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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 $\frac{E C}{E B}$.
$ 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 |
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 |
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 |
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 |
Amanda drew the following figure:
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:
 In the figure below, there are three squares with sides 9, 6, and $x$. Determine the value of $x$.
b) Marcelo continues the drawing above and draws more squares (many!). Since these becam... | Solution
a) Observe the figure:
By similarity of triangles, we have
$$
\frac{6-x}{x}=\frac{3}{6}
$$
Therefore, $x=4$.
b) We have that the squares are similar in a ratio of $\frac{2}{3}$.... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $A B C D$ and $E F G H$ be squares with side lengths 33 and 12, respectively, with $E F$ lying on side $D C$ (as shown in the figure below). Let $X$ be the intersection point of segments $H B$ and $D C$. Suppose that $\overline{D E}=18$.
 Denote $\overline{E X}=x$. We have that $|\overline{C X}|=33-18-x=15-x$.
Now note that the triangles $E X H$ and $C X B$ are similar, so:
$$
\frac{|\overline{E H}|}{|\overline{... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A group of boys and girls went out to eat pizza for two consecutive days. At the restaurant they went to, the pizzas are cut into twelve equal slices. Maria observed that on the first day, each boy ate 7 slices, and each girl 3 slices. On the second day, each boy ate 6 slices, and each girl 2 slices. Interestingly, on ... | Solution
Let $x$ and $y$ be the number of boys and girls, respectively. We know that the total number of slices consumed was at least 49 (4 pizzas and one slice of the last pizza) and at most 59 (4 pizzas plus 11 slices, remember that at least one slice of the last pizza was left). On the other hand,
$$
\begin{aligne... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In a class, there are 70 students, such that:
I) 14 boys passed in Mathematics;
II) 12 boys passed in Physics;
III) 10 boys and 16 girls did not pass in Mathematics or Physics;
IV) 32 are boys;
V) 10 passed in both subjects;
VI) 22 passed only in Mathematics.
How many girls passed only in Physics? | Solution
To solve the problem, we will use the diagram below, where the upper rectangle represents the quantities of boys in each case and the lower one the quantities of girls; in the left circle, the number of students who passed in mathematics, while in the right one, the number who passed in physics, and in the in... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Using 5 non-zero digits, we can form 120 numbers, without repeating a digit within the same number. Let $S$ be the sum of all these numbers. Determine the sum of the digits of $S$, where:
a) 1, 3, 5, 7, and 9 are the 5 digits;
b) 0, 2, 4, 6, and 8 are the 5 digits, remembering that 02468 is a number with 4 digits and... | Solution
a) There are 120 numbers in total, with all possible combinations. Thus, in each of the positions (units, tens, hundreds, thousands, ten thousands), each digit appears the same number of times, that is, $\frac{120}{5}=24$. For example, in the units place, the digit 1 appears 24 times, as does 3, 5, 7, and 9. ... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
On a $4 \times 4$ board, the numbers from 1 to 16 must be placed in the cells without repetition, such that the sum of the numbers in each row, column, and diagonal is the same. We call this sum the Magic Sum.
a) What is the Magic Sum of this board?
b) If the sum of the cells marked with $X$ in the board below is 34, ... | Solution
a) Since there are 4 rows (as well as 4 columns), the Magic Sum is:
$$
\frac{1+2+3+\ldots+16}{4}=34
$$
b) If we add the two diagonals, we will have exactly the sum of the cells marked with $X$ and with $Y$. Thus, the sum of the cells marked with $Y$ is $2 \cdot 34-34=34$.
c) We have:
$$
\begin{aligned}
\f... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The numbers $a$ and $b$ are positive integers and satisfy $96 a^{2}=b^{3}$. What is the smallest value of $a$? | Factoring 96 we have: $2^{5} \times 3 \times a^{2}=b^{3}$. For $2^{5} \times 3 \times a^{2}$ to be a cube, the number $a$ must be of the form: $a=2^{n} \times 3^{m}$. Thus,
$$
2^{5} \times 3 \times a^{2}=2^{5} \times 3 \times\left(2^{n} \times 3^{m}\right)^{2}=2^{5+2 n} \times 3^{1+2 m}=b^{3}
$$
Therefore, $5+2 n$ an... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Twelve points are marked on a sheet of graph paper, as shown in the figure. What is the maximum number of squares that can be formed by connecting four of these points?
 | In total, we have 11 possible squares as shown below.
 | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the map of Figure 21.1, the curve XY is one of the boundaries. Countries like I and II share a common border. Point $Y$ is not considered a border, that is, countries like I and V do not share a common border. You must color the map such that countries with a common border have different colors.
 At a minimum, two colors are necessary, as shown in Figure 21.2.
(b) Figures 21.3 and 21.4 display two maps with six countries that require at least four colors to be painted. | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The cities of Coco da Selva and Quixajuba are connected by a bus line. Buses leave Coco da Selva for Quixajuba every hour, with the first one departing at midnight on the dot. Buses leave Quixajuba for Coco da Selva every hour, with the first one departing at half past midnight. The bus journey takes exactly 5 hours.
... | Let's observe that the bus departing from Coco da Selva to Quixajuba meets the buses that, at the time of its departure, are
Suggestion: Divide the coins into three groups of 16 coins. on the way from Quixajuba to Coco da Selva and the buses that depart in the next five hours.
The buses that are on the road are those... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
pongue
Some students from the seventh and eighth grades of a school participate in a ping pong tournament, where each student plays against all the others exactly once, receiving 1 point for a win and 0 points for a loss. There are ten times as many eighth-grade students as seventh-grade students. The total score of t... | pongue - Solution
a) Each of the $k$ students will play $k-1$ times. Summing the number of games each one plays, we get $k(k-1)$. However, we will have counted each game twice, once for each of the participants in the match. Therefore, the number of games is $\frac{k(k-1)}{2}$.
b) Let $n$ be the number of seventh-gra... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A group of 10 students participates in a mathematics competition formed by teams of 4 students. We know that any two of the teams have exactly one student in common.
a) What is the maximum number of teams a student can participate in? Provide an example of a distribution of 10 students where this maximum number can be... | Solution
a) Consider a student $A$ who is part of the most teams and say that he is in a team with three other students $B, C$ and $D$. Any other team that also includes $A$ must contain three other students who are not in the set $\{B, C, D\}$. Since there are only $10-1=9$ students different from $A$, the maximum nu... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Sérgio chooses two positive integers $a$ and $b$. He writes 4 numbers in his notebook: $a, a+2, b$ and $b+2$. Then, all 6 products of two of these numbers are written on the board. Let $Q$ be the number of perfect squares written on it, determine the maximum value of $Q$.
# | Solution
Initially, we will prove that the product $a(a+2)$ is not a perfect square for any choice of $a$. We have two cases to consider:
i) If $a$ is odd, then no prime that divides $a$ can divide $a+2$. Therefore, $a$ and $a+2$ must each be a perfect square. This clearly has no solution for $a \geq 1$, as the diffe... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A chessboard is an $8 \times 8$ square where the squares are arranged in 8 rows and 8 columns.
A rook on a chessboard attacks all pieces that are in its row or column. Knowing this, determin... | Solution
a) Below, we have an example with 8 towers, none of which are attacking each other.
If we place 9 or more towers, since there are only 8 rows, there will be two in the same row, a... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the number $6 a 78 b$, $a$ is the digit of the thousands place and $b$ is the digit of the units place. 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 number 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, we have that $6+a+7+8+0=21+a$ or $6+a+7+8+5=26+a$ are multiples of 9. Hence, $a=6$ or $a=1$, respectively. From this, we get: $a+b... | 6 | Number Theory | MCQ | Yes | Yes | olympiads | false |
In 13 boxes, 74 pencils were packed. If the maximum capacity of each box is 6 pencils, what is the minimum number of pencils that can be in a box?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 6 | 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$, which is greater than the number of pencils 74. In 12 boxes we would have: $12 \times 6=72$. Thus, there would be one box with $74-72=2$ pencils. Therefore, the correct o... | 2 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
Imagine as 2015 fractions:
$$
\frac{2}{2016}, \frac{3}{2015}, \frac{4}{2014}, \ldots, \frac{2014}{4}, \frac{2015}{3}, \frac{2016}{2}
$$
Is it possible to choose three of these fractions with a product equal to 1? | Solution
Yes. See that each fraction is of the form $\frac{x}{2018-x}$. Thus, for $x=1009$, the fraction $\frac{1009}{1009}=1$ is part of the list. It then suffices to multiply the fractions:
$$
\frac{2}{2016} \cdot \frac{1009}{1009} \cdot \frac{2016}{2}=1
$$ | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Carlinhos likes to write numbers in his notebook. One day he wrote the numbers from 1 to 999, one after the other, to form the giant number:
$$
123456789101112 \ldots 997998999
$$
About this number, the following questions are asked:
(a) How many digits were written?
(b) How many times does the digit 1 appear?
(c)... | Solution
(a) Observe that we have 9 numbers with one digit each, $99-9=90$ numbers with two digits each, and $999-99=900$ numbers with three digits each. Therefore, the number of digits written is:
$$
9 \cdot 1 + 90 \cdot 2 + 900 \cdot 3 = 2889
$$
(b) Note that the digit 1 appears once among the one-digit numbers. A... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The figure below shows a "staircase" formed by two squares, one with a side of $8 \mathrm{~cm}$ and one with a side of $6 \mathrm{~cm}$. The task is to cut the figure into three pieces and reassemble them to form a square without any gaps.
 Since the square should not have any holes, the final area must be equal to the original area. If we call $L$ the side of the square, we have:
$$
\begin{aligned}
L^{2} & =8^{2}+6^{2} \\
L^{2} & =64+36 \\
L^{2} & =100 \\
L & =10
\end{aligned}
$$
(b) By the Pythagor... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The 9 squares of a $3 \times 3$ board, as shown in the figure below, must be painted in such a way that in each row, each column, and each of its two diagonals, there are no squares of the same color. What is the minimum number of colors needed for this painting?
 Suppose it is possible to have 0 adults. Thus, the sum of the quantities of young people and children is 100. Consequently, if each child pays 1.70 more to match the contribution of the young people, we will have $2 \cdot 100=200... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the board below, in each row and in each column, exactly one number from the set $\{1,2,3,4\}$ is written. Which number is in the square with the symbol $\star$?
| 1 | 2 | | |
| :--- | :--- | :--- | :--- |
| | 1 | 2 | |
| 2 | | $\star$ | 1 |
| 3 | | 1 | | | Solution
Consider the positions indicated by the letters: $A, B, C$ and $D$. Whenever a row or column already has three numbers written, its fourth element is completely determined and is equal to the number that has not yet appeared in it. Thus, $C=4$. Analyzing the numbers in the second row, we have $D=3$. The cell ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Around a circular table, 18 girls are sitting, 11 dressed in blue and 7 dressed in red. Each of them is asked if the girl to their right is dressed in blue, and each one answers yes or no. It is known that a girl tells the truth only when her two neighbors, the one on the right and the one on the left, are wearing clot... | Solution
Let's analyze the possibilities. If the two neighbors of a certain girl are wearing blue, then she answers yes, and if they both wear red, she answers no, because in these cases she tells the truth. If the neighbor on the right is wearing red and the one on the left is wearing blue, the answer is yes, and if ... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A round table has a diameter of $1.40 \, \text{m}$. For a party, the table is expanded by adding three planks, each $40 \, \text{cm}$ wide, as shown in the figure. If each person at the table should have a space of $60 \, \text{cm}$, how many guests can sit at the table?
The perimeter of the enlarged table is
$$
140 \times \pi + 40 \times 6 \simeq 140 \times 3.14 + 240 = 679.60 \mathrm{~cm}
$$
If each guest needs $60 \mathrm{~cm}$ to sit around the table and... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A surveillance service is to be installed in a park in the form of a network of stations. The stations must be connected by telephone lines so that any of the stations can communicate with all the others, either by a direct connection or through at most one other station. Each station can be directly connected by a cab... | The example shows that we can connect at least 7 stations under the proposed conditions. We start with a particular station, and think of it as the base of the network. It can be connected to 1, 2, or 3 stations as shown in the diagram.
 In how many ways can the number 105 be written as the difference of two perfect squares?
(b) Show that it is not possible to write the number 106 as the difference of two perfect squares.
## | (a) Let $x$ and $y$ be two positive integers such that the difference between their squares is equal to 105, that is, $x^{2}-y^{2}=105$. Factoring, we get $(x-y)(x+y)=105$, and therefore, $x+y$ and $x-y$ must be divisors of 105, with $x+y > x-y$. Note that $1 \cdot 105 = 3 \cdot 35 = 5 \cdot 21 = 7 \cdot 15$ are all th... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
If $(x+y)^{2}-(x-y)^{2}=20$, then $x y$ is equal to:
(a) 0
(b) 1
(c) 2
(d) 5
(e) 10 | As $(x+y)^{2}=x^{2}+2 x y+y^{2}$ and $(x-y)^{2}=$ $x^{2}-2 x y+y^{2}$, we have:
$$
(x+y)^{2}-(x-y)^{2}=x^{2}+2 x y+y^{2}-x^{2}+2 x y-y^{2}=4 x y=20
$$
it follows that $x y=5$. The correct option is (d). | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
feira 13 - What is the maximum number of Friday the 13ths that can occur in a non-leap year? In this case, what is the $10^{\text{th}}$ day of the year? | Feira 13 - Given that the days of the week repeat every 7 days, the difference between the days of the week is given by the remainder when dividing by 7 the number of days that have passed.
$\mathrm{In}$ the following table:
- in the first row, the number of days between the 13th of one month and the 13th of the next... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
How many triangles exist whose sides are integers and the perimeter is 12?
(a) 1
(b) 3
(c) 5
(d) 7
(e) 9 | For three numbers $a, b, c$ to be the lengths of the sides of a triangle, each of them must be greater than the difference and less than the sum of the other two.
Let $a \leq b \leq c$ be the lengths of the sides of the triangle. Thus, $c < a + b$.
Now, adding $c$ to both sides, we have: $2c < a + b + c = 12$, that i... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
Archaeologists found a gold necklace made of plates in the shape of regular pentagons. Each of these plates is connected to two other plates, as illustrated in the figure.
Figure 51.1
How m... | The internal angle of a regular pentagon measures $108^{\circ}$. Therefore, the internal angle of the polygon determined by the necklace measures $360^{\circ}-108^{\circ}-108^{\circ}=144^{\circ}$. We must then find $n$ such that
$$
\frac{180^{\circ}(n-2)}{n}=144^{\circ}
$$
Solving this equation, we get $n=10$. Theref... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The sum of 3 numbers is 100, two are prime and one is the sum of the other two.
(a) What is the largest of the 3 numbers?
(b) Give an example of these 3 numbers.
(c) How many solutions exist for this problem? | (a) Initially observe that:
- the largest number is the sum of the other two;
- the largest number cannot exceed 50, otherwise the sum of the three would be greater than 100;
- the largest number cannot be less than 50, otherwise the sum of the three would be less than 100.
Therefore, the largest number can only be 5... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the digit of the $1997^{\mathrm{th}}$ decimal place of:
(a) $\frac{1}{22}$
(b) $\frac{1}{27}$
## Level 1 Solutions
## List 1
# | (a) Dividing 1 by 22 we get: $\frac{1}{22}=0.0454545 \ldots$ We observe that the digit 4 is in the even positions: $2,4,6, \ldots$ and the digit 5 in the odd positions: $3,5,7 \ldots$ Since 1997 is an odd number, the digit in the $1997^{\text{th}}$ decimal place is 5.
(b) Dividing 1 by 27 we get: $\frac{1}{27}=0.03703... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In how many ways can the 6 faces of a cube be colored black or white? Two colorings are the same if one can be obtained from the other by a rotation. | Let's observe that it suffices to count how many colorings exist that have exactly 0, 1, 2, and 3 black faces, because the other cases are symmetric. With one or no black face, there is a unique coloring for each case. When we have two black faces, we have two possible colorings: when these faces are opposite and when ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Each person in a group of ten people calculates the sum of the ages of the other nine members of the group. The ten sums obtained are $82, 83, 84, 85, 87, 89, 90, 90, 91$ and 92.
Determine the age of the youngest person. | Observe that the age of each person appears as a term in 9 of the 10 numbers. Thus, if we sum the 10 numbers, we will get nine times the sum of all the ages. Therefore, the sum of the ages of the ten people is
$$
\frac{82+83+84+85+87+89+90+90+91+92}{9}=\frac{873}{9}=97
$$
The youngest person obtained the highest sum,... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The unit digit of the number $1 \times 3 \times 5 \times 79 \times 97 \times 113$ is
(a) 1
(b) 3
(c) 5
(d) 7
(e) 9 | The correct option is (c).
The given product has one of its factors equal to 5, therefore, it is a multiple of 5, which always has the unit digit equal to 0 or 5. Moreover, since all the factors are odd numbers, the product is an odd number. Thus, its unit digit is 5. | 5 | Number Theory | MCQ | Yes | Yes | olympiads | false |
See the promotions of two supermarkets:
| Supermarket A | Supermarket B |
| :---: | :---: |
| 6 cans of 3 liters of QUOTE ice cream | QUOTE ice cream - can of 3 liters |
| $\mathrm{R} \$ 24.00$ | 4 cans - only $\mathrm{R} \$ 14.00$ |
Joana wants to buy 12 cans of ice cream for her birthday party. In which supermarket... | The correct option is (d).
If Joana buys at supermarket A, she will spend $2 \times 24=48$ reais. If she buys at supermarket B, she will spend $3 \times 14=42$ reais. Therefore, at supermarket B, she will save 6 reais compared to A. | 6 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
Seven teams, divided into two groups, participated in the football tournament in my neighborhood. Group 1 consisted of the teams Avaqui, Botágua, and Corinense. Group 2 consisted of the teams Dinossaurs, Esquisitos, Flurinthians, and Guaraná.
$\mathrm{In}$ the first round of the tournament, each team played against ea... | Let's denote the seven teams by their initial letter.
(a) In the first round of Group 1, three matches were played: $\mathrm{A} \times \mathrm{B}, \mathrm{B} \times \mathrm{C}$, and $\mathrm{C} \times \mathrm{A}$.
(b) In the first round of Group 2, six matches were played: $\mathrm{D} \times \mathrm{E}, \mathrm{D} \ti... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Have you seen a numeric trick? Here are the steps of a numeric trick:
(i) Choose any number.
(ii) Multiply it by 6.
(iii) Subtract 21 from the result.
(iv) Divide this new result by 3.
(v) Subtract twice the number you chose from this last result.
(a) Try this sequence of five steps three times, starting each tim... | (a) Let's do the experiment with the numbers 0, 5, and -4.
$$
\begin{aligned}
& 0 \xrightarrow[\times 6]{ } 0 \xrightarrow[-21]{ }-21 \xrightarrow[\div 3]{ }-7 \xrightarrow[-(0 \times 2)=0]{ }-7 \\
& 5 \xrightarrow[\times 6]{ } 30 \xrightarrow[-21]{ } 9 \xrightarrow[\div 3]{ } 3 \underset{-(5 \times 2)=-10}{\longright... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The blank squares in the figure must be filled with numbers in such a way that each number, starting from the second row, is equal to the sum of the two adjacent numbers in the immediately preceding row. For example, the number in the first cell of the second row is 11, because $11=5+6$. What number will appear in the ... | The correct option is (e).
Filling the board according to the rules of the problem, it follows that $60=(x+17)+(2x+13)=3x+30$, from which $x=10$.
 | 10 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
Camila and Lara each have a $4 \times 4$ board. Starting with both boards blank, they play a game with the following sequence of events.
- Camila, hidden from Lara, paints some squares on her board black.
- Still on her board, Camila writes in each square the number of adjacent squares that are painted black (two dist... | The correct option is (b).
First, we note that if a cell has the digit 0, then none of its neighboring cells can be painted. Therefore, the cells marked with an $\times$ in the figure on the right were not painted.
 11
(b) 20
(c) 21
(d) 31
(e) 41 | The correct option is (a).
Dividing 237 by 37, we get $237=7 \times 31+20$. Therefore, 237 is not divisible by 31. This means the teacher really needs to buy more candies so that all the students can receive the same number of candies. We need to add to the expression $7 \times 31+20$ the smallest positive integer $x$... | 11 | Number Theory | MCQ | Yes | Yes | olympiads | false |
If $m$ and $n$ are integers greater than zero and $m < n$, we define $m \nabla n$ as the sum of the integers between $m$ and $n$, including $m$ and $n$. For example, $5 \nabla 8 = 5 + 6 + 7 + 8 = 26$. What is the value of $\frac{22 \nabla 26}{4 \nabla 6}$?
(a) 4
(b) 6
(c) 8
(d) 10
(e) 12 | The correct option is (c).
By definition, we obtain $\frac{22 \nabla 26}{4 \nabla 6}=\frac{22+23+24+25+26}{4+5+6}=\frac{120}{15}=8$. | 8 | Algebra | MCQ | Yes | Yes | olympiads | false |
Which is the greatest among the given numbers?
(a) $2 \times 0 \times 2006$
(c) $2+0 \times 2006$
(e) $2006 \times 0+0 \times 6$
(b) $2 \times 0+6$
(d) $2 \times(0+6)$ | The correct option is (d).
Remember that if one of the factors in a product is zero, then the product is also zero. We have $2 \times 0 \times 2006=0$, $2 \times 0+6=0+6=6$, $2+0 \times 2006=2+0=2$, $2 \times(0+6)=2 \times 6=12$ and $2006 \times 0+0 \times 6=0+0=0$. Therefore, the largest number is $2 \times(0+6)=12$. | 12 | Algebra | MCQ | Yes | Yes | olympiads | false |
If two sides of a triangle measure 5 and 7 cm, then the third side cannot measure how many centimeters?
(a) 11
(b) 10
(c) 6
(d) 3
(e) 1 | The correct option is (e).
Remember that, in a triangle, the sum of any two sides must be greater than the third side. Since $1+5$ is not greater than 7, the third side cannot measure $1 \mathrm{~cm}$. | 1 | Geometry | MCQ | Yes | Yes | olympiads | false |
If I give two chocolate bars to Tião, he will lend me his bicycle for 3 hours. If I give him 12 chocolates, he will lend me the bicycle for 2 hours. Tomorrow, I will give him one chocolate bar and 3 chocolates. For how many hours will he lend me the bicycle?
(a) $1 / 2$
(b) 1
(c) 2
(d) 3
(e) 4 | The correct option is (c).
Given $\left\{\begin{array}{l}2 \text { bars give } 3 \text { hours} \\ 12 \text { chocolates give } 2 \text { hours}\end{array}\right.$, it follows that $\left\{\begin{array}{l}1 \text { bar gives } 1.5 \text { hours }=1 \text { hour } 30 \text { minutes} \\ 3 \text { chocolates give } 0.5 ... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
How many positive integers satisfy the double inequality $2000<\sqrt{n(n-1)}<2005$?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5 | The correct option is (e).
Since the numbers that appear are all positive, we can square them while maintaining the direction of the inequalities, obtaining
$$
2000 \times 2000=2000^{2}<n(n+1)<2005^{2}=2005 \times 2005
$$
Notice that $n$ and $n+1$ are consecutive integers, so the only options are the following.
- $... | 5 | Inequalities | MCQ | Yes | Yes | olympiads | false |
If $(x+y)^{2}-(x-y)^{2}=20$, then $x y$ is equal to:
(a) 0 ;
(b) 1 ;
(c) 2 ;
(d) 5 ;
(e) 10 . | The correct option is (d).
Since $(x+y)^{2}=x^{2}+2 x y+y^{2}$ and $(x-y)^{2}=x^{2}-2 x y+y^{2}$, we have
$$
20=(x+y)^{2}-(x-y)^{2}=x^{2}+2 x y+y^{2}-x^{2}+2 x y-y^{2}=4 x y
$$
therefore $x y=5$. | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
feira treze - What is the maximum number of Friday the 13ths that can occur in a non-leap year? In this case, what day of the week does the tenth day of the year fall on? | feira treze - Since the days of the week repeat every 7 days, the difference between the days of the week is given by the remainder when the number of days that have passed is divided by 7. In the table below,
(a) in the first row, the number of days between the 13th of one month and the 13th of the following month;
... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
How many triangles exist whose sides are integers and the perimeter measures 12 units?
(a) 1
(b) 3
(c) 5
(d) 7
(e) 9 | The correct option is (b).
For three numbers $a, b$, and $c$ to be the lengths of the sides of a triangle, each of them must be greater than the difference and less than the sum of the other two. Let $a \leq b \leq c$ be the lengths of the sides of the triangle, so that $c < a + b$. Now, adding $c$ to both sides, we h... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
The number $(\sqrt{6}+\sqrt{2})(\sqrt{3}-2) \sqrt{\sqrt{3}+2}$ is equal to:
(a) $-\sqrt{3}$;
(b) $-\sqrt{2}$;
(c) -2 ;
(d) 1 ;
(e) 2. | The correct option is (c).
Observe that, denoting by $A$ the given expression, we have
$$
\begin{aligned}
A^{2} & =[(\sqrt{6}+\sqrt{2})(\sqrt{3}-2) \sqrt{\sqrt{3}+2}]^{2} \\
& =(\sqrt{6}+\sqrt{2})^{2}(\sqrt{3}-2)^{2}(\sqrt{\sqrt{3}+2})^{2} \\
& =(\sqrt{6}+\sqrt{2})^{2}(\sqrt{3}-2)^{2}(\sqrt{3}+2) \\
& =(\sqrt{6}+\sqr... | -2 | Algebra | MCQ | Yes | Yes | olympiads | false |
Five points lie on the same line. When we list the 10 distances between any two of these points, from smallest to largest, we find $2,4,5$, $7,8, k, 13,15,17$ and 19. What is the value of $k$? | Solution 1: Since the greatest distance between two points is 19 and the smallest is 2, we draw a number line with the two points 0 and 19 at the ends and the point 2 two units from 0, obtaining the first three points in the figure.
^{2}+2 \times 2^{4} \times 2^{6}+\left(2^{\frac{n}{2}}\right)^{2}$. Therefore, for $n=12$, we have $2^{8}+2^{11}+2^{12}=\left(2^{4}+2^{6}\right)^{2}$. Thus, $n=12$ is a solution.
Solution 2: If $2^{8}+2^{11}+2^{n}=k^{2}$, then
$$
\begin{aligned}
2^{8}+2^{... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The reverse of a two-digit integer is the number obtained by reversing the order of its digits. For example, 34 is the reverse of 43. How many numbers exist that, when added to their reverse, give a perfect square? | Let's recall that two-digit numbers $ab$, where $a$ is the tens digit and $b$ is the units digit, are given by $ab = a \times 10 + b$. For example, $47 = 4 \times 10 + 7$.
If $ab$ is a two-digit number, then its reverse is $ba$. We have that
$$
ab + ba = a \times 10 + b + b \times 10 + a = (a + b) \times 11
$$
On th... | 8 | Number Theory | math-word-problem | 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 $2 \%$ and the price of apples increased by $10 \%$. 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, set 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 10 apples also cost 100. Since the price of ... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
The numbers $a$ and $b$ are positive integers that satisfy $96 a^{2}=b^{3}$. What is the smallest possible value of $a$? | Factoring 96, we have $2^{5} \times 3 \times a^{2}=b^{3}$. For $2^{5} \times 3 \times a^{2}$ to be a cube, the number $a$ must have, at least, the factorization $2^{n} \times 3^{n}$. To find the smallest value of $a$, we take $a=2^{n} \times 3^{n}$, and thus,
$$
2^{5} \times 3 \times a^{2}=2^{5} \times 3 \times\left(2... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Twelve points are marked on a sheet of graph paper, as shown in the figure. What is the maximum number of squares that can be formed by connecting four of these points?
 | In total, we have 11 possible squares, shown in the following figures.
 | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Professor Newton divided his students into groups of 4 and 2 were left over. He divided his students into groups of 5 and one student was left out. If 15 students are women and there are more women than men, the number of male students is:
(a) 7 ;
(b) 8 ;
(c) 9 ;
(d) 10 ;
(e) 11 . | The correct option is (c).
Since the number of male students is less than 15 and the number of female students is 15, we have $15<$ male students + female students $<15+15=30$, which means the total number of students is between 15 and 30.
Solution 1: When we divide the number of students by 4, there are 2 students l... | 9 | Number Theory | MCQ | Yes | Yes | olympiads | false |
The surface of the Earth consists of $70\%$ water and $30\%$ land. Two fifths of the land are deserts or covered by ice and one third of the land is pasture, forest, or mountain; the rest of the land is cultivated. What is the percentage of the total surface of the Earth that is cultivated? | The fraction of land that is cultivated is
$$
1-\frac{2}{5}-\frac{1}{3}=\frac{15-6-5}{15}=\frac{4}{15}
$$
Since land occupies 3/10 of the total surface area of the Earth, it follows that the cultivated area is $\frac{4}{15} \times \frac{3}{10}=\frac{2}{25}$, that is, $\frac{2}{25}=\frac{2}{25} \times \frac{4}{4}=\fra... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A store was selling each unit of a toy for $R \$ 13.00$. To sell its entire stock, which was not more than 100 units, the store management decided to lower the price by an integer number of reais, thus obtaining $\mathrm{R} \$ 781.00$ for the entire stock. What was the price reduction, per unit? | If $x$ denotes the discount in reais and $y$ the total number of pieces, then $(13-x) \times y=781$. Thus, $(13-x)$ and $y$ are divisors of 781, and since $781=11 \times 71$, the only divisors of 781 are $1, 11, 71$, and 781. The divisor $13-x$ of 781 cannot be equal to 1, as we know that $y \leq 100$. The only option,... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
If the price of a product increased from 5.00 to 5.55 reais, what was the percentage increase? | In reais, the increase was $5.55 - 5 = 0.55$ and, therefore, the percentage increase was
$$
\frac{0.55}{5}=\frac{0.55 \times 20}{5 \times 20}=\frac{11}{100}=11 \%
$$ | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A factory produced an original calculator that performs two operations,
(a) the usual addition, denoted by $+\mathrm{e}$
(b) an operation denoted by $\circledast$.
We know that, for any natural number $a$, the following hold
$$
\text { (i) } \quad a \circledast a=a \quad \text { and } \quad \text { (ii) } \quad a \... | To calculate $(2+3) \circledast(0+3)$, we will use property (iii), obtaining $(2+3) \circledast(0+3)=(2 \circledast 0)+(3 \circledast 3)$. Now, by (ii), we have $2 \circledast 0=2 \times 2=4$ and, by (i), we have $3 \circledast 3=3$. Therefore, $(2+3) \circledast(0+3)=4+3=7$.
To calculate $1024 \circledast 48$ we will... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Joana wrote a sequence in 10 lines using the digits from 0 to 9, following the pattern below.
$$
\begin{array}{lllllllllll}
0 & & & & & & & & & \\
1 & 1 & 0 & & & & & & & \\
2 & 2 & 2 & 1 & 1 & 0 & & & & \\
3 & 3 & 3 & 3 & 2 & 2 & 2 & 1 & 1 & 0
\end{array}
$$
Which digit was used the most? How many times was this dig... | According to the pattern of the sequence, we have
Thus,
one digit 0 in each line gives \(1 \times 10 = 10\) digits 0 in total; two digits 1 in nine lines give \(2 \times 9 = 18\) digits 1... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The measures of the sides of a rectangle are even numbers. How many rectangles of this type exist with an area equal to 96? | If $a$ and $b$ denote the length and width of the rectangle, we have $a \times b=96$. Therefore, $a$ and $b$ are even divisors of 96, and thus we have four rectangles satisfying the given conditions, namely, rectangles with sides measuring 2 and 48; 4 and $24 ; 6$ and 16 and 8 and 12. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine the nearest integer to
(a) $\frac{19}{15}+\frac{19}{3}$
(b) $\frac{85}{42}+\frac{43}{21}+\frac{29}{14}+\frac{15}{7}$
(c) $-\frac{11}{10}-\frac{1}{2}-\frac{7}{5}+\frac{2}{3}$ | (a) We have:
$$
\frac{19}{15}+\frac{19}{3}=1+\frac{4}{15}+6+\frac{1}{3}=7+\frac{9}{15}=7+\frac{3}{5}
$$
Thus, the given sum is between 7 and 8. Since $\frac{3}{5}>\frac{1}{2}$, the nearest integer is 8.
Would it be possible for her to make a path passing through nine edg... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a promotion, Joana bought blouses for $\mathrm{R} \$ 15.00$ each and pants for $\mathrm{R} \$ 17.00$ each, spending a total of $\mathrm{R} \$ 143.00$. How many blouses and pants did Joana buy? | Let $b$ and $c$ be the number of blouses and pants bought, respectively. Therefore, we have $15 b + 17 c = 143$, where $b$ and $c$ are positive integers. Note that $b < 10$ and $c < 9$, because both $15 \times 10$ and $17 \times 9$ are greater than 143. From this point, we present two possible solutions.
Solution 1: W... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the figure, $O$ is the center of the semicircle with diameter $P Q, R$ is a point on the semicircle, and $R M$ is perpendicular to $P Q$. If the measure of arc $\widehat{P R}$ is twice the measure of arc $\overparen{R Q}$, what is the ratio between $P M$ and $M Q$?
 Since the circles with radii of 1 and $3 \mathrm{~cm}$ are concentric, the new circles tangent to the original ones must also have a radius of $1 \mathrm{~cm}$.
(b) The centers of the three circles with a radius of $1 \mathrm{~cm}$ shown in the figure form an equilateral triangle with a side length of 2 $\mathrm{c... | 6 | Geometry | 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 $E C / B E$.
 Show that triangle CEA is isosceles.
(b) Knowing that the length of the angle bisector of angle BÂC is 2, determine $\mathrm{BC}-\mathrm{AB}... | (a) We have $C \hat{A B}=180^{\circ}-20^{\circ}-40^{\circ}=120^{\circ}$. Since triangle $A B E$ is isosceles, it follows that
$$
A \hat{E} B=E \hat{A} B=\frac{180^{\circ}-20^{\circ}}{2}=80^{\circ}
$$
Thus, $\mathrm{CA} E=120^{\circ}-80^{\circ}=40^{\circ}$ and triangle ACE has two angles of $40^{\circ}$, and therefore... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The square of 13 is 169, which has the digit 6 in the tens place. The square of another number has the digit 7 in the tens place. What are the possible values for the digit in the units place of this square? | Suppose the number is $10a + b$, with $b$ being a digit. When we square it, we get
$$
(10a + b)^2 = 100a^2 + 20ab + b^2
$$
which has three terms: $100a^2$, $20ab$, and $b^2$.
The first term ends in 00, while the second term ends in an even number followed by zero. For the tens digit to be 7, which is odd, it is nece... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The fractions of the form $\frac{n}{n+1}$, with $n$ a positive integer are:
$$
\underbrace{\frac{1}{2}}_{n=1} ; \quad \underbrace{\frac{2}{3}}_{n=2} ; \quad \underbrace{\frac{3}{4}}_{n=3} ; \quad \underbrace{\frac{4}{5}}_{n=4} ; \quad \underbrace{\frac{5}{6}}_{n=5} \cdots
$$
Observe that this sequence of fractions is... | 2 - Converting to decimal numbers we have: $7 / 9=0.777 \ldots$ and $1 / 2=0.5$; $2 / 3=0.666 \ldots ; 3 / 4=0.75 ; 4 / 5=0.8 ; 5 / 6=0.8333 \ldots$
Therefore, the sequence is increasing and only $1 / 2=0.5 ; 2 / 3=0.666 \ldots ; 3 / 4=0.75$ are less than $7 / 9=0.777 \ldots$
Solution 3 - If $\frac{n}{n+1}<\frac{7}{9... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Comment: Note that:
the units digit of $\left(9867^{3}-9867^{2}\right)=$ the units digit of $\left(7^{3}-7^{2}\right)$ | 2: $n^{3}-n^{2}=n^{2}(n-1)$. Thus, $n^{2}=(9867)^{2}$ ends in 9 and $n-1=9866$ ends in 6. Since, $9 \times 6=54$, the last digit of the result is 4. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
As $96 \div 8=12$, we have $8 \times 12=96$.
Notice that the solution is equivalent to solving the equation $8 x=96$, whose root is $x=\frac{96}{8}=12$. | 2 - We must find among the list of five options which number, when multiplied by 8, gives 96. The unit digit of this number can only be 2 or 7.
Therefore, it can only be the number 12. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The figures show that the volumes occupied by the liquids correspond, approximately to more than half in bottle A, half in bottle B, and less than half in bottle C.
The only group of fractions that corresponds to these estimates is: $\frac{2}{3}$ (more than half); $\frac{1}{2}$ (half); $\frac{1}{4}$ (less than half). | 2 - The figures show that the volumes occupied by the liquids are decreasing numbers. The only possible options are B and E. Since $\frac{3}{3}=1$ and none of the flasks are full, the answer is B. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the following figure, the circle with center $B$ is tangent to the circle with center $A$ at $X$. The circle with center $C$ is tangent to the circle with center $A$ at $Y$. Additionally, the circles with centers $B$ and $C$ are also tangent. If $A B=6, A C=5$ and $B C=9$, what is the measure of $A X$?
 João arranged 13 sticks in the shape of a rectangular fence $1 \times 4$ as shown in the figure below. Each stick is the side of a $1 \times 1$ square, and inside each of these squares, he placed an ant. What is the minimum number of sticks we need to remove to ensure that all 4 ants can escape and return to their a... | Solution
a) It is possible to free all the ants by removing 4 matchsticks as indicated in the following figure.
Since each matchstick is shared by at most two squares and each square must ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the figure, line $t$ is parallel to segment $E F$ and tangent to the circle. If $A E=12, A F=10$ and $F C=14$, determine the length of $E B$.
 | Solution
Since the line $t$ is tangent to the circle, we have $\angle X A E=\angle A C B$. Furthermore, since $t$ and $E F$ are parallel, we have $\angle X A E=\angle A E F$. Similarly, we have $\angle Y A F=\angle A F E=\angle A B C$. Therefore, $\triangle A F E \simeq \triangle A B C$. Thus,
$$
\begin{aligned}
\fra... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the figure below, triangle $A B C$ is a right triangle at $C$ and both $B C D E$ and $C A F G$ are squares. If the product of the areas of triangles $E A B$ and $B F A$ is 64, determine the area of triangle $A B C$.
 See that the sides of the lake squared result in 74, 116, and 370. The conditions that must be met are the equations derived from three applications of the Pythagorean Theorem:
$$
\begin{aligned}
a^{2}+c^{2} & =74 \\
b^{2}+d^{2} & =116 \\
(a+b)^{2}+(c+d)^{2} & =370
\end{aligned}
$$
We can limit the test... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
We know that
$$
\frac{8^{x}}{2^{x+y}}=64 \text{ and } \frac{9^{x+y}}{3^{4 y}}=243
$$
Determine the value of $2 x y$.
# | Solution
Since $8=2^{3}$ and $9=3^{2}$, we have
$$
\begin{aligned}
64 & =\frac{8^{x}}{2^{x+y}} \\
2^{6} & =2^{3 x-(x+y)} \\
& =2^{2 x-y}
\end{aligned}
$$
$$
\begin{aligned}
243 & =\frac{9^{x+y}}{3^{4 y}} \\
3^{5} & =3^{(2 x+2 y)-4 y} \\
& =3^{2 x-2 y}
\end{aligned}
$$
Thus, we have the following system:
$$
\left\{... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
triminos and I-triminós
We want to cover a grid board with certain pieces without overlap and so that no part of them is outside the board. We will use pieces, formed by squares, called L-triminós and I-triminós, which can be rotated into the positions described in the following figure.
 The figure below shows a way to cover the $3 \times 4$ board using only L-triminós.
b) Consider the $3 \times 5$ board below with the 6 shaded squares.
... | 7 | Combinatorics | proof | Yes | Yes | olympiads | false |
How many fractions of the form $\frac{n}{n+1}$ are less than $7 / 9$, knowing that $n$ is a positive integer?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5 | The correct option is (c).
Solution 1: The fractions of the form $\frac{n}{n+1}$, with $n$ a positive integer, are
Observe that we have $\frac{1}{2} \cdot \frac{35}{45} = \frac{7}{9}$. Th... | 3 | Number Theory | MCQ | Yes | Yes | olympiads | false |
The absolute value $|a|$ of any number $a$ is defined by
$$
|a|=\left\{\begin{array}{cl}
a & \text { if } a>0 \\
0 & \text { if } a=0 \\
-a & \text { if } a<0
\end{array}\right.
$$
For example, $|6|=6,|-4|=4$ and $|0|=0$. What is the value of $N=|5|+|3-8|-|-4|$?
(a) 4
(b) -4
(c) 14
(d) -14
(e) 6 | The correct option is (e).
We have: $|5|=5,|3-8|=|-5|=5$ and $|-4|=4$. Therefore, $N=5+5-4=6$. | 6 | Algebra | MCQ | Yes | Yes | olympiads | false |
The vertices of a cube are numbered from 1 to 8, such that one of the faces has the vertices $\{1,2,6,7\}$ and the other five have the vertices $\{1,4,6,8\},\{1,2,5,8\}$, $\{2,3,5,7\},\{3,4,6,7\}$ and $\{3,4,5,8\}$. Which is the number of the vertex that is farthest from the vertex numbered 6?
(a) 1
(b) 3
(c) 4
(d) 5
(... | The correct option is (d).
Solution 1: By drawing the cube and numbering its vertices according to the question statement, we obtain a figure in which we can see that vertex 5, being diametrically opposite, is the farthest from vertex 6.
 1
(b) 0
(c) 2
(d) -2
(e) -1 | The correct option is (a).
By setting $a=1$ and $b=0$ in $a=a^{2}-a b+b^{2}$, we obtain $1 \times 1^{2}-1 \times 0+0^{2}=1$. | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
In the given figure, $ABCD$ is a rectangle and $\triangle ABE$ and $\triangle CDF$ are right triangles. The area of triangle $\triangle ABE$ is $150 \mathrm{~cm}^{2}$ and the segments $AE$ and $DF$ measure, respectively, 15 and $24 \mathrm{~cm}$. What is the length of the segment $CF$?
^{2}=(C F)^{2}+(F D)^{2}=(C F)^{2}+24^{2}$, and from this, we get $(C F)^{2}=(C D)^{2}-24^{2}$. In other words, to find $C F$, it is enough to know... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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