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7.5. Determine the integer numbers $x$, for which the numbers $n=x^{2}+5 x+1$ and $m=x^{2}+3 x+$ 7 are simultaneously perfect squares. | ## Solution:
$\operatorname{Given} x^{2}+3 x+7=k^{2} \Rightarrow 4 x^{2}+12 x+28=4 k^{2} \Leftrightarrow(2 x+3)^{2}+19=4 k^{2} \Leftrightarrow$ $(2 x+3-2 k)(2 x+3+2 k)=-19$.
a) $\left\{\begin{array}{l}2 x+3-2 k=-1 \\ 2 x+3+2 k=19\end{array} \Rightarrow 4 x+6=18 \Leftrightarrow 4 x=12 \Leftrightarrow x=3\right.$.
b) ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. On a board, all natural numbers $1,2,3, \ldots, 15$ are written. Determine the minimum number of numbers that can be erased from the board so that with the remaining numbers we can form two non-empty sets of numbers that simultaneously satisfy the following properties:
a) they have no common elements;
b) they ha... | ## Solution:
The product of all the numbers initially written on the board is $n=1 \cdot 2 \cdot 3 \cdot 2^{2} \cdot 5 \cdot 2 \cdot 3 \cdot 7 \cdot 2^{3} \cdot 3^{2} \cdot 2 \cdot 5 \cdot 11 \cdot 2^{2} \cdot 3 \cdot 13 \cdot 2 \cdot 7 \cdot 3 \cdot 5=2^{11} \cdot 3^{6} \cdot 5^{3} \cdot 7^{2} \cdot 11 \cdot 13$. To ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.3. For any natural number $m$, we denote by $S(m)$ the sum of the digits of the number $m$. Calculate $S\left(S\left(S\left(2023^{2023}\right)\right)\right)$. | ## Solution.
For any natural number $m$, we denote by $S(m)$ the sum of the digits and by $N(m)$ the number of digits of the number $m$. We have $2023^{2023}<\left(10^{4}\right)^{2023}=10^{8092}$, which implies $N\left(2023^{2023}\right) \leq 8092$. The number $2023^{2023}$ has no more than 8092 digits, and the sum of... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.4. Let the function $f: \mathbf{R} \rightarrow(0,+\infty)$ be differentiable, strictly decreasing, with $f(0)=1$ and $f^{\prime}(0)=-1$. Let the sequence $\left(a_{n}\right)_{n=1}^{\infty}$ be defined by: $a_{1}=1$ and $a_{n+1}=a_{n} f\left(a_{n}\right), \forall n \geq 1$. Show that the limit $\lim _{n \rightarrow \... | Solution. Since $a_{1}=1>0$ and $a_{n+1}=a_{n} f\left(a_{n}\right), \forall n \geq 1$, by mathematical induction it can be shown that $a_{n}>0, \forall n \geq 1$.
From the fact that the function $f$ is strictly decreasing, the inequalities follow: $0 < a_{n+1} = a_{n} f(a_{n}) < a_{n}, \forall n \geq 1$. Therefore, th... | 1 | Calculus | proof | Yes | Yes | olympiads | false |
11.8. A student was supposed to multiply a three-digit number by a four-digit number (the numbers do not start with 0). Instead of doing so, he simply appended the four-digit number to the right of the three-digit number, obtaining a seven-digit number, which is $N$ times ($N \in \mathbb{Z}$) larger than the correct re... | Solution. Let $\overline{abc}$ be the three-digit number and $\overline{defg}$ be the four-digit number. Then the seven-digit number has the form $\overline{abcdefg}$. From the condition of the problem, we have $\overline{\overline{abcdefg}}=N \cdot \overline{abc} \cdot \overline{defg}$, which is equivalent to $\overli... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. Mihai drew several right-angled triangles on the board, such that any two triangles do not have common points. In each of these triangles, Mihai drew the following important lines (segments): all angle bisectors, all medians, and the altitude to the hypotenuse. Counting, the total number of important lines drawn i... | ## Solution:
Right-angled triangles can be scalene or isosceles. In a scalene right-angled triangle, Mihai constructs exactly 7 distinct important segments: 3 angle bisectors, 3 medians, and one altitude. In a right-angled isosceles triangle, Mihai constructs exactly 5 distinct important segments, because the median, ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12.5. Let $I_{n}=\int_{\frac{1}{n}}^{n} \frac{d x}{(x+1)\left(\ln ^{2 n} x+1\right)}, n \geq 2$. Calculate $\lim _{n \rightarrow \infty} I_{n}$. | ## Solution.
$$
I_{n}=\int_{\frac{1}{n}}^{n} \frac{d x}{(x+1)\left(\ln ^{2 n} x+1\right)}=\int_{\frac{1}{n}}^{1} \frac{d x}{(x+1)\left(\ln ^{2 n} x+1\right)}+\int_{1}^{n} \frac{d x}{(x+1)\left(\ln ^{2 n} x+1\right)}
$$
Consider
$$
\begin{aligned}
& \int_{\frac{1}{n}}^{1} \frac{d x}{(x+1)\left(\ln ^{2 n} x+1\right)}=... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
8.3. In triangle $A B C$, point $D$ is the midpoint of side $A C$, and $E$ is an interior point of side $B C$ such that angles $B E A$ and $C E D$ are congruent. Find the numerical value of the ratio $\frac{A E}{D E}$. | Solution. On the extension of the ray $(ED$, we take the point $F$ such that $DE = DF$ (see figure).
Since $AD = DC$, it follows that the quadrilateral $AECF$ is a parallelogram. Then the lines $AE$ and $CF$ are parallel, as are the lines $AF$ and $BC$. Then we obtain
$$
m(\angle BEA) = m(\angle EAF) = \alpha = m(\an... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.4. A rectangular table has 6 rows and 7 columns and includes 42 squares of size $1 \times 1$. In each square, one of the numbers 0 or 1 is written such that for any two different rows, the sums of the numbers written on them are different, and for any two columns, the sums of the numbers written on them are equal. Fi... | Solution. Since the sums of the numbers on the lines are different, it follows that the sum of 7 numbers (0 or 1) written in a line can take only 8 values, namely: $0,1,2,3,4,5,6,7$. Since we have only 6 lines, then two values out of the eight are missing. Let's denote the missing values by $x$ and $y$. Then the sum of... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.3. Solve in the set of real numbers the equation
$$
\cos \frac{\pi}{x}=-x^{6}-4 x^{5}+2 x^{4}+12 x^{3}-9 x^{2}-1
$$ | Solution.
We have
$$
\begin{gathered}
1 \geq -\cos \frac{\pi}{x} = x^{6} + 4 x^{5} - 2 x^{4} - 12 x^{3} + 9 x^{2} + 1 = x^{2}\left(x^{4} + 4 x^{3} - 2 x^{2} - 12 x + 9\right) + 1 = \\
= x^{2}\left[\left(x^{2} + 2 x\right)^{2} - 6 x^{2} - 12 x + 9\right] + 1 = x^{2}\left[\left(x^{2} + 2 x\right)^{2} - 6\left(x^{2} + 2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.7. Solve in $\square$ the equation $16 x+(x-1) \cdot 4^{x+1}=x^{2}\left(4^{x}+8+4^{\frac{1}{x}}\right)$. | Solution: DVA: $x \neq 0$.
The equation can be written in the form $x^{2} \cdot 4^{\frac{1}{x}}+(2-x)^{2} \cdot 4^{x}=8 x(2-x)$.
We observe that $x=2$ is not a solution.
Since $x \neq 2$, the equation can be brought to the form $\frac{x}{2-x} \cdot 4^{\frac{1}{x}}+\frac{2-x}{x} \cdot 4^{x}=8$.
If $\frac{x}{2-x} > 0... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.1. Let $\left(a_{n}\right)_{n=1}^{\infty}$ be a sequence of integers that satisfies the relation $a_{n+1}=a_{n}^{1009}+3^{2017}, \forall n \geq 1$. How many perfect squares can this sequence contain? Justify your answer. | Solution. The possible values of the pairs $\left(a_{n} \bmod 4, a_{n+1} \bmod 4\right)$ are $(0,3),(1,0),(2,3)$ and $(3,2)$. Therefore, regardless of the value of $a_{1}$, all terms $a_{n}, n \geq 3$, are equal to 2 or $3(\bmod 4)$ and, thus, are not perfect squares. In conclusion, we have at most two perfect square t... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.1. Determine all natural numbers $n$ for which the number $\sqrt{n+\sqrt{n-2}}$ is rational. | Solution: Suppose for some value of the number $n, n \geq 2$ we have $m=\sqrt{n+\sqrt{n-2}} \in Q, m \in N^{*}$. The equality $m^{2}=n+\sqrt{n-2}$ holds. Then there exists $k \in N^{*}$, such that $k=\sqrt{n-2}$, i.e., $n=k^{2}+2$. We can write $m^{2}=k^{2}+k+2$.
The inequality $k^{2}<k^{2}+k+2<k^{2}+4 k+4=(k+2)^{2}$ ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.6. Let the sequence $\left(a_{n}\right)_{n=0}^{\infty}$ be defined by the relations $a_{0}=1, a_{1}=\frac{1+\sqrt{3}}{2 \sqrt{2}}$ and $a_{n}=2 a_{1} a_{n-1}-a_{n-2}, \forall n \geq 2$. Calculate the value of $a_{2020}$ and determine $\lim _{n \rightarrow \infty} \frac{a_{n}}{n^{2}}$. | ## Solution.
We have $\quad a_{2}=2 a_{1}^{2}-a_{0}=2\left(\frac{1+\sqrt{3}}{2 \sqrt{2}}\right)^{2}-1=\frac{\sqrt{3}}{2}, \quad a_{3}=2 a_{1} a_{2}-a_{1}=2 \cdot \frac{\sqrt{3}}{2} \cdot\left(\frac{1+\sqrt{3}}{2 \sqrt{2}}\right)-\frac{1+\sqrt{3}}{2 \sqrt{2}}=\frac{\sqrt{2}}{2}$, $a_{4}=2 a_{1} a_{3}-a_{2}=2 \cdot \fra... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.6. Solve in $R$ the equation
$$
\sqrt{2 x^{5}+x^{4}+4 x^{3}+2 x^{2}+2 x+1}+\sqrt{17-2 x+34 x^{2}-4 x^{3}+17 x^{4}-2 x^{5}}=7 x^{2}-8 x+22
$$ | Solution: We will apply the inequality $\frac{a+b}{2} \leq \sqrt{\frac{a^{2}+b^{2}}{2}}$, which is true for any real numbers $a$ and $b$.
On $D V A$, the inequality holds:
$$
\frac{\sqrt{2 x^{5}+x^{4}+4 x^{3}+2 x^{2}+2 x+1}+\sqrt{17-2 x+34 x^{2}-4 x^{3}+17 x^{4}-2 x^{5}}}{2} \leq \sqrt{\frac{18 x^{4}+36 x^{2}+18}{2}}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.8. The natural number $n$ and the prime numbers $p$ and $q$ satisfy the relation $\frac{3}{\sqrt{n}}=\frac{2}{\sqrt{p}}+\frac{1}{\sqrt{q}}$. Determine all possible values of the expression $E=p+q-2 n$. | ## Solution:
From $\frac{3}{\sqrt{n}}=\frac{2}{\sqrt{p}}+\frac{1}{\sqrt{q}}$, we get $\frac{3}{\sqrt{n}}=\frac{2 \sqrt{q}+\sqrt{p}}{\sqrt{p q}}, 3 \sqrt{p q}=\sqrt{n}(2 \sqrt{q}+\sqrt{p})$. Squaring both sides of the last equality, we obtain $9 p q=n(4 q+p+4 \sqrt{p q})$. Since $9 p q \in \mathbb{N}$, it follows that ... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.1. For what real values of the parameter a does the graph of the function $f: \square \rightarrow \square$, $f(x)=x^{4}-8 x^{3}+14 x^{2}+$ ax have an axis of symmetry parallel to the line $x=0$? | ## Solution.
Since the graph of the function $f(x)$ has an axis of symmetry parallel to the line $x=0$, then the axis of symmetry has the equation $x=c$, where $c$ is a constant real value. Substituting $t=x-c$, we obtain that the graph of the function $f(t)$, as a function of $t$, will be symmetric with respect to th... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. On the ceiling of a room, there are 15 lamps, numbered from 1 to 15. Initially, all the lamps are off. In another room, there are 15 switches: one switch for lamp 1 and 2, one switch for lamp 2 and 3, one switch for lamp 3 and 4, and so on, up to a switch for lamp 15 and 1. When the switch for such a pair of la... | Solution. With three rounds of walking, Raymond can never be certain. Indeed, if for each switch you note whether it is in the initial position or the flipped position in the three rounds, then you can get $2^{3}=8$ different patterns. However, there are 15 switches, so there are multiple switches with the same pattern... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Let $n$ be an even positive integer. A sequence of $n$ real numbers is called complete if for every integer $m$ with $1 \leq m \leq n$, the sum of the first $m$ terms or the sum of the last $m$ terms of the sequence is an integer. Determine the minimum number of integers in a complete sequence of $n$ numbers. | Solution. We prove that the minimum number of integers in a complete sequence is 2. First, consider the case $n=2$. Let $a_{1}$ and $a_{2}$ be the numbers in the sequence. Then $a_{1}$ or $a_{2}$ is an integer, say without loss of generality $a_{1}$. Furthermore, $a_{1}+a_{2}$ is an integer, but then $a_{2}$ is also an... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 5. Given a positive integer $n$. Determine all positive real numbers $x$ such that
$$
n x^{2}+\frac{2^{2}}{x+1}+\frac{3^{2}}{x+2}+\ldots+\frac{(n+1)^{2}}{x+n}=n x+\frac{n(n+3)}{2} .
$$ | Solution. For $1 \leq i \leq n$ we have
$\frac{(i+1)^{2}}{x+i}=i+1+\frac{(i+1)^{2}-(i+1)(x+i)}{x+i}=i+1+\frac{i+1-(i+1) x}{x+i}=i+1+\frac{(i+1)(1-x)}{x+i}$,
so we can rewrite the left side of the given equation to
$$
n x^{2}+2+3+\ldots+(n+1)+\frac{2(1-x)}{x+1}+\frac{3(1-x)}{x+2}+\ldots+\frac{(n+1)(1-x)}{x+n}
$$
It ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 4. Let $m \geq 3$ and $n$ be positive integers with $n > m(m-2)$. Find the largest positive integer $d$ such that $d \mid n!$ and $k \nmid d$ for all $k \in \{m, m+1, \ldots, n\}$. | Solution. We will prove that $d=m-1$ is the largest that satisfies. First note that $m-1 \mid n$ ! and for $k \geq m$ we have $k \nmid m-1$, so $d=m-1$ indeed satisfies. Now suppose that for some $d$ we have: $d \mid n$ ! and $k \nmid d$ for all $k \in\{m, m+1, \ldots, n\}$. We will prove that $d \leq m-1$. Write $d=p_... | -1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 2. Determine the largest real number $M$ such that for every infinite sequence $x_{0}, x_{1}, x_{2}, \ldots$ of real numbers that satisfies
a) $x_{0}=1$ and $x_{1}=3$,
b) $x_{0}+x_{1}+\cdots+x_{n-1} \geq 3 x_{n}-x_{n+1}$,
it holds that
$$
\frac{x_{n+1}}{x_{n}}>M
$$
for all $n \geq 0$. | Solution. Answer: the largest possible constant for which this holds is $M=2$.
This problem is a typical example where it is beneficial to take a stronger induction hypothesis than strictly necessary: we show by induction that $x_{n+1}>2 x_{n}>x_{n}+x_{n-1}+\ldots+x_{0}$. For $n=0$, this translates to $x_{1}>2 x_{0}>x... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
## Problem 3.
(a) Let $a$ and $b$ be positive integers such that $M(a, b)=a-\frac{1}{b}+b\left(b+\frac{3}{a}\right)$ is an integer. Prove that $M(a, b)$ is a perfect square.
(b) Find integers $a$ and $b$, both not equal to zero, such that $M(a, b)$ is a positive integer but not a perfect square. | ## Solution.
(a) Since $a+b^{2}$ is an integer, $-\frac{1}{b}+\frac{3 b}{a}$ is also an integer. This can be written as $\frac{-a+3 b^{2}}{a b}$. We see that $a b$ is a divisor of $3 b^{2}-a$. In particular, $b$ is a divisor of $3 b^{2}-a$, and thus $b \mid a$. But this means that $b^{2}$ is a divisor of $a b$ and the... | 7 | Algebra | proof | Yes | Yes | olympiads | false |
Task 2. Determine all positive integers $n$ for which there exist positive integers $a_{1}, a_{2}, \ldots, a_{n}$ such that
$$
a_{1}+2 a_{2}+3 a_{3}+\ldots+n a_{n}=6 n
$$
and
$$
\frac{1}{a_{1}}+\frac{2}{a_{2}}+\frac{3}{a_{3}}+\ldots+\frac{n}{a_{n}}=2+\frac{1}{n}
$$ | Solution. If we apply the inequality of the arithmetic and harmonic mean to $a_{1}$, twice $a_{2}$, three times $a_{3}, \ldots, n$ times $a_{n}$, then we find
$$
\frac{6 n}{\frac{1}{2} n(n+1)}=\frac{a_{1}+2 a_{2}+3 a_{3}+\ldots+n a_{n}}{\frac{1}{2} n(n+1)} \geq \frac{\frac{1}{2} n(n+1)}{\frac{1}{a_{1}}+\frac{2}{a_{2}}... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 3. Find all positive integers $k$ for which the equation
$$
\operatorname{lcm}(m, n)-\operatorname{gcd}(m, n)=k(m-n)
$$
has no positive integer solutions $(m, n)$ with $m \neq n$. | Solution. Let $d=\operatorname{gcd}(m, n)$ and write $m=d a$ and $n=d b$. It holds that $\operatorname{lcm}(m, n) \cdot \operatorname{gcd}(m, n)=m n$, so we can write the given equation as
$$
\frac{d a \cdot d b}{d}-d=k(d a-d b)
$$
or equivalently
$$
a b-1=k(a-b) .
$$
From now on, we will look at the equivalent pro... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 2. The sets $A$ and $B$ are subsets of the positive integers. The sum of any two different elements from $A$ is an element of $B$. The quotient of any two different elements of $B$ (where we divide the larger by the smaller) is an element of $A$. Determine the maximum number of elements in $A \cup B$. | Solution. Suppose $A$ contains at least three elements, say $a<b<c$, so $b-a$ is positive, thus it must hold that $a+c \leq b-a$. This gives $c \leq b-2a$. Therefore, $A$ contains at most two elements.
Suppose $B$ contains at least four elements, say $a<b<c<d$. Then $A$ contains the three distinct elements $\frac{d}{a... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
B3. One of the four gnomes, Anne, Bert, Chris, and Dirk, has stolen gold from the king. Each of the gnomes, who know each other inside and out, makes two statements. If a gnome is a liar, at least one of those two statements is a lie. If a gnome is not a liar, both statements are true.
Anne says: "Bert is a liar." and... | B3. 5
First, consider the case where Anne stole from the king. Then the last two statements of Bert and Chris are true, and those of Anne and Dirk are false. Therefore, Anne and Dirk are liars. Since both statements of Chris are true, Chris is not a liar. From this, we conclude that Bert is a liar, because his first s... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
B1. We consider numbers of two or more digits where none of the digits is 0. We call such a number thirteenish if every two adjacent digits form a multiple of 13. For example, 139 is thirteenish because $13=1 \times 13$ and $39=3 \times 13$.
How many thirteenish numbers of five digits are there? | B1. 6 The multiples of 13 that consist of two non-zero digits are 13, 26, 39, 52, 65, 78, and 91. In a thirteenish number, a 1 can only be followed by a 3, and a 2 can only be followed by a 6. The digit 4 cannot be followed by any digit. In the figure, each digit is marked with the digit that can follow it.
 | B2. 10
We split the black area from the problem into four triangles, two of which we have shaded gray. The two gray triangles both have a base of 2 and together have a height of $7-2=5$, which is the height of the large square minus the height of the small square. The combined area of the two gray triangles is thus $\... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
B3. In a class, there are 23 students, each of whom has chosen exactly one foreign language, either German or French. There are a total of 10 girls in the class, and there are 11 students in total who are taking French. The number of girls who have chosen French plus the number of boys who have chosen German is 16.
Ho... | B3. 7 In total, there are 23 students. From the data it follows:
$$
\begin{aligned}
16+11+10= & (\text { girls with French }+ \text { boys with German })+\text { everyone with French }+ \text { all girls } \\
= & (\text { girls with French }+ \text { boys with German }) \\
& +(\text { girls with French }+ \text { boys... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
C2. We consider a triangle $A B C$ and a point $D$ on the extension of $A B$ towards $B$. The point $E$ lies on the side $A C$ such that angle $\angle D B C$ and angle $\angle D E C$ are equal. The intersection of $D E$ and $B C$ is $F$. Suppose that $|B F|=2,|B D|=3,|A E|=4$ and $|A B|=5$. (Note: the image below is no... | C2. (a) The situation is as shown in the figure below. Since angles $\angle A E C$ and $\angle A B D$ are straight, it follows that
$$
\angle A B C=180^{\circ}-\angle D B C=180^{\circ}-\angle D E C=\angle A E D .
$$
Since angle $A$ is present in both triangles, triangles $\triangle A B C$ and $\triangle A E D$ have t... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
B4. Een parallellogram heeft twee zijden van lengte 4 en twee zijden van lengte 7. Ook heeft een van de diagonalen lengte 7. (Let op: het plaatje hiernaast is niet op schaal.) Hoe lang is de andere diagonaal?
|
B4. 9 We tekenen enkele hulplijnen, zoals in de figuur. Het parallellogram is $A B C D$ met $|A B|=4$ en $|A D|=|B C|=7$. We zijn op zoek naar $|A C|$.
Aangezien $|A D|=|B D|$, is driehoek $\triangle A B D$ een gelijkbenige driehoek. Laat $E$ het punt zijn midden tussen $A$ en $B$. Dan zijn de driehoeken $\triangle A... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
B2. In a top hat, there are a hundred tickets numbered from 1 to 100. You want to have three tickets with the property that each of the three numbers is smaller than the sum of the other two numbers. For example, the three tickets with numbers 10, 15, and 20 would be suitable (since $10<15+20, 15<10+20$, and $20<10+15$... | 2. 11
Suppose we have drawn 11 lottery tickets with numbers $a_{1}<a_{2}<\cdots<a_{11}$ on them such that no three tickets meet our wish. We know first that $a_{1} \geqslant 1$ and $a_{2} \geqslant 2$. Since the triplet $a_{1}, a_{2}$, and $a_{3}$ does not meet our wish, it must be that $a_{3} \geqslant a_{1}+a_{2} \g... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
B3. On each of the twelve edges of a cube, we write the number 1 or -1. We then multiply the four numbers on the edges of each face of the cube and write the result on that face. Finally, we add up the eighteen written numbers.
What is the smallest (most negative) result we can get this way? In the figure, you see an ... | 3. -12
First, we show that the result is always -12 or more. If we write a -1 on each of the twelve edges, we get a 1 in each face. In that case, the result is $-12+6=-6$. For every -1 on an edge that we change to a 1, at most two faces change from a 1 to a -1. The result thus becomes at most 2 lower $(-1+1+1$ becomes... | -12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
B5. Given is the sequence of numbers $a_{0}, a_{1}, a_{2}, \ldots, a_{2020}$. It is given that $a_{0}=0$. Moreover, for every $k=1,2, \ldots, 2020$ it holds that
$$
a_{k}= \begin{cases}a_{k-1} \cdot k & \text { if } k \text { is divisible by } 8, \\ a_{k-1}+k & \text { if } k \text { is not divisible by } 8 .\end{case... | B5. 02 We see that $a_{2000}=a_{1999} \cdot 2000$ ends in three zeros. We can now determine the last digits of $a_{2001}, a_{2002}$ up to $a_{2020}$ using the formulas. It is only important to keep the last two digits, the rest we will ignore.
| $a_{2001}$ | $a_{2002}$ | $a_{2003}$ | $a_{2004}$ | $a_{2005}$ | $a_{2006... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
B4. For the second round of the Math Olympiad, 999 students are invited. Melanie prepares invitation letters in the order of participant numbers: $1,2,3, \ldots$ For some values of $n \geqslant 100$, she notices the following: the number of participant numbers from $1$ to $n$ that end in a 5 is exactly equal to the num... | B4. 9 Let $a, b$ and $c$ be the digits of $n$; that is, $n=100 a+10 b+c$. Then the number of participant numbers from $1 \mathrm{t} / \mathrm{m} n$ that end in 5 is equal to $10 a+b$ if $c<5$ and equal to $10 a+b+1$ if $c \geqslant 5$. The number formed by the last two digits of $n$ is $10 b+c$.
First, consider the ca... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
B2. Consider a circle with diameter $A B$. Point $C$ lies on the line segment $A B$ such that $2 \cdot|A C|=|B C|$. The points $D$ and $E$ lie on the circle so that $C D$ is perpendicular to $A B$ and $D E$ is also a diameter of the circle. Denote the areas of triangles $A B D$ and $C D E$ as $O(A B D)$ and $O(C D E)$.... | B2. Let the center of the circle be $M$. Triangles $C D M$ and $C E M$ have equal areas. Indeed, the two triangles have bases of equal length $|D M|=|E M|$ and the same height. It follows that
$$
O(C D E)=2 \cdot O(C D M)
$$
We have $|A C|=\frac{1}{3}|A B|$ and $|A M|=\frac{1}{2}|A B|$, so $|C M|=|A M|-|A C|=$ $\frac... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
B4. Emile stands in a circle with nine other people. Each of the ten people thinks of an integer (which can also be negative) and whispers this number to both of their neighbors. Then, each person loudly states the average of the two numbers they heard from their neighbors. It turns out that Emile says the number 10, h... | B4. 5
The number that Emile had in mind we call $c_{10}$, the number of his right neighbor we call $c_{9}$, and so on until the left neighbor of Emile, who had the number $c_{1}$ in mind. We then see that
\[
\begin{aligned}
c_{10}+c_{8} & =2 \cdot 9=18, \\
c_{8}+c_{6} & =2 \cdot 7=14, \\
c_{6}+c_{4} & =2 \cdot 5=10, ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
B5. The sum of the number of eyes on opposite sides of a die is always 7. Nine identical dice are glued together in a $3 \times 3$ square. This is done in such a way that two glued sides always have the same number of eyes. In the figure, you see the top view of the result, where the number of eyes is not shown for fiv... | B5. 3 The numbers on opposite sides of a die are called opposite. Together, these numbers always add up to 7. We consider two dice that are touching with the same numbers, but can still rotate relative to each other. Around the two dice, we then see the same numbers, but in reverse cyclic order.
Consider the situation... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
B2. Given is a square $A B C D$ with side lengths of 4. Inside the square, two semicircles with diameters $A B$ and $B C$ are drawn (see figure).
What is the combined area of the two gray regions?
 and on... | B1. 11 Of every three consecutive integers, there is always exactly one that is a multiple of three. Now consider a sequence of $k$ consecutive integers. If $k$ itself is a multiple of three, say $k=3 \ell$, then exactly $\ell$ numbers in the sequence are multiples of three: one in each consecutive group of three numbe... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
B4. A parallelogram $A B C D$ is cut by a line $m$. From the four vertices $A, B, C$ and $D$, we drop perpendiculars to line $m$. The feet of these perpendiculars are $P, Q, R$ and $S$ respectively. Point $S$ is also exactly the intersection of line $m$ and $A B$. The lengths of segments $A P, B Q$ and $D S$ are 6, 7, ... | B4. 12 We will prove that:
$$
|A P|+|B Q|+|C R|=|D S| .
$$
From this, the required length follows:
$$
|C R|=|D S|-|A P|-|B Q|=25-6-7=12 .
$$
Let $T$ be the foot of the perpendicular from $A$ to $D S$. It follows that $A P S T$ is a rectangle, so $|S T|=|A P|$.
 In every match of the tournament, 2 or 3 points are scored. In a tournament with four teams, $\left(\begin{array}{l}4 \\ 2\end{array}\right)$, so six matches are played, resulting in a total of 12 to 18 points scored. For three teams, the maximum is $\left(\begin{array}{l}3 \\ 2\end{array}\right) \times 3$, thus ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ are five different real numbers. The number of different values that the sum $a_{i}+a_{j}$ can take for all $i, j$ with $1 \leq i<j \leq 5$ we call $m$.
Determine the smallest possible value of $m$. | 3. We order the five numbers from smallest to largest and call them $a, b, c, d$, and $e$, so $a<b<c<d<e$. Then it holds that: $a+b<a+c<a+d<a+e<b+e<c+e<d+e$. Therefore, the smallest value of $m$ satisfies: $m \geq 7$. With the example $a=1, b=2, c=3, d=4$ and $e=5$, we find the sums to be $3,4,5,6,7,8$ and 9. Thus, 7 i... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A quadruple $(p, a, b, c)$ of positive integers is called a Leiden quadruple if
- $p$ is an odd prime number,
- $a, b$, and $c$ are distinct, and
- $ab+1, bc+1$, and $ca+1$ are divisible by $p$.
a) Prove that for every Leiden quadruple $(p, a, b, c)$, it holds that $p+2 \leqslant \frac{a+b+c}{3}$.
b) Determine al... | 4. a) Without loss of generality, we assume that $a < b < c$. The numbers $a$ and $c$ are not divisible by $p$, because otherwise $ac + 1$ would be a multiple of $p$ plus 1, and thus not divisible by $p$. Since $bc + 1$ and $ac + 1$ are divisible by $p$, their difference $(bc + 1) - (ac + 1) = (b - a)c$ is also divisib... | 5 | Number Theory | proof | Yes | Yes | olympiads | false |
5. Kira has 3 blocks with the letter A, 3 blocks with the letter B, and 3 blocks with the letter C. She puts these 9 blocks in a sequence. She wants to have as many distinct distances between blocks with the same letter as possible. For example, in the sequence ABCAABCBC the blocks with the letter A have distances 1, ... |
5. We will show that the maximum number of distinct distances is 7 . First we prove that the number of distinct distances cannot be more than 7 , then we will show that there is a sequence of blocks with 7 distances.
The possible distances between two blocks in the sequence are the numbers 1 to 8 . Therefore, there c... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Daan distributes the numbers 1 through 9 over the nine cells of a $3 \times 3$ table (each cell contains exactly one number). Then Daan circles the middle number in terms of size in each row. For example, if the numbers 8, 1, and 2 are in a row, he circles the number 2. He does this for each column and for each of t... | 1. (a) The smallest possible number of circled numbers is 3. That it cannot be done with fewer follows from the fact that at least one number is circled in each row (and these are three different numbers).
A median table with only 3 numbers circled is shown alongside. In the rows, the numbers 7, 5, 3 are circled, in t... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. We consider sequences of numbers $a_{1}, a_{2}, a_{3}, \ldots$ that satisfy the formula $a_{n+1}=\frac{a_{n}+a_{1}}{a_{n}+1}$ for all $n \geqslant 1$.
(a) Suppose that $a_{1}=-3$. Calculate $a_{2020}$.
(b) Suppose that $a_{1}=2$. Prove that for all $n \geqslant 2$ it holds that $\frac{4}{3} \leqslant a_{n} \leqsla... | ## 2. Version for Grade 5 & Grade 4 and lower
(a) If $a_{1}=-3$, then $a_{2}=\frac{a_{1}+a_{1}}{a_{1}+1}=\frac{-6}{-2}=3$. Next, we calculate $a_{3}=\frac{a_{2}+a_{1}}{a_{2}+1}=\frac{0}{4}=0$. Then comes $a_{4}=\frac{a_{3}+a_{1}}{a_{3}+1}=\frac{-3}{1}=-3$. We see that $a_{4}=a_{1}$. Since $a_{n+1}$ depends only on $a_... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. We consider sports tournaments with $n \geqslant 4$ participating teams and where every pair of teams plays against one another at most one time. We call such a tournament balanced if any four participating teams play exactly three matches between themselves. So, not all teams play against one another.
(a) Prove t... | ## 2. Version for klas 5 \& klas 4 and below
(a) Suppose towards a contradiction that we can find three teams in a balanced tournament that all play against each other, say teams A, B and C. Because $n \geqslant 5$ there are two other teams, say D and E. Since A, B and C already play three matches between them, there ... | 5 | Combinatorics | proof | Yes | Yes | olympiads | false |
1. Niek has 16 square cards that are yellow on one side and red on the other. He lays them out in a $4 \times 4$ square. Some cards may lie with the yellow side up, and some with the red side up. For each color pattern, he calculates the monochromaticity of the pattern as follows. For each pair of cards that share a si... | 1. (a) First note that there are $3 \cdot 4=12$ horizontal boundaries between two tiles, and also 12 vertical boundaries. Suppose there are $k$ boundaries that count for -1, then there are $24-k$ boundaries that count for +1. This gives a monochromaticity of $(24-k) \cdot (+1) + k \cdot (-1) = 24 - 2k$. Therefore, the ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. We consider sports tournaments with $n \geqslant 4$ teams where each pair of teams plays against each other at most once. We call such a tournament balanced if every quartet of teams plays exactly three matches among themselves. Not all teams play against each other.
(a) Prove that in a balanced tournament with $n ... | ## 2. Version for Grade 5 & Grade 4 and Lower
(a) Assume, for the sake of contradiction, that in a balanced tournament, there are three teams that all play against each other, say teams A, B, and C. Since \( n \geqslant 5 \), there are two other teams, say teams D and E. Because A, B, and C already play three matches ... | 5 | Combinatorics | proof | Yes | Yes | olympiads | false |
3. In a tournament with six teams, each team plays once against every other team. If a team wins a match, it receives 3 points and the loser gets 0 points. If a match ends in a draw, both teams receive 1 point.
Can the final scores of the teams be exactly six consecutive numbers $a, a+1, \ldots, a+5$? If so, determine... | 3. In total, 15 matches are played. In each match, a total of 2 or 3 points are earned. The sum of the six scores therefore lies between $15 \cdot 2=30$ (all matches end in a draw) and $15 \cdot 3=45$ (no match ends in a draw).
The sum of the six scores is also equal to $a+(a+1)+\cdots+(a+5)=15+6a$. We see that $30 \l... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. The eight points below are the vertices and the midpoints of the sides of a square. We would like to draw a number of circles through the points, in such a way that each pair of points lie on (at least) one of the circles.
Determine the smallest number of circles needed to do this.
|
5. Four of the eight points are coloured black and the other four points are coloured white in the way indicated in the figure on the left. The circle through the four black points is denoted $C_{1}$ and the circle through the four white points is denoted $C_{2}$. If two points lie on a circle $C$, we say that $C$ cov... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Kira has 3 blocks with the letter A, 3 blocks with the letter B, and 3 blocks with the letter C. She places these 9 blocks in a row. She wants to have as many different distances between blocks with the same letter as possible. For example, in the sequence ABCAABCBC, the blocks with the letter A have distances of 1,... | 5. We will show that there can be a maximum of 7 different distances. First, we prove that the number of different distances cannot be more than 7, then we show that there is a sequence of blocks with 7 different distances.
The possible distances between two blocks in a row of nine blocks are 1 through 8. Therefore, t... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In a hockey tournament, six teams participate. Each team plays exactly one match against each other team. A team receives 3 points for a win, 1 point for a draw, and 0 points for a loss. At the end of the tournament, a ranking list is compiled. It turns out that there are no shared positions on the ranking list. It ... | 3. Let the six team scores be equal to $s, s+2, s+4, s+6, s+8$ and $s+10$. Let $T$ be the total number of points scored. Therefore, $T=6s+30$. The total number of points is thus a multiple of six.
In total, $\frac{6 \cdot 5}{2}=15$ matches were played. The number of matches that ended in a draw we call $g$. A draw yie... | 2 | Combinatorics | proof | Yes | Yes | olympiads | false |
5. The eight points below are the vertices and midpoints of the sides of a square. We want to draw a number of circles through the points in such a way that each pair of points lies on (at least) one of the circles.
Find the smallest possible number of circles with which this can be done. | 5. Of the eight points, we color four black and the other four white, as indicated in the left figure. The circle through the four black points we call $C_{1}$ and the circle through the four white points we call $C_{2}$. If a pair of points lies on a circle $C$, we say that $C$ covers that pair.
 and $15 \cdot 3=45$ (no match is a draw).
On the other hand, the sum of the six scores equals $a+(a+1)+\cdots+(a+5)=15+6 a... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. We call a positive integer a shuffle number if the following hold:
(1) All digits are nonzero.
(2) The number is divisible by 11 .
(3) The number is divisible by 12. If you put the digits in any other order, you again have a number that is divisible by 12 .
How many 10-digit shuffle numbers are there?
| ## 1. Version for klas $4 \&$ below
First observe that a shuffle number can only contain the digits 2, 4, 6, and 8 . Indeed, if we place the digits in any order, we obtain an even number (since it is divisible by 12) because of property (3). Since the last digit of an even number is also even, the digit must be $2,4,6... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The numbers 1 to 25 are each coloured blue or red. Determine all possible colourings that satisfy the following rules:
- The number 5 is red.
- If numbers $x$ and $y$ have different colours and $x+y \leqslant 25$, then $x+y$ is blue.
- If numbers $x$ and $y$ have different colours and $x \cdot y \leqslant 25$, the... | ## 2. Version for klas $5 \&$ klas 4 and below
We first consider the case that 1 is red. Then all numbers from 1 to 25 are red. Indeed, suppose that some number $k$ is blue. Then 1 and $k$ have different colours, hence by the third rule the number $1 \cdot k=k$ must be red. But this contradicts the assumption that it ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The numbers 1 to 15 are each coloured blue or red. Determine all possible colourings that satisfy the following rules:
- The number 15 is red.
- If numbers $x$ and $y$ have different colours and $x+y \leqslant 15$, then $x+y$ is blue.
- If numbers $x$ and $y$ have different colours and $x \cdot y \leqslant 15$, th... | ## 2. Version for klas 6
We first consider the case that 1 is red. Then all numbers from 1 to 15 are red. Indeed, suppose that some number $k$ is blue. Then 1 and $k$ have different colours, hence by the third rule the number $1 \cdot k=k$ must be red. But this contradicts the assumption that it was blue. Observe that... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. We make groups of numbers. Each group consists of five distinct numbers. A number may occur in multiple groups. For any two groups, there are exactly four numbers that occur in both groups.
(a) Determine whether it is possible to make 2015 groups.
(b) If all groups together must contain exactly six distinct numbe... |
1. (a) It is possible to make 2015 groups. For example, take the 2015 groups $\{-4,-3,-2,-1, i\}$, where $i$ runs from 1 to 2015. Each group consists of five distinct numbers, as required, and any two groups have exactly four numbers in common: $-4,-3,-2$, and -1 .
(b) Using six available numbers, there are only six ... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Daan distributes the numbers 1 to 9 over the nine squares of a $3 \times 3$-table (each square receives exactly one number). Then, in each row, Daan circles the median number (the number that is neither the smallest nor the largest of the three). For example, if the numbers 8,1 , and 2 are in one row, he circles th... |
1. (a) The smallest possible number of circled numbers is 3 . Fewer than 3 is not possible since in each row at least one number is circled (and these are three different numbers).
On the right, a median table is shown in which only 3 numbers are circled. In the rows, the numbers $7,5,3$ are circled, in the columns t... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. We consider number sequences $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{n+1}=\frac{a_{n}+a_{1}}{a_{n}+1}$ holds for all $n \geqslant 1$.
(a) Suppose that $a_{1}=-3$. Compute $a_{2020}$.
(b) Suppose that $a_{1}=2$. Prove that $\frac{4}{3} \leqslant a_{n} \leqslant \frac{3}{2}$ holds for all $n \geqslant 2$.
| ## 2. Version for klas 5 \& klas 4 and below
(a) If $a_{1}=-3$, we have $a_{2}=\frac{a_{1}+a_{1}}{a_{1}+1}=\frac{-6}{-2}=3$. Next, we find $a_{3}=\frac{a_{2}+a_{1}}{a_{2}+1}=\frac{0}{4}=0$. Then, we have $a_{4}=\frac{a_{3}+a_{1}}{a_{3}+1}=\frac{-3}{1}=-3$. We see that $a_{4}=a_{1}$. Since $a_{n+1}$ only depends on $a_... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (a) Determine all pairs $(x, y)$ of (real) numbers with $0<x<1$ and $0<y<1$ for which $x+3 y$ and $3 x+y$ are both integers. An example of such a pair is $(x, y)=\left(\frac{3}{8}, \frac{7}{8}\right)$, because $x+3 y=\frac{3}{8}+\frac{21}{8}=\frac{24}{8}=3$ and $3 x+y=\frac{9}{8}+\frac{7}{8}=\frac{16}{8}=2$.
(b) Fi... | 4. (a) Suppose $(x, y)$ is such a pair and consider the integers $a=x+3 y$ and $b=3 x+y$. Since $0<x, y<1$, it follows that $0<a, b<4$, or rather $1 \leqslant a, b \leqslant 3$.
Now, conversely, let $a$ and $b$ be two integers with $1 \leqslant a, b \leqslant 3$. There is exactly one pair of numbers $(x, y)$ for which... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In a table consisting of $n$ by $n$ small squares some squares are coloured black and the other squares are coloured white. For each pair of columns and each pair of rows the four squares on the intersections of these rows and columns must not all be of the same colour.
What is the largest possible value of $n$ ?
|
1. We will prove that $n=4$ is the largest possible $n$ for which an $n \times n$-table can be coloured according to the rules. The following figure shows a valid colouring for $n=4$.
Now we... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Call a positive integer a hussel number if:
(1) All digits are not equal to 0.
(2) The number is divisible by 11.
(3) The number is divisible by 12. If you rearrange the digits in any other random order, you always get a number that is divisible by 12.
How many 5-digit hussel numbers are there? | ## 1. Version for Class 4 and Lower
To begin with, we note that a hussel number can only contain the digits 2, 4, 6, and 8. Indeed, if we arrange the digits in any random order, we get an even number (divisible by 12) due to property (3). Since the last digit of an even number is always even, that digit must be 2, 4, ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The numbers 1 through 25 are each colored blue or red. Determine all possible colorings that satisfy the following rules:
- The number 5 is red.
- If the numbers $x$ and $y$ have different colors and $x+y \leqslant 25$, then $x+y$ is blue.
- If the numbers $x$ and $y$ have different colors and $x \cdot y \leqslant ... | ## 2. Version for class $5 \&$ class 4 and lower
First, we consider the case where 1 is red. Then all numbers from 1 to 25 must be red. Suppose a certain number $k$ is blue. Then 1 and $k$ have different colors, so according to the third rule, the number $1 \cdot k = k$ must be red. But this number was blue, which is ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The numbers 1 through 15 are each colored blue or red. Determine all possible colorings that satisfy the following rules:
- The number 15 is red.
- If the numbers $x$ and $y$ have different colors and $x+y \leqslant 15$, then $x+y$ is blue.
- If the numbers $x$ and $y$ have different colors and $x \cdot y \leqslant... | ## 2. Version for Grade 6
First, let's consider the case where 1 is red. Then all numbers from 1 to 15 must be red. Suppose a certain number $k$ is blue. Then 1 and $k$ have different colors, so according to the third rule, the number $k=1 \cdot k$ must be red. But this number was blue, which is a contradiction. Note ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
B2 The point $S$ lies on the chord $A B$ of a circle such that $S A=3$ and $S B=5$. The radius of the circle from the center $M$ through $S$ intersects the circle at $C$. Given $C S=1$.
Calculate the length of the radius of the circle.
^{2}$
(1) - (2) gives $15=r^{2}-(r-1)^{2... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. We call a positive integer squareful if every two adjacent digits in the number form a square. The number 364 is, for example, squareful, because both 36 and 64 are squares.
How many digits does the largest squareful number have?
A) 2
B) 3
C) 4
D) 5
E) 6 | A1. D) 5 The squares of two digits are $16, 25, 36, 49, 64$, and 81. We start with each of these squares and look for which digit we can append so that the last two digits also form a square. For example: after 16, a 4 can be appended because of 64 (and no other digit can follow the 6), then comes a 9 to form 49, and t... | 5 | Number Theory | MCQ | Yes | Yes | olympiads | false |
4. We place the numbers 1 through 9 one by one in a $3 \times 3$ grid. The number 1 can be placed in any cell; each subsequent number must be placed in a cell that is horizontally or vertically adjacent to the cell containing the previous number. See the example picture below. We call such a grid a snake grid. The scor... | A4. E) 5
We color the cells of the $3 \times 3$ grid alternately gray and white and assign each cell a letter, as shown in the figure below. A snake always alternates between the white and gray cells, and thus alternates between even and odd numbers. Since there are five odd numbers and five white cells, the odd numbe... | 5 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
4. Pjotr has a bathtub with a plug at the bottom that he can fill with two identical taps. The water flows from both taps at the same constant speed, and the bathtub drains (if the plug is not in) at a constant speed.
On Monday, Pjotr lets the bathtub fill to the brim by turning on one tap fully, after which he pulls ... | B4. 10 Given that water flows from a tap at a rate of $v$ liters per minute and the bathtub drains at a rate of $w$ liters per minute when the taps are off. Assume the bathtub can hold $b$ liters of water. We want to know how long it takes for a full bathtub to drain, which is $\frac{b}{w}$.
On Monday, the bathtub fir... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Xander draws five points and a number of infinitely long lines on an infinite sheet of paper, in such a way that at least two of those points lie on each line and that the lines intersect only at points that Xander has drawn.
What is the maximum number of lines that Xander can have drawn?
A) 3
B) 4
C) 5
D) 6
E) 7 | A5. D) 6 We can get 6 lines as follows. We take the extended sides of a square together with the two diagonals. The 5 points are now the four corners of the square and the center, see also the figure below.
We will show that 7 (or more) lines are not possible. We do this by contradiction: assume that a situation with ... | 6 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
1. Arthur writes down five different positive integers less than $10 \mathrm{on}$. If you add any two of these five numbers, the sum is never 10.
Which number has Arthur certainly written down?
A) 1
B) 2
C) 3
D) 4
E) 5 | A1. E) 5 Of the numbers 1 and 9, Arthur could have written down only one, because no two written numbers add up to 10. The same applies to 2 and 8, for 3 and 7, and also for 4 and 6. Of the eight numbers 1, 2, 3, 4, 6, 7, 8, and 9, Arthur has thus written down at most four. This means that he must have written down the... | 5 | Number Theory | MCQ | Yes | Yes | olympiads | false |
4. Er zijn 13 verschillende veelvouden van 7 die uit twee cijfers bestaan. Je wilt een zo lang mogelijke keten maken van deze veelvouden, waarbij twee veelvouden alleen naast elkaar mogen staan als het laatste cijfer van het linker veelvoud gelijk is aan het eerste cijfer van het rechter veelvoud. Je mag elk veelvoud ... |
A4. B) 7 De dertien veelvouden zijn: $14,21,28,35,42,49,56,63,70,77,84,91$ en 98.
In de veelvouden 70 en 77 komen alleen de cijfers 0 en 7 voor. Bovendien zijn het de enige veelvouden met een 0 of 7 . Met 70 of 77 kan de keten dus hooguit lengte 2 hebben.
Op dezelfde manier zijn de veelvouden 35, 56 en 63 de enige v... | 7 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
4. Alice has a number of cards. On each card, three of the letters A through I are written. For any two letters you choose, there is at least one card where those two letters appear together.
What is the smallest number of cards Alice can have? | B4. 12 There are 9 letters, namely $A$ through $I$. The number of possible pairs of letters is $\frac{9 \times 8}{2}=36$. Indeed, each of the 9 letters can form a pair with 8 other letters, but this way we count each pair twice (pair AE is the same as pair EA).
Each card has three different letters that form 3 pairs. ... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. By replacing every $*$ in the expression $1 * 2 * 3 * 4 * 5 * \cdots * 2019 * 2020$ with a plus or minus sign ( + or - ), a long arithmetic expression is formed. Place the plus and minus signs in such a way that the result is the smallest possible positive number (greater than 0).
What is this result? | B1. 2 We look at the expression from back to front. By taking $2017-2018-2019+2020$, we get exactly zero. Before that, we choose $2013-2014-2015+2016$ and so on until $5-6-7+8$. At the beginning, we take $1+2+3-4$ and thus we get a result of 2.
Now we only need to show that the result is never equal to 1, making 2 the... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Annemiek and Bart each wrote three different positive integers on a piece of paper. It turns out that there is exactly one number that appears on both of their papers. Furthermore, if you take two different numbers from Annemiek's paper and add them together, the result is always a number on Bart's paper. One of the... | B3. Suppose we call the three numbers on Annemieke's note $a, b$, and $c$ and assume that $a<b<c$. Then the three numbers on Bart's note, in order of size, are $a+b, a+c$, and $b+c$. The last two numbers are greater than $c$ and thus greater than any number on Annemieke's note. The number that appears on both notes mus... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. We consider rows of 2020 coins. Each coin has a value of 1, 2, or 3. Between two coins with value 1, there is always at least one other coin. Between two coins with value 2, there are always at least two other coins. Between two coins with value 3, there are always at least three other coins.
How many different row... | B4. 10 Consider a continuous segment where no coin of value 3 is present. In this segment, there can be at most one coin of value 2, because otherwise we would have two coins of value 2 with only coins of value 1 in between, which is not allowed. Since there can be at most one coin of value 2, there can also be at most... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On a $4 \times 4$ board, there are 16 grasshoppers, each on its own square. At a certain moment, each grasshopper jumps to an adjacent square: a square up, down, to the right, or to the left, but not diagonally and not off the board.
What is the maximum number of squares that can be empty afterward?
A) 8
B) 9
C) 10... | A5. C) 10 We color the squares of the $4 \times 4$ board in a checkerboard pattern and name the squares from a1 to d4 as shown in the figure. We see that the grasshoppers on white squares all jump to a black square, and the grasshoppers on black squares jump to a white square.
. Also, each of the three columns (read from top to bottom) is a correct arithmetic operation. However, the digits in the table have been replaced by letters. Different letters stand for different digits, an... | A6. E) 9
From the first row, it follows that
$$
\mathrm{ADF}+\mathrm{F}=\mathrm{ABC} .
$$
The digit $\mathrm{C}$ is therefore even and, moreover, not 0. The possibilities are thus: 2, 4, 6, and 8. From the middle row, it follows that
$$
\mathrm{C} \cdot \mathrm{GC}=\mathrm{ADD} .
$$
Since $2 \cdot 2=4, 4 \cdot 4=1... | 9 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
8. Lieneke makes beaded bracelets. Each bracelet has six beads: two white, two gray, and two black beads. Some bracelets may seem different at first glance, but they are not: if you rotate or flip one, it turns out to be the same as the other. For example, the three bracelets below are the same.
 11 We divide the bracelets into three groups: (I) the black beads are next to each other, (II) there is one bead between the two black beads, and (III) the black beads are opposite each other. In each case, we count how many different ways we can place the remaining beads.
 6
(B) 8 (C) 10
(D) 12
(E) 15
 | A2. In the left triangle, the third angle is equal to 180-7 $x$ degrees. In the right triangle, the third angle is equal to 180-13 $x$ degrees. The two base angles of the middle triangle are each equal to one of those angles.
So it must be true that: $5 x+(180-7 x)+(180-13 x)=180$ thus $15 x=180$ and therefore $x=12$. | 12 | Geometry | MCQ | Yes | Yes | olympiads | false |
A5. In a magic square, the sum of the three numbers in each row, the sum of the three numbers in each column, and the sum of the three numbers in each of the two diagonals is always the same number.
In the magic square below, four of the nine numbers are given.
What is the value of $N$?
| | $\boldsymbol{N}$ | |
| ... | A5. Compare the left column with the diagonal in which 10 is located. In the middle box (below the $N$), 13 must be placed because it must hold that $13+10=12+11$. In the middle below, 17 must be placed because $(11+13+15)-(12+10)=17$. Therefore, $N=39-13-17=9$. | 9 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
1. An archipelago consists of a large, a medium, and a small island. The total area of the three islands is $23 \mathrm{~km}^{2}$. The difference between the areas of the large and the medium island turns out to be exactly $1 \mathrm{~km}^{2}$ more than the area of the small island. How many $\mathrm{km}^{2}$ is the ar... | A1. C) 12 The difference between the areas of the large and the medium island is 1 $\mathrm{km}^{2}$ more than the area of the small island. In other words, the area of the medium island plus the area of the small island is $1 \mathrm{~km}^{2}$ less than the area of the large island. Therefore, the sum of the areas of ... | 12 | Algebra | MCQ | Yes | Yes | olympiads | false |
8. Michael prints the result below twice on cardboard and makes two identical dice such that the visible dots are on the outside of the dice. He stacks them to form a tower. The front side of the bottom die shows 3 dots. The total number of dots on the two touching sides of the dice is 9. The total number of dots on th... | A8. A) 1 On the bottom die, we see 3 eyes on the front, so there is a 4 on the back. On the right side of the bottom die, there can still be 1, 2, 5, or 6. With 5 or 6, the total number of eyes on the back must be at least 15 or 18, and that is not possible. If there is a 1 on the right side, then there is a 2 on the t... | 1 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
A4. How many positive integers $N$ of two digits are there with the following property: The sum of $N$ and the reverse of $N$ is a square of an integer (the reverse of $N$ is the number you get by writing $N$ backwards)? | A4 If you write the number $N$ as $a b$ then $N=10 a+b$. The reverse of $N$ is then equal to $10 b+a$. Together this gives $11 a+11 b=11(a+b)$. Since the latter must be a square, it must hold that $(a+b)$ is equal to 11 or a square times 11. $(a+b)=11$ gives the solutions $29,38,47,56,65,74,83$ and 92. If $(a+b)$ must ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In a classroom, there are chairs and stools. A child is sitting on each chair and each stool. Each chair has 4 legs, each stool has 3 legs, and each child has 2 legs. Together, this results in a total of 39 legs and feet.
How many chairs are there in the classroom?
A) 3
B) 4
C) 5
D) 6
E) 9 | A1. B) 4
A chair with a child on it has a total of 6 legs and feet. A stool with a child on it has a total of 5 legs and feet. With 7 chairs, we already have $7 \times 6 = 42$ legs and feet, which is too many. We create a table with the number of chairs and the number of legs and feet:
| Number of chairs: | 0 | 1 | 2... | 4 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
3. If you continue the chain of squares and regular pentagons in the same way, does it close precisely upon itself after going all the way around? If so, how many pentagons does the chain contain in total?
A) 9
B) 10
C) 11
D) 12
E) It does not close.
 10 The angles of a regular $n$-gon are equal to $180-\frac{360}{n}$ degrees. Indeed, if you walk around the $n$-gon, you change direction $n$ times, after which you have turned a total of 360 degrees. At each vertex, your direction changes by $\frac{360}{n}$ degrees, and the corresponding angle of the $n$-gon is... | 10 | Geometry | MCQ | Yes | Yes | olympiads | false |
4. Julian wants to make a list of numbers as long as possible. Each number on the list must consist of one or more of the digits 1 through 9. Additionally, it must be true that
- each of the digits 1 through 9 is used exactly once;
- no number on the list is divisible by any other number on the list.
What is the maxi... | A4. D) 7 The only list of 9 numbers that meets the first requirement consists of the nine digits each as a separate number. This list does not meet the second property (since 2, for example, is divisible by 1).
A list of 8 numbers that meets the first requirement must consist of seven single-digit numbers and one two-... | 7 | Number Theory | MCQ | Yes | Yes | olympiads | false |
5. At a party, there are nine people. Upon entering, some people greeted each other by shaking hands. Quintijn is present at the party and asks everyone else how many people they have shaken hands with. He receives eight different answers.
How many people did he shake hands with?
A) 0
B) 1
C) 2
D) 3
E) 4 | A5. E) 4 Of the eight people Quintijn meets at the party, let's call them A, B, C, D, E, F, G, and H in ascending order of the number of hands they shake. Since there are a total of 9 people present, the number of hands someone shakes is always a number from 0 to 8. If someone shakes 8 hands, then he shakes hands with ... | 4 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
A1. The numbers 1, 2, 3, 4, 5, 6, 7, and 8 must be distributed over the circles in the figure such that the sum of the three numbers on each of the four sides of the square is the same.
Provide a solution where this sum is minimized.
$ of positive integers with $a+b<100$ satisfy the equation: $a+\frac{1}{b}=13 \times\left(b+\frac{1}{a}\right)$?
(A) 5
(B) 7
(C) 9
(D) 13
(E) 28 | A4. $\frac{a b+1}{b}=13 \times \frac{b a+1}{a}$. Thus $a=13 b$ and hence the seven pairs $(13,1),(26,2), \cdots,(91,7)$. (B) | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
A6. A sequence of numbers is constructed as follows:
The first number is 2. The second number is 2.
Each subsequent number is the product of its two predecessors.
The sequence thus becomes: $2, 2, 4, 8, 32, \ldots$
What is the last digit of the $2007^{\text{th}}$ number in this sequence?
(A) 0
(B) 2
(C) 4
(D) 6
(E)... | A6. We continue the row with only the last digits of the numbers: we find
$2,2,4,8, \cdot 2, \cdot \cdot 6, \cdot 2, \cdot \cdot 2, \cdot 4, \cdot 8, \cdot 2, \cdot 6, \cdot 2$, etc. We see that a group of six digits repeats itself. Since $2007=334 \times 6+3$, the last digit of the 2007th number is the same as that o... | 4 | Number Theory | MCQ | Yes | Yes | olympiads | false |
2. In a toy store, at the beginning of the day, there are 20 sticker sheets (0.30 euros each), 18 footballs (3 euros each), 5 teddy bears (5 euros each), and 8 water guns (15 euros each) in stock. The new cashier makes a mess of the administration and reports only the total amount of 75.80 euros at the end of the day f... | A2. D) 4 Because three of the four items have a price in whole euros, the sticker sheets must have either 6 sold (for 1.80 euros) or 16 (for 4.80 euros). More is not possible, as there were only 20 in stock. For the other three items, there is thus 74 or 71 euros left. First, assume that there is 74 euros left. Since t... | 4 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
3. A chessboard consists of $8 \times 8$ squares. In a corner square stands a knight. The knight makes jumps of either two squares horizontally and one square vertically, or two squares vertically and one square horizontally. In each square of the chessboard, we write down how many jumps the knight needs to reach it.
... | A3. C) 4
In each field of the chessboard, we write down how many moves a knight needs to get there. We start with 0 in the corner, then write a 1 in the (empty) squares that can be reached from the 0 in one move, then write a 2 in the (empty) squares that can be reached from a 1 in one move, and so on. Eventually, we ... | 4 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
4. We have a triangle $\triangle A B C$ and two points $D$ and $E$ on the line segment $B C$, such that $\angle B A D=\angle D A E=\angle E A C$. Furthermore, $|A B|=6,|B D|=4,|D E|=2$ and $|E A|=3$. See the picture below: note that it is not drawn to scale!
What is the length of the line segment $C E$?
 3 Name the three equal angles $\alpha$. We see that $|E B|=|E D|+|D B|=2+4=$ $6=|A B|$, and thus triangle $\triangle A B E$ is an isosceles triangle. From this it follows that $\angle A E B=$ $\angle B A E=2 \alpha$ and due to the straight angle at $E$ we find that $\angle A E C=180^{\circ}-2 \alpha$. Since $\an... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
7. How many pairs of positive integers $a$ and $b$ with $a<b$ are there such that
$$
\frac{1}{a}+\frac{1}{b}=\frac{4}{15} ?
$$
A) 1
B) 2
C) 3
D) 4
E) 5 | A7. C) 3 Because $a\frac{1}{b}$. In particular, $\frac{1}{a}$ must be more than half of $\frac{4}{15}$: $\frac{1}{a}>\frac{2}{15}$. From this, it follows that $a<\frac{15}{2}<8$. Moreover, $a$ must be at least 4, otherwise $\frac{1}{a}$ would be greater than $\frac{4}{15}$. So we have four possibilities for $a$: 4, 5, ... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
1. For integers $a, b, c$ and $d$, it is given that the difference between $a$ and $b$ is 2, the difference between $b$ and $c$ is 3, and the difference between $c$ and $d$ is 4.
Which of the following values cannot be the difference between $a$ and $d$?
A) 1
B) 3
C) 5
D) 7
E) 9 | A1. D) 7
From the data in the problem, we deduce that $a-b= \pm 2, b-c= \pm 3$ and $c-d= \pm 4$.
We then get
$$
a-d=(a-b)+(b-c)+(c-d)= \pm 2 \pm 3 \pm 4
$$
We can now go through all eight possibilities for the plus and minus signs:
$$
\begin{array}{rrrr}
2+3+4=9, & 2+3-4=1, & 2-3+4=3, & 2-3-4=-5, \\
-2-3-4=-9, & -... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
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