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We have 32 stones, all with different weights. Describe a process to show that we can find the two heaviest stones with 35 weighings on a balance scale. | We divide the stones into 16 pairs, weigh each pair, and take the 16 heaviest. We repeat the process with the 16 stones, obtaining 8 stones with eight more weighings, 4 stones with four more weighings, 2 stones with two more weighings, and the heaviest stone with the final weighing.
Up to this point, 16 + 8 + 4 + 2 + ... | 35 | Combinatorics | proof | Yes | Yes | olympiads | false |
How many five-digit natural numbers have the product of their digits equal to 2000? | Initially, observe that $2000=2^{4} \times 5^{3}$. Since the digits of the number are less than 10, each factor of 5 must be a digit of this number. Additionally, the product of the other digits must be $2^{4}=16$. Thus, we have two cases:
- The missing digits are 2 and 8. In this case, there are five possibilities fo... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
On a $123 \times 123$ board, each cell is painted purple or blue according to the following conditions:
- Each purple cell that is not on the edge of the board has exactly 5 blue cells among its 8 neighbors.
- Each blue cell that is not on the edge of the board has exactly 4 purple cells among its 8 neighbors.
Note: ... | (a) Observing a $3 \times 3$ board, we can clearly see that its center is not on the edge of the board. The center cell can:
- Be painted purple. In this case, among its 8 neighbors, there are 5 blue and 3 purple. In total, there are 4 purple cells and 5 blue cells on this board.
- Be painted blue. In this case, among... | 6724 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the smallest multiple of 9 that does not have any odd digits. | As the number is divisible by 9, the sum of the digits is divisible by 9.
On the other hand, since all the digits are even, the sum of the digits is also even. Therefore, the sum of the digits is at least 18. The smallest multiple of 9 with the sum of the digits equal to 18 is 99, but its digits are odd. This implies ... | 288 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A box has the shape of a rectangular block with dimensions $102 \mathrm{~cm}, 255 \mathrm{~cm}$, and $170 \mathrm{~cm}$. We want to store the smallest possible number of small cubes with integer edge lengths in this box, so as to fill the entire box.
Hint: Note that the length of the edge of the cube must be a divisor... | (a) Since the number of blocks is the smallest possible, the edge of the same must be the largest possible. The measure of the edge must be a divisor of 102, 255, and 170. Since we want the largest possible edge, the measure of it must be equal to $\operatorname{gcd}(102,255,170)=17$. Therefore, the edge of the cube me... | 900 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
We say that a natural number is lucky if all of its digits are equal to 7. For example, 7 and 7777 are lucky, but 767 is not. João wrote on a piece of paper the first twenty lucky numbers starting from 7, and then added them up. What is the remainder of the division of this sum by 1000? | Let's observe that if a lucky number has more than 3 digits, the remainder of the division by 1000 is 777.
Thus, the remainder we are looking for is the same remainder of the division of
$$
7+77+\underbrace{777+777+\cdots+777}_{18 \text { times }}
$$
by 1000. But this number is
$$
84+18 \times 777=84+13986=14070
$$... | 70 | Number Theory | 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 product of 50 consecutive integers is zero and the sum of these numbers is positive. What is the smallest value that this sum can assume? | As the product is equal to zero, one of the numbers must be zero. Thus, to minimize the sum, we should have the largest number of negative numbers, but in such a way that the sum is still positive.
Therefore, the number of negative numbers must be less than the number of positive numbers. Thus, among the 49 non-zero n... | 25 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Every term of a sequence, starting from the second, is equal to the sum of the previous term and the sum of its digits. The first elements of the sequence are
$$
1,2,4,8,16,23,28,38,49, \ldots
$$
Is it possible that 793210041 belongs to this sequence? | We know that a number and the sum of its digits leave the same remainder when divided by 3. In each case, if the number leaves a remainder of 1 when divided by 3, then the number plus the sum of its digits leaves a remainder of 2 when divided by 3, and if the number leaves a remainder of 2, then the sum of the number a... | 793210041 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In 1998, the population of Canada was 30.3 million. Which of the options below represents the population of Canada in 1998?
(a) 30300000
(b) 303000000
(c) 30300
(d) 303300
(e) 30300000000 | The correct option is (a).
Since 1 million $=1000000$, we have 30.3 million $=30.3 \times 1000000=30300000$. | 30300000 | Other | MCQ | Yes | Yes | olympiads | false |
A certain machine is capable of producing eight rulers per minute. How many rulers can this machine produce in 15 minutes?
(a) 104
(b) 110
(c) 112
(d) 128
(e) 120 | The correct option is (e).
If the machine produces eight rulers in a minute, in 15 minutes it will produce $8 \times 15=$ 120 rulers. | 120 | Algebra | MCQ | 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 |
In the figure, we have $\widehat{B}=50^{\circ}$, with $A D$ and $C D$ being the bisectors of angles $\widehat{A}$ and $\widehat{C}$, respectively. What is the measure of angle $A \widehat{D} C$?
(a) $90^{\circ}$
(b) $100^{\circ}$
(c) $115^{\circ}$
(d) $122.5^{\circ}$
(e) $125^{\circ}$
.
In this question, we will use an important theorem of Plane Geometry, as follows.
Theorem: The sum of the interior angles of a triangle is always $180^{\circ}$.
By the theorem, we have $\widehat{A}+\widehat{B}+\widehat{C}=180^{\circ}$ and, since $\widehat{B}=50^{\circ}$, it follows that $\... | 115 | Geometry | 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 |
Paulo wants to buy an ice cream with four scoops at an ice cream shop that offers three flavors: açaí, vanilla, and cajá. In how many different ways can he make this purchase?
(a) 6
(b) 9
(c) 12
(d) 15
(e) 18 | The correct option is (d).
Let's denote each flavor of ice cream by its initial letter, that is, a - açaí, $b$ - vanilla, $c$ - cajá. To enumerate all the possibilities of buying an ice cream with four scoops, we must consider the following cases:
- four scoops of the same flavor (1st column on the side);
- three sco... | 15 | Combinatorics | 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 |
In the figure below, we see a checkered billiard table and part of the trajectory of a ball, which is hit from one corner of the table, such that every time the ball hits one of the edges of the table, it continues its movement forming angles of $45^{\circ}$ with the edge.
(a) In which of the four pockets will the bal... | The ball changes the direction of its trajectory each time it hits one of the table edges. Since the trajectory always makes an angle of $45^{\circ}$ with the edge, the trajectory of the ball, hit from a corner, will always follow the diagonals of the squares it crosses. By tracing this trajectory, we conclude that (b)... | 23 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the figure, the triangle $\triangle A B C$ is isosceles, with $B \widehat{A} C=20^{\circ}$.
Knowing that $B C=B D=B E$, determine the measure of the angle $B \widehat{D} E$.
 | By definition, a triangle is isosceles if it has two equal sides. The third side is called the base of the isosceles triangle, and the angles formed between the base and the two equal sides are the base angles.
.
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 |
A soccer ball is made with 32 pieces of leather. Of these pieces, 12 are identical regular pentagons and the other 20 are also regular and identical hexagons. The sides of the pentagons are equal to the sides of the hexagons. To join two sides of two of these pieces, a seam is required. How many seams are needed to mak... | The correct option is (c).
If we add up the number of sides of all the polygons (20 hexagons and 12 pentagons) that make up the surface of the ball, we will get a value that is twice the number of seams, since each seam is a side common to exactly two polygons. Thus, we have $2 \times$ number of seams $=12 \times 5+20... | 90 | Geometry | MCQ | Yes | Yes | olympiads | false |
On a hot summer day, 64 children each ate an ice cream in the morning and another in the afternoon. The ice creams were of four flavors: pineapple, banana, chocolate, and dulce de leche. The table given shows how many children consumed one of these flavors in the morning and another in the afternoon. For example, the s... | The correct option is (c).
Let's first analyze the information contained in the diagonal of the table indicated by the numbers inside the small squares.
| | | AFTERNOON | | | |
| :---: | :---: | :---: | :---: | :---: | :---: |
| | Pineapple | Banana | Chocolate | Dulce de leche | |
| M | Pineapple | 1 | 8 | 0 ... | 60 | Combinatorics | 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.
Jorge places the first card, and then the turns... | The correct option is (b).
The formation of a 6-digit number is illustrated below.
| hundred thousand | ten thousand | unit thousand | hundred | ten | unit |
| :---: | :---: | :---: | :---: | :---: | :---: |
To obtain the smallest number possible, the smallest digits should be placed as far to the left as possibl... | 253416 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
A teacher has 237 candies to give to her 31 students. What is the minimum number of additional candies she needs to ensure that all her students receive the same number of candies, with none left over?
(a) 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 $x+y=8$ and $x y=15$, what is the value of $x^{2}+6 x y+y^{2}$?
(a) 64
(b) 109
(c) 120
(d) 124
(e) 154 | The correct option is (d).
Squaring both sides of the equality $x+y=8$, we obtain $x^{2}+2 x y+y^{2}=(x+y)^{2}=8^{2}=64$. Since $x y=15$, we conclude that
$$
x^{2}+6 x y+y^{2}=\left(x^{2}+2 x y+y^{2}\right)+4 x y=64+4 \times 15=124
$$ | 124 | Algebra | MCQ | Yes | Yes | olympiads | false |
In the figure, the measures of some angles in degrees are indicated as functions of $x$. What is the value of $x$?
(a) $6^{\circ}$
(b) $12^{\circ}$
(c) $18^{\circ}$
(d) $20^{\circ}$
(e) $24^{\circ}$
.
We complete the figure by marking the angles $\alpha$ and $\beta$, remembering that vertically opposite angles are equal. Recall that the sum of the interior angles of a triangle is $180^{\circ}$. Looking at the leftmost triangle, we see that
$$
3 x+4 x+\alpha=180^{\circ} \text {. }
$$
![]... | 18 | Geometry | 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 |
How many numbers between 1 and 601 are multiples of 3 or multiples of 4?
(a) 100
(b) 150
(c) 250
(d) 300
(e) 430 | The correct option is (d).
To find the number of multiples of 3 between 1 and 601, it is enough to use the division algorithm and observe that $601=200 \times 3+1$. This shows that $3 \times 1, 3 \times 2$, ..., $3 \times 200$ are the multiples of 3 between 1 and 601, that is, there are 200 of these multiples. Similar... | 300 | Number Theory | MCQ | Yes | Yes | olympiads | false |
If $x$ and $y$ are positive integers such that $x y z=240, x y+z=46$ and $x+y z=64$, what is the value of $x+y+z$?
(a) 19
(b) 20
(c) 21
(d) 24
(e) 36 | The correct option is (b).
Solution 1: From $x y z=240$, it follows that $x y=\frac{240}{z}$. Substituting in $x y+z=46$, we get $\frac{240}{z}+z=46$, that is, $z^{2}-46 z+240=0$. The roots of this equation are numbers whose sum is 46 and whose product is 240, and it is easy to verify that these roots are 6 and 40. Th... | 20 | Algebra | MCQ | Yes | Yes | olympiads | false |
In a car race, a driver covered three sections: one of $240 \mathrm{~km}$, one of $300 \mathrm{~km}$, and one of $400 \mathrm{~km}$. The driver knows that the average speeds on these sections were $40 \mathrm{~km} / \mathrm{h}$, $75 \mathrm{~km} / \mathrm{h}$, and $80 \mathrm{~km} / \mathrm{h}$, but does not remember w... | The correct option is (d).
The shortest travel time is achieved when the longest stretch is covered at the highest speed and the shortest stretch at the lowest speed. The longest time is achieved when the longest stretch is covered at the lowest speed and the shortest stretch at the highest speed. Thus, the total time... | 15 | Algebra | MCQ | Yes | Yes | olympiads | false |
From square $A B C D$, isosceles triangles were cut out as shown in the figure, leaving rectangle $P Q R S$. Knowing that the total area of what was cut out measures $200 \mathrm{~cm}^{2}$, what is the length of $P R$, in $\mathrm{cm}$?
(a) $\sqrt{200}$
(b) 200
(c) $\sqrt{800}$
(d) 25
(e) 88
.
First, we notice that the triangles $\triangle A P S$ and $\triangle C Q R$ are congruent, as they have three equal angles (one of them being a right angle) and also one of their sides $(P S=Q R)$. Similarly, the triangles $\triangle B P Q$ and $\triangle D R S$ are also congruent. Let $A P=... | 20 | Geometry | MCQ | Yes | Yes | olympiads | false |
Determine the value of $123456123456 \div 10000001$. | Of course, with such large numbers, the purpose of the question is not to perform the division. Instead, we decompose the number into convenient parts.
$$
\begin{aligned}
123456123456 & =123456000000+123456=123456 \times 1000000+123456 \\
& =123456 \times(1000000+1)=123456 \times 1000001
\end{aligned}
$$
Therefore, $... | 123456 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
You only have sticks that are 6 and $7 \mathrm{~cm}$ long. What is the minimum number of these sticks needed to cover a 2-meter line segment?
(a) 29
(b) 30
(c) 31
(d) 32
(e) 33 | The correct option is (a).
The number of sticks is minimal when the number of $7 \mathrm{~cm}$ sticks used is as large as possible. The segment measures $200 \mathrm{~cm}$. Dividing 200 by 7, we get $200 = 28 \times 7 + 4$. Therefore, if we tried to use only $7 \mathrm{~cm}$ sticks, we would need 29 sticks, but there ... | 29 | Number Theory | MCQ | Yes | Yes | olympiads | false |
What is the largest root of the equation $(x-37)^{2}-169=0$?
(a) 39
(b) 43
(c) 47
(d) 50
(e) 53 | The correct option is (d).
Solution 1: Using the factorization $a^{2}-b^{2}=(a-b)(a+b)$, we have $0=(x-37)^{2}-169=(x-37)^{2}-13^{2}=(x-37-13)(x-37+13)=(x-50)(x-24)$.
Therefore, the roots are 24 and 50.
Solution 2: Taking the square root on both sides of $(x-37)^{2}=13^{2}$, we get $x-37=13$, or $x-37=-13$. Thus, $x... | 50 | Algebra | MCQ | Yes | Yes | olympiads | false |
A rectangle $A B C D$ is divided into four smaller rectangles. The areas of three of them are indicated in the given figure. What is the area of rectangle $A B C D$?
(a) 80
(b) 84
(c) 86
(d) 88
(e) 91
.
Solution 1: First, observe that the ratio between the areas of two rectangles that have the same base is equal to the ratio between their heights. Indeed, in the figure on the left, two rectangles are represented that have the same base $b$ and heights $h_{1}$ and $h_{2}$.
 $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 |
The symbol $\odot$ represents a special operation with numbers; some examples are $2 \odot 4=10,3 \odot 8=27,4 \odot 27=112$ and $5 \odot 1=10$. What is the value of $4 \odot(8 \odot 7)$?
(a) 19
(b) 39
(c) 120
(d) 240
(e) 260 | $ A correct option is (e).
We need to figure out what the rule of this operation is. Note that
$$
2 \odot 4=10=2 \times 4+2,3 \odot 8=27=3 \times 8+3,4 \odot 27=112=4 \times 27+4
$$
and $5 \odot 1=10=5 \times 1+5$.
A plausible hypothesis is that the rule defining the operation $\odot$ is $a \odot b=a \times b+a$. A... | 260 | 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 |
The given figure is composed of congruent isosceles right triangles. What is the area, in $\mathrm{cm}^{2}$, of the shaded part?
(a) 20
(d) 45
(b) 25
(e) 50
(c) 35
 | The correct option is (d).
Solution 1: The length of the hypotenuse of each of the five isosceles right triangles in the strip measures $30 \div 5=6 \mathrm{~cm}$. The square formed by four of these triangles has a side length of $6 \mathrm{~cm}$, so its area is $36 \mathrm{~cm}^{2}$. Therefore, each of the triangles ... | 45 | 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 |
The square $S T U V$ is formed of a square bounded by 4 equal rectangles. The perimeter of each rectangle is $40 \mathrm{~cm}$. What is the area, $\mathrm{in} \mathrm{cm}^{2}$, of the square STUV?
(a) 400
(c) 160
(e) 80
(b) 200
(d) 100
.
Let $C$ and $L$ denote the length and width, respectively, of each of the four rectangles. The perimeter of each rectangle is given by $2(C+L)$. Since this perimeter measures $40 \mathrm{~cm}$, we obtain $C+L=20 \mathrm{~cm}$. Notice, in the given figure, that the side of the square $S T U V... | 400 | Geometry | 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 |
In the given figure, $T U=S V$. What is the measure of the angle $S \widehat{V} U$, in degrees?
(a) 30
(d) 65
(b) 50
(e) 70
(c) 55
 | The correct option is (d).
Remember that the sum of the interior angles of a triangle is $180^{\circ}$. From triangle $\triangle S T U$, we have that $T \widehat{S} U=180^{\circ}-\left(75^{\circ}+30^{\circ}\right)=75^{\circ}$. Therefore, this triangle is isosceles (by having two equal angles) and, thus, $T U=S U$. Sin... | 65 | Geometry | MCQ | Yes | Yes | olympiads | false |
In the triangle $\triangle K L M$ we have $K L=K M, K T=K S$ and $L \widehat{K} S=30^{\circ}$. What is the measure of the angle $x=T \widehat{S} M$?
(a) $10^{\circ}$
(b) $15^{\circ}$
(c) $20^{\circ}$
(d) $25^{\circ}$
(e) $30^{\circ}$
.
Let \( T \widehat{S} M = x \), \( S \widehat{K} T = y \), \( K \widehat{L} S = \alpha \), and \( K \widehat{T} S = \beta \). The triangle \( \triangle K L M \) is... | 15 | Geometry | MCQ | Yes | Yes | olympiads | false |
The given figure shows nine squares. The area of square $\mathrm{A}$ is $1 \mathrm{~cm}^{2}$ and that of square B is $81 \mathrm{~cm}^{2}$. What is the area, in $\mathrm{cm}^{2}$, of square I?
(a) 196
(d) 324
(b) 256
(e) 361
(c) 289
.
The side of $A$ measures $\sqrt{1}=1 \mathrm{~cm}$ and the side of $B$ measures $\sqrt{81}=9 \mathrm{~cm}$. Now we have:
- Side of $G=$ side of $B$ - side of $A=9-1=8 \mathrm{~cm}$.
- Side of $C=$ side of $B+$ side of $A=1+9=10 \mathrm{~cm}$.
- Side of $F=$ side of $G$ - side of $A=8-1=7 \m... | 324 | Geometry | 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 |
To celebrate her birthday, Ana is going to prepare pear and apple pies. At the market, an apple weighs $300 \mathrm{~g}$ and a pear, $200 \mathrm{~g}$. Ana's bag can hold a maximum weight of $7 \mathrm{~kg}$. What is the maximum number of fruits she can buy to be able to make pies of both fruits? | Let $m$ denote the number of apples and $p$ the number of pears that Ana buys, so the weight she carries in the bag is $300 m + 200 p$ grams. Since the bag can hold a maximum of 7000 grams, we have $300 m + 200 p \leq 7000$, which is equivalent to $3 m + 2 p \leq 70$. As pears weigh less, Ana should take a larger numbe... | 34 | Logic and Puzzles | math-word-problem | 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
$$
\begin{aligned}
A^{2} & =[(\sqrt{6}+\sqrt{2})(\sqrt{3}-2) \sqrt{\sqrt{3}+2}]^{2} \\
& =(\sqrt{6}+\sqrt{2})^{2}(\sqrt{3}-2)^{2}(\sqrt{\sqrt{3}+2})^{2} \\
& =(\sqrt{6}+\sqrt{2})^{2}(\sqrt{3}-2)^{2}(\sqrt{3}+2) \\
& =(\sqrt{6}+\sqr... | -2 | Algebra | MCQ | Yes | Yes | olympiads | false |
Five points lie on the same line. When we list the 10 distances between any two of these points, from smallest to largest, we find $2,4,5$, $7,8, k, 13,15,17$ and 19. What is the value of $k$? | Solution 1: Since the greatest distance between two points is 19 and the smallest is 2, we draw a number line with the two points 0 and 19 at the ends and the point 2 two units from 0, obtaining the first three points in the figure.
(7-n)(7-p)(7-q)=4$, then the sum $m+n+p+q$ is equal to:
(a) 10 ;
(b) 21 ;
(c) 24 ;
(d) 26 ;
(e) 28 . | The correct option is (e).
As $m, n, p$ and $q$ are integers, $7-m, 7-n, 7-p$ and $7-q$ are also integers. Now, $4=1 \times 2 \times 2$ and
$$
4=(-1) \times(-2) \times 1 \times 2
$$
is the only decomposition of 4 into a product of distinct integers. It follows that
$$
(7-m)+(7-n)+(7-p)+(7-q)=(-1)+(-2)+1+2
$$
thus,... | 28 | Number Theory | MCQ | Yes | Yes | olympiads | false |
Determine a value of $n$ for which the number $2^{8}+2^{11}+2^{n}$ is a perfect square. | Solution 1: Observe that $2^{8}+2^{11}+2^{n}=\left(2^{4}\right)^{2}+2 \times 2^{4} \times 2^{6}+\left(2^{\frac{n}{2}}\right)^{2}$. Therefore, for $n=12$, we have $2^{8}+2^{11}+2^{12}=\left(2^{4}+2^{6}\right)^{2}$. Thus, $n=12$ is a solution.
Solution 2: If $2^{8}+2^{11}+2^{n}=k^{2}$, then
$$
\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 |
A game is played with four integers in the following way: add three of these numbers, divide this sum by 3, and then add the result to the fourth number. There are four ways to do this game, yielding the following results: 17, 21, 23, and 29. What is the largest of the four numbers? | Let $a, b, c$ and $d$ be the numbers we are looking for. The given numbers are
$$
\frac{a+b+c}{3}+d, \frac{a+b+d}{3}+c, \frac{a+c+d}{3}+b \quad \text { and } \quad \frac{b+c+d}{3}+a
$$
but we do not know their order. However,
$$
\begin{aligned}
90 & =17+21+23+29 \\
& =\frac{a+b+c}{3}+d+\frac{a+b+d}{3}+c+\frac{a+c+d}... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In a warehouse, a dozen eggs and 10 apples had the same price. After a week, the price of eggs dropped by $2 \%$ and the price of apples increased by $10 \%$. How much more will be spent on the purchase of a dozen eggs and 10 apples?
(a) $2 \%$
(b) $4 \%$
(c) $10 \%$
(d) $12 \%$
(e) $12.2 \%$ | The correct option is (b).
Since the statement and the answer are in percentages, we can, in this case, set any price and any currency unit, and the answer will always be the same. The simplest approach, therefore, is to assume that initially, a dozen eggs cost 100 and that 10 apples also cost 100. Since the price of ... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
The numbers $a$ and $b$ are positive integers that satisfy $96 a^{2}=b^{3}$. What is the smallest possible value of $a$? | Factoring 96, we have $2^{5} \times 3 \times a^{2}=b^{3}$. For $2^{5} \times 3 \times a^{2}$ to be a cube, the number $a$ must have, at least, the factorization $2^{n} \times 3^{n}$. To find the smallest value of $a$, we take $a=2^{n} \times 3^{n}$, and thus,
$$
2^{5} \times 3 \times a^{2}=2^{5} \times 3 \times\left(2... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A factory produces blouses at a cost of $\mathrm{R} \$ 2.00$ per unit, in addition to a fixed cost of $R \$ 500.00$. If each unit produced is sold for $\mathrm{R} \$ 2.50$, from how many units produced does the factory start making a profit?
(a) 250
(b) 500
(c) 1000
(d) 1200
(e) 1500 | The correct option is (c).
Let $x$ be the number of units produced. Then the production cost is $500+2x$ reais. In the sale, the manufacturer is receiving $2.5x$. Thus, he will have a profit when $2.5x > 500+2x$, that is, $0.5x > 500$, or $x > 1000$. | 1000 | Algebra | MCQ | 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 |
The number 119 has the following properties:
(a) division by 2 leaves a remainder of 1;
(b) division by 3 leaves a remainder of 2;
(c) division by 4 leaves a remainder of 3;
(d) division by 5 leaves a remainder of 4;
(e) division by 6 leaves a remainder of 5.
How many positive integers less than 2007 satisfy thes... | Given positive integers $d$ and $r$, we say that $N$ divided by $d$ leaves a remainder $r$ if there exists an integer $n$ such that $N = n d + r$. If $M$ divided by $d$ also leaves the same remainder $r$, then there exists an integer $m$ such that $M = m d + r$. In this case, if $M > N$, it follows that $m = n + p$ for... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Sílvia is going to fill her 10 water jugs at a source that has three taps. One of the jugs takes one minute to fill, another two minutes, another three minutes, and so on. How should Sílvia distribute the 10 jugs among the three taps to spend the least amount of time possible? What is this time? | To simplify, we number the 10 water dispensers according to the respective times they take to fill up, from 1 to 10.
Solution 1: One idea is to use the "remaining time" of one water dispenser to fill another, filling two others simultaneously. The following figures illustrate the solution. In Figure I, the 3 taps take... | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
There are already $N$ people sitting around a circular table with 60 chairs. What is the smallest possible value for $N$ if the next person to sit down will have to sit next to someone? | If the next person to sit down will have to sit next to an occupied chair, this means there are at most 2 consecutive unoccupied chairs. (In the figure, occupied chairs are represented by black squares and unoccupied chairs by white squares.)
![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-072.jpg... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the "Let's Go to the Theater" campaign, 5 tickets can be purchased for the usual price of 3 tickets. Mario bought 5 tickets in this campaign. The savings Mario made represent what percentage of the usual price of the tickets?
(a) $20 \%$
(b) $33 \frac{1}{3} \%$
(c) $40 \%$
(d) $60 \%$
(e) $66 \frac{2}{3} \%$ | The correct option is (c).
Mário paid 3 and took 5, so he only paid $\frac{3}{5}$ of the usual price, thus saving $\frac{2}{5}$. Since $\frac{2}{5}=\frac{40}{100}$, the savings amount to $40 \%$. | 40 | Algebra | MCQ | 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 |
In the figure, rectangle $A B C D$ represents a rectangular plot of land whose width measures $3 / 5$ of the length. Rectangle $A B E F$ represents a rectangular garden whose width also measures $3 / 5$ of the length.
.
From the data of the problem we know that
$$
\frac{3}{5} A D=A B \quad \text { and } \quad \frac{3}{5} A B=A F
$$
Therefore, $A F=\frac{3}{5} A B=\left(\frac{3}{5}\right)^{2} A D=\frac{9}{25} A D$. The area of the land is $A B \times A D$ and the area of the garden is $A B \times A F$, so ... | 36 | Geometry | MCQ | Yes | Yes | olympiads | false |
Three couples dine at the same restaurant every Saturday at the same table. The table is round, and the couples have agreed that
(a) husband and wife never sit next to each other at the table; and
(b) the arrangement of the six at the table is different each Saturday.
Ignoring rotations of the arrangements at the tab... | To simplify, let's denote each couple by a pair of numbers, one number representing the husband and the other the wife. We have, then, the three pairs \((1,2),(3,4),(5,6)\), which cannot be neighbors. We can consider the husband 1's place at the table as fixed, since we are disregarding rotations in the arrangement aro... | 16 | Combinatorics | math-word-problem | 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 |
The set $\{1,2,3, \ldots, 3000\}$ contains a subset of 2000 elements in which no element is double the other? | Let's construct a subset of $\{1,2,3, \ldots, 3000\}$ in which no element is double another. We start by including all the odd numbers, 1, 3, 5, ..., 2999. Thus, we already have 1500 numbers, and none of them is double any other. Now we can add the numbers that are quadruple some odd number, that is, add
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 |
How many integers between 10 and 999 have the sum of their digits equal to 9? | Solution 1: Let's divide it into two cases: two-digit numbers and three-digit numbers. In the case of two-digit numbers, we have $18, 27, 36, 45, 54, 63$, 72, 81, and 90, for a total of 9 numbers. Similarly, we list the three-digit numbers as follows:
$$
\begin{aligned}
108,117,126,135,144,153,162,171,180 & \leadsto 9... | 54 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The measures of the sides of a rectangle are even numbers. How many rectangles of this type exist with an area equal to 96? | If $a$ and $b$ denote the length and width of the rectangle, we have $a \times b=96$. Therefore, $a$ and $b$ are even divisors of 96, and thus we have four rectangles satisfying the given conditions, namely, rectangles with sides measuring 2 and 48; 4 and $24 ; 6$ and 16 and 8 and 12. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
We know that two distinct points determine a unique line. How many lines are determined by any two of the nine points marked on the given grid? | Solution 1: To count the number of lines, we will divide the lines according to their positions.
- lines parallel to the sides of the squares: three horizontal and three vertical;
- lines... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Determine the nearest integer to
(a) $\frac{19}{15}+\frac{19}{3}$
(b) $\frac{85}{42}+\frac{43}{21}+\frac{29}{14}+\frac{15}{7}$
(c) $-\frac{11}{10}-\frac{1}{2}-\frac{7}{5}+\frac{2}{3}$ | (a) We have:
$$
\frac{19}{15}+\frac{19}{3}=1+\frac{4}{15}+6+\frac{1}{3}=7+\frac{9}{15}=7+\frac{3}{5}
$$
Thus, the given sum is between 7 and 8. Since $\frac{3}{5}>\frac{1}{2}$, the nearest integer is 8.
+(3-4)+\cdots+(199-200) \\
& =1000-(\... | 900 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
An ant starts from a vertex of a cube, walking only along the edges, until it returns to the initial vertex, without passing through any vertex twice. What is the longest walk that this ant can make? | In the figure, we have a path consisting of eight edges that the ant can take starting from the vertex identified as 1.
Would it be possible for her to make a path passing through nine edg... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a promotion, Joana bought blouses for $\mathrm{R} \$ 15.00$ each and pants for $\mathrm{R} \$ 17.00$ each, spending a total of $\mathrm{R} \$ 143.00$. How many blouses and pants did Joana buy? | Let $b$ and $c$ be the number of blouses and pants bought, respectively. Therefore, we have $15 b + 17 c = 143$, where $b$ and $c$ are positive integers. Note that $b < 10$ and $c < 9$, because both $15 \times 10$ and $17 \times 9$ are greater than 143. From this point, we present two possible solutions.
Solution 1: W... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A number less than 200 is formed by three different digits, and the double of this number also has all different digits. Moreover, the number and its double do not have any digits in common. What is this number? How many solutions does this problem have? | Initially note that the double of an integer is even, so it ends in $0,2,4,6$ or 8. However, the number we are looking for cannot end in 0, because in that case its double would also end in 0, and both would have the digit 0 in common. Therefore, we have the following cases.
$ are there such that
$$
\frac{x y}{x+y}=144 ?
$$ | The given equation is equivalent to $x y=144(x+y)=144 x+144 y$, so isolating $x$, we get $x=\frac{144 y}{y-144}$. Since $x$ and $y$ must be positive integers, the denominator $y-144$ must be a positive integer, say, $y-144=n$. Substituting this expression into the value of $x$, we get
$$
x=\frac{144(n+144)}{n}=144+\fr... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The ratio between the number of men and women in the city of Campo Verde is 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 $H / M=2 / 3$, so that $M=(3 H) / 2$ and, therefore, the population of Campo Verde is given by
$$
H+M=H+\frac{3}{2} H=\frac{5}{2} H
$$
If the average age of men is 37 years, then
$$
37=\text { average age of } H \text { men }=\frac{\text { sum of the... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the trapezoid of the given figure, $AB$ is parallel to $DC$, $AD=AB=BC=$ $1 \text{ cm}$ and $DC=2 \text{ cm}$. What is the measure of the angle $D \widehat{A} C$?
(a) $30^{\circ}$
(b) $45^{\circ}$
(c) $60^{\circ}$
(d) $90^{\circ}$
(e) $120^{\circ}$
.
Let $P$ be the midpoint of segment $D C$ and draw segments $A P$ and $B P$. The three triangles thus formed, $\triangle A D P, \triangle A P B$, and $\triangle B P C$, are equilateral (why?), so $D \widehat{A} P=60^{\circ}=P \widehat{A} B$. Since segment $A C$ is the bisector of angle $P \wi... | 90 | Geometry | MCQ | Yes | Yes | olympiads | false |
Determine the value of the natural number $a$, knowing that $4 a^{2}$ and $\frac{4}{3} \times a^{3}$ are four-digit integers. | We have $1000 \leq 4 a^{2}<10000$, which implies $250 \leq a^{2}<2500$. But, $15^{2}=225,16^{2}=256$ and $50^{2}=2500$, therefore, since $a$ is a natural number, we get $15<a<50$. We also have $1000 \leq \frac{4}{3} \times a^{3}<10000$, which implies $750 \leq a^{3}<7500$. But, $9^{3}=729,10^{3}=1000,19^{3}=6859$ and $... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
How many pairs $(x, y)$ of positive integers are solutions to the equation $3 x+5 y=501 ?$
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。
（注：最后一句为说明翻译要求，不需要翻译） | As $501-3x=3(167-x)$, the given equation is equivalent to $y=\frac{3}{5}(167-x)$. Since $y$ is a positive integer, $167-x$ must be some positive multiple of 5, that is, $167-x=5k$, for some positive integer $k$ and, therefore, $x=167-5k$ or, still, $x=5 \times 33+2-5k=5(33-k)+2$. Since $x$ is a positive integer, we mus... | 33 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The director of a certain 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 decided to remove one student from each row, placing them... | The diagrams below represent the situation of the problem, where the students who were initially removed are represented in black and the students removed the second time, in gray.
Let $m$... | 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 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 |
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's not possible, you must have entered the number backwards and switched the order of two digits, because I remember that last week it cost less than 50 reais!" The cashier replies: I'm s... | The old price was less than 50 reais and underwent an increase of $20\%$, resulting in a new price that is still a two-digit number, which we represent as $ab$, where $a$ is the tens digit and $b$ is the units digit, i.e., $ab = 10a + b$. The new price is the old price $ba$ with an increase of $20\%$, i.e.,
$$
10a + b... | 54 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In a certain condominium, 29 families live, each of which has either one, three, or five cats. The number of families that have only one cat is the same as the number of families that have five cats. How many cats are there in this condominium? | Let $x$ be the number of families that have only one or exactly five cats, and $y$ be the number of families that have exactly three cats. It follows that $x+y+x=29$ and, therefore, $2 x+y=29$. Since the number of cats is $x+3 y+5 x=6 x+3 y$, we obtain
$$
\text { number of cats }=6 x+3 y=3(2 x+y)=3 \times 29=87 \text ... | 87 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the addition below, the same letters represent the same digit, and different letters represent different digits. Find the number $A B C D E$.
$A B C D E$ $B C D E$ $C D E$ $D E$ $A A A A A$ | Note that $5 \times E$ is a multiple of 5 and, in this case, ends in $A$. Since $A$ cannot be 0, it follows that $A=5$ and $E$ is odd. Observe that $E$ cannot be 1, because in that case, $4 D=5$, which is impossible for digits. Therefore, $E=3, 5, 7$ or 9. Let's analyze each of these possibilities.
If $E=3$, then $4 D... | 52487 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In triangle $\triangle A B C$, point $F$ is on side $A C$ and $F C=2 A F$. If $G$ is the midpoint of segment $B F$ and $E$ is the intersection point of the line passing through $A$ and $G$ with segment $B C$, calculate the ratio $E C / B E$.
=\text { Area }(\triangle B C D) \quad \text { and Area }(\triangle A B C)=\text { Area }(\triangle A C D)
$$
Let's denote the areas of the four regions determined by the diagonals by $X, Y, Z$ and ... | 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 $a b=10 a+b$ be a two-digit number with digits $a$ and $b$ that is divisible by the sum $a+b$ of its digits. Note that, being a two-digit number, necessarily $a \neq 0$ and that, being divisible by the sum of its digits, the difference $(10 a+b)-(a+b)=9 a$ is also divisible by $a+b$ (prove this). Thus, it suffices ... | 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 number that is divisible by the sum of its digits and is also less than 900, we can start our search among numbers with the digit 8 in the hundreds place, since, at a minimum, 800 is divisible by the sum $8+0+0=8$ of its digits, and 899 does not have this property. Thus, we will examine the numbers ... | 888 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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