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Using only the digits 1, 2, 3, 4, and 5, Peri constructed the sequence
$$
1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,1,1,1,1,1,1,2,2,2,2,2,2,2, \ldots
$$
starting with one 1, followed by two 2s, three 3s, four 4s, five 5s, six 1s, seven 2s, and so on. What is the hundredth term of this sequence? | We group the sequence into consecutively numbered blocks, each block formed by consecutive equal terms, as shown below.
$$
\begin{aligned}
& \underbrace{1}_{\text {block } 1}, \underbrace{2,2}_{\text {block } 2}, \underbrace{3,3,3}_{\text {block } 3}, \underbrace{4,4,4,4}_{\text {block } 4}, \underbrace{5,5,5,5,5}_{\t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A point $P$ is at the center of a square with a side length of $10 \mathrm{~cm}$. How many points on the edge of the square are $6 \mathrm{~cm}$ away from $P$?
(a) 1
(b) 2
(c) 4
(d) 6
(e) 8 | The correct option is (e).
The points that are $6 \mathrm{~cm}$ away from point $P$ form a circle with center $P$ and radius $R=6$ $\mathrm{cm}$. If $d$ denotes the diagonal of the square, by the Pythagorean Theorem we have
$$
d=\sqrt{10^{2}+10^{2}}=\sqrt{2 \times 10^{2}}=10 \sqrt{2}
$$
The circle with radius $L / 2... | 8 | Geometry | MCQ | Yes | Yes | olympiads | false |
If $2\left(2^{2 x}\right)=4^{x}+64$, what is the value of $x$?
(a) -2
(b) -1
(c) 1
(d) 2
(e) 3 | The correct option is (e).
Solution 1: We observe that the terms on the right side of the given equation can be written as powers of 2. Indeed, $4^{x}=\left(2^{2}\right)^{x}=2^{2 x}$ and $64=2^{6}$. Thus, the equation becomes $2\left(2^{x}\right)=2^{2 x}+2^{3}$. We then have $2\left(2^{2 x}\right)-2^{2 x}=2^{6}$, whic... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
In the figure, the number 8 was obtained by adding the two numbers directly below its position. By doing the same to fill in the blank spaces, one gets 42 in the indicated position. What is the value of $x$?
(a) 7
(b) 3
(c) 5
(d) 4
(e) 6
.
Using the given rule, we fill in the empty cells starting from the second row from the bottom and obtain the figure. Therefore,
$$
(13+x)+(11+2 x)=42
$$
and thus, $24+3 x=42$, which means $x=6$.
 0
(b) 2
(c) 4
(d) 6
(e) 8 | The correct option is (c).
Solution 1: The last digit of $9867^{3}$ is the same as that of $7^{3}=343$, which is 3. The last digit of $9867^{2}$ is the same as that of $7^{2}=49$, which is 9. If we subtract a number ending in 9 from another ending in 3, the last digit of the result is 4.
Note: Observe that the last d... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
The function $f$ is given by the table
| $x$ | 1 | 2 | 3 | 4 | 5 |
| :---: | :--- | :--- | :--- | :--- | :--- |
| $f(x)$ | 4 | 1 | 3 | 5 | 2 |
For example, $f(2)=1$ and $f(4)=5$. What is the value of $\underbrace{f(f(f(f(\ldots f}_{2004 \text { times }}(4) \ldots)))$ ?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5 | The correct option is (d).
From the table
| $x$ | 1 | 2 | 3 | 4 | 5 |
| :---: | :--- | :--- | :--- | :--- | :--- |
| $f(x)$ | 4 | 1 | 3 | 5 | 2 |
we obtain
$$
\begin{aligned}
& f(4)=5, f(\underbrace{f(4)}_{5})=f(5)=2, f(f(\underbrace{f(4)}_{5}))=f(\underbrace{f(5)}_{2})=f(2)=1 \text{ and} \\
& \underbrace{f(f(f(f(4... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
A bus, a train, and a plane depart from city A to city B at the same time. If I take the bus, which has an average speed of 100 $\mathrm{km} / \mathrm{h}$, I will arrive in city B at 8 PM. If I take the train, which has an average speed of $300 \mathrm{~km} / \mathrm{h}$, I will arrive in city B at 2 PM. What will be t... | Let $d$ be the distance between the two cities and $h$ be the common departure time of the bus, the train, and the airplane. Since distance $=$ speed $\times$ time, we have $d=100 \times(20-h)$ and $d=300 \times(14-h)$. Therefore, $100 \times(20-h)=300 \times(14-h)$, from which $h=11$. Thus, the distance between the tw... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
How many different pairs of positive integers $(a, b)$ are there such that $a+b \leq 100$ and $\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=13$?
(a) 1
(b) 5
(c) 7
(d) 9
(e) 13 | The correct option is (c).
We have
$$
13=\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=\frac{\frac{a b+1}{b}}{\frac{1+a b}{a}}=\frac{(a b+1) \times a}{(1+a b) \times b}=\frac{a}{b}
$$
Thus, $a=13 b$. Since $a+b \leq 100$, it follows that $14 b \leq 100$ and, therefore, $b \leq 7.14$. Since $b$ is an integer, we must have $b \... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
In the figure, the three circles are concentric, and the area of the smallest circle coincides with the area of the largest ring, highlighted in gray. The radius of the smallest circle is $5 \mathrm{~cm}$ and of the largest $13 \mathrm{~cm}$. What is the radius (in cm) of the intermediate circle?
(a) 12
(c) $10 \sqrt{6... | The correct option is (a).
The area of the largest circle is $13^{2} \pi=169 \pi$ and that of the smallest is $5^{2} \pi=25 \pi$, which is also the area of the largest ring. Let $r$ be the radius of the intermediate circle. Then, the area of the largest ring is $169 \pi-\pi r^{2}$. Therefore, $169 \pi-\pi r^{2}=25 \pi... | 12 | Geometry | MCQ | Yes | Yes | olympiads | false |
How many of the numbers $-5,-4,-3,-2,-1,0$, $1,2,3$ satisfy the inequality $-3 x^{2}<-14$?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5 | The correct option is (d).
If $-3 x^{2}14$, or $x^{2}>\frac{1}{3} 14=4 \frac{2}{3}$. Since we are only looking at integer values of $x$, then $x^{2}$ is also an integer. Given that $x^{2}>4 \frac{2}{3}$, we conclude that $x^{2}$ is at least 5. Among the numbers $-5,-4,-3,-2,-1,0,1,2,3$ only four, namely, $-5,-4,-3$ an... | 4 | Inequalities | MCQ | Yes | Yes | olympiads | false |
If $S_{n}=1-2+3-4+5-6+\cdots+(-1)^{n+1} n$ for each positive integer $n$, then $S_{1992}+S_{1993}$ is equal to
(a) -2 ;
(b) -1 ;
(c) 0 ;
(d) 1 ;
(e) 2 . | The correct option is (d).
The expression $(-1)^{n+1}$ in the definition of $S_{n}$ has a value of 1 if $n$ is even and a value of -1 if $n$ is odd.
Solution 1: By associating consecutive terms in pairs, we obtain a sum of several terms equal to -1: $(1-2)+(3-4)+(5-6)+\cdots$. Therefore,
$$
S_{1992}=\underbrace{(1-2... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
How many positive integers $n$ exist such that the quotient $\frac{2 n^{2}+4 n+18}{3 n+3}$ is an integer? | As
$$
\frac{2 n^{2}+4 n+18}{3 n+3}=\frac{2}{3}\left[\frac{\left(n^{2}+2 n+1\right)+8}{n+1}\right]=\frac{1}{3}\left(2 n+2+\frac{16}{n+1}\right)
$$
it follows that the expression in parentheses must be a multiple of 3 and, in particular, $n+1$ must divide 16. Thus, $n$ can be $1, 3, 7$ or 15. From the table below, in e... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Four mayors decide to build a circular highway that passes within the boundaries of their cities. Since the four cities are not on the same circle, the mayors hire a company to develop a project for the construction of a circular highway equidistant from the four cities. What is the maximum number of geographically dis... | The number of highways is equal to the number of points that can be the center of a circle (highway) that is equidistant from four given points (cities). Since no circle passes through all four points, if any circle is equidistant from the four points, this circle cannot leave all four points on the inside or all on th... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Five cards are on a table, and each has a number on one side and a letter on the other. Simone must decide whether the following statement is true: "If a card has a vowel on one side, then it has an even number on the other." What is the minimum number of cards she needs to turn over to make a correct decision?
![](ht... | Simone doesn't need to turn over the card that has the number $\mathbf{2}$ because the other side, whether a vowel or a consonant, will satisfy the condition "If a card has a vowel on one side, then it has an even number on the other."
(a) How many squares can we construct such that their vertice... | (a) The only squares that do not have any of their sides parallel to line $r$ or line $s$ are those of type 1 and type 2 (see figures).
Thus, there are a total of six squares, four of type ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
All natural numbers starting from 1 were written consecutively, forming a sequence of digits, as follows.
$$
1234567891011121314151617181920212223 \ldots
$$
What is the digit that appears in the 206788th position? | The numbers with one digit form the first 9 terms of the sequence. The 90 two-digit numbers form the next 180 terms. Then come the 2700 terms corresponding to the three-digit numbers, followed by the 36000 terms corresponding to the four-digit numbers, and finally, the 450000 terms corresponding to the five-digit numbe... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
How many zeros are there at the end of the number $9^{2007}+1$? | Initially, we verify how the powers of 9 end, that is, we list the last two digits, the tens and units, of the powers $9^{n}$, in order.
| If $n$ is | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
When Isabel was born, her mother was turning 20 years old. If Isabel and her mother live for another 100 years, how many times will their ages be multiples of each other? | When Isabel is $a$ years old, her mother is $20+a$ years old. If $a$ is a divisor of $20+a$, then $(20+a) / a=(20 / a)+1$ is an integer, and consequently, $20 / a$ is also an integer. Therefore, $a$ is a divisor of 20, and thus $a$ can be 1, 2, 4, 5, 10, or 20. So, there are a total of 6 times when their ages are multi... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
An empty swimming pool was filled with water by two faucets A and B, both with constant flow rates. For four hours, both faucets were open and filled $50 \%$ of the pool. Then, faucet B was turned off and, for two hours, faucet A filled $15 \%$ of the pool's volume. After this period, faucet A was turned off and faucet... | Given that taps A and B pour water into the pool at a constant flow rate, the volume of water poured by each tap is proportional to the time it is open. Therefore, if tap A fills $15 \%$ of the pool volume in two hours, then in four hours it will fill $30 \%$ of the pool volume.
However, when taps A and B are both ope... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A triangle has vertices $A=(3,0), B=(0,3)$, and $C$, where $C$ lies on the line with equation $x+y=7$. What is the area of this triangle? | Observe that the height $h$ relative to side $AB$ of all triangles $\triangle ABC$ that have vertex $C$ on the line $x+y=7$ is always the same, since the line $x+y=7$ is parallel to the line $x+y=3$ that passes through $A$ and $B$. Therefore, all these triangles have the same area, namely,
$$
\frac{1}{2}(AB \times h)
... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Three circles, with radii measuring 1, 2, and $3 \mathrm{~cm}$, are pairwise externally tangent, as shown in the given figure. Determine the radius of the circle that is externally tangent to the three circles.
$ the sum of the digits of the number $n$. For example, $s(2345) = 2 + 3 + 4 + 5 = 14$. Observe that:
\[
\begin{gathered}
40 - s(40) = 36 = 9 \times 4; \quad 500 - s(500) = 495 = 9 \times 55 \\
2345 - s(2345) = 2331 = 9 \times 259
\end{gathered}
\]
(a) What can we say about the number $n - s(n)$?... | (a) It is immediate that if $a$ is a digit between 1 and 9, then $s\left(a \cdot 10^{k}\right)=a$, since the number $a \cdot 10^{k}$ is formed by the digit $a$ followed by $k$ zeros. Therefore, we have
$$
a \cdot 10^{k}-s\left(a \cdot 10^{k}\right)=a \cdot 10^{k}-a=a\left(10^{k}-1\right)=a \times \underbrace{9 \cdots ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
If $S_{n}=1-2+3-4+5-6+\ldots+(-1)^{n+1} n$, where $n$ is a positive integer, then $S_{1992}+S_{1993}$ is:
(a) -2
(b) -1
(c) 0
(d) 1
(e) 2 | Solution 1: Remember that
$(-1)^{n+1}= \begin{cases}1 & \text { if } n \text { is odd } \\ -1 & \text { if } n \text { is even }\end{cases}$
Observe that by grouping consecutive terms in pairs,
$$
(1-2)+(3-4)+(5-6)+\cdots
$$
we get a sum of $n$ terms all equal to -1. Therefore,
$S_{1992}=\underbrace{(1-2)+(3-4)+(5-... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
Ester goes to a stationery store to buy notebooks and pens. In this stationery store, all notebooks cost $\mathrm{R} \$ 6.00$. If she buys three notebooks, she will have R \$4.00 left. If, instead, her brother lends her an additional $\mathrm{R} \$ 4.00$, she will be able to buy two notebooks and seven pens, all the sa... | By buying three notebooks at 6 reais each, Ester still has 4 reais left, so the amount she has is $3 \times 6 + 4 = 22$ reais.
(a) If her brother lends her 4 reais, she then has $22 + 4 = 26$ reais and can buy 2 notebooks at 6 reais each, leaving her with $26 - 2 \times 6 = 26 - 12 = 14$ reais for 7 pens. We conclude ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Diamantino put three liters of water and one liter of soft drink in a container. The soft drink is composed of $20 \%$ orange juice and $80 \%$
water. After mixing everything, what percentage of the final volume represents the orange juice?
(a) $5 \%$
(b) $7 \%$
(c) $8 \%$
(d) $20 \%$
(e) $60 \%$ | The correct option is (a).
The refreshment is composed of $20 \%$ of a liter, that is, 0.2 liters of juice and $80 \%$ of a liter, that is, 0.8 liters of water. Therefore, the final mixture has 0.2 liters of juice and $3+0.8=3.8$ liters of water. The percentage of juice in relation to the volume of the mixture is then... | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
What is the value of $2^{6}+2^{6}+2^{6}+2^{6}-4^{4}$?
(a) 0
(b) 2
(c) 4
(d) $4^{2}$
(e) $4^{4}$ | The correct option is (a).
We have $2^{6}+2^{6}+2^{6}+2^{6}-4^{4}=4 \times 2^{6}-4^{4}$. There are several ways to calculate this.
Solution 1: $4 \times 2^{6}-4^{4}=4 \times\left(2^{2}\right)^{3}-4^{4}=4 \times 4^{3}-4^{4}=4^{4}-4^{4}=0$.
Solution 2: $4 \times 2^{6}-4^{4}=4\left(2^{6}-4^{3}\right)=4\left[2^{6}-\left... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
In a year, at most how many months have five Sundays?
(a) 3
(b) 4
(c) 5
(d) 6
(e) 7 | The correct option is (c).
A common year has 365 days and a leap year has 366. From the division of 365 by 7, we get $365=52 \times 7+1$ and from the division of 366 by 7 we get $366=52 \times 7+2$. Therefore,
$$
\begin{aligned}
\text { common year } & =52 \text { weeks }+1 \text { day } \\
\text { leap year } & =52 ... | 5 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
What is the unit digit of the number
$$
1 \times 3 \times 5 \times \cdots \times 97 \times 99 ?
$$
(a) 1
(b) 3
(c) 5
(d) 7
(e) 9 | The correct option is (c).
The last digit of a multiple of 5 is 0 or 5; those ending in 0 are even and those ending in 5 are odd. Since $1 \times 3 \times 5 \times \cdots \times 97 \times 99$ is odd, being a product of odd numbers, and is a multiple of 5, it follows that its units digit is 5. | 5 | Number Theory | MCQ | Yes | Yes | olympiads | false |
The figure shows the map of a country (imaginary) consisting of five states. It is desired to color this map with the colors green, blue, and yellow, so that two adjacent states do not have the same color. In how many different ways can the map be painted?
.
State A can be painted in three ways: green, blue, or yellow. For any neighboring state, for example, state $\mathrm{B}$, we have two possibilities, and the colors of the other states are determined. Therefore, we can color the map in $3 \times 2=6$ ways.
Below we illustrate two of these wa... | 6 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
In a $3 \times 3$ board, nine houses must be painted such that in each column, each row, and each of the two diagonals, there are no two houses of the same color. What is the minimum number of colors needed for this?
(a) 3
(b) 4
(c) 5
(d) 6
(e) 7 | The correct option is (c).
To satisfy the conditions of the problem, the five houses on the diagonals, marked with *, must have different colors. Therefore, we will need at least five distinct colors. Let's denote these five distinct colors by 1, 2, 3, 4, and 5, and determine how we can choose the colors for the remai... | 5 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
If $\frac{n}{24}$ is a number between $\frac{1}{6}$ and $\frac{1}{4}$, who is $n$?
(a) 5
(b) 6
(c) 7
(d) 8
(e) 9 | The correct option is (a).
Since $\frac{1}{6}=\frac{4}{24}$ and $\frac{1}{4}=\frac{6}{24}$, then $n$ can only be equal to 5. | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
What is the smallest natural number $n$ for which $10^{n}-1$ is a multiple of 37?
(a) 6
(b) 5
(c) 4
(d) 3
(e) 2 | The correct option is (d).
Observe that $10^{n}-1$ is a number that has all its digits equal to 9. Note, also, that a multiple of 37, of the form $37 \times n$, only ends in 9 if $n$ ends in 7. Therefore, the smallest multiples of 37 ending in 9 are $37 \times 7=259$, $37 \times 17=629$, and $37 \times 27=999$. Since ... | 3 | Number Theory | MCQ | Yes | Yes | olympiads | false |
What is the value of $2-2\{2-2[2-2(4-2)]\}$?
(a) 0
(b) 2
(c) -2
(d) 4
(e) -10 | The correct option is (e).
The order of precedence for solving an expression is
$$
\underbrace{\text { parentheses }}_{1 \varrho} \rightarrow \underbrace{\text { brackets }}_{2 \varrho} \rightarrow \underbrace{\text { braces }}_{3 \propto}
$$
and
$$
\underbrace{\text { multiplications and divisions }}_{1 \varrho} \... | -10 | Algebra | MCQ | Yes | Yes | olympiads | false |
What is the digit $a$ in $a 000+a 998+a 999=22$ 997? | Performing the addition
| 111 |
| ---: |
| $a 000$ |
| $a 998$ |
| $+a 999$ |
| $\square 997$ |
we find $\square 997=22997$, where $\square=a+a+a+1$. Therefore, $22=a+a+a+1$, and thus, $a=7$. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A game starts with seven coins aligned on a table, all with the crown face up. To win the game, you need to flip some coins in such a way that, in the end, two adjacent coins always have different faces up. The rule of the game is to flip two adjacent coins in each move. What is the minimum number of moves required to ... | Assigning the value 1 to heads and -1 to tails and summing the results after each flip, the game starts with a sum of 7 and we want to reach alternating heads and tails, so that the game ends at 1 or -1. We observe that, in each step of the game, we have the following possibilities: we exchange two heads for two tails ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The price of a kilogram of chicken was $R \$ 1.00$ in January 2000, when it started to triple every 6 months. How long will it take for the price to reach $\mathrm{R} \$ 81.00$?
(a) 1 year
(b) 2 years
(c) $2 \frac{1}{2}$ years
(d) 13 years
(e) $13 \frac{1}{2}$ years | The correct option is (b).
Since $81=3^{4}$, the value of the chicken has tripled four times. The number of months that have passed is $4 \times 6=24$ months, that is, two years, meaning that in January 2002, the chicken will reach a price of R\$ 81.00. | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
In a certain warehouse, a dozen eggs and 10 apples had the same price. After a week, the price of eggs dropped by $10 \%$ and the price of apples increased by $2 \%$ How much more will be spent on the purchase of a dozen eggs and 10 apples?
(a) $2 \%$
(b) $4 \%$
(c) $10 \%$
(d) $12 \%$
(e) $12.2 \%$ | The correct option is (b).
Since the statement and the answer are in percentages, we can, in this case, assume any price and any currency unit, and the answer will always be the same. The simplest approach, therefore, is to assume that initially, a dozen eggs cost 100 and that ten apples also cost 100. Since the price... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
The squares of the natural numbers from 1 to 99 were written one after another, forming the number 14916253649... What is the digit that occupies the 100th position? (The positions are counted from left to right, so the $1^{\underline{a}}$ position is the 1, the $2^{\underline{a}}$ is the 4, and so on.) | Separating the numbers whose squares have 1, 2, and 3 digits, we have,
$$
\begin{array}{ll}
\text { with } 1 \text { digit: } & 1,2,3 \\
\text { with } 2 \text { digits: } & 4,5,6,7,8,9 \\
\text { with } 3 \text { digits: } & 10,11,12, \ldots, 31
\end{array}
$$
Up to $31^{2}$, the number already has $3+12+66=81$ digi... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Seventy-four pencils were packed into 13 boxes. If the maximum capacity of each box is six pencils, what is the minimum number of pencils that can be in a box?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 6 | The correct option is (b).
Let's see in how many boxes we can place the maximum number of pencils, which is 6 per box. In 13 boxes it is not possible, because $13 \times 6=78$ is greater than the total number of pencils, which is 74. In 12 boxes we can have $12 \times 6=72$, leaving one box with $74-72=2$ pencils. | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
In the number $6 a 78 b$, $a$ represents the thousands digit and $b$ represents the units digit. If $6 a 78 b$ is divisible by 45, then the value of $a+b$ is:
(a) 5 ;
(b) 6 ;
(c) 7 ;
(d) 8 ;
(e) 9 . | The correct option is (b).
The number is divisible by $45=5 \times 9$, so it is divisible by 5 and 9. Every number divisible by 5 ends in 0 or 5. Thus, $b=0$ or $b=5$. Every number divisible by 9 has the sum of its digits as a multiple of 9. Therefore, $6+a+7+8+b=21+a+b$ is a multiple of 9. Since $a \leq 9$, and $b=0$... | 6 | Number Theory | MCQ | Yes | Yes | olympiads | false |
Henrique bought chocolate bars for $\mathrm{R} \$ 1.35$ each. He paid with a $\mathrm{R} \$ 10.00$ bill and received change less than $\mathrm{R} \$ 1.00$. How many bars did he buy? | As $8 \times 1.35 = 10.8$ is greater than 10, Henrique bought 7 chocolate bars and received $10 - 7 \times 1.35 = 0.55$ dollars, or 55 cents, in change. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The sum of three numbers is 100, two are prime and one is the sum of the other two.
(a) What is the largest of the three numbers?
(b) Give an example of such three numbers.
(c) How many solutions exist for this problem? | (a) Initially observe that, since the sum of the three numbers is 100 and the largest of them is equal to the sum of the other two, then twice the largest number is 100, that is, the largest number is 50.
(b) Since 50 is not a prime number, the other two numbers are primes and their sum is 50. For example, 3 and 47 ar... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Antônio has a parrot that performs fantastic calculations with integers. When Antônio whispers certain numbers into its ear, the parrot multiplies that number by 5, then adds 14, divides the result by 6, and finally subtracts 1, shouting the result afterward. However, the parrot knows nothing about decimals, so sometim... | (a) We have $8 \xrightarrow{\times 5} 40 \xrightarrow{+14} 54 \xrightarrow{\dot{\circ} 6} 9 \xrightarrow{-1} 8$. Therefore, the parrot shouts 8.
(b) We should perform the inverse operation of what the parrot did, starting from the last operation, that is, add 1 to the number, multiply the number by 6, then subtract 14... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Roberto wants to write the number 111111 as a product of two numbers, neither of which ends in 1. Is this possible? Why? | Factoring 111111, we obtain $111111=3 \times 7 \times 11 \times 13 \times 37$. It follows that it is indeed possible to write the number 111111 as a product of two factors, neither of which ends in 1. For example, $111111=3 \times 37037$. But there are other possibilities, such as, for example, $111111=7 \times 15873$.... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A class has 22 male students and 18 female students. During the holidays, $60 \%$ of the students in this class went to do community work. At a minimum, how many female students participated in this work?
(a) 1
(b) 2
(c) 4
(d) 6
(e) 8 | The correct answer is (b).
The total number of students in this class is $22+18=40$, of which $60\%$ went to do community work, that is, $0.6 \times 40=24$. The minimum number of female students who participated in this work is obtained when the number of male students who participated is maximum, that is, when all 22... | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
The sum on the side is incorrect. To correct it, it is enough to replace a certain digit in all the places where it appears in the equation with another digit. Which is the incorrect digit and what is its correct substitute? | At first inspection, we can admit that the three digits to the right of the numbers are correct, that is, the digits $0,1,3,4,5$, 6 and 8 are correct. Therefore, among the digits 2, 7, and 9, one of them is wrong. The digit 9 is correct, because if we change it, the sum with 2 will not be correct. Thus, only 2 and 7 re... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the luminous mechanism of the figure, each of the eight buttons can light up in green or blue. The mechanism works as follows: when turned on, all buttons light up blue, and if we press a button, that button and its two neighbors change color. If we turn on the mechanism and successively press buttons 1, 3, and 5, h... | The correct answer is (c).
The table shows the color of each button at each step.
| | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ | $\mathbf{5}$ | $\mathbf{6}$ | $\mathbf{7}$ | $\mathbf{8}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| start | blue | blue | blue | blue |... | 5 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
Will the number $3^{444}+4^{333}$ be divisible by 5? | There is a pattern for the unit digit of a power of 3: it has a period of 4, as it repeats every four times. Indeed, we have
$$
\begin{array}{ll}
3 & 3^{5}=243 \\
3^{2}=9 & 3^{6}=\ldots 9 \\
3^{3}=27 & 3^{7}=\ldots 7 \\
3^{4}=81 & 3^{8}=\ldots 1
\end{array}
$$
Since 444 is a multiple of 4, the unit digit of $3^{444}$... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Writing successively the natural numbers, we obtain the sequence
$$
12345678910111213141516171819202122 \ldots
$$
What is the digit that is in the $2009^{th}$ position of this sequence? | Observe that:
- the numbers from 1 to 9 occupy nine positions;
- the numbers from 10 to 99 occupy $2 \times 90=180$ positions;
- the numbers from 100 to 199 occupy $3 \times 100=300$ positions;
- the numbers from 200 to 299 occupy $3 \times 100=300$ positions;
- the numbers from 300 to 399 occupy $3 \times 100=300$ po... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
An ant leaves point $A$, walks $7 \mathrm{~cm}$ to the left, $5 \mathrm{~cm}$ up, $3 \mathrm{~cm}$ to the right, $2 \mathrm{~cm}$ down, $9 \mathrm{~cm}$ to the right, $2 \mathrm{~cm}$ down, $1 \mathrm{~cm}$ to the left, and $1 \mathrm{~cm}$ down, arriving at point $B$. What is the distance, in cm, between $A$ and $B$?
... | The correct answer is (c).
 | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
Elisa has 24 science books and others of mathematics and literature. If Elisa had one more mathematics book, then one ninth of her books would be mathematics and one quarter literature. If Elisa has fewer than 100 books, how many mathematics books does she have? | Let $N$ be the total number of books Elisa has. Since $N+1$ is a multiple of 9 and 4, it follows that $N+1$ is a multiple of 36. Therefore, $N+1$ is 36 or 72, as Elisa has fewer than 100 books. If $N=35$, then the number of math books is $36 \div 9-1=3$ and the number of literature books is $36 \div 4=9$. However, this... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
O dobro de um número dividido por 5 deixa resto 1. Qual é o resto da divisão desse número por 5 ?
The double of a number divided by 5 leaves a remainder of 1. What is the remainder of the division of this number by 5? | Solution 1: The double of the number sought is a multiple of 5 increased by 1. Since multiples of 5 end in 0 or 5, the double ends in 1 or 6. But the double is an even number, so it ends in 6. Thus, the number ends in 3 or 8 and, therefore, when divided by 5, leaves a remainder of 3.
Solution 2: We know that the integ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. $V$ each of the four rooms there are several items. Let $n \geqq 2$ be a natural number. We move one $n$-th of the items from the first room to the second. Then, one $n$-th of the (new) number of items is moved from the second room to the third. Similarly, from the third room to the fourth, and from the fourth back ... | SOLUTION. When analyzing the number of items after individual steps, we will proceed "backwards." First, let's show how the numbers of items in two rooms before the transfer can be determined from the numbers of items in the rooms after the transfer. Let's say that before the transfer from room $A$ to room $B$, there a... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Nela and Jana choose a natural number $k$ and then play a game with a $9 \times 9$ table. Starting with Nela, each player, on their turn, selects an empty cell and writes a zero in it. Jana, on her turn, writes a one in some empty cell. Additionally, after each of Nela's moves, Jana makes $k$ moves. If at any point ... | SOLUTION. Let's first show that in the case $k=3$, Jana wins. We will work with squares $A_{1}, A_{2}$, and $A_{3}$ of size $3 \times 3$ (Fig. 3). A $3 \times 3$ square is considered covered if there is exactly one one in each of its rows and columns. If Jana covers the squares $A_{1}, A_{2}$, and $A_{3}$ without playi... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. For positive real numbers $a, b, c$ it holds that
$$
a b+b c+c a=16, \quad a \geqq 3 .
$$
Find the smallest possible value of the expression $2 a+b+c$. | 3. Let's modify the square of the expression $V=2a+b+c$, which is clearly positive. We will conveniently use the given relationship $ab+bc+ca=16$:
$$
\begin{aligned}
V^{2} & =(2a+b+c)^{2}=4a^{2}+b^{2}+c^{2}+4ab+4ac+2bc= \\
& =4a^{2}+b^{2}-2bc+c^{2}+4(ab+bc+ca)=4a^{2}+(b-c)^{2}+4 \cdot 16 .
\end{aligned}
$$
According ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Determine the largest integer $n$, for which it is possible to fill a square table $n \times n$ with natural numbers from 1 to $n^{2}$ in such a way that in every $3 \times 3$ square part of it, at least one square of an integer is written. | SOLUTION. It will be easy for us to fill the $11 \times 11$ table in the required way; it is enough to choose nine out of the 11 squares of integers $1^{2}, 2^{2}, \ldots, 11^{2}$ and place them on nine cells of the table with coordinates
$$
(3,3),(3,6),(3,9), \quad(6,3),(6,6),(6,9), \quad(9,3),(9,6),(9,9)
$$
and fil... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. On the board, there are several (not necessarily distinct) prime numbers such that their product is $2020$ times greater than their sum. Determine their smallest possible number. (Patrik Bak) | SOLUTION. Let the prime numbers on the board be denoted as $p_{1}, p_{2}, \ldots, p_{n}$. According to the problem statement, we have
$$
2020 \cdot\left(p_{1}+p_{2}+\ldots+p_{n}\right)=p_{1} p_{2} \ldots p_{n}
$$
The left side of equation (1) is divisible by the number $2020=2 \cdot 2 \cdot 5 \cdot 101$. Therefore, t... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. If a, b, c are distinct non-negative real numbers, what is the smallest possible number of distinct numbers among the numbers $a+b, b+c, c+a, a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+a^{2}$? (Patrik Bak) | SOLUTION. The problem is symmetric in the variables $a, b, c$: changing their order only changes the order of the six numbers being examined. In the first part of the solution, we will assume that $a < b < c$ and $a > 1$. Then it will hold that
$$
11$ will satisfy the condition $a^{2}=a+1$. A simple calculation reveal... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Along a circle, 16 real numbers are arranged with a sum of 7.
a) Prove that there exists a segment of five consecutive numbers with a sum of at least 2.
b) Determine the smallest $k$ such that in the described situation, one can always find a segment of $k$ consecutive numbers with a sum of at least 3. | 2. a) Among 16 numbers written along a circle, there are exactly 16 segments of five consecutive numbers (if we select any of the written numbers and mark the numbers along the circle sequentially as the first, second, ..., sixteenth, the first segment will consist of the first to fifth numbers, the second segment will... | 7 | Combinatorics | proof | Yes | Yes | olympiads | false |
1. On the table lie 54 piles of stones with $1,2,3, \ldots, 54$ stones. In each step, we select any pile, say with $k$ stones, and remove it from the table along with $k$ stones from each pile that has at least $k$ stones. For example, after the first step, if we select the pile with 52 stones, the piles remaining on t... | 1. If in each step we choose the pile with the most stones, we will gradually remove piles with $54, 53, 52, \ldots$ stones, and after 53 steps, only one pile with one stone will remain on the table.
We can prove that regardless of the procedure, the last pile will always contain a single stone. We will show that afte... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Given a rectangle $A B C D$ with perimeter $o$. In its plane, find the set of all points whose sum of distances from the lines $A B, B C, C D, D A$ is equal to $\frac{2}{3} o$. | SOLUTION. The required value of the sum of four distances can be written in the form
$$
\frac{2}{3} o=\frac{1}{6} o+\frac{1}{2} o=\frac{1}{6} o+|A B|+|B C| \text {. }
$$
For any point in the strip determined by the lines $A B$ and $C D$, it holds that the sum of its distances from these two parallel lines is equal to... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. An unconventional piece, which we will call the "sick lady," threatens any square in the row and column where it stands, while on the diagonal, it only threatens the adjacent squares. How many "sick ladies" do we need to place on an $8 \times 8$ chessboard so that they threaten all unoccupied squares?
 are called directly threatened if they lie in the column or row where the queen stands; threatened fields that are not directly threatened are called indirectly threatened. Each queen thus threatens 15 fields directly (including the... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. The average of Pažout's grades is exactly 3. If we did not include three of Pažout's fives in the average, the average of his grades would be exactly 2. Determine the maximum number of ones that Pažout could have received. (Possible grades are 1, 2, 3, 4, 5.) | SOLUTION. Let $s$ be the sum of all Pážout's grades and $p$ their count. We do not know either of these numbers, but we know that
$$
\frac{s}{p}=3 \quad \text { or } \quad s=3 p .
$$
According to the second sentence of the problem, we set up another equation for the unknowns $s, p$ and immediately simplify it:
$$
\f... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Consider any $3 \times 3$ table filled with positive integers such that the sum of the numbers $v$ in each row and each column is 10. How many numbers $v$ in such a table can be: a) the same, b) different?
(Ján Mazák) | SOLUTION. a) Six identical numbers can be found, for example, in the following table that meets the conditions of the problem:
| 8 | 1 | 1 |
| :--- | :--- | :--- |
| 1 | 8 | 1 |
| 1 | 1 | 8 |
If the table contained at least seven identical numbers, one of the rows would contain three of these seven identical numbers.... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find all real numbers $p$ for which the system of inequalities
$$
\begin{aligned}
& 25+2 x^{2} \leqq 13 y+10 z-p \\
& 25+3 y^{2} \leqq 6 z+10 x \\
& 25+4 z^{2} \leqq 6 x+5 y+p
\end{aligned}
$$
with unknowns $x, y, z$ has a solution in the set of real numbers. | 1. By adding all three inequalities, we obtain the inequality
$$
75+2 x^{2}+3 y^{2}+4 z^{2} \leqq 16 x+18 y+16 z
$$
from which, after "completing the square," we get
$$
2(x-4)^{2}+3(y-3)^{2}+4(z-2)^{2} \leqq 0 .
$$
This inequality, which is a consequence of the given system of inequalities, clearly holds only when ... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. Find the greatest natural number $d$ that has the property that for any natural number $n$, the value of the expression
$$
V(n)=n^{4}+11 n^{2}-12
$$
is divisible by the number $d$. | SOLUTION. First, let's calculate the values of $V(n)$ for the smallest natural numbers $n$ and write their prime factorizations in a table:
| $n$ | 1 | 2 | 3 | 4 |
| :---: | :---: | :---: | :---: | :---: |
| $V(n)$ | 0 | $48=2^{4} \cdot 3$ | $168=2^{3} \cdot 3 \cdot 7$ | $420=2^{2} \cdot 3 \cdot 5 \cdot 7$ |
From thi... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. On the board, five different positive numbers are written. Determine the maximum number of ways in which pairs can be formed from them, the sum of which equals one of the five numbers written on the board. | 1. Let $a_{1}<a_{2}<a_{3}<a_{4}<a_{5}$ be positive numbers written on the board. The smallest numbers $a_{1}$ and $a_{2}$ clearly cannot be the sum of any two numbers written on the board. The number $a_{3}$ can be obtained as the sum of some pair in at most one way, namely $a_{3}=a_{1}+a_{2}$. The number $a_{4}$ can t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Consider the expression
$$
V(x)=\frac{5 x^{4}-4 x^{2}+5}{x^{4}+1} .
$$
a) Prove that for every real number $x$, $V(x) \geq 3$.
b) Find the maximum value of $V(x)$. | 1. The expression $V$ is apparently defined for all real numbers $x$.
a) Since $x^{4}+1>0$ for every $x$, the inequality $V(x) \geqq 3$ is equivalent to the inequality $5 x^{4}-4 x^{2}+5 \geqq 3\left(x^{4}+1\right)$ or $2 x^{4}-4 x^{2}+2 \geqq 0$. The expression on the left side is equal to $2\left(x^{2}-1\right)^{2}$... | 5 | Algebra | proof | Yes | Yes | olympiads | false |
3. Find the smallest positive number $x$, for which the following holds: If $a, b, c, d$ are any positive numbers whose product is 1, then
$$
a^{x}+b^{x}+c^{x}+d^{x} \geqq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}
$$
(Pavel Novotný) | Solution. Let $a, b, c, d$ be any positive numbers whose product equals 1. According to the inequality between the arithmetic and geometric means of the triplet of numbers $a^{x}, b^{x}, c^{x}$ for any $x>0$, we have
$$
\frac{a^{x}+b^{x}+c^{x}}{3} \geqq \sqrt[3]{a^{x} b^{x} c^{x}}=\sqrt[3]{\frac{1}{d^{x}}}
$$
By choo... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. In a $11 \times 11$ square grid, we sequentially wrote the numbers $1,2, \ldots, 121$ from left to right and from top to bottom. Using a $3 \times 3$ square tile, we covered exactly nine cells in all possible ways. In how many cases was the sum of the nine covered numbers a perfect square of an integer? | 2. Let $n$ be the number covered by the middle cell of the square tile. Then the first row of this tile covers the numbers $n-12, n-11, n-10$, its second row covers the numbers $n-1, n$ and $n+1$, and finally the third row covers the numbers $n+10, n+11$ and $n+12$. The sum of all numbers covered by the tile is thus $9... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Determine all integers greater than 1 by which some fraction of the form
$$
\frac{3 p-q}{5 p+2 q}
$$
can be divided, where $p$ and $q$ are coprime integers.
The written part of the school round in category A takes place
## on Tuesday, December 2, 2008
so that it starts in the morning and the contestants have 4 ... | 3. A fraction can be reduced by an integer $d>1$ if and only if the number $d$ is a common divisor of the numerator and the denominator of the considered fraction. Let us assume, therefore, that $d \mid 3 p-q$ and at the same time $d \mid 5 p+2 q$, where $p$ and $q$ are coprime integers. By adding suitable multiples of... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Find all positive integers $n$ for which the number $n^{2}+6 n$ is a perfect square of an integer. | 2. Clearly $n^{2}+6 n>n^{2}$ and at the same time $n^{2}+6 n<n^{2}+6 n+9=(n+3)^{2}$. In the given range, there are only two squares of integers: $(n+1)^{2}$ and $(n+2)^{2}$.
In the first case, we have $n^{2}+6 n=n^{2}+2 n+1$, thus $4 n=1$, but no integer $n$ satisfies this.
In the second case, we have $n^{2}+6 n=n^{2... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find all prime numbers $p$ for which there exists a natural number $n$ such that $p^{n}+1$ is a cube of some natural number. | SOLUTION. Let's assume that for a natural number $a$ the following holds: $p^{n}+1=a^{3}$ (obviously $a \geq 2$). We will rearrange this equation to make it possible to factor one side:
$$
p^{n}=a^{3}-1=(a-1)\left(a^{2}+a+1\right) \text {. }
$$
From this factorization, it follows that if $a>2$, both numbers $a-1$ and... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Determine the values that the expression
$$
\frac{a+b c}{a+b}+\frac{b+c a}{b+c}+\frac{c+a b}{c+a}
$$
can take if \( a \), \( b \), and \( c \) are positive real numbers with a sum of 1. (Michal Rolínek, Pavel Calábek) | SOLUTION. The fractions in the given expression make sense because their denominators are positive numbers according to the problem statement. Thanks to the condition $a+b+c=1$, for the first fraction we have
$$
\frac{a+b c}{a+b}=\frac{(a+b)+(b c-b)}{a+b}=1-b \cdot \frac{1-c}{a+b}=1-b \cdot \frac{a+b}{a+b}=1-b.
$$
Si... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In the plane, there is a right-angled triangle $ABC$, on whose hypotenuse $AB$ we consider an arbitrary point $K$. The circle constructed over the segment $CK$ as a diameter intersects the legs $BC$ and $CA$ at internal points, which we denote by $L$ and $M$ respectively. Determine for which point $K$ the quadrilate... | 2. Since the angles $KLC$, $KMC$, and $LCM$ are right angles (Fig. 1), the quadrilateral $KLCM$ is a rectangle and triangles $AKM$ and $KBL$ are similar to triangle $ABC$. Let us denote
Fig.... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Consider the 2022 fractions
$$
\frac{0}{2022}, \frac{1}{2021}, \frac{2}{2020}, \ldots, \frac{2021}{1}
$$
in the form of the ratio of two non-negative integers, whose sum for each fraction is equal to 2022. How many of them take integer values?
(Jaroslav Zhouf) | SOLUTION. The denominators of the given fractions are natural numbers from 1 to 2022. The numerator $c$ of the fraction with a given denominator $j$ is determined by the equation $c+j=2022$, i.e., $c=2022-j$. Therefore, our fractions can be expressed, and we will immediately simplify this expression:
$$
\frac{2022-j}{... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In the real domain, consider the system of equations
$$
\begin{aligned}
& x^{4}+y^{2}=\left(a+\frac{1}{a}\right)^{3}, \\
& x^{4}-y^{2}=\left(a-\frac{1}{a}\right)^{3}
\end{aligned}
$$
with a non-zero real parameter $a$.
a) Find all values of $a$ for which the given system has a solution.
b) Prove that for any sol... | SOLUTION. We solve the system as a linear system of equations with unknowns $x^{4}$ and $y^{2}$. By adding both equations and dividing by two, we get
$$
x^{4}=\frac{1}{2}\left(\left(a+\frac{1}{a}\right)^{3}+\left(a-\frac{1}{a}\right)^{3}\right)=a^{3}+\frac{3}{a} .
$$
Similarly, by subtracting the second equation from... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
3. On the board, the numbers $1,2, \ldots, 33$ are written. In one step, we choose several numbers written on the board (at least two), whose product is a square of a natural number, erase the chosen numbers, and write the square root of their product on the board. We continue this process until only such numbers remai... | 3. The product of all numbers written on the board is equal to
$$
S=2^{31} \cdot 3^{15} \cdot 5^{7} \cdot 7^{4} \cdot 11^{3} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 .
$$
The presence of odd exponents means that $S$ is not a perfect square. Therefore, we cannot erase all the numbers in the first step... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. To each vertex of a regular 63-gon, we assign one of the numbers 1 or -1. To each of its sides, we append the product of the numbers at its vertices and sum all the numbers next to the individual sides. Find the smallest possible non-negative value of such a sum. | SOLUTION. Let $S$ be the value we are examining, i.e., the sum of the numbers on the sides of the 63-gon. If we assign the number 1 to each vertex of the 63-gon, we get $S=63$, because each of its sides will be assigned the number 1. It is clear that for any chosen numbering of the vertices, it can be reached by gradua... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Dana wrote a three-digit number on paper, which when divided by seven gives a remainder of 2. By swapping the first two digits, a three-digit number was formed that when divided by seven gives a remainder of 3. The number formed by swapping the last two digits of the original number gives a remainder of 5 when divid... | SOLUTION. Let's denote the digits of Dana's number as $a, b, c$. The information about the remainders when divided by seven from the problem can be rewritten into equations
$$
\begin{aligned}
& 100 a+10 b+c=7 x+2, \\
& 100 b+10 a+c=7 y+3, \\
& 100 a+10 c+b=7 z+5 .
\end{aligned}
$$
The digits $a$ and $b$ cannot be zer... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. We have a certain number of boxes and a certain number of balls. If we put exactly one ball in each box, we will have n balls left. However, if we set aside exactly n boxes, we can distribute all the balls so that there are exactly $n$ balls in each of the remaining boxes. How many boxes and how many balls do we hav... | SOLUTION. If we denote $x$ as the number of boxes and $y$ as the number of balls, the problem leads to the system of equations
$$
x+n=y \quad \text { and } \quad(x-n) \cdot n=y
$$
with unknowns $x$, $y$, and $n$ from the set of natural numbers. By eliminating the unknown $y$, we get the equation $x+n=(x-n) \cdot n$, ... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Construct a trapezoid, given the lengths of its diagonals $9 \mathrm{~cm}$ and $12 \mathrm{~cm}$, the length of the midline $8 \mathrm{~cm}$, and the distance between the midpoints of the diagonals $2 \mathrm{~cm}$. | SOLUTION. Let's choose the notation according to Fig. $1, K P$ is the midline in triangle $A C D$,
Fig. 1
thus $|K P|=\frac{1}{2}|D C|$, similarly $|Q L|=\frac{1}{2}|D C|,|P L|=\frac{1}{2}|A... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find all pairs of natural numbers $a, b$ for which
$$
n(a, b)+D(a, b)=63
$$
where $n(a, b)$ denotes the least common multiple and $D(a, b)$ the greatest common divisor of the numbers $a, b$. | SOLUTION. We will use what we stated in the 1st problem. Let $a=D p, b=D q, n=D p q$, where $D$ is the greatest common divisor, $n$ is the least common multiple of the numbers $a, b$, and the numbers $p, q$ are coprime. According to the problem statement, $D(1+p q)=63$ must hold, so we have the following options (witho... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. On the board, one or several different two-digit natural numbers are written. We will call a digit c on the board good if the sum of those numbers on the board that contain the digit c is equal to the number 71. (Sometimes it may be a "sum" of a single number with the given digit)
a) Which of the digits 0 to 9 can ... | SOLUTION. a) The following examples of one or two numbers written on the board show that the digits 1, 2, 3, 4, 5, 7 can be good:
$$
\begin{array}{ll}
\text { 1: } & \{71\}, \\
\text { 2: } & \{29,42\}, \\
\text { 3: } & \{32,39\}, \\
\text { 4: } & \{24,47\}, \\
\text { 5: } & \{15,56\}, \\
\text { 7: } & \{71\} .
\e... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A $10 \times 10$ table is filled with numbers 1 and -1 such that the sum of the numbers in each row, except one, is equal to 0, and the sum of the numbers in each column, except one, is equal to the same number s. Determine the largest possible value of s and show that it cannot be larger. Also, provide an example o... | SOLUTION. In the first part of the solution, we will show that the value of $s$ never exceeds the number 2. For the value $s=2$, we will provide an example of a suitable table in the second part.
Let us consider any $10 \times 10$ table filled according to the problem statement and, in addition to the number $s$, let ... | 2 | Combinatorics | proof | Yes | Yes | olympiads | false |
2. Let's examine whether it is possible to fill an $n \times n$ square table with natural numbers from 1 to $n^{2}$ such that in every $2 \times 2$ square part, at least one multiple of five is written.
a) Prove that for no even $n$ is this possible.
b) Find the largest odd $n$ for which this is possible. | 2. For even $n=2k$, divide the table into non-overlapping $2 \times 2$ squares. There will be exactly $k^2$ of these, and in each of them, a different multiple of five must be written. For this, we need $k^2$ multiples of five, the smallest of which are the numbers $5, 10, \ldots, 5k^2$. However, the last of these numb... | 9 | Combinatorics | proof | Yes | Yes | olympiads | false |
1. Find the bases $z$ of all number systems in which the four-digit number $(1001)_{z}$ is divisible by the two-digit number $(41)_{z}$. | 1. Since the digit 4 appears in the notation of a two-digit number, it necessarily follows that $z \geq 5$. From the expanded notations $(1001)_z = z^3 + 1$ and $(41)_z = 4z + 1$, we are looking for exactly those natural numbers $z \geq 5$ for which the number $z^3 + 1$ is a multiple of the number $4z + 1$. Using the E... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Prove that there exists a unique digit $c$, for which there is a unique natural number $n$ ending in the digit $c$ and having the property that the number $2n+1$ is the square of a prime number. | SOLUTION. Let the (odd) number $2n+1$ be the square of a prime number $p$, then $p$ is also an odd number. From the relation $p^2 = 2n + 1$, it follows that $n = \frac{1}{2}(p^2 - 1) = \frac{1}{2}(p - 1)(p + 1)$. Let's create a table of the first few odd prime numbers $p$ and their corresponding numbers $n$:
| $p$ | 3... | 2 | Number Theory | proof | Yes | Yes | olympiads | false |
1. Each cell of a $68 \times 68$ table is to be colored with one of three colors (red, blue, white). In how many ways can this be done so that every triplet of adjacent cells in each row and each column contains cells of all three colors?
(Josef Tkadlec) | Solution. Once we determine the colors of some two adjacent fields in a row or column of the table, the coloring of all its other fields is uniquely determined by the requirements of the problem. The colors in each row and column thus alternate regularly with a period of 3.
Let's say that $a b c a b c .$. is the color... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In how many ways can a $3 \times 3$ square table be filled with the numbers $2,2,3,3,3,4,4,4,4$ so that the sum of the numbers in each $2 \times 2$ square of this table is equal to 14? | ANOTHER SOLUTION. Just like in the previous solution, we will show that the number 14 can be obtained from the given numbers as a sum of four numbers only as $4+4+4+2$ or $4+4+3+3$. It follows from this that in any $2 \times 2$ square that contains the number 2 (or 3), no other 2 can be present. In the central square (... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let $a, b, c$ be positive real numbers such that $a b + b c + c a = 1$. Determine the values that the expression
$$
\frac{a\left(b^{2}+1\right)}{a+b}+\frac{b\left(c^{2}+1\right)}{b+c}+\frac{c\left(a^{2}+1\right)}{c+a}
$$
can take.
(Josef Tkadlec \& Patrik Bak) | SOLUTION. Given the condition $a b+b c+c a=1$, we first modify the first of the three addends of the expression under consideration, similarly to problem 4 from the homework round,
$$
\frac{a\left(b^{2}+1\right)}{a+b}=\frac{a\left(b^{2}+a b+b c+c a\right)}{a+b}=\frac{a(b+a)(b+c)}{a+b}=a(b+c)=a b+c a .
$$
Similarly, t... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Determine the largest natural number $k$ for which it is possible to place $k$ rooks and $k$ bishops on an $8 \times 8$ chessboard such that no piece threatens another. (A bishop threatens any square on the same diagonal, and a rook threatens any square on the same row or column.)
(Josef Tkadlec) | Solution. Consider any valid placement of $k$ rooks and $k$ archers. We will prove that the inequality $k \leqq 5$ holds.
If a rook is placed in any row (or column) of the chessboard, it must be the only piece in that row (or column). Therefore, for the number $k$ of placed rooks, it must be true that $k \leqq 8$, and... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let's have an $8 \times 8$ chessboard and for each "edge" that separates two of its squares, write a natural number that indicates the number of ways the entire chessboard can be cut into $2 \times 1$ rectangles such that the given edge is part of the cut. Determine the last digit of the sum of all such written numb... | Solution. There are $7 \cdot 8=56$ vertical edges and the same number of horizontal edges, making a total of $56 \cdot 2=112$. When the chessboard is cut in any way, 32 rectangles of $2 \times 1$ are formed, so each such cut touches exactly $112-32=80$ edges, contributing the number 80 to the total sum. Therefore, the ... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A cinema received 234 visitors. Determine for which $n \geqq 4$ it could have happened that the visitors could be seated in $n$ rows such that each visitor in the $i$-th row knew exactly $s_j$ visitors in the $j$-th row for any $i, j \in\{1,2, \ldots, n\}, i \neq j$. (The relationship of knowing is symmetric.)
(Tom... | Solution. For each $k \in\{1,2, \ldots, n\}$, let $p_{k}$ be the number of viewers in the $k$-th row. The condition of the problem for given $i$ and $j$ means that the number of acquaintances between viewers in the $i$-th and $j$-th rows is equal to $j p_{i}$. By swapping the roles of the numbers $i$ and $j$, we find t... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Determine the maximum possible area of triangle $ABC$, whose medians satisfy the inequalities $t_{a} \leqq 2, t_{b} \leqq 3, t_{c} \leqq 4$.
(Pavel Novotný) | Solution. Let $T$ be the centroid of triangle $ABC$ and $K, L, M$ the midpoints of sides $BC, CA, AB$. The medians divide triangle $ABC$ into six smaller triangles of equal area: For example, triangle $AMT$ has side $|AM|=\frac{1}{2} c$ and its height to side $AM$ has a length of $\frac{1}{3} v_{c}$, so $S_{AMT}=\frac{... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Let's consider 20 statements:
| "I have exactly one sister." | "I have exactly one brother." |
| :--- | :--- |
| "I have exactly two sisters." | "I have exactly two brothers." |
| "I have exactly ten sisters." | "I have exactly ten brothers." |
a) Each of the four siblings made a different one of these 20 statemen... | SOLUTION. a) Yes, it is possible. In the case where the four siblings are two brothers and two sisters, they can truthfully make four mutually different statements. One brother says: "I have exactly one brother" and the other: "I have exactly two sisters," one sister says: "I have exactly one sister" and the other: "I ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 4. Find the smallest value of the fraction
$$
V(n)=\frac{n^{3}-10 n^{2}+17 n-4}{n^{2}-10 n+18}
$$
$where\ n$ is any natural number greater than 2. | SOLUTION. First, let's calculate the values of the expression $V(n)$ for several natural numbers $n \geq 3:$
| $n$ | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $V(n)$ | $5 \frac{1}{3}$ | $5 \frac{1}{3}... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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