problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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|---|---|---|---|---|---|---|---|---|
What is the value of $\square$ in $\frac{6400000}{400}=1.6 \times \square$? | For simplification $\frac{6400000}{400}=16000$, thus:
$$
\frac{6400000}{400}=1.6 \times \square \quad \Longrightarrow \quad 16000=1.6 \times
$$
It follows that $\square=10000$. | 10000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A certain concrete mixture is made of cement, sand, and soil in the ratio $1: 3: 5$ per kilogram. How many kilograms of this mixture can be made with 5 kilograms of cement?
(a) $13 \frac{1}{3}$
(b) 15
(c) 25
(d) 40
(e) 45 | According to the data of the problem, we have:
$$
\begin{array}{cccc}
\text { cement } & \text { sand } & & \text { soil } \\
1 \mathrm{~kg} & \longleftrightarrow 3 \mathrm{~kg} & \longleftrightarrow & 5 \mathrm{~kg}
\end{array}
$$
Therefore, with $5 \mathrm{~kg}$ of cement, we have:
$$
\begin{aligned}
& \text { cem... | 45 | Algebra | MCQ | Yes | Yes | olympiads | false |
In the sum below, the same letters represent the same digits and different letters represent different digits.
$$
\begin{array}{ccc}
& & X \\
+ & & X \\
& Y & Y \\
\hline Z & Z & Z
\end{array}
$$
What digit does the letter $X$ represent?
# | Solution
Initially note that $X+X+Z \leq 2 \cdot 9+8=26$. Therefore, at most 2 units will be added to the tens place. Since $Y \neq Z$, when we add 1 or 2 units to $Y$, we should obtain the two-digit number $\overline{Z Z}$. As $Y+2 \leq 9+2=11$, the only possibility is to have $\overline{Z Z}=11$, thus, $Y=9$. For th... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A regular icosahedron is a geometric solid with 20 faces, which are equilateral triangles. By drawing the numbers from 1 to 20 on the faces of a regular icosahedron, it is transformed into a die. Luísa and Mathias invented a game in which each of them rolls the icosahedral die 5 times and records the numbers drawn in s... | Solution
a) 2,020,190,909. For this to happen, Luísa scored 20 in her 5 throws and Mathias scored 1 in his 5 throws, making the difference equal to 2,020,202,020 - 11,111 = 2,020,190,909.
b) No. The highest possible score would be by scoring 20 on the last throw, resulting in 141,181,920, which is less than 162,012,5... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Starting with any non-zero natural number, it is always possible to form a sequence of numbers that ends in 1, by repeatedly following the instructions below:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Starting with any non-zero natural number, it is always possible to form a sequence of numbers that ends in 1, by repeatedly... | Solution
a)
$$
100 \rightarrow 99 \rightarrow 33 \rightarrow 11 \rightarrow 12 \rightarrow 4 \rightarrow 3 \rightarrow 1
$$
b) $\{27,9,3,1\},\{10,9,3,1\},\{8,9,3,1\},\{12,4,3,1\},\{6,2,3,1\}$. Analyzing the process in reverse, we must always arrive at 1, which will always be the last element of the sequence; to arri... | 21 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Ana, Bia, Cátia, Diana, and Elaine work as street vendors selling sandwiches. Every day, they stop by Mr. Manoel's snack bar and take the same number of sandwiches to sell. One day, Mr. Manoel was sick and left a note explaining why he wasn't there, but asking each of them to take $\frac{1}{5}$ of the sandwiches. Ana a... | Solution
a) $\frac{4}{25}$. Since Bia took $\frac{1}{5}$ of $\frac{4}{5}$, she ended up with $\frac{1}{5} \cdot \frac{4}{5}=\frac{4}{25}$ of the sandwiches.
b) Ana took $\frac{1}{5}=\frac{5}{25}$, and Cátia, Diana, and Elaine divided the remaining fraction, which was $1-\frac{9}{25}=\frac{16}{25}$, among the three, m... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
On a table, there is a certain number $N$ of candies. Aline and Bruna agree that, alternately, each one must eat at least one, but no more than half of the existing quantity. The one who eats the last candy wins the game. Ana always starts the game.
a) For $N=5$, which of the two has the winning position? (A winning p... | Solution
a) Bruna. The number of candies that should be eaten is from 1 to half of the existing amount on the table. Therefore, the winner $(V)$ must find 1 candy on the table in their last move, and for this to happen, there should have been 2 candies on the previous move for the player $(P)$, forcing her to eat only... | 191 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The grid in the figure below is composed of 25 small unit squares. Determine:
a) How many squares with vertices on the points of the figure and sides on the segments of the figure exist?
b) How many pairs of parallel lines, each containing some segment of the figure, exist?
c) How many rectangles with vertices on th... | Solution
a) 55. Let's divide our count: | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Mario completed 30 hours in an extra math course. On the days he had classes, the duration was only 1 hour, which occurred exclusively in the morning or in the afternoon. Additionally, there were 20 afternoons and 18 mornings without classes during the course period.
a) On how many days were there no classes?
b) How ... | Solution
We can construct a diagram with two intertwined sets. Set $T$, representing afternoons without classes, and set $M$, representing mornings without classes. Thus, the intersection $(T \cap M)$, which has $x$ elements, represents the full days without classes.
 On a digital clock, hours are displayed using four digits. For example, when showing 00:00 we know it is midnight, and when showing 23:59 we know it is one minute before midnight. How many times a day do all four digits shown are even?
b) A clock, with hour, minute, and second hands, makes a "plim" every time a han... | Solution
a) Let's say the display is made up of 4 digits. The first digit can only be occupied by the numbers 0, 1, or 2. The second digit depends on the first: when the first is 0 or 1, any digit can appear, and when the first is 2, 0, 1, 2, or 3 can appear. The third digit only allows the digits from 0 to 5. Additio... | 1438 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a chess tournament, all players faced each other twice, scoring 1 point for a win, half a point for a draw, and zero points for a loss. The winner was the one who scored the most points in the competition. Luíza, a curious mathematician, found a piece of paper stating that the total score of all participants was 210... | Solution
a) Let $x$ be the number of chess players. Since in each match exactly one point is contested, the total number of points is equal to the number of games. Each athlete will play $(x-1) \cdot 2$ matches, as they will play against all the others twice. However, the matches $a \times b$ and $b \times a$ will be ... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
King Arthur had to fight the Three-Headed and Three-Tailed Dragon (known as $D T C T C$). His task was made easier when he managed to obtain a magical sword that could deliver the following strikes (one at a time):
A) cut off one head;
C) cut off one tail;
B) cut off two heads;
D) cut off two tails.
Moreover, the Fair... | Solution
a) Yes, it is enough for the King to use, for example, the sequence of strikes $D \rightarrow C \rightarrow C \rightarrow C$.
b) From a practical standpoint, the only strike that reduces the number of heads is $B$, but we need two of them to perform it (the same applies to tails and the strike $D$), unless s... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In each of the 16 circles in the figure, there is a student. A total of 3360 coins are distributed among the 16 students, such that students who are the same distance from the center receive the same number of coins. Five by five, from the outside to the inside, the circles are on the vertices of a regular pentagon, wi... | Solution
Denote each vertex with capital letters and the quantities each one passed by the respective lowercase letter. For example,
The 5 A's sent $4a$ and received $2a$ and $2b$, so $4a ... | 280 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
On a $30 \times 30$ board, the rows and columns are numbered from 1 to 30. In each row, João paints red the cells where the column number is a multiple of the row number. How many cells will be painted red?
# | Solution
The number of houses painted in line $i$ is equal to the number of multiples of $i$ less than or equal to 30. If we divide 30 by $i$ and get quotient $q$ and remainder $r$, with $0 \leq r<i$, then the number of houses painted is $q$, because
$$
i \cdot 1<i \cdot 2<\ldots<i \cdot q \leq 30<i \cdot q+i
$$
The... | 111 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A tournament will take place with 100 competitors, all with different skill levels. The most skilled competitor always wins against the least skilled competitor. Each participant plays exactly twice, with two randomly drawn opponents (once against each). A competitor who wins two matches receives a medal. Determine the... | Solution
Let's denote that a player $A$ is more skilled than a player $B$ by $A \rightarrow B$. Suppose that
$$
L_{1} \rightarrow L_{2} \rightarrow L_{3} \rightarrow \ldots \rightarrow L_{100}
$$
are the 100 players in the tournament. Since $L_{1}$ is the most skilled of all, he will certainly win a medal by winning... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Using the digits 1, 3, and 5, Mônica forms three-digit numbers that are greater than 150. How many numbers can Mônica form?
# | Solution
There are three possibilities for the first digit of the number formed by Mônica.
a) The hundreds digit is 1. For the number to be greater than 150, the tens digit must be 5. There are only three options left for the units digit, namely: 1, 3, or 5. In this case, she can form only 3 numbers.
b) The hundreds... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Professor Antônio discovered an interesting property related to the integer $x$ that represents his age. He told his students that $x^{2}=\overline{a b a c}$ and that $x=\overline{a b}+\overline{a c}$. What is the professor's age?
Note: We are using a bar to distinguish the decimal representation of the four-digit num... | Solution
Given that $x^{2}$ has 4 digits, we have $31<x<100$. Note that $\overline{a b c d}=\overline{a b} \cdot 100+\overline{c d}$, and thus it follows that
$$
\begin{aligned}
x^{2} & =\overline{a b} \cdot 100+\overline{c d} \\
& =99 \overline{a b}+\overline{a b}+\overline{c d} \\
& =99 \cdot \overline{a b}+x
\end{... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Fernando is a cautious person, and the gate of his house has 10 distinct padlocks, each of which can only be opened by its respective key, and each key opens only one padlock. To open the gate, he must have at least one key for each padlock. For security, Fernando has distributed exactly two different keys in the 45 dr... | Solution
a) Notice that the pair $\{a, b\}$ is the same as $\{b, a\}$, since they will open the same locks. Therefore, after choosing key $a$ in 10 ways and key $b$ in 9 ways, we must divide the result by 2, obtaining $\frac{10 \cdot 9}{2}=45$ pairs.
b) Since each key will pair with all the others exactly once, Ferna... | 37 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Consider the set $A=\{1,2,3, \ldots, 2011\}$. How many subsets of $A$ exist such that the sum of their elements is 2023060? | Observe that the sum $1+2+\cdots+2011=\frac{2011 \times 2012}{2}=$ 2023066. Therefore, to obtain a subset of $A$ that has a sum of its elements equal to 2023060, it suffices to remove from $A$ the elements whose sum is 6. The possible cases are:
- Subsets with one element: $\{6\}$.
- Subsets with two elements: $\{2,4\... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
An ant moves one unit per second over the points 0, 1, and 2 in the following figure, starting from point 0.
(a) What are the possible paths the ant can take in 3 seconds?
(b) How many poss... | (a) Up to three seconds, we have two possible paths: $0-1-0-1$ or $0-1-2-1$.
(b) Observe that when the ant is at points 0 and 2, it only has one possibility for the next second, which is to go to 1. When it is at 1, it has two possibilities for the next second, which is to go to 0 or 2. Thus, at odd seconds, the ant i... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A convex polyhedron $\mathcal{P}$ has 26 vertices, 60 edges, and 36 faces. 24 faces are triangular and 12 are quadrilaterals. A spatial diagonal is a line segment connecting two vertices not belonging to the same face. How many spatial diagonals does $\mathcal{P}$ have? | The 26 vertices determine exactly $\binom{26}{2}=26 \times 25 / 2=$ 325 segments. Of these segments, 60 are edges and since each quadrilateral has two diagonals, we have $12 \times 2=24$ diagonals that are not spatial.
Therefore, the number of spatial diagonals is $325-60-24=241$. | 241 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A grid of points with 10 rows and 10 columns is given. Each point is colored red or blue. Whenever two points of the same color are adjacent in the same row or column, they are connected by a segment of the same color as the points. If two points are adjacent but of different colors, they are connected by a green segme... | Initially, observe that there are 9 segments in each row and in each column, so there are $9 \times 10+9 \times 10=180$ segments in total.
Let $A$ be the number of blue segments and $V$ be the number of red segments. Then $A+V+98=180$, so $A+V=82$, since there are 98 green segments.
Notice that from red points, only ... | 37 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A number is said to be TOP if it has 5 digits and the product of the $1^{\circ}$ and the $5^{\circ}$ is equal to the sum of the $2^{\circ}, 3^{\circ} \mathrm{and} 4^{\circ}$. For example, 12,338 is TOP, because it has 5 digits and $1 \cdot 8=2+3+3$.
a) What is the value of $a$ so that $23,4 a 8$ is TOP?
b) How many T... | Solution
a) We have that $2 \cdot 8 = 3 + 4 + a$, so $a = 9$.
b) Let $1 b . c d 2$ be a TOP number. We have that $b + c + d = 2$, and all the possible combinations $(b, c, d)$ are $(0,0,2), (0,1,1), (0,2,0), (1,0,1), (1,1,0), (2,0,0)$, which means there are 6 TOP numbers.
c) Let 9e.fgh be a TOP number. We will analy... | 112 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The segment $AB$ of length $16 \mathrm{~cm}$ is the diameter of a circle with center $O$. A secant line intersects the circle at $C$ and $D$ and the line $AB$ at $P$, as indicated in the following figure. If $OD = DP$ and $\angle APC = 18^{\circ}$, what is the value of the angle $\angle AOC$?
 Starting from the corners, there are 4 possibilities for each ( $4 \cdot 4=16$ in total), for example, starting with 1, we have the passwords 123, 125, 145, 147. Starting from 2, 4, 6, or 8, there are 5 possibilities for each ( $4 \cdot 5=20$ in total), for example, starting with 2, we have the passwords 2... | 44 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The 2,019 lockers of 2,019 students in a school are numbered with the squares of the first 2,019 positive natural numbers, that is, the first locker has the number $1^{2}=1$, the second locker has the number $2^{2}=4$, the third locker has the number $3^{2}=9$, and so on until the last locker which has the number $2,01... | Solution
a) Let's divide into groups by the number of digits:
I) 1 digit: 3 lockers $\left(1^{2}, 2^{2}, 3^{2}\right)$;
II) 2 digits: $9-3=6$ lockers ( $4^{2}$ to $9^{2}$ );
III) 3 digits: $31-9=22$ lockers $\left(10^{2}\right.$ to $\left.31^{2}\right)$;
IV) 4 digits: $99-31=68$ lockers $\left(32^{2}\right.$ to $\l... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a parking lot, there are motorcycles, cars, buses, and trucks, totaling 80 vehicles and 540 wheels. Each motorcycle has 2 wheels, each car has 4, each bus has 6, and each truck has 8. The number of cars is the sum of the number of motorcycles and the number of buses. How many trucks are there in this parking lot, if... | Solution
Let the quantities of motorcycles, cars, buses, and trucks be $m, k, o$, and $c$ respectively. Organizing the information, we have:
$$
\left\{\begin{array}{cccc}
m+k+o+c & = & 80 & (I) \\
2 m+4 k+6 o+8 c & = & 540 & (II) \\
k & = & m+o & (III)
\end{array}\right.
$$
Substituting (III) into (II) and (I), we a... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
At the edge of a circular lake, there are stones numbered from 1 to 10, in a clockwise direction. Frog starts from stone 1 and jumps only on these 10 stones in a clockwise direction.
a) If Frog jumps 2 stones at a time, that is, from stone 1 to stone 3, from stone 3 to stone 5, and so on, on which stone will Frog be a... | Solution
a) After 5 jumps, Frog returns to stone 1 and starts the same sequence. Since 100 is a multiple of 5, on the $100^{\text{th}}$ jump, he goes to stone 1.
b) On the $1^{\text{st}}$ jump, he moves 1 stone; on the $2^{\text{nd}}$, 2 stones; on the $3^{\text{rd}}$, 3 stones, and so on until the last jump when he ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On a line $r$, points $A$ and $B$ are marked, and on a line $s$, parallel to $r$, points $C$ and $D$ are marked, such that $A B C D$ is a square. Point $E$ is also marked on the segment $C D$.
 Let $2 k$ be the area of square $A B C D$, then the area of triangle $A B E$ is equal to $k$ and the area of triangle $B C D$ is also equal to $k$, therefore, the ratio between the areas is 1.
b) If $\frac{A_{B F E}}{A_{D F E}}=2$, then $\frac{B F}{F D}=2$. If $r / / s$, then $\triangle A B F \sim \triang... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Two positive integers $x$ and $y$ are such that:
$$
\frac{2010}{2011}<\frac{x}{y}<\frac{2011}{2012}
$$
Find the smallest possible value for the sum $x+y$.
# | Solution
As $\frac{2011}{2012}4021 d \\
& \geq 12063
\end{aligned}
$$
Consequently, the minimum value of the sum is obtained with $d=2$ and in this case $x+y=8044$.
# | 8044 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $m=999 \ldots 99$ be the number formed by 77 digits all equal to 9 and let $n=777 \ldots 77$ be the number formed by 99 digits all equal to 7. What is the number of digits of $m \cdot n$? | Solution
Since $m+1=10^{77}$, notice that:
$$
\begin{aligned}
m \cdot n & =(m+1) \cdot n-n \\
& =\underbrace{777 \ldots 77}_{99} \underbrace{000 \ldots 00}_{77}-\underbrace{777 \ldots 77}_{99} .
\end{aligned}
$$
Since $\underbrace{777 \ldots 77}_{99} \underbrace{000 \ldots 00}_{77}$ has $99+77$ digits and $\underbra... | 176 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the figure below, all the small squares on the board are equal. What is the value of the angle $\angle A E F$? Justify.
# | Solution
Mark the points $G$ and $H$ as indicated in the figure. The triangles $E F G$ and $A E H$ correspond to half of a $1 \times 3$ rectangle, and consequently $\angle F E G=\angle E A H$ and $\angle E F G=$ $\angle A E H$. Since these triangles are right triangles, we have $\angle E A H+\angle A E H=90^{\circ}$. ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The table shows the final standings of the Quixajuba football championship. In this championship, each team played against each of the other teams four times. Each team earns 3 points for a win, 1 point for a draw, and no points for a loss.
| Team | Points |
| :---: | :---: |
| Bissetriz | 22 |
| Primo | 19 |
| Potênc... | (a) There are 6 possible matches between the four teams: (Bissetriz vs Primo), (Bissetriz vs Potência), (Bissetriz vs MDC), (Primo vs Potência), (Primo vs MDC), and (Potência vs MDC). Each of these matches occurred 4 times, so the number of matches is equal to $4 \times 6 = 24$.
(b) The maximum number of points in the... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Guilherme wrote 0 or 1 in each cell of a $4 \times 4$ board. He placed the numbers in such a way that the sum of the numbers in the neighboring cells of each cell on the board was equal to 1.
For example, in Figure 40.1, considering the cell marked with $\bullet$, the sum of the numbers in the shaded cells is equal to... | Each house can only have one neighbor with a number 1, and the other neighbors must be zeros, since the sum of the neighbors is 1.
Starting from the top left corner, we can assume without loss of generality that we fill the board as shown in Figure 40.2.
In the following steps, the filled cells are the neighbors of t... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A game starts with 7 coins aligned on a table, all with the head side facing up. To win the game, you need to flip some coins so that in the end, two adjacent coins are always with different sides facing up. The rule of the game is: in each move, you have to flip two adjacent coins. What is the minimum number of moves ... | If we assign the value of 1 to heads and -1 to tails and sum the results after each flip, the game starts with a sum of 7 and we need to reach alternating heads and tails, so the game ends at 1 or -1. We observe that at each step of the game, we have the following possibilities: we swap two heads for two tails and the ... | 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 and began to triple every 6 months. When will it reach $R \$ 81.00$?
(a) 1 year
(b) 2 years
(c) $21 / 2$ years
(d) 13 years
(e) $131 / 2$ years | As $81=3^{4}$, then the value of the franc has tripled 4 times, the number of months elapsed is $4 \times 6=24$ months, that is, 2 years, meaning that in January 2002 the chicken will reach the proposed price. The correct option is (b). | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
In 2005, a travel agency organized a tour to Foz do Iguaçu, distributing the people in buses with 27 seats, and it was necessary to form an incomplete bus with 19 seats. In 2006, the number of participants increased by 53, and they continued to use buses with 27 seats. How many more buses were needed and how many peopl... | We have a bus with $27-19=8$ free seats and still need to accommodate $53-8=45$ participants in buses with 27 seats. It is clear that one bus is not enough, so we need 2 buses and there will be $2 \times 27-45=9$ free seats in the last bus. 18 people remained in the incomplete bus. | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the unit digit of the product
$$
(5+1)\left(5^{3}+1\right)\left(5^{6}+1\right)\left(5^{12}+1\right) ?
$$
(a) 0
(b) 1
(c) 2
(d) 5
(e) 6 | The unit digit of any power of 5 is 5, so the unit digit of each factor of the product is $5+1=6$. But, $6 \times 6=36$, that is, the product of two numbers ending in 6 is also a number that ends in 6. Therefore, the unit digit of this product is 6. The correct option is (e). | 6 | Number Theory | MCQ | Yes | Yes | olympiads | false |
In a warehouse, a dozen eggs and 10 apples had the same price. After a week, the price of eggs dropped by $10 \%$ and the price of apples increased by $2 \%$. How much more will be spent on the purchase of a dozen eggs and 10 apples?
(a) $2 \%$
(b) $4 \%$
(c) $10 \%$
(d) $12 \%$
(e) $12.2 \%$
## Solutions to List 3 | Suppose, initially, that a dozen eggs cost $R \$ 1.00$. Thus, 10 apples also cost $R \$ 1.00$. Since the price of eggs increased by $10 \%$, the new price of eggs is $R \$ 1.10$. The price of apples decreased by $2 \%$, so the new price of apples is $R \$ 0.98$.
Thus, before it cost 2 reais to buy 1 dozen eggs and 10 ... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
What is the digit $a \mathrm{in}$
$$
a 000+a 998+a 999=22997 ?
$$ | Performing the addition
$$
\begin{array}{r}
a \\
a \\
a 998 \\
+\quad a 999 \\
\hline \square 997
\end{array}
$$
we find $\square 997=22997$, where $\square=a+a+a+1$.
Therefore, $22=a+a+a+1$. Thus, $a=7$. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A number is called ascending if each of its digits is greater than the digit to its left. For example, 2568 is ascending and 175 is not. How many ascending numbers are there between 400 and 600? | The numbers we are looking for are greater than 400 and less than 600, so the hundreds digit is 4 or 5. Since they are ascending numbers, the tens digit must be less than the units digit. Let's see how to choose the tens and hundreds digits.
$$
4\left\{\begin{array}{l}
56 \\
57 \\
58 \\
59
\end{array} \quad ; \quad 4\... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The opposite of a two-digit number, both different from zero, is the number obtained by swapping the order of its digits. For example, the opposite of 25 is 52 and the opposite of 79 is 97. Which of the numbers below is not the sum of a two-digit number and its opposite?
A) 44
B) 99
C) 121
D) 165
E) 181 | Solution
## ALTERNATIVE E
Let $n$ be a two-digit number, with $a$ being its tens digit and $b$ its units digit; then $n=10a+b$. If $a$ and $b$ are both different from zero, the reverse of $n$ is $10b+a$. Thus, the sum of $n$ and its reverse is:
$$
(10a+b)+(10b+a)=11a+11b=11(a+b)
$$
and, therefore, the sum of a numb... | 181 | Number Theory | MCQ | Yes | Yes | olympiads | false |
The number abcde has five distinct digits, none of which are zero, each represented by one of the letters $a, b, c, d, e$. Multiplying this number by 4 results in a five-digit number edcba. What is the value of $a+b+c+d+e$?
A) 22
B) 23
C) 24
D) 25
E) 27 | Solution
## ALTERNATIVE E
The multiplication can be schematized as
The solution is based on the following observations:
- The digit $a$ can only be 1 or 2, because if $a \geq 3$, then $... | 27 | Number Theory | MCQ | Yes | Yes | olympiads | false |
Write the digits from 0 to 9 in a line, in the order you choose. On the line below, join the neighbors, forming nine new numbers, and add these numbers as in the example:
| 2 | | 1 | | 3 | | 7 | | 4 | | 9 | | 5 | | 8 | | 0 | | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | ... | Solution
## ALTERNATIVE B
For any arrangement of the digits, the sum of the "joined" neighbors will always have nine terms, without repetition of digits in the units or in the tens. The only digit that does not appear in the units is the first, and the only one that does not appear in the tens is the last. For the su... | 494 | Number Theory | MCQ | Yes | Yes | olympiads | false |
Alberto, Bernardo, and Carlos competed in a race, during which each of them ran at a constant speed throughout the course. When Alberto crossed the finish line, Bernardo and Carlos were 36 and 46 meters behind him, respectively. When Bernardo crossed the finish line, Carlos was 16 meters behind him. What is the length ... | Solution
## ALTERNATIVE A
Let $x$ be the length in meters of the track. The distance between Bernardo and Carlos was 10 meters when Alberto crossed the finish line, and it was 16 meters when Bernardo crossed the finish line. Thus, during the time interval in which Alberto and Bernardo completed the race, Bernardo ran... | 96 | Algebra | MCQ | Yes | Yes | olympiads | false |
The square in Figure I is called special because:
- it is divided into 16 equal squares;
- in each row and in each column, the digits 1, 2, 3, and 4 appear;
- in each of the squares \(A, B, C\), and \(D\) (as in Figure II), the digits 1, 2, 3, and 4 appear.
| 4 | 2 | 1 | 3 |
| :--- | :--- | :--- | :--- |
| 1 | 3 | 2 ... | Solution
a) The solution is presented in the figure below:
| $\mathbf{1}$ | 2 | $\mathbf{4}$ | $\mathbf{3}$ |
| :--- | :--- | :--- | :--- |
| 3 | 4 | 2 | 1 |
| $\mathbf{2}$ | $\mathbf{3}$ | 1 | $\mathbf{4}$ |
| $\mathbf{4}$ | $\mathbf{1}$ | $\mathbf{3}$ | 2 |
b) No. Since the small squares in the last column of squa... | 288 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The circles in the figure below have been filled with the numbers 1 to 7, such that all arrows point from a smaller number to a larger one. In this case, we say that the figure is well filled.
 There is only one way to fill in the diagram, as we show below.
- The number 9 cannot be placed below any number, so it must be at the top.
- Above the number 7, we can only place the 9 or the 8. Since the 9 is already at the top, the 8 will be above the 7.
- The number 6 cannot be placed below the 5 or t... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The six triangles in the figure are right-angled and their angles with vertex at point $A$ are equal. In addition, $A B=24 \mathrm{~cm}$ and $A C=54 \mathrm{~cm}$. What is the length of $A D$?
A) $110^{\circ}$
B) $115^{\circ}$
C) $120^... | Solution
## ALTERNATIVE C
Consider the triangle $ABC$ in the figure below. It is a right triangle with $AB=1 \text{ cm}$ and $BC=2 \text{ cm}$, that is, one leg is half the hypotenuse. It follows that $D \hat{C} B = A \hat{C} B = 30^{\circ}$, and similarly, $C \hat{B} D = 30^{\circ}$. Since the sum of the interior an... | 120 | Geometry | MCQ | Yes | Yes | olympiads | false |
On a sheet of paper, we mark points equally spaced horizontally and vertically, so that square $A$ has an area of $1 \mathrm{~cm}^{2}$, as shown in the figure. We say that a square is legal if its vertices are four of these points; for example, squares $A$ and $B$ are legal.
 $1^{\text {st }}$ solution: Observing the figure below, we see that square $B$ can be inscribed in a square consisting of 9 small squares.
 The ... | Solution
a) The sum of all odd numbers from 1 to 17 is 81. Since there are three columns in the square, and all columns (and rows) have the same sum, the sum in each column must be $81 / 3 = 27$. From this, we deduce that the missing number in the third column (the one with the numbers 13 and 3) is the number 11, beca... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Cyclists have an aversion to the number zero (because it is oval) and the number eight (because that's how the wheels look after accidents). How many members can register in a cycling club if each must have a three-digit identification number, without using the digit zero or the digit eight?
# | Solution
Since the cyclists do not use the digit 0 or 8, the remaining digits are $1,2,3,4,5,6,7$ and 9. Thus, there are 8 possibilities for the choice of each digit. We need to choose three-digit numbers. Therefore, we have 8 options for the first digit, 8 options for the second digit, and 8 options for the third dig... | 512 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the city of Cordisburgo, a bicycle race was held on a circular circuit, in which three cyclists, Guimarães, Rosa, and João, participated. In the first hour of the race, Guimarães completed exactly 230 full laps, João completed exactly 111 full laps, but it is not known how many laps Rosa completed. It is only known ... | Solution
Let $x$ be the number of laps Rosa completed in the first hour of the race. Since Guimarães completed more laps than Rosa, and both ran at a constant speed, the number of times Guimarães lapped Rosa was
$$
230-x
$$
Since Rosa completed more laps than João, and both ran at a constant speed, the number of tim... | 238 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Write, in ascending order, each of the multiples of 3 whose sum with 1 is a perfect square:
$$
3,15,24,48, \ldots
$$
a) What is the next number that will appear in this sequence after 48?
b) What is the eighth number in this sequence?
c) What is the number that will appear in this sequence in the $2013^{th}$ positi... | Solution
a) Note that 48 is a term in the sequence that is a multiple of 3 and satisfies $48+1=49=7^{2}$. Thus, to find its successor, we will try to obtain a multiple of three that, when increased by one unit, equals $8^{2}=64$. Subtracting one unit from 64, we get 63, which is indeed a multiple of 3.
In other words... | 9120399 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A mathematical worm starts from point $A$ and reaches point $B$ in the figure below.
B
This mathematical worm always moves along the black lines in the drawing above, and never passes over ... | Solution
a) To make a path contained in the segments of the figure, the mathematical worm must choose, every two steps, one up and one to the right, or one to the right and one up. For example, if the worm takes the path illustrated below (in the thicker line)
. Knowing that Benito Juárez turned $x$ years old in the year $x^{2}$, what was the year of his birth?
# | Solution
The perfect squares that are closest to 1801-1900 are:
$$
\begin{aligned}
& 42 \times 42=1764 \\
& 43 \times 43=1849 \\
& 44 \times 44=1936
\end{aligned}
$$
Let $x$ be the age of Benito Juárez in the year $x^{2}$. The number $x$ cannot be 42, because in this case Benito would not have been born in the 19th ... | 1806 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $a$ be a positive integer such that there are exactly 10 perfect squares greater than $a$ and less than $2a$.
a) Find the smallest possible value of $a$.
b) Find the largest possible value of $a$.
# | Solution
a) Since there are exactly ten perfect squares greater than $a$ and less than $2a$, these ten perfect squares must be consecutive. Therefore, there must exist a positive integer $x$ such that
$$
(x-1)^{2} \leq a\frac{961}{2}>480
$$
which gives us
$$
a \geq 481
$$
Now, we observe that between 481 and $2(48... | 684 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Consider the illustrated diagram below:
Augusto likes to count paths starting from some point, reaching point $A$ and never passing through the same vertex twice. To do this, he represents ... | Solution
a) The unusual paths starting from $D$ and arriving at $A$ are $D C B A$, $D B A$, and $D C A$.
b) An unusual path starting from $E$ must visit one of the points $D$ or $C$ next. In the case where it visits point $D$, its continuation to point $A$ can be any of the unusual paths starting from $D$ and arrivin... | 89 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Arnaldo, Bernaldo, Cernaldo, Dernaldo, and Ernaldo are students from different parts of Brazil who were chosen to represent their country in international olympiads. After several weeks of training, some friendships were formed. We then asked each of them how many friends they had made in the group. Arnaldo, Bernaldo, ... | Solution
Suppose Ernaldo has $x$ friends within the group. Since Dernaldo has 4 friends and the group has 5 members, everyone is a friend of Dernaldo. Let's remove Dernaldo from the group. Thus, Arnaldo, Bernaldo, Cernaldo, and Ernaldo now have $0, 1, 2$, and $x-1$ friends within the subgroup, respectively. Since Arna... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Little Abel received as a gift a $2 \times n$ board and $n$ tiles of size $2 \times 1$. For example, the following figure shows the case where $n=10$, that is, when Abel has a $2 \times 10$ board and 10 tiles of size $2 \times 1$.
 We can easily count the first few cases and observe that $a_{1}=1, a_{2}=2 \text{ and } a_{3}=3$, as shown in the figure below.
b) When starting to fill his $2 \times 10$ board,... | 89 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
On the sides $A B$ and $B C$ of an equilateral triangle $A B C$, two points $D$ and $E$ are fixed, respectively, such that $\overline{A D}=\overline{B E}$.
If the segments $A E$ and $C D$ in... | Solution
Consider the triangles $D A C$ and $E B A$. It is known that $\overline{D A}=\overline{E B}$. Furthermore, since triangle $A B C$ is equilateral, then $\overline{A C}=\overline{B A}$. Additionally, in an equilateral triangle, all internal angles measure $60^{\circ}$. Therefore, $\measuredangle D A C=\measured... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On a $5 \times 5$ board, the numbers $1, 2, 3, \ldots, 25$ have been distributed such that each cell is occupied by a single number and that two consecutive numbers are always placed in adjacent cells. The following figure shows an example of how to distribute these numbers.
| 3 | 4 | 5 | 24 | 23 |
| :---: | :---: | :... | Solution
a) Let's color the squares of the board white and black as indicated in the following figure. In this way, the squares corresponding to two consecutive numbers will always be painted in different colors.
 The number 978563412 is, clearly, a zigzag number. Let's show that if $N$ is a zigzag number, then
$$
N \leq 978563412
$$
The maximum number of digits a zigzag number can have is 9. Therefore, we can assume that $N$ is of the form
$$
N=\overline{a b c d e f g h i}
$$
since otherwise, the inequality (.1... | 978563412 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a class, $40 \%$ of the students have poor eyesight. Of these, $70 \%$ wear glasses and the remaining $30 \%$ wear contact lenses. Knowing that 21 students wear glasses, how many students are there in this class? | Let $A$ be the total number of students in the class. Thus, $\frac{40}{100} \times A$ do not see well. Therefore, $\frac{70}{100} \times \frac{40}{100} \times A$ wear glasses. Consequently, we have:
$$
\frac{70}{100} \times \frac{40}{100} \times A=21 \Rightarrow A=\frac{21 \times 100}{7 \times 4}=3 \times 25=75
$$ | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The ratio between the number of men and women in the city of Campo Verde is $\frac{2}{3}$. The average age of men is 37 years and that of women is 42 years. What is the average age of the inhabitants of Campo Verde? | If $H$ denotes the number of men and $M$ the number of women, then:
$$
\frac{H}{M}=\frac{2}{3} \quad \Rightarrow \quad M=\frac{3 H}{2}
$$
The average age of the population is:
$$
\frac{37 H+42 M}{H+M}=\frac{37 H+42 \frac{3 H}{2}}{H+\frac{3 H}{2}}=\frac{100 H}{\frac{5 H}{2}}=\frac{100 \times 2}{5}=40 \text { years. }... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the trapezoid in the figure below, $AB$ is parallel to $CD$, $AD = AB = BC = 1 \, \text{cm}$, and $DC = 2 \, \text{cm}$. What is the measure of the angle $\angle CAD$?
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $90^{\circ}$
(E) $120^{\circ}$
.
Let $P$ be the midpoint of segment $C D$ and draw segments $A P$ and $B P$. The three triangles formed $\triangle A D P$, $\triangle A B P$, and $\triangle B C P$ are equilateral (why?). Therefore, the angles $D \widehat{A} P=60^{\circ}=P \widehat{A B}$. Since segment $A C$ is the bisector o... | 90 | Geometry | MCQ | Yes | Yes | olympiads | false |
How many times does the word BRAZIL appear in the figure on the side? Only count the word by joining letters that are written in adjacent squares.
| | | | | $\bar{B}$ | R |
| :---: | :---: | :---: | :---: | :---: | :---: |
| | | | $\bar{B}$ | $\bar{R}$ | $\bar{A}$ |
| | | $B$ | $\bar{R}$ | $\bar{A}$ | $\bar{S... | To read the word BRASIL, we must follow a path that starts at a letter B and ends at a letter L. Observe that the path to be followed is composed successively of horizontal moves to the right and vertical moves downward. Thus, let us represent these paths by sequences of letters $\mathrm{H}$ (meaning move to the right)... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Show that $M=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$ is an integer. | Let $a=\sqrt[3]{\sqrt{5}+2}$ and $b=\sqrt[3]{\sqrt{5}-2}$. Thus, $M=a-b$ and we have:
$$
M^{3}=(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)
$$
We know that $a^{3}-b^{3}=4$ and $a b=1$. Therefore, $M^{3}+3 M-4=0$, which means the number $M$ is a root of the polynomial $x^{3}+3 x-4$.
In turn, the number 1 is a root of the polynom... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
The squared rectangle in the figure is made of 31 segments of $0.5 \mathrm{~cm}$, and comprises 12 squares. Rosa drew on a rectangular sheet of $21 \mathrm{~cm}$ by $29.7 \mathrm{~cm}$, quadriculated with squares of side $0.5 \mathrm{~cm}$, a large squared rectangle made with 1997 segments. How many squares does this r... | Let $m$ and $n$ be, respectively, the number of segments of $0.5 \, \text{cm}$ on two consecutive sides of the rectangle. We know that the total number of segments of $0.5 \, \text{cm}$ in the division of the rectangle into $m \times n$ squares of side $0.5 \, \text{cm}$ is: $m(n+1) + n(m+1)$ (prove this). Thus,
$$
m(... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Knowing that $a$ is a natural number, and that $4 a^{2}$ and $\frac{4}{3} \times a^{3}$ are natural numbers with 4 digits, determine $a$.
## List 4 | We have $1000 \leq 4 a^{2}<10000$ and also $1000 \leq \frac{4}{3} \times a^{3}<10000$.
From $1000 \leq 4 a^{2}<10000$ it follows that $250 \leq a^{2}<2500$. Since $a$ is an integer and $15^{2}=225,16^{2}=256$ and $50^{2}=2500$, we have that $15<a<50$.
From $1000 \leq \frac{4}{3} a^{3}<10000$ we get $750 \leq a^{3}<75... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
How many pairs of positive integers $(x, y)$ are solutions to the equation $3 x+5 y=501 ?$
untranslated text has been directly translated while preserving the formatting and structure. | The given equation is equivalent to $y=\frac{3(167-x)}{5}$. Since $y$ is a positive integer, $167-x$ must be a positive multiple of 5, that is:
$167-x=5 k \quad \Rightarrow \quad x=167-5 k \quad \Rightarrow \quad x=5 \times 33+2-5 k \quad \Rightarrow \quad x=5(33-k)+2$
where $k$ is a positive integer. Since $x$ is po... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The director of the school decided to take a photo of the 2008 graduates. He arranged the students in parallel rows, all with the same number of students, but this arrangement was too wide for the camera's field of view. To solve this problem, the director noticed that it was enough to remove one student from each row ... | The figures below represent the situation of the problem, where in black are represented the students who were initially removed and in gray the students removed for the second time.
Let $n... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Draw two circles with the same center, one with a radius of $1 \mathrm{~cm}$ and the other with a radius of $3 \mathrm{~cm}$. In the region outside the circle with a radius of $1 \mathrm{~cm}$ and inside the circle with a radius of $3 \mathrm{~cm}$, draw circles that are simultaneously tangent to both circles, as shown... | (a) Since the circles with radii $1 \mathrm{~cm}$ and $3 \mathrm{~cm}$ are concentric, the other circles shown in the figure must have a radius equal to $1 \mathrm{~cm}$.
(b) The centers of the 3 circles with radius $1 \mathrm{~cm}$ shown in the figure form an equilateral triangle with side length $2 \mathrm{~cm}$. Th... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Márcia is in a store buying a recorder that she has wanted for a long time. When the cashier registers the price, she exclaims: "That can't be right, you entered the number backwards, you switched the order of two digits, I remember it cost less than 50 reais last week!" The cashier replies: I'm sorry, but all of our i... | The old price was less than 50 reais and increased by $20 \%$. Therefore, the new price is still a two-digit number. Let's represent it as $a b$, where $a$ is the tens digit and $b$ is the units digit. Thus, the new price is $b a$, and we have:
$$
10 b+a=1.2(10 a+b) \Rightarrow 10 b-1.2 b=12 a-a
$$
Therefore:
$$
8.8... | 54 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a condominium, 29 families live, each of them has either 1 cat or 3 cats or 5 cats. The number of families that have only 1 cat is the same as the number of families that have 5 cats. How many cats are there in this condominium? | Let:
$$
\begin{aligned}
& x=\text { number of families that have only } 1 \text { cat; } \\
& y=\text { number of families that have exactly } 3 \text { cats; } \\
& z=\text { number of families that have } 5 \text { cats. }
\end{aligned}
$$
It follows that $x+y+z=29$ and $x=z$. Therefore, $2 x+y=29$. On the other ha... | 87 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In triangle $\triangle A B C$, point $F$ is on side $A C$ and $F C=2 A F$. If $G$ is the midpoint of segment $B F$ and $E$ is the intersection point of the line passing through $A$ and $G$ with segment $B C$, calculate the ratio $\frac{E C}{E B}$.
=\text { Area }(\triangle B C D) \text { and Area }(\triangle A B C)=\text { Area }(\triangle A C D) \text {. }
$$
But,
$$
\begin{aligned}
\text { Area }(\triangle A B D) & =X+W \\
\text { Area }(... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The number 81 has the following property: it is divisible by the sum of its digits $8+1=9$. How many two-digit numbers satisfy this property? | Let $ab$ be such a number. By hypothesis, $ab = 10a + b$ is divisible by $a + b$. Therefore, the difference $(10a + b) - (a + b) = 9a$ is also divisible by $a + b$. Moreover, we know that $10a + b$ is divisible by $a + b$ if and only if $(10a + b) - (a + b) = 9a$ is divisible by $a + b$ (prove this).
Before proceeding... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The sum $1+1+4$ of the digits of the number 114 divides the number itself. What is the largest number, less than 900, that satisfies this property? | To find the largest possible number less than 900, we start with the digit 8 in the hundreds place. We observe that the number 800 satisfies the property. Therefore, the number we are looking for is greater than or equal to 800.
We need to find $a$ and $b$ such that $8+a+b$ divides $8ab = 800 + 10a + b$. We remember t... | 888 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The base $A D$ of a trapezoid $A B C D$ measures $30 \mathrm{~cm}$. Suppose there is a point $E$ on $A D$ such that the triangles $\triangle A B E, \triangle B C E$ and $\triangle C D E$ have equal perimeters. Determine the length of $B C$. | Suppose that $A E$ is greater than $B C$, and let $A^{\prime}$ be a point on $A E$ such that $A^{\prime} E = B C$.
Since $A^{\prime} E$ and $B C$ are parallel, $A^{\prime} B C E$ is a paral... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the star $A B C D E$ in the following figure, we know that $\measuredangle G B F=20^{\circ}$ and $\measuredangle G H I=130^{\circ}$. What is the value of the angle $\measuredangle J E I$ ?
$ is divisible by 9, which means $a + b$ is divisible by 9. If $a + b = 9$, we have the 9 solutions:
$a=1$ and $b... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the sequence $1,3,2, \ldots$ each term after the first two is equal to the preceding term subtracted from the term that precedes it, that is: if $n>2$ then $a_{n}=a_{n-1}-a_{n-2}$. What is the sum of the first 100 terms of this sequence? | Let's initially write some terms:
$$
1,3,2,-1,-3,-2,1,3,2, \ldots
$$
The 7th and 80th terms are, respectively, equal to the 10th and 2nd terms. This means that the sequence repeats every 6 terms. The sum of the first 6 terms is $1+3+2-1-3-2=0$, and therefore, the sum of the first 96 terms is also 0. Thus, the sum of ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The bisector of an angle is a ray originating from the vertex of the angle that divides it into two other congruent angles. For example, in the drawing below, the ray $O C$ is the bisector of the angle $\angle A O B$.
 Let $\angle B A D=2 x$ and $\angle B A C=2 y$ be the adjacent angles.
The angle between the angle bisectors is
$$
\begin{aligned}
\angle E A F & =\angle E A B-\angle F A B \\
&... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a group of 200 people, only $1 \%$ are women. Determine the number of men who must leave the group so that $98 \%$ of the remaining people are male.
# | Solution
The number of women is $200 \cdot \frac{1}{100}=2$. For this number to represent $2\% = 100\% - 98\%$ of the new total number of people $x$, we must have $2 = x \cdot \frac{2}{100}$, that is, $x = 100$. Therefore, $198 - 98 = 100$ men must leave the group. | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the drawing below, points $E$ and $F$ belong to the sides $A B$ and $B D$ of triangle $\triangle A B D$ in such a way that $A E=A C$ and $C D=F D$. If $\angle A B D=60^{\circ}$, determine the measure of the angle $\angle E C F$.
 Joana, a teacher f... | Solution
a) If Carlão follows Joana's suggestion, item $c$ will be worth 5 points and items $a$ and $b$ must sum to another 5 points. We will then have four possible divisions of items $(a, b, c):(1,4,5),(2,3,5)$, $(3,2,5)$, and $(4,1,5)$.
b) Once the scores for items $a$ and $b$ are defined, item $c$ will be worth $1... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Consider two drums of sufficiently large capacity, one of them empty and the other full of liquid.
a) Determine if it is possible to place exactly one liter of liquid from the full drum into the empty one, using two buckets, one with a capacity of 5 liters and the other with a capacity of 7 liters.
b) Determine if it... | Solution
a) It is enough to fill the empty drum with 15 liters ($3 \times 5$ liters) using the 5-liter bucket three times, and then remove 14 liters ($2 \times 7$ liters) using the 7-liter bucket twice. This way, we transport $3 \times 5 - 2 \times 7 = 1$ liter.
b) The amount $a$ that we can transport from the full dr... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
João works selling astrological prediction packages. To boost sales of his predictions, he offers discounts if people of the same sign want to hire his services. In the Greek Horoscope, since there are exactly 12 signs, therefore, in a group of 13 people, at least two of them will have the same sign and may be interest... | Solution
a) The minimum is 25. If in a group of 24 people each zodiac sign appears at most twice, we will have at most $2 \cdot 12=24$ people. Since $24<25$, this shows that at least one of the zodiac signs must appear three times. Indeed, this is the minimum where such a property occurs, because if we consider 24 peo... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Um decorador distribuirá flores em oito pontos ao redor de um arranjo quadrado de flores, como indicado na figura abaixo. Ele quer fazer isso de modo tal que, em cada lado do arranjo, as pessoas vejam sempre a mesma quantidade de flores. No exemplo abaixo, temos o total de 11 flores e em cada um dos 4 lados do quadrado... | Solução
a) A soma das flores vistas nos lados é $4 \cdot 9=36$. Como as flores nos cantos são vistas por dois lados e as flores no meio dos lados são vistas apenas uma vez, podemos escrever:
$$
2 C+M=36
$$
onde $C$ e $M$ indicam as quantidades de flores nos cantos e no meio. Consequentemente, $C+M=36-C \leq 36$ e se... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Some Christmas lights are arranged using magical wires. Each light can be green or yellow. Each wire is connected to two lights and has a magical property: when someone touches a wire connecting two lights, each of them changes color from green to yellow or from yellow to green.
a) In the arrangement below, each point... | Solution
a) Each wire we touch changes the color of exactly two lamps. Since there are 16 yellow lamps, we must touch at least 8 wires. The figure below shows an example of wire choices that make this possible:
 What is the minimum number of squares we need to mark on this board so that each of its $3 \times 3$ subboards has at least one marked square?
b) What is the minimum number of squares we need to mark on this board so that each of its $3 \times 3$ subboards has at least three marked squ... | Solution
a) Consider the figure below.
Each of the four $3 \times 3$ sub-boards marked in the figure must have at least one cell marked. Additionally, with the four marked cells in the fig... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
a) Show that it is not possible to separate the numbers in the set $A=\{1,2,3, \ldots, 10\}$ into two sets such that the product of the numbers in each of them is the same.
b) What is the minimum number of elements that need to be removed from the set $A$ so that the remaining elements can be divided into two sets who... | Solution
a) It suffices to look at the number 7. Since it is the only number in $A$ with a factor of 7, it is not possible to divide them into two with the same product of their elements, because one of these products would be a multiple of 7 and the other would not.
b) By removing only the number 7, we will show that... | 720 | Number Theory | proof | Yes | Yes | olympiads | false |
João managed to paint the squares of an $n \times n$ board in black and white such that the intersections of any two rows and any two columns were not made up of squares of the same color. What is the maximum value of $n$?
# | Solution
An example $\operatorname{with} n=4$ is given in the figure below:
We want to show now that, if $n \geq 5$, such a coloring is not possible. Consider then an $n \times n$ board wi... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In this problem, we will learn and use the famous Beak Theorem, which is named so because the figure formed really resembles the head and beak of a bird.
a) The Beak Theorem states that the distances from an external point to the points where its tangents touch the circle are equal. In the following figure, $A P$ and ... | Solution
a) Draw $O A$. Observe that the triangles $\triangle O P A$ and $\triangle O Q A$ are congruent because they are right triangles with the same hypotenuse and one of the legs of the same length. Therefore, $A P=A Q$.
b) Each of the sides is divided by the point of tangency into two segments, as shown in the f... | 4 | Geometry | proof | Yes | Yes | olympiads | false |
A common mistake many students make is thinking that two quadrilaterals are congruent if they have their respective sides equal. This is not true. In this problem, we will see that quadrilaterals can have corresponding sides equal, but different areas.
a) Show that the maximum possible area for a quadrilateral that ha... | Solution
a) There are two ways to assemble the quadrilateral with pairs of equal sides: either they are together or they are separated. In both cases, the quadrilateral can be divided into two triangles that will be congruent by the (S.S.S.) case. See the figure below.
Observe that the sum along any side of the triangle above is always the same, as we can verify,
$$
1+3+6=6+2+2=1+7+2
$$
a) Complete the numbers that are ... | Solution
a) In the bottom row, the sum is $2+3+5=10$. Since the sums along any side are equal, the missing number in the top right corner of the square must be equal to 2, as shown in the following figure:
 Find a five-digit number $A B C D E$... | Solution
a) We are looking for a five-digit number $A B C D E$ such that
$$
A B C D E \times 9=E D C B A
$$
Adding $A B C D E$ to both sides of this equality, we get that $A B C D E \times 10 = E D C B A + A B C D E$. But the number $A B C D E \times 10$ can also be represented by the digits $A B C D E 0$. Thus, we ... | 1099989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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