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A boy tried to align 480 coins in the shape of a triangle, with one coin in the first row, two coins in the second row, and so on. At the end of the attempt, 15 coins were left over. How many rows does this triangle have? | Suppose the triangle is formed by $n$ rows, then $1+2+3+\cdots+n$ coins were used, that is,
$$
\frac{1}{2} n(n+1)=1+2+\cdots+n=480-15=465
$$
which gives $n^{2}+n-930=0$. Solving this equation, we get
$$
n=\frac{-1 \pm \sqrt{1+4 \times 930}}{2}=\frac{-1 \pm 61}{2}
$$
Since $n=30$ is the only positive solution to thi... | 30 | Number Theory | 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 ... | The example shows that we can connect at least seven 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 one, two, or three stations, as shown in the first of the two diagrams below. Stations A, B, and C still have two unused lin... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The lines of the first given table are all arithmetic progressions of the same ratio, and the columns of this table are all arithmetic progressions of the same ratio. In the second given table, the same formation rule was used, but someone erased some numbers, leaving only three. What is the number that was in the posi... | We see that in the first table, each row is an arithmetic progression with a common difference of 3, and each column is an arithmetic progression with a common difference of 7.
| 5 | 8 | 11 | 14 | 17 |
| :---: | :---: | :---: | :---: | :---: |
| 12 | 15 | 18 | 21 | 24 |
| 19 | 22 | 25 | 28 | 31 |
| 26 | 29 | 32 | 35... | 102 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In a class at school, all students are the same age, except for seven of them who are 1 year younger and two of them who are 2 years older. The sum of the ages of all the students in this class is 330. How many students are there in this class? | Let's denote by $a$ the common age of the students and by $n$ the number of students in this class. We have seven students who are $a-1$ years old, two who are $a+2$ years old, and the rest, that is, $n-9$ students, are $a$ years old. Therefore, the sum of the ages of all the students, which is 330, can be broken down ... | 37 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A round table has a diameter of 1.40 meters.
For a party, the table is expanded by adding three planks, each 40 cm wide, as shown in the figure. If each person at the table should have 60 cm of space, how many guests can sit at the table?
 | When we add a new triangle to a figure composed of triangles, it intersects each of the sides of the existing triangles at, at most, two points. Initially, starting with a single triangle, we have no intersection points. By adding a second triangle, we introduce, at most, \(2 \times 3 = 6\) intersection points. Similar... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Two types of candles have the same length but are made of different materials. One of them burns completely in three hours, and the other in four hours, both burning at a uniform rate. How many minutes after 1 PM should the two candles be lit simultaneously so that, at 4 PM, the length of one is double that of the othe... | The correct option is (c).
Since the actual length of the candles is irrelevant, we can assume any size for the candles, and the answer will always be the same. The simplest is to assume that both candles have a length of one unit. Thus, the one that burns in 3 hours burns at a constant rate of $1 / 3$ (candle per hou... | 36 | Algebra | MCQ | Yes | Yes | olympiads | false |
A teacher proposes 80 problems to a student, informing that they will award five points for each problem solved correctly and deduct three points for each problem not solved or solved incorrectly. In the end, the student ends up with eight points. How many problems did he solve correctly? | Let $c$ be the number of problems solved correctly and $e$ be the sum of the number of problems solved incorrectly and problems not solved. Therefore, $c+e=80$ and $5 c-3 e$ is the number of points the student scored in the evaluation. In the present case,
$$
\left\{\begin{aligned}
c+e & =80 \\
5 c-3 e & =8
\end{align... | 31 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the given figure, we have 16 points forming a square grid and two lines, $r$ and $s$, which are perpendicular to each other.
(a) How many squares can we construct such that their vertice... | (a) The only squares that do not have any of their sides parallel to line $r$ or line $s$ are those of type 1 and type 2 (see figures).
Thus, there are a total of six squares, four of type ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
All natural numbers starting from 1 were written consecutively, forming a sequence of digits, as follows.
$$
1234567891011121314151617181920212223 \ldots
$$
What is the digit that appears in the 206788th position? | The numbers with one digit form the first 9 terms of the sequence. The 90 two-digit numbers form the next 180 terms. Then come the 2700 terms corresponding to the three-digit numbers, followed by the 36000 terms corresponding to the four-digit numbers, and finally, the 450000 terms corresponding to the five-digit numbe... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
How many zeros are there at the end of the number $9^{2007}+1$? | Initially, we verify how the powers of 9 end, that is, we list the last two digits, the tens and units, of the powers $9^{n}$, in order.
| If $n$ is | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 were written (in an unknown order) around a circle. Reading these digits in groups of three in a clockwise direction, nine three-digit numbers are formed. Determine the sum of these nine numbers. | In the following figure, we represent the nine digits written around the circumference.
Reading the digits written around the circumference, in groups of three, in a clockwise direction, we... | 4995 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A certain number leaves a remainder of 1 when divided by 3, leaves a remainder of 2 when divided by 4, leaves a remainder of 3 when divided by 5, and leaves a remainder of 4 when divided by 6. What is the smallest positive integer that satisfies these properties? | Solution 1: Let $x$ be the number sought. Observe that $x+2$ is divisible by $3, 4, 5,$ and $6$. The least common multiple of these numbers is $60$. Therefore, $x+2=60k$, so $x=58$.
Solution 2: Let $x$ be the number sought. The remainder of the division of $x$ by $3$ is $1$. Therefore, $x$ is of the form $x=3a+1$, wit... | 58 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A group of five friends decides to play Secret Santa, each buying a gift for their secret friend. According to the rules of the game, each person gives exactly one gift and receives exactly one gift. In how many ways can the gifts be distributed so that no one gives a gift to themselves? | First, let's observe that the number of ways to distribute the gifts without any restrictions is $5!=5 \times 4 \times 3 \times 2 \times 1=120$. From this, we need to subtract the "bad" cases, that is, the cases where at least one friend ends up with their own gift. These cases to be eliminated can be listed by the num... | 44 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What are the last two digits of the number
$$
8+88+888+\cdots+\overbrace{88 \cdots 88}^{2008} ?
$$ | Solution 1: Since we only want to know the last two digits, it is enough to know the last two columns of this sum (the tens and the units), that is,
$$
8+88 \times 2007=8+\ldots 16
$$
Since $8+16=24$, the last two digits of the number are 24.
 / a=(20 / a)+1$ is an integer, and consequently, $20 / a$ is also an integer. Therefore, $a$ is a divisor of 20, and thus $a$ can be 1, 2, 4, 5, 10, or 20. So, there are a total of 6 times when their ages are multi... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Ana has a cube with a side length of $10 \mathrm{~cm}$. She cut the cube into smaller cubes with a side length of $1 \mathrm{~cm}$ and plays by forming other rectangular blocks with these smaller cubes, without any cubes left over. For example, she formed a block of $10 \times 20 \times 5$. In total, how many different... | The volume of the cube is $10 \times 10 \times 10=1000 \mathrm{~cm}^{3}$. The volume $V$ of a block is the product of its three dimensions, height $(=a)$, width $(=l)$, and length $(=c)$.
To construct each block, Ana must use all the small blocks, so the volume of each block is
$$
V=\text { height } \times \text { wi... | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Joana wrote the numbers from 1 to 10000 on the blackboard and then erased all multiples of 7 and 11. What number remained in the 2008th position? | Initially observe that, from 1 to 77, Joana erased 11 multiples of 7 and 7 multiples of 11. Since 77 is a multiple of 7 and 11, she erased $11+7-1=17$ numbers, leaving $77-17=60$ numbers. Now, grouping the first 10000 numbers into groups of 77 consecutive numbers, this reasoning applies to each of the lines below, that... | 2577 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the maximum number of elements of a subset of $\{1,2, \ldots, 100\}$ such that none of its elements is a multiple of any other? | Initially, observe that the set $\{51,52,53, \ldots, 100\}$ has 50 elements and none of its elements is a multiple of another. Thus, the subset with the largest number of elements that satisfies the required property has at least 50 elements. To conclude that 50 is the largest possible number of elements of a subset th... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A chicken coop with $240 \mathrm{~m}^{2}$ of area must house chickens and chicks, with a desirable free space of $4 \mathrm{~m}^{2}$ for each chicken and $2 \mathrm{~m}^{2}$ for each chick. Additionally, each chick consumes $40 \mathrm{~g}$ of feed per day and each chicken consumes $160 \mathrm{~g}$ per day, with a max... | Let $x$ and $y$ be, respectively, the number of hens and chicks in the henhouse.
(a) We have $4 x+2 y=240$, which simplifies to $2 x+y=120$. Since $8 \text{ kg} = 8000 \text{ g}$, we have $160 x+40 y \leq 8000$. Thus, $4 x+y \leq 200$.
In summary, the numbers $x$ of hens and $y$ of chicks satisfy
$$
(*)\left\{\begin... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Starting from her house to get to school, Júlia must walk eight blocks to the right and five blocks up, as indicated in the given figure.
She knows that there are many different ways to mak... | Regardless of the route Júlia takes from her house to school, if she wishes to follow a shorter path, she must travel exactly eight blocks to the right and five blocks up. A shorter path connecting her house to the school is, then, a sequence of "block crossings," with eight of them in the horizontal direction (to the ... | 1287 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A square board of three rows by three columns contains nine cells. In how many different ways can we write the three letters A, B, and $\mathbf{C}$ in three different cells, so that in each row exactly one of these three letters is written? | Starting with the letter A, it can be written in any of the nine spaces on the board. Once the letter $\mathbf{A}$ is written, there are six spaces left where the letter B can be written. Once the letters $\mathbf{A}$ and $\mathbf{B}$ are written on the board, there are three spaces left for the letter $\mathbf{C}$ to ... | 162 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
If in the fraction $x / y$ we decrease the numerator $x$ by $40 \%$ and the denominator $y$ by $60 \%$, then the fraction $x / y$
(a)decreases by $20 \%$;
(b)increases by $20 \%$;
(c)decreases by 50\%;
(d)increases by $50 \%$. | The correct option is (d).
If a number $x$ is decreased by $40\%$, it becomes $60\%$ of $x$, or $0.6 x$. Similarly, when a number $y$ is decreased by $60\%$, it becomes $0.4 y$. Therefore, the fraction $x / y$ becomes
$$
\frac{0.6 x}{0.4 y}=\frac{6}{4} \frac{x}{y}=1.5 \times \frac{x}{y}
$$
which means that the fract... | 50 | Algebra | MCQ | Yes | Yes | olympiads | false |
An empty swimming pool was filled with water by two faucets A and B, both with constant flow rates. For four hours, both faucets were open and filled $50 \%$ of the pool. Then, faucet B was turned off and, for two hours, faucet A filled $15 \%$ of the pool's volume. After this period, faucet A was turned off and faucet... | Given that taps A and B pour water into the pool at a constant flow rate, the volume of water poured by each tap is proportional to the time it is open. Therefore, if tap A fills $15 \%$ of the pool volume in two hours, then in four hours it will fill $30 \%$ of the pool volume.
However, when taps A and B are both ope... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
An isosceles triangle has a base of $10 \mathrm{~cm}$ and two equal sides measuring $13 \mathrm{~cm}$. Is it possible to cut this triangle into two other triangles in such a way that, by joining these triangles in another manner, we obtain another isosceles triangle (obviously with the same area)?
 Who will e... | Those who had a monthly grade of $x$ will receive a discount of $x \%$ on that grade, meaning they will lose
$$
x \% \text { of } x=\frac{x}{100} \times x=\frac{x^{2}}{100}
$$
Thus, a grade initially of $x$, after the penalty, becomes $x-\frac{x^{2}}{100}$. Let's consider this "grade after the penalty" function, give... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the quadrilateral $A B C D$ given in the figure, we have $A B=5, B C=17$, $C D=5$ and $D A=9$. Determine $D B$, knowing that its measure is an integer.
 | Remember that any side of a triangle is greater than the difference and less than the sum of the other two sides. In triangle $A D B$, we have $A D - A B < B D < A D + A B$ and in triangle $C B D$, we have $B C - C D < B D < B C + C D$.
^{2}}{1,001^{k+1}}, \ldots
$$
is maximum. We want to compare the sizes of any two terms of this sequence. The simplest way is to ... | 2001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A triangle has vertices $A=(3,0), B=(0,3)$, and $C$, where $C$ lies on the line with equation $x+y=7$. What is the area of this triangle? | Observe that the height $h$ relative to side $AB$ of all triangles $\triangle ABC$ that have vertex $C$ on the line $x+y=7$ is always the same, since the line $x+y=7$ is parallel to the line $x+y=3$ that passes through $A$ and $B$. Therefore, all these triangles have the same area, namely,
$$
\frac{1}{2}(AB \times h)
... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Three circles, with radii measuring 1, 2, and $3 \mathrm{~cm}$, are pairwise externally tangent, as shown in the given figure. Determine the radius of the circle that is externally tangent to the three circles.
(\sqrt{2}-1)=3-2 \sqrt{2},(\sqrt{2}+1)(\sqrt{2}+1)=3+2 \sqrt{2}$ and $(\sqrt{2}-1)(\sqrt{2}+1)=1$. Suppose that $a$ products are equal to $3-2 \sqrt{2}$, $b$ products are equal to $3+2 \sqrt{2}$ and $1002-a-b$ pr... | 502 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $a$ be a positive integer that is a multiple of 5 and such that $a+1$ is a multiple of 7, $a+2$ is a multiple of 9, and $a+3$ is a multiple of 11. Determine the smallest possible value of $a$. | Solution 1: Note that if a positive integer $m$ is a multiple of an integer $p$, then $2m - p$ is a multiple of $p$. Indeed, if $m = np$, then $2m - p = 2np - p = (2n - 1)p$. In particular, taking $m$ equal to $a, a+1, a+2$, and $a+3$, we establish that $2a - 5$ is a multiple of 5, $2(a+1) - 7$ is a multiple of 7, $2(a... | 1735 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $A B C D$ be a right trapezoid with bases $A B$ and $C D$, and right angles at $A$ and $D$. Given that the shorter diagonal $B D$ is perpendicular to the side $B C$, determine the smallest possible value for the ratio $C D / A D$. | Let $A \widehat{B} D = B \widehat{D} C = \alpha$, as shown in the given figure. Then we have $C D = \frac{B D}{\cos \alpha}$ and $A D = B D \sin \alpha$, hence
$$
\begin{aligned}
\frac{C D}{A D} & = \frac{\frac{B D}{\cos \alpha}}{B D \sin \alpha} = \frac{1}{\sin \alpha \cos \alpha} \\
& = \frac{2}{\sin 2 \alpha} \geq ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The twelve students in an olympiad class went out to play soccer every day after their math class, forming two teams of six players each and playing against each other. Each day they formed two different teams from those formed in previous days. By the end of the year, they found that every group of five students had p... | For every group of five students, there is a unique team formed that contains them. Therefore, we count
$$
C_{12}^{5}=\frac{12 \times 11 \times 10 \times 9 \times 8}{5!}=792
$$
teams for each five students chosen. On the other hand, in each team of six players, there are $C_{6}^{5}=6$ ways to choose five players, tha... | 132 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let's denote by $s(n)$ the sum of the digits of the number $n$. For example, $s(2345) = 2 + 3 + 4 + 5 = 14$. Observe that:
\[
\begin{gathered}
40 - s(40) = 36 = 9 \times 4; \quad 500 - s(500) = 495 = 9 \times 55 \\
2345 - s(2345) = 2331 = 9 \times 259
\end{gathered}
\]
(a) What can we say about the number $n - s(n)$?... | (a) It is immediate that if $a$ is a digit between 1 and 9, then $s\left(a \cdot 10^{k}\right)=a$, since the number $a \cdot 10^{k}$ is formed by the digit $a$ followed by $k$ zeros. Therefore, we have
$$
a \cdot 10^{k}-s\left(a \cdot 10^{k}\right)=a \cdot 10^{k}-a=a\left(10^{k}-1\right)=a \times \underbrace{9 \cdots ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
If $S_{n}=1-2+3-4+5-6+\ldots+(-1)^{n+1} n$, where $n$ is a positive integer, then $S_{1992}+S_{1993}$ is:
(a) -2
(b) -1
(c) 0
(d) 1
(e) 2 | Solution 1: Remember that
$(-1)^{n+1}= \begin{cases}1 & \text { if } n \text { is odd } \\ -1 & \text { if } n \text { is even }\end{cases}$
Observe that by grouping consecutive terms in pairs,
$$
(1-2)+(3-4)+(5-6)+\cdots
$$
we get a sum of $n$ terms all equal to -1. Therefore,
$S_{1992}=\underbrace{(1-2)+(3-4)+(5-... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
An arc of a circle measures $300^{\circ}$ and its length is $2 \mathrm{~km}$. What is the nearest integer to the measure of the radius in meters?
(a) 157
(b) 284
(c) 382
(d) 628
(e) 764 | Solution 1: If the radius is $r$, then the length of an arc of $\theta$ degrees is $2 \pi \frac{\theta}{360} r$. Therefore, in the given problem, we have
$$
2 \pi \frac{300}{360} r=2000 m \Longrightarrow r=2000 \times \frac{3}{5 \pi} \simeq 382.17 m
$$
Thus, the correct option is (c).
Solution 2: Since the circumfer... | 382 | Geometry | MCQ | Yes | Yes | olympiads | false |
A mason is able to lay 8 meters of wall per day. How many meters of wall can this mason lay in 15 days?
(a) 104
(b) 110
(c) 120
(d) 128
(e) 112 | The correct option is (c).
If the mason lays 8 meters per day, in 15 days he will lay $15 \times 8=120$ meters. | 120 | Algebra | MCQ | Yes | Yes | olympiads | false |
Consider two natural numbers, each with three different digits. The larger one only has even digits, and the smaller one only has odd digits. If the difference between them is as large as possible, what is this difference?
(a) 997
(b) 777
(c) 507
(d) 531
(e) 729 | The correct option is (e).
To make the difference as large as possible, we should choose the largest number with three different even digits and the smallest number with three different odd digits. The largest number with three different even digits is 864 and the smallest number with three different odd digits is 135... | 729 | Number Theory | MCQ | Yes | Yes | olympiads | false |
A pharmacy offers a $30 \%$ discount on the list price of all medications it sells. When purchasing a medication with a list price of $\mathrm{R} \$ 120.00$, how many reais will a person pay?
(a) 36
(b) 84
(c) 64
(d) More than 116
(e) 94 | The correct option is (b).
Solution 1: The person will pay 120 reais minus the discount, which is 30% of 120, that is, $0.3 \times 120=36$ reais. Thus, the person pays $120-36=84$ reais.
Solution 2: Since the discount is 30%, the person will pay 70% of 120, that is, $0.7 \times 120=84$ reais. | 84 | Algebra | MCQ | Yes | Yes | olympiads | false |
A liter of alcohol costs $\mathrm{R} \$ 0,75$. Maria's car travels $25 \mathrm{~km}$ with 3 liters of alcohol. How many reais will Maria spend on the alcohol needed to travel $600 \mathrm{~km}$?
(a) 54
(b) 72
(c) 50
(d) 52
(e) 45 | The correct option is (a).
Solution 1: If in a journey of $25 \mathrm{~km}$ she spends 3 liters, then to travel 100 km Maria will spend $4 \times 3=12$ liters. Therefore, to travel $600 \mathrm{~km}$, the car will spend $6 \times 12=72$ liters. Since each liter costs 0.75 reais, 72 liters will cost $0.75 \times 72=54$... | 54 | Algebra | MCQ | Yes | Yes | olympiads | false |
The four cities $A, B, C$ and $D$ were built along a straight highway, as illustrated.
The distance between $A$ and $C$ is $50 \mathrm{~km}$ and the distance between $B$ and $D$ is $45 \mat... | The correct option is (a).
Solution 1: From the problem statement, we have $A C=50, B D=45$ and $A D=80$. From the figure, it follows that
$$
C D=A D-A C=80-50=30
$$
Therefore,
$$
B C=... | 15 | Geometry | MCQ | Yes | Yes | olympiads | false |
In a warehouse, some boxes were stacked to form the pile shown in the figure. If each box weighs $25 \mathrm{~kg}$, how many kilograms does the pile weigh with all the boxes?
(a) 300
(b) 325
(c) 350
(d) 375
(e) 400 | The correct option is (e).
$\mathrm{Na}$ In the figure, we can see one column with three boxes, four columns with two boxes, and three columns with one box. Therefore, the total number of boxes is
$$
1 \times 3 + 4 \times 2 + 3 \times 1 = 14
$$
Since each box weighs $25 \mathrm{~kg}$, the weight of the stack of boxe... | 350 | Number Theory | MCQ | Yes | Yes | olympiads | false |
A book of one hundred pages has its pages numbered from 1 to 100. How many sheets of this book have the digit 5 in their numbering? (NOTE: a sheet has two pages.)
(a) 13
(b) 14
(c) 15
(d) 16
(e) 17 | The correct option is (c).
Between 1 and 100, the digit 5 appears in the numbers $5,15,25,35,45,50,51,52,53$, $54,55,56,57,58,59,65,75,85$ and 95. The first sheet contains pages 1 and 2, the second sheet contains pages 3 and 4, the third sheet contains pages 5 and 6, and so on. Therefore, the two pages that make up ea... | 15 | Number Theory | MCQ | Yes | Yes | olympiads | false |
Write the numbers from 0 to 9 in the circles next to them, so that they increase in a counterclockwise direction. Then, subtract one unit from the odd numbers and add one unit to the even numbers. By choosing three consecutive circles, what is the highest sum you can obtain?
(a) 19
(b) 21
(c) 23
(d) 24
(e) 25
.
Starting from any circle, we initially get the sequence $0,1,2,3,4,5,6,7,8,9$. By subtracting one from the odd numbers and adding one to the even numbers, the sequence becomes $1,0,3,2,5,4,7,6,9,8$. It is now easy to verify that the largest possible sum of three consecutive numbers is $8+9+6... | 23 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
The city hall of a certain city launched a campaign that allows exchanging four 1-liter empty bottles for one 1-liter full bottle of milk. How many liters of milk can a person obtain who has 43 empty 1-liter bottles by making several of these exchanges?
(a) 11
(b) 12
(c) 13
(d) 14
(e) 15 | The correct option is (d).
Since $43=10 \times 4+3$, in the first round, the 43 empty bottles can be exchanged for 10 full bottles, leaving 3 empty bottles. Now, after consuming the milk from these 10 bottles, we have 13 empty bottles, which can be exchanged for 3 full bottles, leaving 1 empty bottle. Finally, after c... | 14 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
The given figure represents a rectangular lawn on which seven squares numbered from 1 to 7 have been marked. If the area of the smallest of these squares is $1 \mathrm{~m}^{2}$, the total area of the lawn, in $\mathrm{m}^{2}$, is equal to
| 1 | 2 | 5 | |
| :---: | :---: | :---: | :---: |
| | 3 | | |
| | 4 | | |... | The correct option is (c).
Solution 1: Since the smaller squares have an area of $1 \mathrm{~m}^{2}$, each of them has a side length of $1 \mathrm{~m}$. From the figure, we conclude that $B C=2 \mathrm{~m}$. Since $A B C D$ is a square, it follows that $B C=C D=A D=2 \mathrm{~m}$. Since $C D E F$ is also a square, we ... | 45 | Geometry | MCQ | Yes | Yes | olympiads | false |
A sequence of square mosaics is constructed with black and white square tiles, all of the same size, with the first being formed by a white tile surrounded by black tiles, the second by four white tiles surrounded by black tiles, and so on, as indicated in the figure. If in a sequence of mosaics formed according to thi... | The correct option is (a).
In the first mosaic, we have $3+3+1+1=8$ black tiles, in the second, we have $4+4+2+2=12$, in the third, we have $5+5+3+3=16$, and it is not difficult to see (and verify) that the next mosaics have 20 and 24 black tiles, as each new mosaic uses four more black tiles, one on each side. Since ... | 55 | Geometry | MCQ | Yes | Yes | olympiads | false |
Ester goes to a stationery store to buy notebooks and pens. In this stationery store, all notebooks cost $\mathrm{R} \$ 6.00$. If she buys three notebooks, she will have R \$4.00 left. If, instead, her brother lends her an additional $\mathrm{R} \$ 4.00$, she will be able to buy two notebooks and seven pens, all the sa... | By buying three notebooks at 6 reais each, Ester still has 4 reais left, so the amount she has is $3 \times 6 + 4 = 22$ reais.
(a) If her brother lends her 4 reais, she then has $22 + 4 = 26$ reais and can buy 2 notebooks at 6 reais each, leaving her with $26 - 2 \times 6 = 26 - 12 = 14$ reais for 7 pens. We conclude ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The famous Greek mathematician Pythagoras referred to the numbers obtained by the sum of the first positive integers as triangular numbers. For example, 1, 3, 6, and 10 are triangular numbers.
$$
\begin{aligned}
1 & =1 \\
3 & =1+2 \\
6 & =1+2+3 \\
10 & =1+2+3+4
\end{aligned}
$$
^{2}=$ $C A B$, that is, the two-digit number $A B$ squared gives the three-digit number $C A B$. Determine the value of $A+B+C$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | Starting from the equality $(A B)^{2}=C A B$ and denoting the two-digit number $A B$ by $x$, we have $x^{2}=(A B)^{2}=C A B=C \cdot 100+x$, that is, $x^{2}-x=C \cdot 100$. Therefore, the product $x(x-1)=x^{2}-x$ is divisible by 100. Considering the factorization $100=2^{2} \cdot 5^{2}$, we divide the solution into thre... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
João has, in his garden, a cistern where he stores rainwater and takes water to water his flowers. At midnight on December 31, 2005, the cistern contained 156 liters of water. João has the habit of writing on a board, every day, the number of liters of water he spends to water the flowers and the amount of rainwater co... | On January 1st, the cistern starts with 156 liters of water, and from then on, the cistern receives rainwater and loses water for watering flowers. Since there was no change in the amount of water in the cistern on the 8th, the number of liters of water in the cistern on the 8th is
$156+$ rainwater from day 1 to day 7... | 132 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
From the equality $9174532 \times 13=119268916$ it can be concluded that one of the following numbers is divisible by 13. Which is that number?
(a) 119268903
(c) 119268911
(e) 119268923
(b) 119268907
(d) 119268913 | The correct option is (a).
Since 119268916 is divisible by 13, as $9174532 \times 13=119268916$, we can conclude that numbers divisible by 13 are those obtained by adding or subtracting multiples of 13 from the number 119268916. Among the numbers presented, the number
$$
119268916-13=119268903
$$
is the only one div... | 119268903 | Number Theory | MCQ | Yes | Yes | olympiads | false |
Arnaldo stated that a billion is the same as a million million. Professor Piraldo corrected him and correctly said that a billion is the same as a thousand million. What is the difference between the correct value of a billion and Arnaldo's statement?
(a) 1000
(b) 999000
(c) 1000000
(d) 999000000
(e) 999000000000 | The correct option is (e).
Arnaldo said that 1 billion $=1000000 \times 1000000=1000000000000=10^{12}$. Professor Piraldo corrected him, saying that 1 billion $=1000 \times 1000000=1000000000=$ $10^{9}$. The difference is
$$
1000000000000-1000000000=999000000000
$$ | 999000000000 | Number Theory | MCQ | Yes | Yes | olympiads | false |
With the energy provided by one liter of honey, a bee can fly 7000 kilometers. How many bees could fly one kilometer each, with the energy provided by 10 liters of honey?
(a) 7000
(b) 70000
(c) 700000
(d) 7000000
(e) 70000000 | The correct option is (b).
The energy spent by a bee to fly 7000 kilometers is the same as 7000 bees spend to fly 1 kilometer each. Since the number of liters of honey was multiplied by 10, we have enough energy for 10 times that number of bees to fly 1 kilometer each, that is, 70000 bees. | 70000 | Algebra | MCQ | Yes | Yes | olympiads | false |
Diamantino put three liters of water and one liter of soft drink in a container. The soft drink is composed of $20 \%$ orange juice and $80 \%$
water. After mixing everything, what percentage of the final volume represents the orange juice?
(a) $5 \%$
(b) $7 \%$
(c) $8 \%$
(d) $20 \%$
(e) $60 \%$ | The correct option is (a).
The refreshment is composed of $20 \%$ of a liter, that is, 0.2 liters of juice and $80 \%$ of a liter, that is, 0.8 liters of water. Therefore, the final mixture has 0.2 liters of juice and $3+0.8=3.8$ liters of water. The percentage of juice in relation to the volume of the mixture is then... | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
What is the value of $2^{6}+2^{6}+2^{6}+2^{6}-4^{4}$?
(a) 0
(b) 2
(c) 4
(d) $4^{2}$
(e) $4^{4}$ | The correct option is (a).
We have $2^{6}+2^{6}+2^{6}+2^{6}-4^{4}=4 \times 2^{6}-4^{4}$. There are several ways to calculate this.
Solution 1: $4 \times 2^{6}-4^{4}=4 \times\left(2^{2}\right)^{3}-4^{4}=4 \times 4^{3}-4^{4}=4^{4}-4^{4}=0$.
Solution 2: $4 \times 2^{6}-4^{4}=4\left(2^{6}-4^{3}\right)=4\left[2^{6}-\left... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
With six identical rectangles, we form a larger rectangle, with one of the sides measuring $21 \mathrm{~cm}$, as shown in the figure. What is the area of the larger rectangle, in $\mathrm{cm}^{2}$?
(a) 210
(b) 280
(c) 430
(d) 504
(e) 588
.
From the figure, we see that the length $a$ of the smaller rectangles is twice their width $b$, that is, $a=2 b$. Therefore,
$$
a+b=2 b+b=3 b=21
$$
that is, $b=7 \text{ cm}$ and $a=14 \text{ cm}$. Therefore, the length of the larger rectangle is $4 b=28$ and its area is $21 \times 28=588 \... | 588 | Geometry | MCQ | Yes | Yes | olympiads | false |
Three years ago, the population of Pirajussaraí was equal to the population that Tucupira has today. Since then, the population of Pirajussaraí has not changed, but the population of Tucupira has grown by $50 \%$. Today, the sum of the populations of the two cities is 9000 inhabitants. What was the sum of these two pop... | The correct option is (e).
Let $p$ be the population of Tucupira three years ago. Since this population grew by $50 \%$, Tucupira currently has $p+50 \%$ of $p$ inhabitants, that is,
$$
p+\frac{50}{100} p=p+0.5 p=1.5 p \quad \text { inhabitants. }
$$
Since the population of Pirajussaraí did not grow in these 3 years... | 7200 | Algebra | MCQ | Yes | Yes | olympiads | false |
In a year, at most how many months have five Sundays?
(a) 3
(b) 4
(c) 5
(d) 6
(e) 7 | The correct option is (c).
A common year has 365 days and a leap year has 366. From the division of 365 by 7, we get $365=52 \times 7+1$ and from the division of 366 by 7 we get $366=52 \times 7+2$. Therefore,
$$
\begin{aligned}
\text { common year } & =52 \text { weeks }+1 \text { day } \\
\text { leap year } & =52 ... | 5 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
Students from a school participated in a field trip, for which two buses were hired. When the buses arrived, 57 students entered the first bus and only 31 in the second. How many students should transfer from the first to the second bus so that the same number of students are transported in both buses?
(a) 8
(b) 13
(c)... | The correct option is (b).
The total number of students in the two buses is $57+31=88$ and $\frac{1}{2} 88=44$. For each bus to have the same number of students, $57-44=13$ students must move from the first bus to the second bus. | 13 | Algebra | MCQ | Yes | Yes | olympiads | false |
What is the unit digit of the number
$$
1 \times 3 \times 5 \times \cdots \times 97 \times 99 ?
$$
(a) 1
(b) 3
(c) 5
(d) 7
(e) 9 | The correct option is (c).
The last digit of a multiple of 5 is 0 or 5; those ending in 0 are even and those ending in 5 are odd. Since $1 \times 3 \times 5 \times \cdots \times 97 \times 99$ is odd, being a product of odd numbers, and is a multiple of 5, it follows that its units digit is 5. | 5 | Number Theory | MCQ | Yes | Yes | olympiads | false |
The figure shows the map of a country (imaginary) consisting of five states. It is desired to color this map with the colors green, blue, and yellow, so that two adjacent states do not have the same color. In how many different ways can the map be painted?
.
State A can be painted in three ways: green, blue, or yellow. For any neighboring state, for example, state $\mathrm{B}$, we have two possibilities, and the colors of the other states are determined. Therefore, we can color the map in $3 \times 2=6$ ways.
Below we illustrate two of these wa... | 6 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
In a $3 \times 3$ board, nine houses must be painted such that in each column, each row, and each of the two diagonals, there are no two houses of the same color. What is the minimum number of colors needed for this?
(a) 3
(b) 4
(c) 5
(d) 6
(e) 7 | The correct option is (c).
To satisfy the conditions of the problem, the five houses on the diagonals, marked with *, must have different colors. Therefore, we will need at least five distinct colors. Let's denote these five distinct colors by 1, 2, 3, 4, and 5, and determine how we can choose the colors for the remai... | 5 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
On the same side of a street, six adjacent houses will be built. The houses can be brick or wooden, but as a safety measure against fire, two wooden houses cannot be adjacent. In how many ways can the construction of these houses be planned? | As the houses are neighbors, we can think of them as a row of houses with six positions. Let's divide the count into cases, according to the number of wooden houses that can be built.
(a) No wooden houses: here there is only one way to build the houses, that is, all of them are masonry.
(b) One wooden house: here we h... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Which is the largest of the given numbers?
(a) $1000+0.01$
(c) $1000 / 0.01$
(e) $1000-0.01$
(b) $1000 \times 0.01$
(d) $0.01 / 1000$ | The correct option is (c).
We have $1000+0.01=1000.01$ and $1000 \times 0.01=1000 \times \frac{1}{100}=10$, as well as
$$
\frac{1000}{0.01}=\frac{1000}{\frac{1}{100}}=1000 \times 100=100000
$$
and $0.01 / 1000=0.00001$. Finally, $1000-0.01$ is less than 1000 (no need to perform the calculation to reach this conclusi... | 100000 | Algebra | MCQ | Yes | Yes | olympiads | false |
What is the largest six-digit number that can be found by removing nine digits from the number 778157260669103, without changing the order of its digits?
(a) 778152
(b) 781569
(c) 879103
(d) 986103
(e) 987776 | The correct option is (c).
Solution 1: For it to be the largest possible, the number must start with the largest digit. To have six digits without changing the order, the largest is 8 and, after that, 7. Now, we need four more digits to complete the number, so we choose 9103. Therefore, the number is 879103.
Solution... | 879103 | Number Theory | MCQ | Yes | Yes | olympiads | false |
If $\frac{n}{24}$ is a number between $\frac{1}{6}$ and $\frac{1}{4}$, who is $n$?
(a) 5
(b) 6
(c) 7
(d) 8
(e) 9 | The correct option is (a).
Since $\frac{1}{6}=\frac{4}{24}$ and $\frac{1}{4}=\frac{6}{24}$, then $n$ can only be equal to 5. | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
Miguel chose a three-digit number and another two-digit number. What is the sum of these numbers if their difference is 989?
(a) 1000
(b) 1001
(c) 1009
(d) 1010
(e) 2005 | The correct option is (c).
Since the difference is 989 and the smaller number has two digits (thus being greater than 9), the three-digit number must be greater than $989+9=998$, so the only option is 999. Therefore, the two-digit number is 10 and the sum of the two is $999+10=1009$. | 1009 | Algebra | MCQ | Yes | Yes | olympiads | false |
What is the smallest natural number $n$ for which $10^{n}-1$ is a multiple of 37?
(a) 6
(b) 5
(c) 4
(d) 3
(e) 2 | The correct option is (d).
Observe that $10^{n}-1$ is a number that has all its digits equal to 9. Note, also, that a multiple of 37, of the form $37 \times n$, only ends in 9 if $n$ ends in 7. Therefore, the smallest multiples of 37 ending in 9 are $37 \times 7=259$, $37 \times 17=629$, and $37 \times 27=999$. Since ... | 3 | Number Theory | MCQ | Yes | Yes | olympiads | false |
In a certain country with 14 million inhabitants, $0.15 \%$ of the population caught a certain flu. How many inhabitants did not catch this flu?
(a) 13979000
(b) 1397900
(c) 139790
(d) 13979
(e) 139790000 | The correct option is (a).
15% of 14000000 contracted the flu, that is,
$$
\frac{0.15}{100} \times 140000000=0.0015 \times 14000000=21000
$$
people. Therefore, the number of people who did not contract the flu is $14000000-21000=13979000$ people. | 13979000 | Algebra | MCQ | Yes | Yes | olympiads | false |
The secret code of a group of students is a three-digit number with distinct non-zero digits. Discover the code using the following information.
$\begin{array}{llll}1 & 2 & 3 & \text { No digit is correct. }\end{array}$
$4 \quad 5 \quad 6 \quad$ Only one digit is correct and in the right position.
$\begin{array}{lll... | The correct option is (b).
The code can only be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9.
From the first piece of information, we know that 1, 2, and 3 are not part of the code (numbers that are not part of the code are underlined in the tables). From the third piece of information, we conclude that 6 is ... | 876 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
What is the value of $2-2\{2-2[2-2(4-2)]\}$?
(a) 0
(b) 2
(c) -2
(d) 4
(e) -10 | The correct option is (e).
The order of precedence for solving an expression is
$$
\underbrace{\text { parentheses }}_{1 \varrho} \rightarrow \underbrace{\text { brackets }}_{2 \varrho} \rightarrow \underbrace{\text { braces }}_{3 \propto}
$$
and
$$
\underbrace{\text { multiplications and divisions }}_{1 \varrho} \... | -10 | Algebra | MCQ | Yes | Yes | olympiads | false |
How many three-digit numbers greater than 200 can be written using only the digits 1, 3, and 5?
(a) 10
(b) 12
(c) 14
(d) 15
(e) 18 | The correct option is (e).
Since they are greater than 200, their hundreds digits can only be 3 or 5.
Starting with 3, we have 315 and 351 (which do not repeat digits) and 311, 313, 331, 335, 353, 333, and 355 (repeating digits), that is, nine numbers.
Starting with 5, it is enough to swap the 3 with the 5 in the nu... | 18 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
Five students each wrote a number on a piece of paper. The numbers could only be 1 or 2 or 4. What could be the product of the five numbers written?
(a) 100
(b) 120
(c) 256
(d) 768
(e) 2048 | The correct answer is (c).
If all the students wrote the number 1, the product would be 1, which is not among the options. Therefore, 2 or 4 are factors of the product, and thus the product must be a power of 2. The largest possible product would be obtained if all 5 students wrote the number 4, and the product would ... | 256 | Number Theory | MCQ | Yes | Yes | olympiads | false |
If $m$ is a natural number such that $3^{m}=81$, what is the value of $m^{3}$?
(a) $81^{3}$
(b) $3^{81}$
(c) 64
(d) 24
(e) 48 | The correct option is (c).
We have $3^{m}=81=3^{4}$, hence $m=4$. Therefore, $m^{3}=4^{3}=4 \times 4 \times 4=64$. | 64 | Algebra | MCQ | Yes | Yes | olympiads | false |
Célia wants to trade stickers with Guilherme from an album about Brazilian animals. Célia wants to trade four butterfly stickers, five shark stickers, three snake stickers, six parakeet stickers, and six monkey stickers. All of Guilherme's stickers are of spiders. They know that
(a) one butterfly sticker is worth thre... | The currency of exchange for Guilherme is spider stickers, so we calculate the spider value of the stickers Célia wants to trade, using the given information.
- 4 butterfly $\stackrel{(\mathrm{a})}{=} \underbrace{12 \text { shark }}_{4 \times 3} \stackrel{(\mathrm{e})}{=} \underbrace{24 \text { parrot }}_{12 \times 2}... | 171 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A house catches fire. A firefighter remains on the middle step of a ladder, spraying water on the fire. The flames subside and he climbs up five steps. The wind blows and the firefighter descends seven steps. A little later, he climbs up eight steps and stays there until the fire is out. Then, he climbs the last seven ... | The correct option is (c).
The up-and-down movement of the firefighter from the middle step until reaching the last step is given by
$$
\overbrace{+5}^{\text {up }} \underbrace{-7}_{\text {down }} \overbrace{+8}^{\text {up }} \overbrace{+7}^{\text {up }}
$$
such that the firefighter climbs $8+5=13$ steps above the m... | 27 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
If $a, b$ and $c$ are positive integers such that $3a=4b=7c$, what is the smallest possible value of $a+b+c$?
(a) 84
(b) 36
(c) 61
(d) 56
(e) 42 | The correct option is (c).
Let $N$ be the number given by $N=3a=4b=7c$. Then, the number $N$ is a multiple of 3, 4, and 7. Therefore, when we factorize the number $N$ into prime factors, at least the factors 2, 3, and 7 appear, with the first having an exponent of at least 2. It follows that $N$ is a multiple of $2^{2... | 61 | Number Theory | MCQ | Yes | Yes | olympiads | false |
A number is a perfect square if it is equal to an integer raised to the square. For example, $25=5^{2}, 49=7^{2}$ and $125=25^{2}$ are perfect squares. What is the smallest number by which we must multiply 120 to obtain a perfect square?
(a) 10
(b) 15
(c) 20
(d) 30
(e) 35 | The correct option is (d).
Factoring 120 we get $120=2^{3} \times 3 \times 5$. To obtain a perfect square, all the exponents in this factorization must be even, so it is enough to multiply 120 by
$$
2 \times 3 \times 5=30
$$
Indeed, we have, $120 \times 30=2^{3} \times 3 \times 5 \times 2 \times 3 \times 5=2^{4} \ti... | 30 | Number Theory | MCQ | Yes | Yes | olympiads | false |
The machine that records the number of visitors to a museum marks 1879564. Note that this number has all distinct digits. What is the smallest number of additional visitors needed for the machine to record another number that also has all its digits distinct?
(a) 35
(b) 36
(c) 38
(d) 47
(e) 52 | The correct option is (c).
Notice that the only digits that do not appear in the number 1879564 are 0, 2, and 3. The next number with all distinct digits will occur when the hundreds digit changes and we have 18796 _ _. Therefore, the smallest number will be 1879602, and there are still $1879602-1879564=38$ visitors r... | 38 | Number Theory | MCQ | Yes | Yes | olympiads | false |
Find the smallest positive multiple of 9 that can be written using only the digits: (a) 0 and 1 ; $\quad$ (b) 1 and 2 . | A number is a multiple of 9 if the sum of its digits is a multiple of 9.
(a) The number must have 9 digits all equal to 1, that is, 111111111.
(b) We should use the largest possible number of digits equal to 2, all placed in the rightmost positions. Thus, the smallest number is 12222. | 111111111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the digit $a$ in $a 000+a 998+a 999=22$ 997? | Performing the addition
| 111 |
| ---: |
| $a 000$ |
| $a 998$ |
| $+a 999$ |
| $\square 997$ |
we find $\square 997=22997$, where $\square=a+a+a+1$. Therefore, $22=a+a+a+1$, and thus, $a=7$. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
We say that a number is ascending if each of its digits is greater than the digit to its left. For example, 2568 is ascending and 175 is not. How many ascending numbers are there between 400 and 600? | The numbers we are looking for are greater than 400 and less than 600, so the hundreds digit can only be 4 or 5. Since they are ascending numbers, the tens digit is less than the units digit. Let's see how to choose the tens and units digits.
$$
\begin{gathered}
4\left\{\begin{array}{l}
56 \\
57 \\
58 \\
59
\end{array... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A game starts with seven coins aligned on a table, all with the crown face up. To win the game, you need to flip some coins in such a way that, in the end, two adjacent coins always have different faces up. The rule of the game is to flip two adjacent coins in each move. What is the minimum number of moves required to ... | Assigning the value 1 to heads and -1 to tails and summing the results after each flip, the game starts with a sum of 7 and we want to reach alternating heads and tails, so that the game ends at 1 or -1. We observe that, in each step of the game, we have the following possibilities: we exchange two heads for two tails ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The price of a kilogram of chicken was $R \$ 1.00$ in January 2000, when it started to triple every 6 months. How long will it take for the price to reach $\mathrm{R} \$ 81.00$?
(a) 1 year
(b) 2 years
(c) $2 \frac{1}{2}$ years
(d) 13 years
(e) $13 \frac{1}{2}$ years | The correct option is (b).
Since $81=3^{4}$, the value of the chicken has tripled four times. The number of months that have passed is $4 \times 6=24$ months, that is, two years, meaning that in January 2002, the chicken will reach a price of R\$ 81.00. | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
In a certain warehouse, a dozen eggs and 10 apples had the same price. After a week, the price of eggs dropped by $10 \%$ and the price of apples increased by $2 \%$ How much more will be spent on the purchase of a dozen eggs and 10 apples?
(a) $2 \%$
(b) $4 \%$
(c) $10 \%$
(d) $12 \%$
(e) $12.2 \%$ | The correct option is (b).
Since the statement and the answer are in percentages, we can, in this case, assume any price and any currency unit, and the answer will always be the same. The simplest approach, therefore, is to assume that initially, a dozen eggs cost 100 and that ten apples also cost 100. Since the price... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
Jorge walks along a rectangular path, where 12 trees are arranged, playing a game of touching each tree during his walk. First, he touches the tree at the corner, marked with $P$ in the figure, and walks 32 meters in the same direction of the path; then he turns back 18 meters and returns to the initial direction for a... | Remembering that the distance between the trees along the path is $5 \mathrm{~m}$, we illustrate the direction of Jorge's journey in the figures.
(a) Walking initially $32 \mathrm{~m}$, he ... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The squares of the natural numbers from 1 to 99 were written one after another, forming the number 14916253649... What is the digit that occupies the 100th position? (The positions are counted from left to right, so the $1^{\underline{a}}$ position is the 1, the $2^{\underline{a}}$ is the 4, and so on.) | Separating the numbers whose squares have 1, 2, and 3 digits, we have,
$$
\begin{array}{ll}
\text { with } 1 \text { digit: } & 1,2,3 \\
\text { with } 2 \text { digits: } & 4,5,6,7,8,9 \\
\text { with } 3 \text { digits: } & 10,11,12, \ldots, 31
\end{array}
$$
Up to $31^{2}$, the number already has $3+12+66=81$ digi... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The library of a school bought 140 new books, ending up with $\frac{27}{25}$ of the number of books it had before the purchase. The number of books before this purchase was:
(a) 1750;
(b) 2500;
(c) 2780;
(d) 2140;
(e) 1140. | When buying 140 books, the library ended up with $\frac{27}{25}$ of the number of books, therefore, 140 corresponds to $\frac{2}{25}$ of the books in the library. If $\frac{2}{25}$ corresponds to 140 books, $\frac{1}{25}$ corresponds to $140 \div 2=70$ books and $\frac{25}{25}$ to $70 \times 25=1750$ books. The correct... | 1750 | Algebra | MCQ | Yes | Yes | olympiads | false |
How many fractions less than 1 are there, in which the numerator and the denominator are single-digit positive integers? | For a fraction to be less than 1, the numerator must be less than the denominator. Eliminating repetitions, we obtain the following list.
(a) 1 fraction with denominator 2: $\frac{1}{2}$
(b) 2 fractions with denominator $3: \frac{1}{3}$ and $\frac{2}{3}$
(c) 2 fractions with denominator 4: $\frac{1}{4}, \underbrace{... | 27 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Seventy-four pencils were packed into 13 boxes. If the maximum capacity of each box is six pencils, what is the minimum number of pencils that can be in a box?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 6 | The correct option is (b).
Let's see in how many boxes we can place the maximum number of pencils, which is 6 per box. In 13 boxes it is not possible, because $13 \times 6=78$ is greater than the total number of pencils, which is 74. In 12 boxes we can have $12 \times 6=72$, leaving one box with $74-72=2$ pencils. | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
When I went to receive the gold medal I won in the OBMEP, the following information appeared on the passenger cabin screens of my flight to Recife:
$$
\begin{aligned}
\text { Average speed: } & 864 \mathrm{~km} / \mathrm{h} \\
\text { Distance from the departure location: } & 1222 \mathrm{~km} \\
\text { Arrival time ... | The correct option is (b).
At the moment the information was given, the remaining flight time was $1 \mathrm{~h} 20 \mathrm{~min}$, or $4 / 3$ of an hour. Therefore, at that moment, the distance to Recife was $864 \times \frac{4}{3}=1152$ $\mathrm{km}$. Since we were $1222 \mathrm{~km}$ from the departure city, the di... | 2400 | Algebra | MCQ | Yes | Yes | olympiads | false |
The figures $\triangle, \boldsymbol{\Lambda}, \diamond, \boldsymbol{\uparrow}, \odot$ and $\square$ are repeated indefinitely in the sequence
$$
\triangle, \boldsymbol{\leftrightarrow}, \diamond, \boldsymbol{\uparrow}, \odot, \square, \triangle, \boldsymbol{\leftrightarrow}, \diamond, \boldsymbol{\phi}, \odot, \square... | The figures repeat in a group of six, always ending with $\square$, both the 1st and the $166^{\circ}$ group. Since $996=6 \times 166$, the last figure of the $166^{\text{th}}$ group, that is, the 996th figure, is $\square$.
 5 ;
(b) 6 ;
(c) 7 ;
(d) 8 ;
(e) 9 . | The correct option is (b).
The number is divisible by $45=5 \times 9$, so it is divisible by 5 and 9. Every number divisible by 5 ends in 0 or 5. Thus, $b=0$ or $b=5$. Every number divisible by 9 has the sum of its digits as a multiple of 9. Therefore, $6+a+7+8+b=21+a+b$ is a multiple of 9. Since $a \leq 9$, and $b=0$... | 6 | Number Theory | MCQ | Yes | Yes | olympiads | false |
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