problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
4. In each cell of the square field below, a building of height 1, 2, 3, 4, or 5 is placed so that the following rules are met:
- In each (horizontal) row and (vertical) column, each building height appears exactly once.
- The numbers on the sides of the square are the sums of the heights of the visible buildings. The... | A4. B) 2 First note that the building of 5 stories is always visible. We first look at the middle column, where the 6 is written. It must hold that $6=1+5$: the first building you see is 1 story high, and behind it you see the building of 5 stories. No building can stand between them, because otherwise we would also se... | 2 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
7. On the board are the integers from 1 to $n$. One of the numbers is erased. The average of the remaining numbers is $11 \frac{1}{4}$.
Which number was erased?
A) 6
B) 7
C) 11
D) 12
E) 21 | A7. A) 6 The average of the numbers from 1 to $n$ is exactly $\frac{n+1}{2}$. We can divide the numbers into pairs, with one number remaining if $n$ is odd: 1 paired with $n$, 2 paired with $n-1$, and so on. The average of each pair is then $\frac{n+1}{2}$.
If we remove a number, the smallest possible average that rem... | 6 | Algebra | MCQ | Yes | Yes | olympiads | false |
A8. Consider all four-digit numbers in which each of the digits $3,4,6$ and 7 appears exactly once. How many of these numbers are divisible by 44?
(A) 2
(B) 4
(C) 6
(D) 8
(E) 12
B-questions | A8. Consider all four-digit numbers in which each of the digits 3, 4, 6, and 7 appears exactly once. How many of these numbers are divisible by 44? (A) 2
Solution Let the number $n$ consisting of the digits $a, b, c$, and $d (n=1000a+100b+10c+d)$ satisfy the condition, so it is divisible by 44. It must then certainly ... | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
B2. How many positive integers $a$ are there, for which it is true that: when you divide 2216 by $a$, the remainder is 29. | B2. 4 Dat 2216 bij deling door $a$ rest 29 geeft, betekent precies dat $2216-29=2187$ deelbaar is door $a$ en dat $a$ groter is dan 29 (de rest is altijd kleiner dan de deler $a$). The divisors of $2187=3^{7}$ that are greater than 29 are 81, 243, 729, and 2187. There are thus 4 possibilities in total. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A5. If the number $\left(\left(\left(\left(7^{6}\right)^{5}\right)^{4}\right)^{3}\right)^{2}$ is written out, what is the last digit?
A) 1
B) 3
C) 5
D) 7
E) 9 | A5. A) 1 The product of two numbers that end in 1 also ends in 1. Since the last digit of $7^{4}=2401$ is 1, this also holds for any power of $7^{4}$. In particular, the last digit of $\left.\left(\left(\left(7^{6}\right)^{5}\right)^{4}\right)^{3}\right)^{2}=7^{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}=\left(7^{4}\right)^{180... | 1 | Number Theory | MCQ | Yes | Yes | olympiads | false |
A6. Calculate $\left((\sqrt{2}+1)^{7}+(\sqrt{2}-1)^{7}\right)^{2}-\left((\sqrt{2}+1)^{7}-(\sqrt{2}-1)^{7}\right)^{2}$.
A) 2
B) 4
C) $8 \sqrt{2}$
D) 128
E) 512 | A6. B) 4 Given $a=(\sqrt{2}+1)^{7}$ and $b=(\sqrt{2}-1)^{7}$. Then the given expression is equal to
$$
(a+b)^{2}-(a-b)^{2}=4 a b=4(\sqrt{2}+1)^{7}(\sqrt{2}-1)^{7}=4((\sqrt{2}+1)(\sqrt{2}-1))^{7}=4 \cdot 1^{7}=4
$$ | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
A1. In the places of the asterisks, there are positive integers arranged in such a way that the multiplication table below is correct.
What is the largest number that appears more than once in the $5 \times 5$ table?
A) 6
B) 8
C) 9
D) 12
E) 18
| $\times$ | $*$ | $*$ | $*$ | 7 |
| :---: | :---: | :---: | :---: | :---:... | A1. D) 12 Look at the left picture. At the place of the star, a number must stand that both 27 and 6 are divisible by. This gives two possibilities: 1 or 3. The first possibility is ruled out, because then the number 27 would stand at the place of the double star, while 36 is not divisible by it.
| $\times$ | | $* *$... | 12 | Algebra | MCQ | Yes | Yes | olympiads | false |
A6. On side $C D$ of a square $A B C D$ lies a point $E$. Line segment $A E$ is divided into four equal parts by the points $P, R$ and $T$. Line segment $B E$ is divided into four equal parts by the points $Q, S$ and $U$. Given that $|P Q|=3$.
What is the area of quadrilateral $P Q U T$?
A) $\frac{15}{4}$
B) 4
C) $\fr... | A6. B) 4 Triangle $E A B$, triangle $E P Q$, and triangle $E T U$ are similar because $|A E|:|P E|:|T E|=4: 3: 1=|B E|:|Q E|:|U E|$ and $\angle A E B=\angle P E Q=\angle T E U$ (SAS). This implies that $|A B|:|P Q|:|T U|=$ $4: 3: 1$. From this, it follows that $|A B|=4$, so the area of triangle $A B E$ is $\frac{1}{2} ... | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
A7. Carry has six cards. On each card, a positive integer is written. She chooses three cards and calculates the sum of the numbers on those cards. She does this for all 20 possible combinations of three cards. Ten times Carry gets a result of 16 and ten times a result of 18. What is the smallest number that appears on... | A7. D) 4 Because there are only two different outcomes, only two different numbers can appear on the cards. Indeed, if three cards bear different numbers, then in combination with a pair of the remaining three cards, they would each yield a different result. Let the two numbers that appear be $a$ and $b$. We can assume... | 4 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
1. Frank has two integers that add up to 26. Kees adds two more integers and gets 41. Pieter adds two more integers and gets 58. At least how many of the six added numbers are even?
A) 0
B) 1
C) 2
D) 3
E) 4 | A1. C) 2 The two numbers Frank has are either both even or both odd; otherwise, their sum would not be even. The two numbers Kees has add up to $41-26=15$, so one of his numbers must be even and the other odd, because otherwise the sum would not be odd. Pieter's numbers add up to $58-41=17$, so one of his numbers must ... | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
3. A positive integer is called fully divisible if it is divisible by each of its digits. These digits must also all be different (and not equal to 0). For example, 162 is fully divisible, because it is divisible by 1, 6, and 2.
How many fully divisible two-digit numbers are there?
A) 4
B) 5
C) 6
D) 7
E) 8 | A3. B) 5
First, look at the numbers that start with the digit 1. A number is always divisible by 1, so this condition is always met. Now we will check which of the numbers 12, 13, 14, 15, 16, 17, 18, and 19 are also divisible by their last digit, and we see that 12 and 15 are.
Now look at the numbers starting with 2.... | 5 | Number Theory | MCQ | Yes | Yes | olympiads | false |
7. A square board is divided into $4 \times 4$ squares. Initially, all the squares are white. We want to color as many squares blue as possible, in such a way that each blue square has exactly one white neighbor (two squares are neighbors if they share a side). What is the maximum number of squares we can color blue?
A... | A7. D) 12 A filling with 12 blue squares where each blue square has exactly one white neighboring square can be seen in the figure. A coloring with more than 12 blue squares does not exist. With a coloring of 13 or more blue squares, there are at most 3 white squares. These white squares together have no more than $3 \... | 12 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
A1. We have a field of $4 \times 4$ square cells. Of these 16 cells, we want to color exactly four black. This must be done in such a way that each row and each column gets exactly one black cell, and no two black cells are diagonally (with a corner) adjacent.
In how many ways can we choose the four cells to color?
A)... | A1. B) 2 Suppose we color field B2 black. Then the 8 surrounding fields may not be colored black anymore: the fields above, below, to the left, and to the right of B2 are in the same column or row as B2, and the other four fields touch B2 diagonally. This leaves only row 4 and column D, and in each, we can only color o... | 2 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
A4. A square paper ring has height 1. The sides have length 4. The ring is depicted in the left figure. By pressing it flat on a table, we get the right figure. In this figure, quadrilateral $A B C D$ is a square.
 3 We look at the top edge of the paper ring. This edge has a length of $4 \times 4=16$. In the folded figure, the top edge follows the rectangle $E F G H$. Since $|A E|=|B F|=|C G|=$ $|D H|=1$, we see that $|A B|+|F G|+|C D|+|E H|=16-4=12$. These four lengths are equal to the lengths of the sides of the square $... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
B4. We consider $5 \times 5$ tables with 25 numbers written in them. The same number can appear multiple times, but in no row or column does the same number appear five times. We call such a table beautiful if in each row the middle number is the average of the numbers in that row, and in each column the middle number ... | B4. 3
We first show that every beautiful table has a score of at least 3. Consider such a table and let the number exactly in the middle be $a$. The five numbers in the middle row have $a$ as their average and are not all equal to $a$. At least one of these five numbers must therefore be smaller than $a$. Similarly, (... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A8. A motorboat moves relative to the water at a constant speed of 25 kilometers per hour. It sails from Arnhem to Zwolle with the constant current. At a certain moment, it has covered $42\%$ of the distance. From that point, it takes the same amount of time to continue sailing to Zwolle as it does to sail back to Arnh... | A8. B) 4 From the mentioned point, we can travel $42\%$ of the distance upstream and $58\%$ of the distance downstream in the same time. This means that the boat goes $\frac{58}{42}$ times faster with the current than against it. If we denote the current speed as $v$, we then find that $\frac{25+v}{25-v}=\frac{58}{42}$... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
A1. A traffic light alternates between green and red at regular intervals. The green and red periods last the same amount of time, always 1, 2, or 3 minutes. There are four color combinations for the light at the times 12:08 and 12:09: red-red, red-green, green-red, and green-green. How many of the four combinations ar... | A1. B) 2
Given that the traffic light is red at 12:05. For a traffic light of period 1, the colors for the times 12:05 to 12:12 are fixed, alternating between red and green. For period 2, there are two possibilities, and for period 3, there are three possibilities:
| period | 12:05 | 12:06 | 12:07 | 12:08 | 12:09 | $... | 2 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
A4. Nine lamps are arranged in a square. Each lamp can be on or off. If you press a lamp, that lamp and the lamps in the same row or column change state: from on to off or vice versa. Initially, all lamps are on.
What is the smallest number of presses needed to turn all the lamps off?
 3 The order in which the lights are pressed does not matter for the final result. By pressing the three lights in the top row, all lights change from on to off. The lights in the top row change state three times, and the other lights change state exactly once.
It is not possible to turn off all the lights by pr... | 3 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
A2. In a magic square, the three row sums, the three column sums, and the two diagonal sums are equal to each other. (A row sum is the sum of the numbers in a row, etc.) For the $3 \times 3$ magic square depicted here, three numbers have been filled in.
Which number must be placed in the position of the question mark?... | A2. B) $4 \quad$ See figure. From $F+10+3=F+D+7$ it follows that $D=6$. From $7+E+3=C+D+E=C+6+E$ it follows that $C=7+3-6=4$. | 4 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
B2. From a set of 50 different numbers from the collection $\{1,2,3, \ldots, 100\}$, the sum is equal to 2900. What is the smallest possible number of even numbers among these 50 numbers? | B2. 6 If you take all 50 odd numbers from the set $\{1,2,3, \ldots, 100\}$, their sum is $\frac{1}{2} \times 50 \times(1+99)=2500$. That is 400 too little. Therefore, replace the smallest odd numbers with the largest possible even numbers, each time in pairs because 400 is even. If we replace 1 and 3 with 100 and 98, w... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A8. How many different (real) solutions does the equation $\left(\left(x^{2}-2\right)^{2}-5\right)^{2}=1 \quad$ have?
A) 4
B) 5
C) 6
D) 7
E) 8
## B-questions | A8. B) 5 The equation is equivalent to
$$
\left(x^{2}-2\right)^{2}-5=1 \quad \text { or } \quad\left(x^{2}-2\right)^{2}-5=-1 .
$$
The first part is equivalent to $x^{2}-2=\sqrt{6}$ or $x^{2}-2=-\sqrt{6}$, with 2 and 0 solutions respectively (the latter due to $-\sqrt{6}+2<0$).
The second part is equivalent to $x^{2}... | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
A1. The fields of a $4 \times 4$ board are colored white or black. Next to each row and below each column, it is indicated how many fields in that row or column must be black.
In how many ways can the board be colored?
A) 0
B) 1
C) 4
D) 5
E) 8
 5 Note first that all boxes in the second row and in the second column must be colored white. We consider two cases, depending on the color of the box in the top left corner.
If this box is white, the last two boxes in the first row and column must be black. This fixes the coloring. See the top figure.
If this... | 5 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
B1. For the number $x$, it holds that: $x=\frac{1}{1+x}$. Calculate $x-\frac{1}{x}$. Simplify your answer as much as possible. | B1. -1 Given that $x=\frac{1}{1+x}$. We see that $x \neq 0$, because $0 \neq \frac{1}{1}$. We can therefore invert the fraction on both sides. This gives $\frac{1}{x}=1+x$. From this, it follows that $x-\frac{1}{x}=x-(1+x)=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A6. The number $a=11 \ldots 111$ consists of exactly 2011 ones.
What is the remainder of $a$ when divided by 37?
A) 0
B) 1
C) 3
D) 7
E) 11 | A6. B) 1 Door with a division of $a=11 \cdots 11$ (2011 ones) to be divided by 37, it quickly becomes apparent that 111 is divisible by 37. We will use this fact. We now see that the number $1110 \cdots 0$ is divisible by 37, regardless of the number of zeros at the end. In particular
 5
B) 6
C) $\frac{15}{2}$
D) 8
E) 12
 5 If you add the perimeters of the two rectangles, you get the perimeter of the square plus twice the length of the line segment in the middle. That line segment in the middle is exactly as long as the side of the square. The sum of the perimeters of the two rectangles is therefore equal to six times the length ... | 5 | Geometry | MCQ | Yes | Yes | olympiads | false |
A5. Jan has a wooden cube. Each of the faces he divides into a $2 \times 2$ grid of squares, which he then paints in a black-and-white pattern: two diagonally adjacent squares are black, and the other two are white. At each corner of the cube, three squares come together. If two or three of these are black, we call the... | A5. C) 2 On the cube, after distribution, there are a total of 24 squares: 12 white and 12 black. The number of dark vertices cannot be zero, because then the number of black squares would not exceed \(8 \times 1=8\) (one per vertex). The number of dark vertices also cannot be one, because then the number of black squa... | 2 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
A6. In how many ways can you make the number 100 by choosing a sequence of consecutive numbers from 1 to 99 and adding them together?
A) 1
B) 2
C) 3
D) 4
E) 5 | A6. B) 2
Let the number of numbers we add together be $n$. There are two cases.
$n$ is odd In this case, there is a middle number. Let this number be $k$. The sum of the $n$ numbers is then $n \times k=100$. Since $n$ is an odd divisor of 100 (and greater than 1), $n$ must be 5 or 25. In the first case, we find $k=\f... | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
3. In a criminal investigation, five suspects have been arrested. They all make one statement:
Eva: "We are all innocent."
Fatima: "Exactly one of us is innocent."
Kees: "Exactly one of us is guilty."
Manon: "At least two of us are innocent."
Mustafa: "At least two of us are guilty."
It turns out that only the gu... | A3. C) 3 At most one of the statements made by Eva, Fatima, and Kees can be true. Therefore, at least two of them are lying and are guilty. From this, we can already conclude that Eva and Kees are lying and thus guilty. Therefore, Mustafa is telling the truth and is innocent.
Fatima cannot be telling the truth, becaus... | 3 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
6. A $100 \times 100$ board is filled with numbers. In the bottom-left cell, there is a 0. From any other cell $V$, we consider a path from the bottom-left cell to cell $V$, where you can only move right or up, not diagonally. If you add up all the numbers you encounter along the way and then add 1 for each step you ha... | A6. $\quad$ B) 3 We consider a route from the bottom-left square to the top-right square. This route consists of $99+99=198$ steps. The numbers on the squares along this route we call $a_{0}, a_{1}, a_{2}, \ldots, a_{198}$.
We know that $a_{0}=0, a_{1}=1$ and $a_{2}=3$. To calculate the next number, $a_{3}$, we need t... | 3 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
## Problem 3.
Let $a_{n}=1+n^{3}$ be the sequence $\{2,9,28,65, \ldots\}$ and $\delta_{n}=\operatorname{gcd}\left(a_{n+1}, a_{n}\right)$. Find the maximum value that $\delta_{n}$ can take. | Solution: $\delta_{n}$ divides $a_{n+1}$ and $a_{n}$, and therefore their difference $b_{n}=a_{n+1}-a_{n}=3 n^{2}+3 n+1$.
It also divides $c_{n}=3 a_{n}-n b_{n}=3-n-3 n^{2}$ and the sum $d_{n}=b_{n}+c_{n}=4+2 n$. But then $\delta_{n}$ also divides $e_{n}=2 b_{n}-3 n d_{n}=2-6 n$. Finally, it divides $3 d_{n}+e_{n}=14$... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Let $x, y, z$ be positive real numbers such that $x+y+z=3$. Find the maximum value reached by
$$
\sqrt{x}+\sqrt{2 y+2}+\sqrt{3 z+6}
$$
For what values of $x, y, z$ is this maximum achieved? | Solution 1. Consider the vectors $(\sqrt{x}, \sqrt{y+1}, \sqrt{z+2})$ and $(\sqrt{1}, \sqrt{2}, \sqrt{3})$, whose coordinates are all real and positive, whose respective magnitudes are $\sqrt{x+y+z+3}=$ $\sqrt{6}$ and $\sqrt{1+2+3}=\sqrt{6}$, and whose dot product is the expression whose maximum is to be found. By the ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem 6.
It is known that the polynomial $\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-\mathrm{x}+\mathrm{k}$ has three roots that are integers.
Determine the number k. | For $\mathrm{k}=0$ we have $\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-\mathrm{x}=\mathrm{x}(\mathrm{x}-1)(\mathrm{x}+1)$, which has roots $0, -1 \text{ and } 1$.
It is shown that this is the only value of $k$ for which $\mathrm{p}(\mathrm{x})$ has three integer roots. Indeed, if $a, b, c$ are integers, and $p(x)=(x-a)(x-b... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. We have 50 chips numbered from 1 to 50, and they need to be colored red or blue. We know that chip 5 is blue. For the coloring of the rest of the chips, the following rules are followed:
a) If the chip with the number $x$ and the chip with the number $y$ are of different colors, then the chip with the number $|x-y|... | Solution. Let's observe that two numbers differing by 5 have the same color. Indeed, if they were of different colors, their difference should be red, according to rule a). But their difference is 5, which is blue. Therefore, it is enough to know the color of the first 4 numbers. Here, we distinguish two cases: 1) the ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5 Solve the exponential equation
$$
2^{x} \cdot 3^{5^{-x}}+\frac{3^{5^{x}}}{2^{x}}=6
$$ | Solution. Applying the arithmetic and geometric means inequality and, subsequently, one of its most well-known consequences (the sum of a positive real number and its inverse is always greater than or equal to 2, and equality only holds for the number 1), we have,
$$
6=2^{x} 3^{5^{-x}}+2^{-x} 3^{5^{x}} \geq 2 \sqrt{2^... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In a triangle $ABC$, the angle bisector from $A$, the median from $B$, and the altitude from $C$ are concurrent, and moreover, the angle bisector from $A$ and the median from $B$ are perpendicular. If the side $AB$ measures one unit, find the lengths of the other two sides. | Solution. We provide two different solutions.
Solution 1. Let $P, M$, and $Q$ be the feet of the angle bisector from $A$, the median from $B$, and the altitude from $C$, respectively, which intersect at point $X$. In triangle $ABM$, the angle bisector from $A, AX$, is perpendicular to $BM$ (since by hypothesis the med... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem II - 5
The natural numbers 22, 23, and 24 have the following property: the exponents of the prime factors in their factorization are all odd:
$$
22=2^{1} \cdot 11^{1} ; \quad 23=23^{1} ; \quad 24=2^{3} \cdot 3^{1}
$$
What is the largest number of consecutive natural numbers that can have this property? Ju... | ## Solution
We will prove that it is impossible to find 8 consecutive numbers with this property. Suppose, for the sake of contradiction, that such 8 consecutive numbers exist. One of them, which we will call $n$, is divisible by 8.
Among the 8 numbers, there must be either $n+4$ or $n-4$. Both are divisible by 4, bu... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Let $p \geq 3$ be a prime number, and consider the right triangle with the larger leg $p^{2}-1$ and the smaller leg $2p$. We inscribe a semicircle in the triangle whose diameter lies on the larger leg of the triangle and is tangent to the hypotenuse and the smaller leg of the triangle. Find the values of $p$ for whi... | Solution. In the right triangle $ABC$, we consider $\overline{AB}=p^{2}-1$, $\overline{AC}=2p$.
By the Pythagorean Theorem, we have $\overline{BC}^{2}=\overline{AC}^{2}+\overline{AB}^{2}$, s... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Calculate the maximum number of distinct real roots that a polynomial $P$ can have, which satisfies the following property: the product of two distinct roots of $P$ is still a root of $P$. | Solution. The answer is 4. The polynomial $x(x-1 / 2)(x-1)(x-2)$, with roots 0, 1/2, 1, and 2, satisfies the given bound. Suppose there was a polynomial with at least 5 distinct roots. It is important to note that a polynomial has a finite number of roots, which allows us to take maximums and minimums.
If a polynomial... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem 6
Find all positive integers $n$ such that $3^{n}+5^{n}$ is a multiple of $3^{n-1}+5^{n-1}$. | ## Solution:
For such an $n$, since
$$
3\left(3^{n-1}+5^{n-1}\right)<3^{n}+5^{n}<5\left(3^{n-1}+5^{n-1}\right)
$$
it follows that
$$
3^{n}+5^{n}=4\left(3^{n-1}+5^{n-1}\right)
$$
which reduces to
$$
5^{n-1}=3^{n-1}
$$
implying $n=1$.
Since $n=1$ is a solution (because 8 is a multiple of 2), it is concluded that ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Determine all positive integers $x$, such that $2x+1$ is a perfect square, but among the numbers $2x+2, 2x+3, \cdots, 3x+2$, there is no perfect square. | Solution. Let $n$ be an integer such that $2 x+1=n^{2}$ and $n^{2} \leq 3 x+2<(n+1)^{2}$. From the first equation, we get $x=\left(n^{2}-1\right) / 2$, and substituting this value into the double inequality, we obtain
$$
n^{2} \leq \frac{3 n^{2}+1}{2}<n^{2}+2 n+1 \Leftrightarrow 2 n^{2} \leq 3 n^{2}+1<2 n^{2}+4 n+2
$$... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## - Problem 1
A pucelana sequence is a strictly increasing sequence of sixteen consecutive positive odd numbers, whose sum is a perfect cube. How many pucelana sequences consist only of three-digit numbers? | ## Solution:
Let the sequence be $n, n+2, \ldots, n+30$. Then the sum is $\frac{1}{2} \cdot 16(2 n+30)=8(2 n+30)$. Therefore, it is necessary that $2 n+30$ be a perfect cube. Now we need to count the number of such $n$ that are odd and satisfy $101 \leq n \leq 969$. The even cubes between 232 and 1968 are 512, 1000, a... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Each rational number is painted with one of two colors, white and red. Such a coloring is called sanferminera when for any two rational numbers $x, y$, with $x \neq y$, if one of the following three conditions is met:
a) $x y=1$,
b) $x+y=0$,
c) $x+y=1$,
then $x$ and $y$ are painted in different colors. How many san... | If a coloring is sanferminera, we can find another sanferminera coloring by simultaneously swapping the color of each rational, from red to white and from white to red; if in the initial coloring two rationals have different colors, they will also have different colors in the resulting one. Let's then find the number o... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Let $n$ be a natural number and a convex polygon with $n$ vertices. Each of the sides and diagonals of this polygon is colored either blue or red. Determine the smallest $n$ such that, for any such coloring, there exist three vertices of the observed polygon that are mutually connected by segments of the same color. | 5. For $n=4$, we color the sides in one color, for example, blue, and the diagonals in another - red. It is then obvious that there does not exist a triangle whose vertices are the vertices of this quadrilateral, and all sides are colored in the same color.
\(=2022 \cdot(35-34+32-33) \quad 4\) POINTS
\(=2022 \cdot 0 \quad 1\) POINT
\(=0 \quad 1\) POINT
TOTAL 6 POINTS
``` | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. After conducting a survey among 96 sixth-grade students at his school, Leo drew a bar chart as shown in the image, but he did not highlight the numbers on the vertical axis or the names of the sports on the horizontal axis. He remembered that the number of students who swim is five times the number of students who p... | First solution.
From the condition of the problem that "five times more students swim than play tennis" and the ratio of the heights of the third and first bars, we conclude that the third ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. After conducting a survey among 96 sixth-grade students at his school, Leo drew a bar chart as shown in the image, but he did not highlight the numbers on the vertical axis or the names of the sports on the horizontal axis. He remembered that there are five times more students who swim than those who play tennis, an... | Another solution.
From the condition of the problem that "five times more students swim than play tennis" and the ratio of the height of the third and the first column, we conclude that the third column represents swimmers, and the first column represents tennis players.
Let the number of students who play tennis be ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. A polygon has 5 fewer sides than another polygon and 50 fewer diagonals than that other polygon. What is the polygon (the one with fewer sides)? | 4. Let $n$ be the number of sides (vertices) of the sought polygon.
From one vertex of this polygon, $n-3$ diagonals can be drawn,
and the total number of diagonals of this polygon is $\frac{n(n-3)}{2}$.
In a polygon with $n+5$ sides,
from one vertex, $n+5-3=n+2$ diagonals can be drawn,
and the total number of dia... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. The perimeter of a parallelogram is $30 \mathrm{~cm}$. The sum of the areas of the squares constructed on two adjacent sides is $113 \mathrm{~cm}^{2}$. What are the lengths of those sides?
The use of a pocket calculator or any reference materials is not allowed. | First method:
Sketch or description (Let $a$ be the length of the longer side, and $b$ be the length of the shorter side.)
Then we have $2a + 2b = 30$ or $a + b = 15$
and $a^2 + b^2 = 113$... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Determine the smallest possible value of the expression $4 x^{2}+4 x y+4 y^{2}+12 x+8$. For which $x$ and $y$ will this expression have the smallest value?
Each task is scored out of 10 points.
The use of a pocket calculator or any reference materials is not allowed. | 5. Let's rewrite the given expression in another form, using the formula for the square of a sum:
$$
\begin{aligned}
4 x^{2}+4 x y+4 y^{2}+12 x+8 & =x^{2}+4 x y+4 y^{2}+3 x^{2}+12 x+12-4 \\
& =(x+2 y)^{2}+3\left(x^{2}+4 x+4\right)-4 \\
& =(x+2 y)^{2}+3(x+2)^{2}-4
\end{aligned}
$$
Since the square of a number is alway... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In a certain store, 2400 kilograms of flour were ordered. They planned to repackage it into 5 kg bags and sell each bag for 14 kn. When the goods arrived, they noticed that 300 kg of flour had been damaged during transport. By how much should the price of the 5 kg packaging be increased so that the planned profit re... | 4. From the ordered quantity of flour, $2400: 5=480$ packages of $5 \mathrm{~kg}$ can be obtained.
1 POINT
The profit would be $480 \cdot 14=6720 \mathrm{kn}$.
1 POINT
$300 \mathrm{~kg}$ was destroyed, leaving $2100 \mathrm{~kg}$.
1 POINT
From this, $2100: 5=420$ packages of $5 \mathrm{~kg}$ can be made.
1 POINT... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A cat can catch 5 mice in 6 days. How many days will it take for 2 cats to catch 3 mice? | ## 2. 5 cats catch 5 mice in 6 days.
Five times fewer cats catch five times fewer mice in the same time:
1 cat catches 1 mouse in 6 days.
Twice as many cats catch twice as many mice in the same time:
2 cats catch 2 mice in 6 days.
The same number of cats catch half as many mice in half the time:
2 cats catch 1 mo... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The length of the side of the square is 1. Prove that for any line $p$ passing through the intersection of the diagonals, which does not pass through any vertex of the square and is not parallel to any of its sides, the following statement holds: "The sum of the squares of the distances from all four vertices of the... | 5.
Let point $S$ be the intersection of the diagonals of square $ABCD$. Through point $S$, any line $m$ is drawn. Let points $M, N, P,$ and $R$ be the feet of the perpendiculars drawn fro... | 1 | Geometry | proof | Yes | Yes | olympiads | false |
1. Andrija read a book over three days. While reading, his reading speed was constant. On the first day, he spent $\frac{1}{32}$ of a day reading, and each subsequent day he read for one-third less time than the previous day. What percentage of the book does Andrija still need to read if it takes him 1 hour and 40 minu... | 1. Andrija read a book over three days. While reading, his reading speed was constant. On the first day, he spent $\frac{1}{32}$ of a day reading, and each subsequent day he read for one-third less time than the previous day. What percentage of the book does Andrija still need to read if it takes him 1 hour and 40 minu... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The area of a rectangle is $\overline{3 x x 3}$, and the lengths of its adjacent sides are $\overline{x x}$ and $x^{2}$. Determine the digit $x$. | First solution.
The lengths of adjacent sides of the rectangle are $\overline{x x}$ and $x^{2}$, so $x \neq 0$, i.e., $x \in\{1,2, \ldots, 9\}$.
The area of the rectangle is $\overline{x x} \cdot x^{2}=\overline{3 x x 3}$.
1 POINT
The product $\overline{3 x x 3}$ is an odd number, so it is sufficient to consider $x... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In September 2014, when they started the first grade, all students in 5.a grade together had 153 years, and in September 2018, at the beginning of the fifth grade, those same students had a total of 245 years. How many of them started school at six, and how many at seven years old?
Each task is scored out of 10 poi... | First method:
From September 2014 to September 2018, 4 years have passed.
1 POINT
During this time, the total number of years of all students increased by
$245-153=92$ years.
2 POINTS
Since each student aged 4 years, there are
$92: 4=23$ students in the class.
2 POINTS
If all students started the first grade a... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Determine all natural numbers $n$ for which the number $\left|n^{2}-100\right|$ is prime.
## Problems worth 10 points: | 5. Let $m=\left|n^{2}-100\right|$.
We have $m=\left|n^{2}-100\right|=|(n-10)(n+10)|$.
The number $m$ is prime if and only if one of the numbers $|n-10|$ and $|n+10|$ is equal to 1, and the other is a prime number.
$1 \text{ POINT}$
Notice that there are no natural numbers $n$ for which $|n+10|=1$.
Let's find all n... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Ivana chose two numbers $a$ and $b$ from the set $\{1,2,3,4, \ldots, 25,26\}$. The product $a \cdot b$ is equal to the sum of the remaining 24 numbers in that set of numbers. What is the value of the expression $|a-b|$? | First method:
Assume $a \geq b$.
Since $a$ and $b$ are from the set $\{1,2,3,4, \ldots, 25,26\}$ and their product is equal to the sum of the remaining 24 numbers, we have $a+b+a \cdot b=1+2+3+4+\ldots+25+26$ or $a+b+a \cdot b=351$.
Adding 1 to both sides of the equation, we get $a+b+a \cdot b+1=352$.
By factoring ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. A circle with center $O$ and diameter $|A B|=4 \text{ cm}$ is given. Draw three tangents to this circle, two at points $A$ and $B$, and the third at a point such that the segment $\overline{C D}$ between the first two tangents is $5 \text{ cm}$ long. Explain and describe the procedure. | First method: Algebraic approach.
It is necessary to determine the distance of point $C$ from line $A B$.
Sketch:
Draw a parallel line with the diameter $\overline{A B}$ through $C$ and le... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. The boy thought of a number. He added 5 to that number, then divided the sum by 2, multiplied the quotient by 9, subtracted 6 from the product, divided the difference by 7, and got the number 3. What number did the boy think of? | 3. We calculate backwards:
$3 \cdot 7=21$
1 POINT
$21+6=27$ 1 POINT
$27: 9=3$ 1 POINT
$3 \cdot 2=6$ 1 POINT
$6-5=1$
1 POINT
The desired number is 1.
1 POINT
(Note: If the student has written that the solution is the number 1 and has only verified the correctness of this solution, they should receive 2 POINTS.) | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let the expressions $A$ and $B$ be given as:
$$
A=2 x+13+11 x-4-5 x+6 x+12 \text { and } \quad B=9+3 y+14 y-4-5 y+7-y+5
$$
If $x=6$, calculate $y$ so that the values of the expressions $A$ and $B$ are equal. | First method:
Simplify each expression:
| $A=2 x+13+11 x-4-5 x+6 x+12=14 x+21$ | 1 POINT |
| :--- | :--- |
| $B=9+3 y+14 y-4-5 y+7-y+5=11 y+17$ | 1 POINT |
| For $x=6, A=14 \cdot 6+21=84+21=105$. | 1 POINT |
| We are looking for $y$ for which the expressions $A$ and $B$ will be equal: | |
| $11 y+17=105$ | 1 POINT |... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. What is $3 x+2 y$, if $x^{2}+y^{2}+x y=28$ and $8 x-12 y+x y=80$?
The use of a pocket calculator or any manuals is not allowed. | 7. From the second equation, we have $x y=80-8 x+12 y$.
By substituting into the first equation, we get:
$x^{2}+y^{2}+80-8 x+12 y=28$, or $x^{2}+y^{2}-8 x+12 y+52=0$.
3 POINTS
Then we complete the square:
$x^{2}-8 x+16+y^{2}+12 y+36+52-16-36=0$
$(x-4)^{2}+(y+6)^{2}=0$
3 POINTS
The sum of two squares is zero if ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate $2015-5 \cdot(25 \cdot 13+13+3 \cdot 25-10)$. | First method:
\(2015-5 \cdot(25 \cdot 13+13+3 \cdot 25-10)=2015-5 \cdot(325+13+75-10)=\quad 1\) POINT
\(=2015-5 \cdot(338+75-10)=\quad 1\) POINT
\(=2015-5 \cdot(413-10)=\quad 1\) POINT
\(=2015-5 \cdot 403=\quad 1\) POINT
\(=2015-2015=\quad 1\) POINT
\(=0 \quad 1\) POINT
TOTAL 6 POINTS | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The annual set of the magazine Blue Swallow consists of 6 issues. Each issue does not have to have the same number of pages, but it is known that each has 40 or 44 pages. Can the annual set have a total of 260 pages? | First method:
If each number had 40 pages, the complete set would have $6 \cdot 40=240$ pages. 2 POINTS
To have a total of 260 pages in the set, we would need to add $260-240=20$ pages.
2 POINTS
Since $44-40=4$ and $20 \div 4=5$, the annual set will have 260 pages if 5 numbers have 44 pages each and 1 number has 40... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate: $1325: 5+15: 5-(99-9: 9) \cdot 2-35-30$. | 1. Calculate: $1325: 5+15: 5-(99-9: 9) \cdot 2-35-30$.
## Solution.
$$
\begin{aligned}
& 1325: 5+15: 5-(99-9: 9) \cdot 2-35-30 \\
& =265+3-(99-9: 9) \cdot 2-35-30 \quad 1 \text { POINT } \\
& =265+3-(99-1) \cdot 2-35-30 \quad 1 \text { POINT } \\
& =268-98 \cdot 2-35-30 \quad 1 \text { POINT } \\
& =268-196-35-30 \qu... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Four elephants and eight zebras eat a ton of food daily. An elephant eats $214 \mathrm{~kg}$ more food daily than a zebra. If a zebra needs 24 minutes to eat $1 \mathrm{~kg}$ of food, how much time does it need to eat its daily amount of food? Express the obtained time in hours and minutes. | 4. Four elephants and eight zebras eat a ton of food daily. An elephant eats $214 \mathrm{~kg}$ more food daily than a zebra. If a zebra needs 24 minutes to eat $1 \mathrm{~kg}$ of food, how much time does it need to eat its daily amount of food? Express the obtained time in hours and minutes.
## Solution.
An elephan... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In 6th grade class a, there are 24 students. Half of the students in this class play an instrument. Also, $\frac{5}{8}$ of the students in this class sing in the choir. If it is known that more than a third, but less than $\frac{5}{12}$ of the students in 6th grade class a both play an instrument and sing in the cho... | 2. In 6th grade, there are 24 students. Half of the students in this class play an instrument. Also, $\frac{5}{8}$ of the students in this class sing in the choir. If it is known that more than a third, but less than $\frac{5}{12}$ of the students in 6th grade class 6a both play an instrument and sing in the choir, how... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In a rectangular coordinate system in the plane, a point $S(2,3)$ and a square $A B C D$ are given, with three known vertices $A(3,1), C(-2,6)$, and $D(-2,1)$. The square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is the central symmetric image of the square $A B C D$ with respect to the point $S$. Determine the ... | 3. In a rectangular coordinate system in the plane, a point $S(2,3)$ and a square $A B C D$ are given, with three known vertices $A(3,1), C(-2,6)$, and $D(-2,1)$. The square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is the central symmetric image of the square $A B C D$ with respect to the point $S$. Determine the ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. At the entrance to the cinema hall, among 30 visitors, 58 candies were distributed. Each girl received six candies, each boy received four candies, and each adult received one candy. How many girls, how many boys, and how many adults could there be in the cinema hall?
## Ministry of Science and Education of the Rep... | First solution.
After each visitor eats 1 candy, there are $58-30=28$ candies left in the hall.
Then the adults do not have any candies.
\[\begin{array}{ll}\text { Each boy has } 3 \text { candies left, so the number of candies with boys is a multiple of 3. } & 1 \text { POINT } \\ \text { Each girl has } 5 \text { c... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. In the expression $2: 2: 2: 2: 2$ place parentheses so that the result is equal to 2. Find two different ways! | 1. $2:(2: 2):(2: 2)=2: 1: 1=2: 1=2$. 2 points
$2:(2: 2:(2: 2))=2:(2: 2: 1)=2:(1: 1)=2: 1=2$. 2 points
ILI
$2:(2:(2: 2): 2)=2:(2: 1: 2)=2:(2: 2)=2: 1=2$
.$U K U P N O 4$ POINTS | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Write all two-digit numbers that can be written using the digits 3, 4, and 9. How many such numbers are there? | 2. The sought numbers are $33,34,43,39,93,44,49,94,99$.
5 POINTS
There are 9 of these numbers.
Note: For writing 1 or 2 numbers, award 1 point, for 3 or 4 award 2 points, for 5 or 6 award 3 points, for 7 or 8 award 4 points, and for all 9 award 5 points. Grade the answer separately. | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Determine the prime number $p$ such that $\frac{4}{23}<\frac{1}{p}<\frac{8}{19}$. | First method:
If we expand the fractions so that their numerators are 8, then $\frac{8}{46}<\frac{8}{8 p}<\frac{8}{19}$. 2 POINTS Further, it follows that $19<8 p<46$.
2 POINTS
This means that $8 p \in\{24,32,40\} \quad 2$ POINTS
or $p \in\{3,4,5\}$. 2 POINTS
Since $p$ is a prime number, then $p \in\{3,5\}$. 2 POI... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Determine the unknown number $\mathrm{x}$ from the proportion $\frac{5}{8}:\left(0.4-2-\frac{1}{2} x\right)=1.25:\left(3.8-2 \frac{1}{4} \cdot x+\frac{1}{2}\right)$. | First method:
$$
\begin{array}{ll}
\frac{5}{8}:\left(0.4-2-\frac{1}{2} x\right)=1.25:\left(3.8-2 \frac{1}{4} \cdot x+\frac{1}{2}\right) & \\
\frac{5}{8} \cdot\left(3.8-2 \frac{1}{4} \cdot x+\frac{1}{2}\right)=1.25 \cdot\left(0.4-2-\frac{1}{2} x\right) & 1 \text{ POINT} \\
\frac{5}{8} \cdot\left(\frac{19}{5}-\frac{9}{4... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Micek is packing candies into bags. He has three types of candies: caramels, chocolate kisses, and gummy candies. If each bag must contain exactly 6 candies and at least 1 candy of each type, how many different bags can he make? | First method:
Let $k$ denote a caramel, $p$ denote a chocolate chip, and $g$ denote a gummy candy.
The problem can be solved by writing six-element sets of the form $\{k, p, g, -, -, -\}$.
The missing elements in the positions should be filled with the following subsets:
| $\{k, k, k\},\{p, p, p\},\{g, g, g\}$ | 1 ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Calculate the sum of the expressions $S_{2015}$ and $S_{2016}$ if $S_{2015}=1-2+3-4+5-\ldots-2014+2015$, and $S_{2016}=1-2+3-4+5-\ldots-2014+2015-2016$.
Problems worth 10 points: | First method:
```
\(S_{2015}=(1-2)+(3-4)+\ldots+(2013-2014)+2015=(2014: 2) \cdot(-1)+2015=-1007+2015=\)
\(=1008 \quad 3\) POINTS
\(S_{2016}=(1-2)+(3-4)+\ldots+(2015-2016)=(2016: 2) \cdot(-1)=-1008 \quad 2\) POINTS
\(S_{2015}+S_{2016}=1008+(-1008)=0\)
1 POINT
```
TOTAL 6 POINTS | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. A rectangle $A B C D$ with a diagonal of length $20 \text{ cm}$ is circumscribed by a circle. The side $\overline{C D}$ of the rectangle $A B C D$ is the base of an isosceles triangle whose third vertex $E$ is on the shorter arc determined by the chord $\overline{C D}$ of the circle circumscribed around the rectangl... | First method:
Let $|AB| = |CD| = a$, $|BC| = |AD| = b$, and let $v_a$ be the length of the height to the base of triangle $DCE$.
The radius of the circle circumscribed around the rectangle... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. In a mathematics competition, 23 fourth-grade students participated. Some of them were supposed to have four classes that day, while others were supposed to have five. How many had four, and how many had five classes if all together they missed a total of 101 classes?
The use of a pocket calculator or any reference... | First solution.
| $\mathbf{4}$ hours of classes | $\mathbf{5}$ hours of classes | Total hours of classes |
| :---: | :---: | :---: |
| 5 students $\ldots . .20$ hours | $23-5=18$ students $\ldots 90$ hours | 110 hours |
| 6 students $\ldots .24$ hours | $23-6=17$ students $\ldots 85$ hours | 109 hours |
| 7 students $... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate:
$$
\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}
$$ | First solution.
$$
\begin{aligned}
\frac{1}{\sqrt{2}-1} & -\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}= \\
& =\frac{\sqrt{2}+1-(\sqrt{2}-1)}{(\sqrt{2}-1)(\sqrt{2}+1)}+\frac{\sqrt{3}+1-(\sqrt{3}-1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \\
& =\frac{\sqrt{2}+1-\sqrt{2}+1}{(\sqrt{2})^{2}-1^{2}}+\frac{\sqrt{3}+1-\s... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Joško added the largest three-digit number divisible by 9 and the smallest three-digit number not divisible by 9, while Fran added the largest three-digit number not divisible by 9 and the smallest three-digit number divisible by 9. Which sum is greater and by how much? | 2. The largest three-digit number divisible by 9 is 999, and the smallest three-digit number not divisible by 9 is 100.
Joško added $999+100=1099$. 1 point The largest three-digit number not divisible by 9 is 998, and the smallest three-digit number divisible by 9 is 108.
Fran added $998+108=1106$. 1 point
Fran's su... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. What is the 2012th digit in the sequence 012343210012343210012...? | 4. In the sequence 012343210012343210012....the "subsequence" 012343210 which has 9 digits repeats.
1 point
$2012: 9=223$
21
32
5
Up to the 2012th digit, the "subsequence" 012343210 repeats 223 times, and there are still five digits remaining.
2 points
This means that the 2012th digit in the given sequence is t... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Determine the unknown number $x$ and explain.
| 9 | 22 | 13 | 16 |
| :---: | :---: | :---: | :---: |
| 144 | 176 | 143 | 192 |
| 16 | 8 | 11 | $x$ |
Tasks worth 10 points | 5. Since $144: 9=16$, $176: 22=8$ and $143: 13=11$, then it must also be so in the last column.
2 points
Therefore, $x=192$ : 16, which means $x=12$.
2 points
TOTAL 4 POINTS | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. The lengths of the sides of a rectangle are expressed as natural numbers (in cm), and the area is $196 \mathrm{~cm}^{2}$. How many such rectangles are there, and which one has the largest perimeter? | 6. If we factorize the number 196 into prime factors, we have that $196=2 \cdot 2 \cdot 7 \cdot 7$.
3 points
From the prime factorization, it is clear that there are five different pairs of natural numbers whose product is 196, so we have five different rectangles with an area of $196 \mathrm{~cm}^{2}$, and these are... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. How many three-digit natural numbers have the product of their digits equal to 28? Write them down! | 7. Since $28=2 \cdot 2 \cdot 7$, the corresponding triples of digits are:
a) 2, 2, 7;
b) $1,4,7$.
2 points
The required numbers are:
a) $227,272,722$;
2 points
b) $147,174,417,471,714,741$.
4 points
There are 9 numbers with the required properties.
2 points
TOTAL 10 POINTS | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Frane has a total of $108 \mathrm{kn}$ in his piggy bank in $5 \mathrm{kn}$, $2 \mathrm{kn}$, and $1 \mathrm{kn}$ coins. The value of the $5 \mathrm{kn}$ and $2 \mathrm{kn}$ coins is the same. The number of $1 \mathrm{kn}$ coins is equal to the number of $5 \mathrm{kn}$ and $2 \mathrm{kn}$ coins combined. How many c... | First solution.
A $1 \mathrm{kn}$ coin has exactly as many as a $5 \mathrm{kn}$ coin and a $2 \mathrm{kn}$ coin together, so we can place a $1 \mathrm{kn}$ coin on each $5 \mathrm{kn}$ coin to form a "packet" of $6 \mathrm{kn}$, and place a $1 \mathrm{kn}$ coin on each $2 \mathrm{kn}$ coin to form a "packet" of $3 \ma... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Iva and Maja received an order to make 72 greeting cards. Each of them made 36 cards, and each of their cards took the same amount of time to make. Iva made 6 cards in the same time it took Maja to make 5 cards. Iva finished making her 36 cards 1 hour and 48 minutes before Maja finished all of hers. How many cards d... | First solution.
Let $x$ be the time it took for Iva to make 6 cards, and for Maja to make 5 cards.
Then, it took Iva $\frac{x}{6}$ hours to make one card, and Maja $\frac{x}{5}$ hours.
This means that Iva made her 36 cards in $\frac{36}{6} x=6 x$ hours, and Maja in $\frac{36}{5} x$ hours.
Convert 1 hour and 48 minu... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. A sequence of digits consists of the first 222 natural numbers written in a row. In this sequence, we cross out the digits that are in odd positions. After that, we again cross out the digits that are in (new) odd positions. We repeat this procedure until only one digit remains. Which digit will it be?
## Ministry ... | 5. The sequence of digits consists of the first 222 natural numbers written in order. In this sequence, we cross out the digits that are in odd positions. After that, we again cross out the digits that are in (new) odd positions. We repeat this process until only one digit remains. Which digit will it be?
## Solution.... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. A car left city A heading towards city B, covering $12 \mathrm{~km}$ in 10 minutes. At the same time, a truck left city B heading towards city A, covering 10 $\mathrm{km}$ in 12 minutes. How many kilometers apart will the car and the truck be after 2 hours and 30 minutes of driving if the distance between cities A a... | 5. A car travels $12 \cdot(60: 10)=72 \text{ km}$ in 1 hour (60 minutes).
A car travels $36 \text{ km}$ in 30 minutes.
A car travels $2 \cdot 72+36=144+36=180 \text{ km}$ in 2 hours and 30 minutes.
A truck travels $10 \cdot(60: 12)=50 \text{ km}$ in 1 hour.
A truck travels $25 \text{ km}$ in 30 minutes.
A truck tr... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. What is the value of the digit in the units place of the number $1+9^{2014}$? | 3. The powers of the number 9 are $9^{1}=9,9^{2}=81,9^{3}=729,9^{4}=6561, \ldots$
2 POINTS
It is easy to notice that powers with odd exponents have the units digit 9, while powers with even exponents have the units digit 1.
2 POINTS
The number $9^{2014}$ has an even exponent, so its units digit is 1.
1 POINT
Ther... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation $6(x-1)(x+2)-4(x-3)(x+4)=2(x-5)(x-6)$. | 4. $6\left(x^{2}+2 x-x-2\right)-4\left(x^{2}+4 x-3 x-12\right)=2\left(x^{2}-6 x-5 x+30\right) \quad 1$ BOD
$6\left(x^{2}+x-2\right)-4\left(x^{2}+x-12\right)=2\left(x^{2}-11 x+30\right) \quad 1$ BOD
$6 x^{2}+6 x-12-4 x^{2}-4 x+48=2 x^{2}-22 x+60 \quad$ 1 BOD
$6 x^{2}-4 x^{2}-2 x^{2}+6 x-4 x+22 x=60-48+12 \quad 1$ BOD... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Determine the unknown digit $a$ of the number $\overline{401512 a}$ so that the remainders when this number is divided by 3 and 5 are equal. | First method:
Possible remainders when dividing by 3 are 0, 1, and 2.
If the remainder when dividing by 3 is 0, then the number $\overline{401512 a}$ is divisible by 3, and the sum of its digits must be divisible by 3.
The sum of the digits of the number $\overline{401512 a}$ is $13+a$, so the digit a can be 2, 5, a... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Let $n>1$ be a natural number. An equilateral triangle with side length $n$ is divided by lines parallel to its sides into congruent equilateral smaller triangles with side length 1. The number of smaller triangles that have at least one side on the side of the original triangle is 1 less than the number of all othe... | First method:
For $n=2$
the number of small triangles with at least one side lying on the side of the original triangle is 3, and the number of all remaining small triangles is 1, which me... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. How many three-digit numbers less than 200 can be expressed in the form $a^{2} \cdot b$, where $a$ and $b$ are prime numbers? | 4. If $a=2$, then $a^{2}=4$. For the product to be a three-digit number, $b$ must be a prime number greater than or equal to 25. For the product to be less than 200, $b$ must be a prime number less than 50.
Then $b \in\{29,31,37,41,43,47\}$.
1 POINT
If $a=3$, then $a^{2}=9$, so $b$ needs to be a prime number greater... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Mama, tata, brat i sestra kupili su ukupno 10 kuglica sladoleda. Na koliko različitih načina mogu međusobno raspodijeliti sladoled, pod uvjetom da svatko dobije barem dvije kuglice? Ispiši sve mogućnosti.
4. Mama, tata, brat i sestra bought a total of 10 scoops of ice cream. In how many different ways can they dist... | 4. Everyone must receive at least two scoops. That's a total of eight scoops, so there are two more to be distributed.
2 POINTS
Both scoops can belong to one of them (mom, dad, brother, or sister)
3 POINTS
or two people can each get one scoop (mom/dad, mom/brother, mom/sister, dad/brother, dad/sister, or
brother/s... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. To the summer math camp, 713 students traveled in 25 buses, some of which have 33 seats, and some have 26 seats. If it is known that the students filled all the seats in the buses, how many buses had 33 seats, and how many buses had 26 seats? | First method:
Assume that each of the 25 buses has 26 seats. 1 POINT
Then, they can accommodate $25 \cdot 26=\quad 1 \text{ POINT}$
$=650$ students. 1 POINT
Then, $713-650=\quad 1 \text{ POINT}$
$=63$ students would be left without seats. 1 POINT
Since $33-26=7$, a larger bus can accommodate seven more children t... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Roko and Marko have equally long steps. They are 1600 steps apart from each other. In one minute, Roko takes 80 steps, and Marko takes 60 steps. Who of the two needs to start earlier and how much earlier should they start to meet exactly halfway while walking towards each other? Express the solution in minutes and s... | First method:
For half the distance, 800 steps are required. 1 POINT
It takes Roku $800: 80=10$ minutes to do this. 1 POINT
In 10 minutes, Marko walks $10 \cdot 60=600$ steps. 1 POINT
$\begin{array}{ll}\text { Marko needs to start earlier. } & 1 \text { POINT }\end{array}$
Marko needs to start earlier by the amoun... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. What is the remainder when the number $1+2+2^{2}+2^{3}+\ldots+2^{2021}$ is divided by 9? | First method:
The number 1 when divided by 9 gives a remainder of 1.
The number 2 when divided by 9 gives a remainder of 2.
The number $2^{2}$ when divided by 9 gives a remainder of 4.
The number $2^{3}$ when divided by 9 gives a remainder of 8.
The number $2^{4}$ when divided by 9 gives a remainder of 7.
The num... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Students sit in the classroom in pairs. There are more girls in the class, so four fifths of the total number of students sit in mixed pairs (boy + girl), and there are three pairs that are exclusively female. How many boys, and how many girls are there in this class? | First method: In mixed pairs, $\frac{4}{5}$ of the class sits, so in female pairs, $\frac{1}{5}$ of the class sits.
$\mathrm{Here} \frac{1}{5}$ of the class consists of 6 girls sitting together,
1 POINT
which means there are $5 \cdot 6=30$ students in the class.
1 POINT
Of these 30 students, 24 students sit in mix... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. The average value of twelve numbers is 4.7. By adding two new numbers, the average value changes to 5.6. What is the average value of the two new numbers? | 2. From the conditions of the problem, it is possible to write the equality $x_{1}+x_{2} \ldots+x_{12}=56.4$. ..... 2 POINTS
After adding two new numbers, the equality becomes $x_{1}+x_{2} \ldots+x_{12}+x_{13}+x_{14}=78.4$. ..... 2 POINTS
By subtracting the first equality from the second, we get $x_{13}+x_{14}=22$. ...... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In a drum, there are balls on which all three-digit numbers are written (each one once). One ball is drawn. What is the probability that the sum of the digits of the drawn number is 2 or 5? | First method:
There are 900 three-digit numbers, so there are as many balls in the drum. ..... $1 \text{ POINT}$
The sum of 2 can be obtained by drawing three balls (200, 110, 101). ..... $1 \text{ POINT}$
The sum of 5 can be obtained by drawing 15 balls (500, 410, 401, 320, 302, 311, 230, 203, 221, 212,
140, 104, 131,... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Find the smallest natural number $n$ such that for every set of $n$ points with integer coordinates, of which no three lie on the same line, there exists a triangle with vertices from this set for which the midpoints of its sides also have integer coordinates.
No use of a pocket calculator or any manuals is allowed... | 5. Find the smallest natural number $n$ such that for any set of $n$ points with integer coordinates, of which no three lie on the same line, there exists a triangle with vertices from this set for which the midpoints of its sides also have integer coordinates.
## Solution.
If the points are $A\left(x_{1}, y_{1}\righ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. One pipe can fill the pool in 5 hours, and the second pipe in 3 hours. How long (in hours, minutes, and seconds) would it take for both pipes to fill the pool if they fill it together? | 2. The first pipe fills $\frac{1}{5}$ of the pool in 1 hour, and the second pipe fills $\frac{1}{3}$ of the pool in 1 hour. 1 point For 1 hour, these pipes together fill $\frac{1}{5}+\frac{1}{3}=\frac{8}{15}$ of the pool. The pool will be full in $\frac{15}{8}=1 \frac{7}{8}$ hours, which is 1 hour 52 minutes and 30 sec... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.