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a) In the figure below, there are three squares with sides 9, 6, and $x$. Determine the value of $x$.
b) Marcelo continues the drawing above and draws more squares (many!). Since these becam... | Solution
a) Observe the figure:
By similarity of triangles, we have
$$
\frac{6-x}{x}=\frac{3}{6}
$$
Therefore, $x=4$.
b) We have that the squares are similar in a ratio of $\frac{2}{3}$.... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Renato has thirty watermelons, Leandro has eighteen watermelons, and Marcelo has twenty-four jackfruits. Unlike Leandro and Renato, Marcelo does not like jackfruit. On the other hand, all three like watermelons. The three then make an agreement: Marcelo gives his twenty-four jackfruits to Leandro and Renato, and the wa... | Solution
The idea is to determine the value of watermelons in terms of jackfruits. Since there are $18+30=$ 48 watermelons and 24 jackfruits, we have the proportion
$$
\begin{array}{ccc}
48 \text { watermelons } & - & 24 \text { jackfruits } \\
1 \text { watermelon } & - & x \text { jackfruits }
\end{array}
$$
which... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In a classroom, there is a class of ten students. A committee of three students needs to be chosen to represent this class, with the committee consisting of: a spokesperson, a director of arts, and a technical advisor. No student can hold more than one position.
a) In how many ways can this committee be formed?
b) Ho... | Solution
a) To choose the spokesperson, we have 10 possibilities, since there are ten students. Once the spokesperson is chosen, we have 9 possibilities left to choose the student who will be the director of arts. Finally, for choosing the technical advisor, 8 possibilities remain. Therefore, we have
$$
10 \times 9 \... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Five pirates found a treasure chest full of gold coins and divided them among themselves. It is known that:
- What the first pirate received is equivalent to half of what the other four received in total.
- What the second pirate received is equivalent to a third of what the other four received in total.
- What the th... | Solution
Let $a, b, c, d$, $e$ be the number of coins received by the five pirates. The total number of coins is $S=a+b+c+d+e$. The first pirate received half of what the other four received in total, that is, $a=(b+c+d+e) / 2=(S-a) / 2$. Therefore, $a=S / 3$. The second pirate received a third of what the other four ... | 1800 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A four-digit number $a b c d$ is called balanced if
$$
a+b=c+d
$$
Calculate the following quantities:
a) How many numbers $a b c d$ are such that $a+b=c+d=8$ ?
b) How many numbers abcd are such that $a+b=c+d=16$ ?
c) How many balanced numbers exist? | Solution
a) Let's first count the possible values for the pair $(a, b)$. Note that $a$ cannot be equal to zero, as it is the first digit in $abcd$. However, $a$ can take any value in
$$
\{1,2,3,4,5,6,7,8\}
$$
because for each of these values, the number 8 - $a$ results in an appropriate value for $b$. Thus, there ar... | 615 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The following figure shows a regular dodecagon.
Answer the following questions:
a) How many equilateral triangles can be formed such that their three vertices are vertices of the dodecagon... | Solution
We can inscribe the regular dodecagon in a circle as shown in the following figure:
Then the arcs corresponding to each side of the dodecagon must measure $(360 / 12)^{\circ}=30^{... | 168 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A positive integer is called "balanced" if it has four digits, and one of these digits is equal to the average of the other three. For example: the number 2631 is balanced because 3 is the average of 2, 6, and 1; 4444 is also balanced because 4 is the average of 4, 4, and 4.
a) Find the three smallest balanced numbers... | Solution
A four-digit number is balanced when one of its digits, say $a$, is the average of the other three digits, say $b, c$ and $d$, that is, when $(b+c+d) / 3=a$ or, equivalently, when $(a+b+c+d) / 4=a$. We see thus that a four-digit number $N$ is balanced when it satisfies the following two conditions:
(i) The s... | 90 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Pedrinho makes a list of all the 5-digit numbers with distinct digits that can be formed with the digits $1,2,3,4,5$. In this list, the numbers are arranged in ascending order.
a) What is the number that occupies the 10th position in the list?
b) What is the number that occupies the 85th position in the list? | Solution
a) We start by writing the first numbers in the list:
12345, 12354, 12435, 12453, 12534, 12543, 13245, 13254, 13425, 13452.
Thus, the tenth number is 13452.
b) To find the number that occupies position 85, we notice that whenever a 5-digit number starts with the digit 1, this number is smaller than a 5-dig... | 43125 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $A B C D$ and $E F G H$ be squares with side lengths 33 and 12, respectively, with $E F$ lying on side $D C$ (as shown in the figure below). Let $X$ be the intersection point of segments $H B$ and $D C$. Suppose that $\overline{D E}=18$.
 Denote $\overline{E X}=x$. We have that $|\overline{C X}|=33-18-x=15-x$.
Now note that the triangles $E X H$ and $C X B$ are similar, so:
$$
\frac{|\overline{E H}|}{|\overline{... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A group of boys and girls went out to eat pizza for two consecutive days. At the restaurant they went to, the pizzas are cut into twelve equal slices. Maria observed that on the first day, each boy ate 7 slices, and each girl 3 slices. On the second day, each boy ate 6 slices, and each girl 2 slices. Interestingly, on ... | Solution
Let $x$ and $y$ be the number of boys and girls, respectively. We know that the total number of slices consumed was at least 49 (4 pizzas and one slice of the last pizza) and at most 59 (4 pizzas plus 11 slices, remember that at least one slice of the last pizza was left). On the other hand,
$$
\begin{aligne... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In a class, there are 70 students, such that:
I) 14 boys passed in Mathematics;
II) 12 boys passed in Physics;
III) 10 boys and 16 girls did not pass in Mathematics or Physics;
IV) 32 are boys;
V) 10 passed in both subjects;
VI) 22 passed only in Mathematics.
How many girls passed only in Physics? | Solution
To solve the problem, we will use the diagram below, where the upper rectangle represents the quantities of boys in each case and the lower one the quantities of girls; in the left circle, the number of students who passed in mathematics, while in the right one, the number who passed in physics, and in the in... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Using 5 non-zero digits, we can form 120 numbers, without repeating a digit within the same number. Let $S$ be the sum of all these numbers. Determine the sum of the digits of $S$, where:
a) 1, 3, 5, 7, and 9 are the 5 digits;
b) 0, 2, 4, 6, and 8 are the 5 digits, remembering that 02468 is a number with 4 digits and... | Solution
a) There are 120 numbers in total, with all possible combinations. Thus, in each of the positions (units, tens, hundreds, thousands, ten thousands), each digit appears the same number of times, that is, $\frac{120}{5}=24$. For example, in the units place, the digit 1 appears 24 times, as does 3, 5, 7, and 9. ... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the figure, $\angle A B C=100^{\circ}, \angle F A C=3 \angle E C B$ and $\angle G C A=3 \angle D A B$. Determine the measure of the acute angle at the intersection of the internal angle bisectors of triangles $\triangle A D B$ and $\triangle C E B$ relative to angles $D$ and $E$.
By the sum o... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On a $4 \times 4$ board, the numbers from 1 to 16 must be placed in the cells without repetition, such that the sum of the numbers in each row, column, and diagonal is the same. We call this sum the Magic Sum.
a) What is the Magic Sum of this board?
b) If the sum of the cells marked with $X$ in the board below is 34, ... | Solution
a) Since there are 4 rows (as well as 4 columns), the Magic Sum is:
$$
\frac{1+2+3+\ldots+16}{4}=34
$$
b) If we add the two diagonals, we will have exactly the sum of the cells marked with $X$ and with $Y$. Thus, the sum of the cells marked with $Y$ is $2 \cdot 34-34=34$.
c) We have:
$$
\begin{aligned}
\f... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A photographer must take photos of a party with 10 members of the same family. Each of the 45 possible pairs of people in this family must appear together in exactly one photo. Additionally, there are only two types of photos: those with 2 or 3 people.
a) Verify that each person in the family must appear in at least o... | Solution
a) Each person must appear with 9 other family members exactly once in some photo. In photos with 3 people, one person appears in exactly 2 pairs, and since 9 is odd, it is necessary that each person appears in at least one photo with exactly 2 people.
b) Let $x$ be the number of photos with 3 people and $y$... | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a chocolate store, there are boxes with 8, 9, and 10 chocolates. Note that some quantities of chocolates cannot be bought exactly, such as, for example, 12 chocolates.
a) Find another quantity of chocolates that cannot be bought.
b) Verify that every number greater than 56 can be written in the form $8x + 9y$ with... | Solution
a) It is not possible to buy 15 chocolates, because $15>10$ and the sum of the quantities of any two boxes is greater than 15
b) Initially note that the numbers from 57 to 64 can be written in the form $8 x+9 y$:
| $x$ | $y$ | $8 x+9 y$ |
| :---: | :---: | :---: |
| 9 | 1 | 57 |
| 5 | 2 | 58 |
| 4 | 3 | 59 ... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Three squares are attached to each other by their vertices and to two vertical rods, as shown in the figure. Determine the measure of angle $x$.
# | Solution
In the drawing below, where $AB$ is parallel to $CD$, we will show that the sum of the white angles is equal to the sum of the measures of the gray angles. This result holds for any number of "peaks" in the drawing and is popularly known as the "Theorem of Peaks".
^{2}=2^{5+2 n} \times 3^{1+2 m}=b^{3}
$$
Therefore, $5+2 n$ an... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A factory produces blouses at a cost of $R \$ 2.00$ per unit in addition to a fixed cost of $R \$ 500.00$. If each unit produced is sold for $R \$ 2.50$, from how many units produced does the factory start making a profit?
(a) 250
(b) 500
(c) 1000
(d) 1200
(e) 1500 | Let's denote by $x$ the number of units produced. Thus, the production cost is $500+2 x$ reais. From the sales, the manufacturer is receiving $2.5 x$. Therefore, he will have a profit when
$$
2.5 x > 500 + 2 x
$$
that is, $0.5 x > 500$. Therefore, $x > 1000$. Hence, the correct option is (c). | 1000 | Algebra | MCQ | Yes | Yes | olympiads | false |
Twelve points are marked on a sheet of graph paper, as shown in the figure. What is the maximum number of squares that can be formed by connecting four of these points?
 | In total, we have 11 possible squares as shown below.
 | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The number 119 has the following property:
- division by 2 leaves a remainder of 1;
- division by 3 leaves a remainder of 2;
- division by 4 leaves a remainder of 3;
- division by 5 leaves a remainder of 4;
- division by 6 leaves a remainder of 5.
How many positive integers less than 2007 satisfy this property? | Initially note that if $N$ divided by $d$ leaves a remainder $r$, then adding a multiple of $d$ to $N$ does not change the remainder, that is:
$$
\frac{(N+\text { multiple of } d)}{d} \text { also leaves a remainder of } r
$$
For example: 38 divided by 3 leaves a remainder of 2, so the remainder of the division of $(... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Sílvia goes to a fountain that has three taps to fill her ten barrels. One of the barrels takes one minute to fill, another two minutes, another three minutes, and so on. How should Sílvia distribute the barrels among the taps to spend the least amount of time possible? What is this time? | Solution 1: For simplicity, we number the water jugs according to the respective times they take to fill up. The idea is to use the "remaining time" of one jug to fill another jug, filling multiple jugs simultaneously. The figures illustrate the solution.
 At a minimum, two colors are necessary, as shown in Figure 21.2.
(b) Figures 21.3 and 21.4 display two maps with six countries that require at least four colors to be painted. | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The cities of Coco da Selva and Quixajuba are connected by a bus line. Buses leave Coco da Selva for Quixajuba every hour, with the first one departing at midnight on the dot. Buses leave Quixajuba for Coco da Selva every hour, with the first one departing at half past midnight. The bus journey takes exactly 5 hours.
... | Let's observe that the bus departing from Coco da Selva to Quixajuba meets the buses that, at the time of its departure, are
Suggestion: Divide the coins into three groups of 16 coins. on the way from Quixajuba to Coco da Selva and the buses that depart in the next five hours.
The buses that are on the road are those... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
João has a deck of 52 cards numbered from 1 to 52. A set of three cards is called lucky if the sum of the digits on each card is the same. What is the minimum number of cards João has to take from the deck, without looking, so that among the cards he takes there are necessarily three cards that form a set of lucky card... | First, observe that the sum of the digits of the cards is at most $4+9=13$, which only happens with the card 49. For sums that are between 1 and 12, there are at least two cards that satisfy each sum. Thus, by taking the card 49 plus two cards for each sum between 1 and 12, that is, $2 \times 12+1=25$ cards, we still d... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Ninety-nine apples are distributed among some boys in such a way that all receive different quantities of apples.
(a) What is the maximum number of boys that can be in this group?
(b) If there are ten boys, what is the maximum number of apples that the boy who received the fewest apples can receive?
## | (a) To maximize the number of boys, we need to minimize the number of apples each one can receive. In this case, the first natural numbers $1,2,3,4, \ldots$, correspond to the quantities of apples each boy should receive, except for the last boy. Since
$$
1+2+3+\cdots+12+13=91
$$
and
$$
1+2+3+\cdots+13+14=105
$$
th... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Consider a board with 11 rows and 11 columns.
Figure 29.1
(a) How many cells form this board?
(b) The diagonal whose cells are shaded separates the board into two regions: one above and one... | (a) Since there are 11 squares in each row of the board and it has 11 rows, the total number of squares is $11 \times 11=121$.
(b) Since there is one square of the diagonal in each row of the board and it has 11 rows, the total number of squares on the diagonal is 11. On the other hand, the diagonal is an axis of symm... | 5050 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A three-digit number and its sextuple are formed by the same digits. The sum of the digits of this number is 17 and that of its sextuple is 21. What is this number? Is there more than one number with these properties? | Answer: 746 is the only solution. | 746 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Carlos wrote consecutively all the numbers from 1 to 60, that is,
$$
1234567891011121314 \cdots 57585960
$$
Then he erased 100 digits in such a way that the number formed with the digits that were not erased was the largest possible, without changing the initial order in which the digits were written. What is this nu... | Answer: 99999585960. | 99999585960 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Sure, here is the translated text:
```
In this exercise, the letters represent digits. Determine each of the addends of the given sum.
$
\begin{array}{r}
a b c d e f \\
a b c d e f \\
+\quad g h i j \\
\hline d e f h j f
\end{array}
$
``` | Three solutions:
| 231468 |
| ---: |
| 231468 |
| $+\quad 5972$ |
| 468908 |$\quad$| 264538 |
| ---: |
| 264538 |
| $\quad 9102$ | | 468908 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Let $O$ and $H$ be the circumcenter and orthocenter of triangle $\triangle ABC$, respectively. Let $O_{1}$ and $O_{2}$ be such that $AC$ is the perpendicular bisector of $OO_{1}$ and $BC$ is the perpendicular bisector of $OO_{2}$, respectively.
a) Verify that $\angle BAH = \angle OAC$.
b) If $M$ is the midpoint of $B... | Solution
a) Since the triangle $A B F$ is a right triangle, we have $\angle B A H=90^{\circ}-\angle A B C$. In the isosceles triangle $A O C$, the base angle $\angle O A C$ measures $\frac{180^{\circ}-\angle A O C}{2}=90^{\circ}-\angle A B C$. Therefore, $\angle B A H=\angle O A C$.
b) Let $D$ and $G$ be the intersec... | 80 | Geometry | proof | Yes | Yes | olympiads | false |
A deck has 32 cards divided into 4 types, each with 8 cards. In how many ways can we choose 6 cards so that all four types of cards are included among them? | Solution
Let's divide the appropriate choices of cards into two groups:
a) Group $S_{1}$: Two types are represented by two cards each, and the other two remaining types by only one card each.
b) Group $S_{2}$: One type is represented by three cards, and the other three remaining types by only one card each.
To coun... | 415744 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
a) Verify that $(x-1)(x+1)+1=x^{2}$.
b) Find the value of $\sqrt{1+2014 \sqrt{1+2015 \sqrt{1+2016 \cdot 2018}}}$. | Solution
a) $(x-1)(x+1)+1=\left(x^{2}-1\right)+1=x^{2}$.
b) Using the previous item several times, we obtain:
$$
\begin{aligned}
\sqrt{1+2014 \sqrt{1+2015 \sqrt{1+2016 \cdot 2018}}} & = \\
\sqrt{1+2014 \sqrt{1+2015 \sqrt{2017^{2}}}} & = \\
\sqrt{1+2014 \sqrt{1+2015 \cdot 2017}} & = \\
\sqrt{1+2014 \sqrt{2016^{2}}} &... | 2015 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
An urn contains $k$ balls marked with $k$, for all $k=1,2, \ldots, 2016$. What is the minimum number of balls we must withdraw, without replacement and without looking at the balls, to be sure that we have 12 balls with the same number? | Solution
Let's sum the maximum number of balls that can be drawn of each type without obtaining 12 balls of each color:
$$
1+2+3+4+5+6+7+8+9+10+11+\underbrace{11+11+\ldots+11}_{2005 \text { times }}=22121
$$
Thus, it is possible that we have bad luck and draw such a number of balls without obtaining 12 balls of each... | 22122 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Consider the table of numbers below. The first row contains the numbers from 1 to $n$. The second row contains the numbers from 1 to $n$, each multiplied by 2. The rows follow this pattern until the last row, which presents each number from 1 to $n$ multiplied by $n$.
 The numbers in row $t$ are the numbers from 1 to $n$ multiplied by $t$. The sum of them is
$$
\begin{aligned}
t+2 t+3 t+\ldots+n t & =t(1+2+3+\ldots+n) \\
& =t \frac{n(n+1)}{2}
\end{aligned}
$$
The sum of the numbers in the table can be calculated using the sum of all rows. Since $\frac{n(n+1)}{2}$ is a... | 25502500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In a certain lottery, there are 60 distinct numbers and 6 of them are drawn without replacement. Each ticket has 6 distinct numbers among the 60 possible ones. The top prize, known as the "goal-in-the-corner," is awarded to the player who has the ticket with the same 6 numbers that were drawn. In this lottery, there is... | Solution
(a) Take the 6 smallest numbers $\{1,2,3,4,5,6\}$. In this case, we have only 6 bolana-trave tickets, which are formed by choosing 6 numbers from the set $\{1,2,3,4,5,6,7\}$. In principle, there would be 7 tickets, but since a gol-no-ângulo ticket is not a bolana-trave, the number of bolana-trave tickets asso... | 376 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
pongue
Some students from the seventh and eighth grades of a school participate in a ping pong tournament, where each student plays against all the others exactly once, receiving 1 point for a win and 0 points for a loss. There are ten times as many eighth-grade students as seventh-grade students. The total score of t... | pongue - Solution
a) Each of the $k$ students will play $k-1$ times. Summing the number of games each one plays, we get $k(k-1)$. However, we will have counted each game twice, once for each of the participants in the match. Therefore, the number of games is $\frac{k(k-1)}{2}$.
b) Let $n$ be the number of seventh-gra... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a tournament, any two players play against each other. Each player gets one point for a win, 1/2 for a draw, and 0 points for a loss. Let $S$ be the set of the 10 lowest scores. We know that each player obtained half of their score playing against players from $S$.
a) What is the sum of the scores of the players in... | Solution
a) The players of $S$, in matches played only among themselves, scored $\frac{10(10-1)}{2}=45$ points. Since the points they scored playing among themselves correspond to half of the points each one scored in the tournament, we can conclude that the sum of the points of the players of $S$ is $45+45=90$.
b) L... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The multiplicative content of a set is the product of its elements. If the set has only one element, its multiplicative content is this single element, and if the set is empty, its multiplicative content is 1. For example, the multiplicative content of $\{1,2,3\}$ is $1 \cdot 2 \cdot 3=6$.
a) Determine the sum of the ... | Solution
a) Observe that in the product $(1+a)(1+b)$, when we apply the distributive property of multiplication of real numbers, the products $1 \cdot 1, 1 \cdot b, a \cdot 1$, and $a \cdot b$ appear. These products coincide with the multiplicative contents of the sets $\varnothing, \{b\}, \{a\}$, and $\{a, b\}$, resp... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What is the maximum number of integers we can choose from the set $\{1,2,3, \ldots, 2017\}$ such that the difference between any two of them is not a prime number? | Solution
The largest quantity is $\frac{2017-1}{4}+1=505$, which can be obtained by choosing the elements that leave a remainder of 1 when divided by 4: $\{1,5,9, \ldots, 2017\}$. The difference between any two of them is always a multiple of 4 and consequently is not a prime number. We will now show that it is not po... | 505 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A group of 10 students participates in a mathematics competition formed by teams of 4 students. We know that any two of the teams have exactly one student in common.
a) What is the maximum number of teams a student can participate in? Provide an example of a distribution of 10 students where this maximum number can be... | Solution
a) Consider a student $A$ who is part of the most teams and say that he is in a team with three other students $B, C$ and $D$. Any other team that also includes $A$ must contain three other students who are not in the set $\{B, C, D\}$. Since there are only $10-1=9$ students different from $A$, the maximum nu... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Sérgio chooses two positive integers $a$ and $b$. He writes 4 numbers in his notebook: $a, a+2, b$ and $b+2$. Then, all 6 products of two of these numbers are written on the board. Let $Q$ be the number of perfect squares written on it, determine the maximum value of $Q$.
# | Solution
Initially, we will prove that the product $a(a+2)$ is not a perfect square for any choice of $a$. We have two cases to consider:
i) If $a$ is odd, then no prime that divides $a$ can divide $a+2$. Therefore, $a$ and $a+2$ must each be a perfect square. This clearly has no solution for $a \geq 1$, as the diffe... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A chessboard is an $8 \times 8$ square where the squares are arranged in 8 rows and 8 columns.
A rook on a chessboard attacks all pieces that are in its row or column. Knowing this, determin... | Solution
a) Below, we have an example with 8 towers, none of which are attacking each other.
If we place 9 or more towers, since there are only 8 rows, there will be two in the same row, a... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Consider three collinear points $B, C$, and $D$ such that $C$ is between $B$ and $D$. Let $A$ be a point that does not lie on the line $BD$ such that $AB = AC = CD$.
(a) If $\angle BAC = 36^... | Solution
(a) Since $\angle B A C=36^{\circ}$ and triangle $A B C$ is isosceles, we have $\angle A B C=\angle A C B=72^{\circ}$. Additionally, since $\angle A C D=108^{\circ}$ and triangle $A C D$ is also isosceles, it follows that $\angle C A D=$ $\angle C D A=36^{\circ}$.
. Explain why at least 17 children have either the same grandfather or the same grandmother in this family. | Solution
For each child, associate two labels, each with the name of one of their grandparents. If a person is the grandfather or grandmother of all the children, then the statement in the question is clearly true. Let's consider the case where not all children are grandchildren of the same person. Suppose a child has... | 17 | Combinatorics | proof | Yes | Yes | olympiads | false |
How many ordered pairs $(a, b)$, with $a$ and $b$ positive integers, satisfy
$$
a+b+\gcd(a, b)=33 ?
$$ | Solution
Let $d=\gcd(a, b)$. We can rewrite the equation as:
$$
\frac{a}{d}+\frac{b}{d}+1=\frac{33}{d}
$$
Since the left side is a sum of integers, it follows that $d$ divides 33. Moreover,
$$
\begin{aligned}
\gcd\left(\frac{a}{d}, \frac{b}{d}\right) & =\gcd\left(\frac{a}{d}, \frac{33}{d}-1\right) \\
& =\gcd\left(\... | 21 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the number $6 a 78 b$, $a$ is the digit of the thousands place and $b$ is the digit of the units place. If $6 a 78 b$ is divisible by 45, then the value of $a+b$ is:
(a) 5
(b) 6
(c) 7
(d) 8
(e) 9 | The number is divisible by 5 and 9.
Every number divisible by 5 ends in 0 or 5. Thus, $b=0$ or $b=5$.
Every number divisible by 9 has the sum of its digits as a multiple of 9.
Therefore, we have that $6+a+7+8+0=21+a$ or $6+a+7+8+5=26+a$ are multiples of 9. Hence, $a=6$ or $a=1$, respectively. From this, we get: $a+b... | 6 | Number Theory | MCQ | Yes | Yes | olympiads | false |
The manager of a store went to check what had been the selling price in 2006 of a television from the VejoTudo brand. He found a faded invoice, which read: "lot of 72 VejoTudo TVs sold for $R \$ \ldots 679 \ldots$ reais", where the digits in the units and ten-thousands place were illegible. What was the selling price i... | Let $a$ be the ten-thousands digit and $b$ be the units digit. Since the number is divisible by $72=8 \times 9$, we have that $79b$ is an even number divisible by 8. Testing the values of $b=0,2,4,6$ and 8, we see that $b=2$. A number is divisible by 9 if the sum of its digits is a multiple of 9. Therefore, $a+6+7+9+2=... | 511 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In 13 boxes, 74 pencils were packed. If the maximum capacity of each box is 6 pencils, what is the minimum number of pencils that can be in a box?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 6 | Let's see in how many boxes we can place the maximum number of pencils, which is 6 per box. In 13 boxes it is not possible, because $13 \times 6=78$, which is greater than the number of pencils 74. In 12 boxes we would have: $12 \times 6=72$. Thus, there would be one box with $74-72=2$ pencils. Therefore, the correct o... | 2 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
If the digit 1 appears 171 times in the page numbering of a book, how many pages does the book have? | Every 10 pages, 1 appears in the units, and every 100 pages, the number 1 appears 10 times in the tens.
Counting the number of pages that contain the digit 1 in each range below, we have:
- 20 pages between 1-99:
1,11,21,31,41,51,61,71,81,91: 10 (1 in the unit)
10,11,12,13,14,15,16,17, 18,19: 10 (1 in the ten)
- 1... | 318 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On my flight to Recife, to receive the gold medal I won in the OBMEP, the following information appeared on the cabin passenger screen:
Average speed: $864 \mathrm{~km} / \mathrm{h}$
Distance from the departure location: $1222 \mathrm{~km}$
Time of arrival in Recife: 1 h $20 \mathrm{~min}$
If the plane maintained t... | At the moment the information was given, the remaining flight time was $1 \text{ h } 20 \text{ min}$, or $4/3 \text{ h}$. Therefore, at this moment, the distance to Recife was $864 \times \frac{4}{3} = 1152 \text{ km}$. Since we were $1222 \text{ km}$ from the city of departure, the distance between this city and Recif... | 2400 | Algebra | MCQ | Yes | Yes | olympiads | false |
Maria and João take a walk around the square together, counting the houses around it. They started counting the houses from different points. The fifth house of Maria is the twelfth of João and the fifth house of João is the thirtieth of Maria. How many houses are there around the square? | As the $5^{\text{th}}$ house from Maria is the $12^{\text{th}}$ house from João, the difference in the counts is 7 houses. Thus, the $1^{\text{st}}$ house from Maria is the $8^{\text{th}}$ house from João, and the $5^{\text{th}}$ house from João corresponds to two houses before the house where Maria started counting. H... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The figures $\triangle, \boldsymbol{\Lambda}, \diamond, \uparrow, \odot, \square$ are repeated in the sequence
$$
\triangle, \boldsymbol{\phi}, \diamond, \boldsymbol{\phi}, \odot, \square, \triangle, \boldsymbol{\phi}, \diamond, \boldsymbol{\phi}, \odot, \square, \ldots
$$
(a) Which figure will appear in the $1000^{\... | The figures repeat every 6. Dividing 1000 by 6 we get: $1000=6 \times 166+4$.
(a) The figure that is in the $1000^{\text{th}}$ place is
(b) The first $\diamond$ is in the $3^{\text{rd}}$ po... | 5997 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Simão needs to discover a number that is the code of the Treasure Ark hidden in the table.
| 5 | 9 | 4 | 9 | 4 | 1 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 6 | 3 | 7 | 3 | 4 | 8 |
| 8 | 2 | 4 | 2 | 5 | 5 |
| 7 | 4 | 5 | 7 | 5 | 2 |
| 2 | 7 | 6 | 1 | 2 | 8 |
| 5 | 2 | 3 | 6 | 7 | 1 |
To discover the code, he ha... | In the following two tables, we show only the numbers whose sum is 14, horizontally and vertically, respectively.
| | | | 9 | 4 | 1 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| | | 7 | 3 | 4 | |
| 8 | 2 | 4 | | | |
| | | | 7 | 5 | 2 |
| | 7 | 6 | 1 | | |
| | | | 6 | 7 | 1 |
| | 9 | | | | 1 |... | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Colorado Jones must solve a great riddle to survive. He must remove only one of the five pots in front of him, as indicated in the figure below, to open the door of the secret chamber. He knows that each pot contains only one type of coin, gold or silver, and that each number written on them represents the number of co... | Solution
The removal of one of the pots should allow Colorado Jones to separate the remaining ones into two groups, where in one of these groups there is double the number of coins of the other group. If the number of coins in the smaller group is $x$, the sum of the coins in the two groups is $x+2x=3x$. Therefore, we... | 37 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Zé Roberto has five children, two are twins and the other three are triplets. It is known that today Zé's age is equal to the sum of the ages of his five children. In 15 years, if we add up the ages of the five children, we will have double the age that Zé will have at the same time and the sum of the ages of the twins... | Solution
(a) Let $z$ be the current age of Zé, $g$ the current age of the twins, and $t$ the current age of the triplets. The information from the problem can be translated into three equations:
$$
\begin{aligned}
z & =2 g+3 t \\
2(z+15) & =2(g+15)+3(t+15) \\
2(g+15) & =3(t+15)
\end{aligned}
$$
To determine $z$, it ... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Juca has fewer than 800 marbles. He likes to separate the marbles into groups with the same number of marbles. He noticed that if he forms groups of 3 marbles each, exactly 2 marbles are left over. If he forms groups of 4 marbles, 3 marbles are left over. If he forms groups of 5 marbles, 4 marbles are left over. Finall... | Solution
(a) Let $B$ be the number of marbles Juca has. Note that the number of marbles left over when forming a group is equal to the remainder of the division of $B$ by the size of the groups. To determine the remainder in the division by 20, one should use the remainders in the division by 4 and by 5, since $20=4 \... | 419 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the drawing below, $C$ is the intersection point of $A E$ and $B F$, $A B=B C$, and $C E=C F$. If $\angle C E F=50^{\circ}$, determine the angle $\angle A B C$.
 | Solution
Since triangle $CEF$ is isosceles, we have $\angle CEF = \angle CFE = 50^{\circ}$ and
$$
\begin{aligned}
\angle FCE & =180^{\circ}-\angle CEF-\angle CFE \\
& =180^{\circ}-50^{\circ}-50^{\circ} \\
& =80^{\circ}
\end{aligned}
$$
We also have $\angle BCA = \angle FCE$, as they are vertically opposite angles. F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Imagine as 2015 fractions:
$$
\frac{2}{2016}, \frac{3}{2015}, \frac{4}{2014}, \ldots, \frac{2014}{4}, \frac{2015}{3}, \frac{2016}{2}
$$
Is it possible to choose three of these fractions with a product equal to 1? | Solution
Yes. See that each fraction is of the form $\frac{x}{2018-x}$. Thus, for $x=1009$, the fraction $\frac{1009}{1009}=1$ is part of the list. It then suffices to multiply the fractions:
$$
\frac{2}{2016} \cdot \frac{1009}{1009} \cdot \frac{2016}{2}=1
$$ | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Consider the following sequence of pieces, where the piece numbered 1 is a small square.
(a) How many small squares form the piece numbered 50?
(b) How many small squares are there in the u... | Solution
(a) See that with each increment of one unit in the piece number, the number of squares increases by 2 units. Therefore, piece 50 will have $1+2 \cdot 49=99$ squares.
(b) See that the pieces can be juxtaposed to form a larger square divided into squares. What determines the side of this square is the number o... | 5050 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Carlinhos likes to write numbers in his notebook. One day he wrote the numbers from 1 to 999, one after the other, to form the giant number:
$$
123456789101112 \ldots 997998999
$$
About this number, the following questions are asked:
(a) How many digits were written?
(b) How many times does the digit 1 appear?
(c)... | Solution
(a) Observe that we have 9 numbers with one digit each, $99-9=90$ numbers with two digits each, and $999-99=900$ numbers with three digits each. Therefore, the number of digits written is:
$$
9 \cdot 1 + 90 \cdot 2 + 900 \cdot 3 = 2889
$$
(b) Note that the digit 1 appears once among the one-digit numbers. A... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The figure below shows a "staircase" formed by two squares, one with a side of $8 \mathrm{~cm}$ and one with a side of $6 \mathrm{~cm}$. The task is to cut the figure into three pieces and reassemble them to form a square without any gaps.
 Since the square should not have any holes, the final area must be equal to the original area. If we call $L$ the side of the square, we have:
$$
\begin{aligned}
L^{2} & =8^{2}+6^{2} \\
L^{2} & =64+36 \\
L^{2} & =100 \\
L & =10
\end{aligned}
$$
(b) By the Pythagor... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Is it possible to multiply the number 101001000100001 by another integer so that the result has no digits equal to zero?
# | Solution
Initially consider the number 101, which has a "hole" with only one zero. Multiplying by 11 eliminates the "hole," as shown in the diagram below of the multiplication algorithm:
| | 1 | 0 | 1 |
| :---: | :---: | :---: | :---: |
| $\times$ | 1 | 1 | |
| | 1 | 0 | 1 |
| 1 | 0 | 1 | |
| 1 | 1 | 1 | 1. |
No... | 11111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the drawing below, each square in the figure has a side length of $1 \mathrm{~m}$. Determine the sum of the angles $\angle P B C+\angle Q A P+\angle Q D C$.
# | Solution
By rotating the square $90^{\circ}$ counterclockwise, we can conclude that triangle $A B Q$ is equal to triangle $B P C$, and consequently, $\angle B A Q=\angle P B C$. Furthermore, by rotating the square $90^{\circ}$ clockwise, we can also conclude that triangle $A D P$ is equal to triangle $D Q C$ and that ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
a) A cube $3 \times 3 \times 3$ was constructed with 27 smaller cubes $1 \times 1 \times 1$. For each pair of touching faces of two cubes, one drop of glue is used. How many drops of glue were used in total?
b) A cube $10 \times 10 \times 10$ was constructed with 1000 smaller cubes $1 \times 1 \times 1$. For each pair... | Solution
a) From the figure below, to form a face $1 \times 3 \times 3$, 12 drops of glue are needed, corresponding to the 12 contact faces between the 9 cubes.
To place one face on anothe... | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Consider the following statements:
1) The number $N$ is divisible by 2.
2) The number $N$ is divisible by 4.
3) The number $N$ is divisible by 12.
4) The number $N$ is divisible by 24.
Three of these statements are true and one is false. Which one is the false one? | Solution
Notice that the numbers in the set $\{2,4,12,24\}$ are all divisors of 24. Therefore, if the last statement is true, all the other three will also be true. Thus, the false statement is the last one. An example of a number that satisfies the first three conditions of the problem but not the last one is the num... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The 9 squares of a $3 \times 3$ board, as shown in the figure below, must be painted in such a way that in each row, each column, and each of its two diagonals, there are no squares of the same color. What is the minimum number of colors needed for this painting?
 Suppose it is possible to have 0 adults. Thus, the sum of the quantities of young people and children is 100. Consequently, if each child pays 1.70 more to match the contribution of the young people, we will have $2 \cdot 100=200... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the board below, in each row and in each column, exactly one number from the set $\{1,2,3,4\}$ is written. Which number is in the square with the symbol $\star$?
| 1 | 2 | | |
| :--- | :--- | :--- | :--- |
| | 1 | 2 | |
| 2 | | $\star$ | 1 |
| 3 | | 1 | | | Solution
Consider the positions indicated by the letters: $A, B, C$ and $D$. Whenever a row or column already has three numbers written, its fourth element is completely determined and is equal to the number that has not yet appeared in it. Thus, $C=4$. Analyzing the numbers in the second row, we have $D=3$. The cell ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The figures below show a way to cut a cube into smaller cubes.
Notice that we start with a single cube, and after the cuts, we end up with 8 smaller cubes. If we choose one of these cubes a... | Solution
(a) Observe that at each step we have a surplus of 7 in the number of cubes. Therefore, after repeating the process $x$ times, we will have exactly $7 x+1$ cubes. Consequently, to find out the number of steps required to obtain 99 cubes, we just need to solve the equation:
$$
\begin{aligned}
7 x+1 & =99 \\
7... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
We say that a positive three-digit integer is three stars if it is the result of the product of three distinct prime numbers. For example, $286=2 \cdot 11 \cdot 13$ is a three-star number, but $30=2 \cdot 3 \cdot 5$ and $275=5 \cdot 5 \cdot 11$ are not three-star numbers, as the first has only two digits and the second... | Solution
(a) We can factor the smallest three-digit numbers $100=2 \cdot 2 \cdot 5 \cdot 5, 101=101$ (it is prime) and $102=2 \cdot 3 \cdot 17$. Since the first two are not three-star numbers, we can conclude that 102 is the smallest three-star number.
(b) If we take a number that is the product of three distinct pri... | 102 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Around a circular table, 18 girls are sitting, 11 dressed in blue and 7 dressed in red. Each of them is asked if the girl to their right is dressed in blue, and each one answers yes or no. It is known that a girl tells the truth only when her two neighbors, the one on the right and the one on the left, are wearing clot... | Solution
Let's analyze the possibilities. If the two neighbors of a certain girl are wearing blue, then she answers yes, and if they both wear red, she answers no, because in these cases she tells the truth. If the neighbor on the right is wearing red and the one on the left is wearing blue, the answer is yes, and if ... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The two tables below were formed according to the same rule, but in the second one, we indicate only three numbers. What number should be placed in the cell with $\star$?
| 5 | 8 | 11 | 14 | 17 |
| :---: | :---: | :---: | :---: | :---: |
| 12 | 15 | 18 | 21 | 24 |
| 19 | 22 | 25 | 28 | 31 |
| 26 | 29 | 32 | 35 | 38 ... | | 5 | 8 | 11 | 14 | 17 |
| :---: | :---: | :---: | :---: | :---: |
| 12 | 15 | 18 | 21 | 24 |
| 19 | 22 | 25 | 28 | 31 |
| 26 | 29 | 32 | 35 | 38 |
| 33 | 36 | 39 | 42 | 45 |
Let's observe ... | 102 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In a class at school, all students are the same age, except for seven who are 1 year younger and two who are 2 years older.
The sum of the ages of all the students in this class is 330. How many students are in this class? | Let $a$ denote the common age of the students and $n$ the number of students, so we have 7 students aged $a-1$, 2 aged $a+2$, and the rest, that is, $n-9$ aged $a$. Thus, the sum of the ages is
$$
7(a-1)+2(a+2)+(n-9) a=n a-3=330
$$
$\log o$
$$
n a=333=9 \times 37
$$
Since the class has more than 9 students, then $a... | 37 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A round table has a diameter of $1.40 \, \text{m}$. For a party, the table is expanded by adding three planks, each $40 \, \text{cm}$ wide, as shown in the figure. If each person at the table should have a space of $60 \, \text{cm}$, how many guests can sit at the table?
The perimeter of the enlarged table is
$$
140 \times \pi + 40 \times 6 \simeq 140 \times 3.14 + 240 = 679.60 \mathrm{~cm}
$$
If each guest needs $60 \mathrm{~cm}$ to sit around the table and... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A surveillance service is to be installed in a park in the form of a network of stations. The stations must be connected by telephone lines so that any of the stations can communicate with all the others, either by a direct connection or through at most one other station. Each station can be directly connected by a cab... | The example shows that we can connect at least 7 stations under the proposed conditions. We start with a particular station, and think of it as the base of the network. It can be connected to 1, 2, or 3 stations as shown in the diagram.
 Which is the larger number: $\mathrm{R}$ or $S$?
(b) Calculate the difference between the larger and the smaller. | (a) Each term of $S$ is of the form $n \times(n+10)=n^{2}+10 n$ and each term of $R$ is of the form $(n+2) \times(n+8)=n^{2}+10 n+16$ with $\mathrm{n}=\{1,2, \ldots, 2001\}$ in both cases. Thus, for every n, each term of $R$ is greater than the corresponding one in $S$, making $\mathrm{R}>\mathrm{S}$.
(b) The differen... | 32016 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(a) In how many ways can the number 105 be written as the difference of two perfect squares?
(b) Show that it is not possible to write the number 106 as the difference of two perfect squares.
## | (a) Let $x$ and $y$ be two positive integers such that the difference between their squares is equal to 105, that is, $x^{2}-y^{2}=105$. Factoring, we get $(x-y)(x+y)=105$, and therefore, $x+y$ and $x-y$ must be divisors of 105, with $x+y > x-y$. Note that $1 \cdot 105 = 3 \cdot 35 = 5 \cdot 21 = 7 \cdot 15$ are all th... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
If $(x+y)^{2}-(x-y)^{2}=20$, then $x y$ is equal to:
(a) 0
(b) 1
(c) 2
(d) 5
(e) 10 | As $(x+y)^{2}=x^{2}+2 x y+y^{2}$ and $(x-y)^{2}=$ $x^{2}-2 x y+y^{2}$, we have:
$$
(x+y)^{2}-(x-y)^{2}=x^{2}+2 x y+y^{2}-x^{2}+2 x y-y^{2}=4 x y=20
$$
it follows that $x y=5$. The correct option is (d). | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
feira 13 - What is the maximum number of Friday the 13ths that can occur in a non-leap year? In this case, what is the $10^{\text{th}}$ day of the year? | Feira 13 - Given that the days of the week repeat every 7 days, the difference between the days of the week is given by the remainder when dividing by 7 the number of days that have passed.
$\mathrm{In}$ the following table:
- in the first row, the number of days between the 13th of one month and the 13th of the next... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
How many triangles exist whose sides are integers and the perimeter is 12?
(a) 1
(b) 3
(c) 5
(d) 7
(e) 9 | For three numbers $a, b, c$ to be the lengths of the sides of a triangle, each of them must be greater than the difference and less than the sum of the other two.
Let $a \leq b \leq c$ be the lengths of the sides of the triangle. Thus, $c < a + b$.
Now, adding $c$ to both sides, we have: $2c < a + b + c = 12$, that i... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
To celebrate her birthday, Ana is going to prepare pear pies 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 types? | Let $m$ denote the number of apples and $p$ the number of pears that Ana bought, 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
$$
300 m+200 p \leq 7000, \text { which is equivalent to } 3 m+2 p \leq 70
$$
Since pears weigh less, Ana has to carry t... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Archaeologists found a gold necklace made of plates in the shape of regular pentagons. Each of these plates is connected to two other plates, as illustrated in the figure.
Figure 51.1
How m... | The internal angle of a regular pentagon measures $108^{\circ}$. Therefore, the internal angle of the polygon determined by the necklace measures $360^{\circ}-108^{\circ}-108^{\circ}=144^{\circ}$. We must then find $n$ such that
$$
\frac{180^{\circ}(n-2)}{n}=144^{\circ}
$$
Solving this equation, we get $n=10$. Theref... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $A B C$ be a triangle with $A B=13, B C=15$ and $A C=9$. Let $r$ be the line parallel to $B C$ drawn through $A$. The bisector of angle $A \widehat{B C}$ intersects the line $r$ at $E$ and the bisector of angle $A \widehat{C} B$ intersects $r$ at $F$. Calculate the measure of the segment $EF$.
## | Since the line EF is parallel to the side $\mathrm{BC}$, the alternate interior angles formed by the transversal CF are equal, that is, FĈB $=C \hat{F} A$. On the other hand, since CF is the angle bisector, we have FCEB $=F \hat{C A}$, and thus, $F \hat{C A}=C \hat{F} A$, which means triangle CAF is isosceles with base... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Suggestion: Show that triangle BEC is isosceles.
Facts that Help: The sum of the measures of the interior angles of a triangle is equal to $180^{\circ}$.
Inside a triangle $ABC$, a point $E... | Knowing that the sum of the internal angles of a triangle is $180^{\circ}$, we obtain
$$
\left\{\begin{array}{l}
A \hat{E} B=180^{\circ}-(\alpha+\alpha)=180^{\circ}-2 \alpha \\
A \hat{E} C=180^{\circ}-(\alpha+2 \alpha)=180^{\circ}-3 \alpha
\end{array}\right.
$$
Thus, we have
$$
\begin{aligned}
\mathrm{C} \hat{\mathr... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The ants Maricota and Nandinha are strolling on a balcony whose floor is made up of rectangular tiles measuring $4 \mathrm{~cm}$ in width by $6 \mathrm{~cm}$ in length. Maricota starts from point $M$ and Nandinha from $N$, both walking only along the sides of the rectangles, following the path indicated in the figure.
... | (a) The journey from $M$ to $N$ consists of 14 lengths and 12 widths of the tiles, so its length is:
$$
14 \times 6 + 12 \times 4 = 84 + 48 = 132 \text{ cm}
$$
Since the ants travel the same distance, each one must walk $132 \div 2 = 66 \text{ cm}$.
(b) Let's follow the path taken by Maricota from the beginning, unt... | 66 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The sum of 3 numbers is 100, two are prime and one is the sum of the other two.
(a) What is the largest of the 3 numbers?
(b) Give an example of these 3 numbers.
(c) How many solutions exist for this problem? | (a) Initially observe that:
- the largest number is the sum of the other two;
- the largest number cannot exceed 50, otherwise the sum of the three would be greater than 100;
- the largest number cannot be less than 50, otherwise the sum of the three would be less than 100.
Therefore, the largest number can only be 5... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A postal service uses short and long bars to represent the Postal Addressing Code - CEP. The short bar corresponds to zero and the long bar to 1. The first and last bar are not part of the code. The conversion table for the code is shown below.
$$
\begin{array}{ll}
11000=0 & 01100=5 \\
00011=1 & 10100=6 \\
01010=2 & 0... | (a) First, we write the ZIP code in the form of 0's and 1's:
We can now write the barcode for this ZIP code:
 How many students are there in the school?
(b) How many students play only football?
(c) How many students play football?
(d) How many students practice one of the 2 spor... | (a)
The total number of students in the school is given by the fraction 12/12, which we can graphically represent by a rectangle divided into 12 equal parts.
We will denote by V, F, and NE... | 900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What is the digit of the $1997^{\mathrm{th}}$ decimal place of:
(a) $\frac{1}{22}$
(b) $\frac{1}{27}$
## Level 1 Solutions
## List 1
# | (a) Dividing 1 by 22 we get: $\frac{1}{22}=0.0454545 \ldots$ We observe that the digit 4 is in the even positions: $2,4,6, \ldots$ and the digit 5 in the odd positions: $3,5,7 \ldots$ Since 1997 is an odd number, the digit in the $1997^{\text{th}}$ decimal place is 5.
(b) Dividing 1 by 27 we get: $\frac{1}{27}=0.03703... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In how many ways can the 6 faces of a cube be colored black or white? Two colorings are the same if one can be obtained from the other by a rotation. | Let's observe that it suffices to count how many colorings exist that have exactly 0, 1, 2, and 3 black faces, because the other cases are symmetric. With one or no black face, there is a unique coloring for each case. When we have two black faces, we have two possible colorings: when these faces are opposite and when ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Ten points are marked around a circle, as illustrated in the figure.
(a) How many chords can be formed by connecting any two of these points? (A chord is a line segment connecting two points... | (a) From each point, 9 strings come out and we have 10 points. But each string is counted twice (a string $A B$ is counted for coming out of $A$ and for coming out of $B$), so we have $9 \times 10 / 2 = 45$ strings.
(b) Each string is a side of 8 triangles (it is enough to choose a point that is not an endpoint of the... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
O esqueleto de um cubo $6 \times 6 \times 6$, formado por cubinhos $1 \times 1 \times 1$ é mostrado na figura.
Figura 64.1
(a) Quantos cubinhos formam este esqueleto?
(b) É dado um cubo $7 ... |
(a) O esqueleto do cubo é formado por uma camada superior e uma inferior com 20 cubinhos cada e quatro colunas com 4 cubinhos cada.
Assim, o total de cubinhos é
$$
2 \times 20+4 \times 4=56
$$
(b) Do cubo $7 \times 7 \times 7$ foi retirado um cubo central $5 \times 5 \times 5$ e em cada uma das faces foram retirad... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Each of the bicycle plates in Quixajuba contains three letters. The first letter is chosen from the set $\mathcal{A}=\{\mathrm{G}, \mathrm{H}, \mathrm{L}, \mathrm{P}, \mathrm{R}\}$, the second letter is chosen from the set $\mathcal{B}=\{\mathrm{M}, \mathrm{I}, \mathrm{O}\}$, and the third letter is chosen from the set... | Initially, it is possible to register $5 \times 3 \times 4=$ 60 bicycles. Let's analyze the two possible situations:
- We increase two letters in one of the sets. With this, we can have
Suggestion: Calculate the initial number of plates that can be made with the elements of sets $\mathcal{A}, \mathcal{B}$, and $\mathc... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a tennis tournament, each player advances to the next round only in case of victory. If it is not possible for an even number of players to always advance to the next round, the tournament organizers decide which rounds certain players should play. For example, a seeded player can, at the discretion of the organizer... | (a) In the first round, 32 matches are played, from which 32 players advance to the next phase. Then 16 matches are played, qualifying 16 for the next round, and so on. Thus, the number of matches in the tournament is
$$
32+16+8+4+2+1=63
$$
(b) Since in each match one player is eliminated, the number of matches is eq... | 2010 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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