geq 3$.
b) Find the maximum value of $V(x)$. | 1. The expression $V$ is apparently defined for all real numbers $x$.
a) Since $x^{4}+1>0$ for every $x$, the inequality $V(x) \geqq 3$ is equivalent to the inequality $5 x^{4}-4 x^{2}+5 \geqq 3\left(x^{4}+1\right)$ or $2 x^{4}-4 x^{2}+2 \geqq 0$. The expression on the left side is equal to $2\left(x^{2}-1\right)^{2}$... | 5 | Algebra | proof | Yes | Yes | olympiads | false |
3. From the set $\{1,2,3, \ldots, 99\}$, several different numbers are chosen such that the sum of any three of them is not a multiple of nine.
a) Prove that among the chosen numbers, there are at most four divisible by three.
b) Show that the number of chosen numbers can be 26. | 3. According to the remainders when divided by nine, we will divide all 99 considered numbers into nine eleven-element classes $T_{0}, T_{1}, \ldots, T_{8}$ (class $T_{i}$ contains all numbers with a remainder of $i$):
$$
\begin{aligned}
T_{0}= & \{9,18,27, \ldots, 99\}, \\
T_{1}= & \{1,10,19, \ldots, 91\}, \\
T_{2}= ... | 26 | Combinatorics | proof | Yes | Yes | olympiads | false |
3. Find the smallest positive number $x$, for which the following holds: If $a, b, c, d$ are any positive numbers whose product is 1, then
$$
a^{x}+b^{x}+c^{x}+d^{x} \geqq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}
$$
(Pavel Novotný) | Solution. Let $a, b, c, d$ be any positive numbers whose product equals 1. According to the inequality between the arithmetic and geometric means of the triplet of numbers $a^{x}, b^{x}, c^{x}$ for any $x>0$, we have
$$
\frac{a^{x}+b^{x}+c^{x}}{3} \geqq \sqrt[3]{a^{x} b^{x} c^{x}}=\sqrt[3]{\frac{1}{d^{x}}}
$$
By choo... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Given a triangle $ABC$. The angle bisector at vertex $A$ intersects side $BC$ at point $D$. Let $E$, $F$ be the centers of the circumcircles of triangles $ABD$, $ACD$. What can be the measure of angle $BAC$ if the center of the circumcircle of triangle $AEF$ lies on the line $BC$?
(Patrik Bak) | Solution. Let $\alpha$ be the size of the angle $BAC$ under investigation, and $O$ the center of the circumcircle of triangle $AEF$. Since angles $BAD$ and $CAD$ are acute, both points $E$ and $F$ lie in the half-plane $BCA$, and thus for the corresponding central and inscribed angles of the chords $BD$ and $CD$ of the... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find the smallest natural number $n$ such that for any coloring of the numbers 1, 2, 3, ..., $n$ with three colors, there exist two numbers of the same color whose difference is a square of a natural number.
(Vojtech Bálint, Michal Rolínek, Josef Tkadlec) | Solution. We will show that the sought natural number is $n=29$. First, we will show that no matter how we color the first 29 natural numbers, there will always be two numbers of the same color that differ by a square of a natural number. Then, we will provide an example of a suitable coloring of 28 numbers, which will... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Determine the number of all triples of distinct three-digit natural numbers, the sum of which is divisible by each of the three addends. | SOLUTION. Let $x, y, z$ be a triplet of distinct natural numbers such that each of them divides their sum $x+y+z$, meaning $x$ divides $y+z$, $y$ divides $x+z$, and $z$ divides $x+y$. Without loss of generality, we can assume $x<y<z$. Therefore, $x+y=kz$ for some natural number $k$. Since $x+y<2z$, it must be that $k=1... | 234 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The number $a_{n}$ is formed by writing down the first $n$ squares of consecutive natural numbers in sequence. For example, $a_{11}=149162536496481100$ 121. Determine how many numbers divisible by twelve are among the numbers $a_{1}, a_{2}, \ldots, a_{100000}$. | 1. As we know, every natural number $k$ gives the same remainder when divided by three as the number $S(k)$ equal to the sum of the digits of the original number $k$. The number $a_{n}$, therefore, gives the same remainder when divided by three as the sum $S\left(1^{2}\right)+S\left(2^{2}\right)+\ldots+S\left(n^{2}\rig... | 16667 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In a $11 \times 11$ square grid, we sequentially wrote the numbers $1,2, \ldots, 121$ from left to right and from top to bottom. Using a $3 \times 3$ square tile, we covered exactly nine cells in all possible ways. In how many cases was the sum of the nine covered numbers a perfect square of an integer? | 2. Let $n$ be the number covered by the middle cell of the square tile. Then the first row of this tile covers the numbers $n-12, n-11, n-10$, its second row covers the numbers $n-1, n$ and $n+1$, and finally the third row covers the numbers $n+10, n+11$ and $n+12$. The sum of all numbers covered by the tile is thus $9... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Determine all integers greater than 1 by which some fraction of the form
$$
\frac{3 p-q}{5 p+2 q}
$$
can be divided, where $p$ and $q$ are coprime integers.
The written part of the school round in category A takes place
## on Tuesday, December 2, 2008
so that it starts in the morning and the contestants have 4 ... | 3. A fraction can be reduced by an integer $d>1$ if and only if the number $d$ is a common divisor of the numerator and the denominator of the considered fraction. Let us assume, therefore, that $d \mid 3 p-q$ and at the same time $d \mid 5 p+2 q$, where $p$ and $q$ are coprime integers. By adding suitable multiples of... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Determine the number of all five-digit palindromes that are divisible by 37. (A palindrome is a number that reads the same backward as forward in decimal notation.) | 1. Every five-digit palindrome $p$ can be written in the form $p=\overline{a b c b a}$, where $a, b, c$ are digits in the decimal system, $a \neq 0$. From the expression
$$
p=10001 a+1010 b+100 c=37(270 a+27 b+3 c)+11(a+b-c)
$$
it follows that $p$ is divisible by 37 if and only if the number $a+b-c$ is divisible by 3... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Adam and Bohouš participated in a tournament played in a round-robin system, where each player was supposed to play one match per day. However, Adam and Bohouš were the only ones who did not complete the tournament due to illness. Bohouš withdrew five days earlier than Adam. In total, 350 matches were played. How ma... | SOLUTION. Every day, the players were divided into pairs, in which they were to play their match. The total number of players was therefore even, let's denote it as $2n$. Every day, $n$ matches were to be played, and the tournament was scheduled for $2n-1$ days. Let's assume that Adam withdrew $d$ days before the end o... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Find all positive integers $n$ for which the number $n^{2}+6 n$ is a perfect square of an integer. | 2. Clearly $n^{2}+6 n>n^{2}$ and at the same time $n^{2}+6 n<n^{2}+6 n+9=(n+3)^{2}$. In the given range, there are only two squares of integers: $(n+1)^{2}$ and $(n+2)^{2}$.
In the first case, we have $n^{2}+6 n=n^{2}+2 n+1$, thus $4 n=1$, but no integer $n$ satisfies this.
In the second case, we have $n^{2}+6 n=n^{2... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find all prime numbers $p$ for which there exists a natural number $n$ such that $p^{n}+1$ is a cube of some natural number. | SOLUTION. Let's assume that for a natural number $a$ the following holds: $p^{n}+1=a^{3}$ (obviously $a \geq 2$). We will rearrange this equation to make it possible to factor one side:
$$
p^{n}=a^{3}-1=(a-1)\left(a^{2}+a+1\right) \text {. }
$$
From this factorization, it follows that if $a>2$, both numbers $a-1$ and... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. We will divide the integers from 1 to 9 into three groups of three and then multiply the numbers within each group.
a) Determine these three products, given that two of them are equal and smaller than the third product.
b) Assume that one of the three products, denoted as $S$, is smaller than the other two product... | 1. First, we express the product of all nine numbers using its prime factorization:
$$
1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9=2^{7} \cdot 3^{4} \cdot 5 \cdot 7
$$
a) Let's denote two of the considered (different) products as $S$ and $Q$, with $S<Q$. From the equality
$$
S \cdot S \cdot Q=2... | 70 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Given arithmetic sequences $\left(x_{i}\right)_{i=1}^{\infty}$ and $\left(y_{i}\right)_{i=1}^{\infty}$ have the same first term and the following property: there exists an index $k(k>1)$, for which the equalities
$$
x_{k}^{2}-y_{k}^{2}=53, \quad x_{k-1}^{2}-y_{k-1}^{2}=78, \quad x_{k+1}^{2}-y_{k+1}^{2}=27 .
$$
hol... | SOLUTION. Let $c, d$ be the differences of the first and second given arithmetic sequences, respectively. Since according to the problem statement $y_{1}=x_{1}$, the terms of both sequences have the general form
$$
x_{i}=x_{1}+(i-1) c \quad \text { and } \quad y_{i}=x_{1}+(i-1) d
$$
for any index $i$. The difference ... | 54 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. From the set $\{1,2,3, \ldots, 99\}$, select the largest number of elements such that the sum of no two selected numbers is a multiple of eleven. (Explain why the chosen selection has the required property and why no selection with a larger number of elements satisfies the condition.) | SOLUTION. We will divide the numbers from 1 to 99 into eleven groups $T_{0}, T_{1}, \ldots, T_{10}$ based on their remainder when divided by 11:
$$
\begin{aligned}
T_{0}= & \{11,22,33, \ldots, 99\}, \\
T_{1}= & \{1,12,23, \ldots, 89\}, \\
T_{2}= & \{2,13,24, \ldots, 90\}, \\
& \vdots \\
T_{10}= & \{10,21,32, \ldots, 9... | 46 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Two athletes are running around a track, each at a constant speed. If they run in opposite directions, they meet every 10 minutes; if they run in the same direction, they meet every 40 minutes. How long does it take the faster athlete to run the circuit? | 1. Let the speeds of the runners be $v_{1}$ and $v_{2}$ such that $v_{1}>v_{2}$ (speeds are given in laps per minute). Suppose the athletes start from the same point but in opposite directions. At the moment of their next meeting after 10 minutes, the sum of the lengths of the two segments they have run will exactly co... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. From a piece of paper, an isosceles trapezoid $C_{1} A B_{2} C_{2}$ with the shorter base $B_{2} C_{2}$ was cut out. The foot of the perpendicular from the midpoint $D$ of the leg $C_{1} C_{2}$ to the base $A C_{1}$ is denoted as $B_{1}$. After folding the paper along the segments $D B_{1}, A D$ and $A C_{2}$, the p... | Solution. From the equality of segments in the described network that correspond to the same edges of the resulting tetrahedron $ABCD$, we get that $\left|AB_{1}\right|=\left|AB_{2}\right|=|AB|=b, \left|B_{1}C_{1}\right|=\left|B_{2}C_{2}\right|=|BC|=c$. Let $S$ be the midpoint of segment $AB_{1}$ and $B_{3}$ the foot o... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Determine the values that the expression
$$
\frac{a+b c}{a+b}+\frac{b+c a}{b+c}+\frac{c+a b}{c+a}
$$
can take if \( a \), \( b \), and \( c \) are positive real numbers with a sum of 1. (Michal Rolínek, Pavel Calábek) | SOLUTION. The fractions in the given expression make sense because their denominators are positive numbers according to the problem statement. Thanks to the condition $a+b+c=1$, for the first fraction we have
$$
\frac{a+b c}{a+b}=\frac{(a+b)+(b c-b)}{a+b}=1-b \cdot \frac{1-c}{a+b}=1-b \cdot \frac{a+b}{a+b}=1-b.
$$
Si... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. On the board, there are several natural numbers from 1 to 100, none of which is divisible by a two-digit prime number, and the product of any two of them is not a square of a natural number.
(a) Determine the largest possible number of numbers on the board.
(b) Determine the largest possible sum of the numbers on ... | SOLUTION. Each number written on the board has a unique prime factorization $n=2^{a} \cdot 3^{b} \cdot 5^{c} \cdot 7^{d}$, where $a, b, c, d$ are suitable non-negative integers. According to the parity of the numbers in the quadruple $(a, b, c, d)$, we introduce the "type" of the number $n$ as a quadruple $(A, B, C, D)... | 979 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. a) Decide whether there exists a natural number $n$ such that $2n$ is the square of a natural number and $3n$ is the cube of a natural number.
b) Decide whether there exists a natural number $n$ that satisfies both conditions from part a) and additionally $4n$ is the fourth power of a natural number.
(Josef Tkadle... | SOLUTION. a) Such a number exists. For example, the number $n=72$ satisfies both conditions, since for it $2 n=144=12^{2}$ and $3 n=216=6^{3}$.
However, let's also see how we arrive at (the smallest suitable) number 72. Consider the prime factorization of the sought number $n$ in the form
$$
n=2^{a_{2}} \cdot 3^{a_{3... | 72 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Let a regular n-gon be denoted as $A_{1} A_{2} \ldots A_{n}$. For which $n \geqq 5$ does the image of point $A_{3}$ under the axial symmetry with respect to the line $A_{1} A_{2}$ lie on the line $A_{4} A_{5}$? (Josef Tkadlec) | SOLUTION. Let $S$ be the center of the regular $n$-gon $A_{1} A_{2} \ldots A_{n}$ and $P$ the intersection of the rays $A_{1} A_{2}$ and $A_{5} A_{4}$. The point $P$ will exist for every $n>6$, the remaining cases $n=5$ and $n=6$ will be discussed at the end of the solution. Now, let us assume that $n>6$.
 | SOLUTION. Let's first consider what the largest possible sum in a row of the table could be if it is composed of the numbers $-4, 3$, and 10 and must be at most 0. Notice that all three numbers -4, 3, and 10 give a remainder of 3 when divided by seven. Therefore, the sum of ten numbers in the same row will give the sam... | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Determine the greatest number of consecutive five-digit natural numbers, among which there is no palindrome, i.e., a number that reads the same forwards and backwards. | Among 109 consecutive five-digit numbers
$$
10902,10903, \ldots, 10999,11000, \ldots, 11009,11010
$$
there is no palindrome (it is possible to provide other suitable examples of 109 five-digit numbers, we have listed the smallest group of them).
The smallest and largest five-digit palindromes are the numbers 10001 a... | 109 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In the plane, there is a right-angled triangle $ABC$, on whose hypotenuse $AB$ we consider an arbitrary point $K$. The circle constructed over the segment $CK$ as a diameter intersects the legs $BC$ and $CA$ at internal points, which we denote by $L$ and $M$ respectively. Determine for which point $K$ the quadrilate... | 2. Since the angles $KLC$, $KMC$, and $LCM$ are right angles (Fig. 1), the quadrilateral $KLCM$ is a rectangle and triangles $AKM$ and $KBL$ are similar to triangle $ABC$. Let us denote
Fig.... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Consider the 2022 fractions
$$
\frac{0}{2022}, \frac{1}{2021}, \frac{2}{2020}, \ldots, \frac{2021}{1}
$$
in the form of the ratio of two non-negative integers, whose sum for each fraction is equal to 2022. How many of them take integer values?
(Jaroslav Zhouf) | SOLUTION. The denominators of the given fractions are natural numbers from 1 to 2022. The numerator $c$ of the fraction with a given denominator $j$ is determined by the equation $c+j=2022$, i.e., $c=2022-j$. Therefore, our fractions can be expressed, and we will immediately simplify this expression:
$$
\frac{2022-j}{... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Šebestová has an average grade of exactly 1.12 from her five-minute tests. Prove that she has at least 22 ones. (Possible grades are 1, 2, 3, 4, 5.) (Josef Tkadlec) | SOLUTION. First, we explain why more than half of Šebestová's grades are ones. We start from the obvious observation that if we replace a grade with a better one, the average of the grades will also improve. Therefore, if ones were at most half, then by replacing grades $3, 4, 5$ (if any exist) with better twos, we wou... | 22 | Number Theory | proof | Yes | Yes | olympiads | false |
6. Determine the largest natural number $n \geqq 10$ such that for any 10 different numbers $z$ from the set $\{1,2, \ldots, n\}$, the following statement holds: If none of these 10 numbers is a prime number, then the sum of some two of them is a prime number.
(Ján Mazák) | SOLUTION. We will show that the sought maximum $n$ is equal to 21.
First, we verify that no natural number $n \geq 22$ has the required property. For each such $n$, the set $\{1,2, \ldots, n\}$ contains even numbers $4,6,8, \ldots, 22$, which are not prime numbers (we excluded the smallest even number 2) and the requi... | 21 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the smallest possible value of the expression
$$
3 x^{2}-12 x y+y^{4} \text {, }
$$
where $x$ and $y$ are any non-negative integers. | 1. Let's denote the given expression by $V$ and transform it using the method called completing the square twice:
$$
V=3 x^{2}-12 x y+y^{4}=3(x-2 y)^{2}-12 y^{2}+y^{4}=3(x-2 y)^{2}+\left(y^{2}-6\right)^{2}-36 .
$$
Clearly, $(x-2 y)^{2} \geq 0$, so the minimum value of the expression $V$ for a fixed $y$ is obtained wh... | -32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In the real domain, consider the system of equations
$$
\begin{aligned}
& x^{4}+y^{2}=\left(a+\frac{1}{a}\right)^{3}, \\
& x^{4}-y^{2}=\left(a-\frac{1}{a}\right)^{3}
\end{aligned}
$$
with a non-zero real parameter $a$.
a) Find all values of $a$ for which the given system has a solution.
b) Prove that for any sol... | SOLUTION. We solve the system as a linear system of equations with unknowns $x^{4}$ and $y^{2}$. By adding both equations and dividing by two, we get
$$
x^{4}=\frac{1}{2}\left(\left(a+\frac{1}{a}\right)^{3}+\left(a-\frac{1}{a}\right)^{3}\right)=a^{3}+\frac{3}{a} .
$$
Similarly, by subtracting the second equation from... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
6. The figure of the rook threatens any square on the diagonal on which the rook stands on the chessboard. However, if there is a tower on some square of the diagonal, the rook no longer threatens the squares behind it. Determine the maximum number of rooks that can be placed together with four towers on an $8 \times 8... | SOLUTION. Each white field lies on one of the seven diagonals parallel to the main white diagonal and marked on the left image with arrows. Similarly, black fields lie on one of the seven diagonals parallel to the main black diagonal. On each of these diagonals without rooks, there can stand at most one archer. Therefo... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In how many ways can the set $\{1,2, \ldots, 12\}$ be divided into six disjoint two-element subsets such that each of them contains coprime numbers (i.e., numbers that do not have a common divisor greater than one)? | 3. No two even numbers can be in the same of the six two-element subsets (let's call them "pairs"), and we can thus limit ourselves to such partitions of the set $\{1,2, \ldots, 12\}$ (let's call them even-odd), in which pairs are always composed of one even and one odd number. Another restriction is that the numbers 6... | 252 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. A village has 100 inhabitants. We know that each of them has exactly three acquaintances in the village. (Acquaintance is a mutual relationship.)
a) Prove that there exists a group of 25 people in the village, none of whom know each other.
b) Find the smallest natural number $n$ with the property that in any group... | 2. a) Assume there exists a group $N$ of $k$ people, in which there is no pair of acquaintances. (Certainly, such a group exists for $k=1$.) We place all people who have at least one acquaintance in group $N$ into group $O$. In group $O$, there are at most $3k$ people. Therefore, the total number of people in both grou... | 51 | Combinatorics | proof | Yes | Yes | olympiads | false |
4. Let M be a set of six distinct positive integers whose sum is 60. We will write all of them on the faces of a cube, with exactly one on each face. In one step, we choose any three faces of the cube that share a common vertex and increase each of the numbers on these three faces by 1. Determine the number of all such... | 4. Let the faces of the cube be denoted by $S_{1}, S_{2}, \ldots, S_{6}$ such that face $S_{1}$ is opposite to face $S_{6}$, face $S_{2}$ is opposite to $S_{5}$, and $S_{3}$ is opposite to $S_{4}$. Let the number on face $S_{i}$ be denoted by $c_{i}$. Clearly, any vertex of the cube belongs to exactly one pair of oppos... | 84 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. On the board, the numbers $1,2, \ldots, 33$ are written. In one step, we choose several numbers written on the board (at least two), whose product is a square of a natural number, erase the chosen numbers, and write the square root of their product on the board. We continue this process until only such numbers remai... | 3. The product of all numbers written on the board is equal to
$$
S=2^{31} \cdot 3^{15} \cdot 5^{7} \cdot 7^{4} \cdot 11^{3} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 .
$$
The presence of odd exponents means that $S$ is not a perfect square. Therefore, we cannot erase all the numbers in the first step... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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