problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
In the right isosceles triangle $A O B$, the points $P, Q$, and $S$ are chosen on the sides $O B, O A$, and $A B$, respectively, such that $P Q R S$ is a square. If the lengths of $O P$ and $O Q$ are $a$ and $b$, respectively, and the area of the square $P Q R S$ is $2 / 5$ of the area of the triangle $A O B$, determin... | Solution
Let $C$ be the foot of the perpendicular from point $S$ to segment $O B$. The triangles $S P C$ and $P Q O$ have the same angles, since
$$
\begin{aligned}
\angle C P S & =\angle 180^{\circ}-\angle S P Q-\angle O P Q \\
& =90^{\circ}-\angle O P Q \\
& =\angle P Q O
\end{aligned}
$$
Since $P S=P Q$, these tri... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
If $n$ is a positive integer, what is the smallest value that the sum of the digits in the decimal representation of $3 n^{2}+n+1$ can take?
# | Solution
If $n=8$, we have that $3 n^{2}+n+1=201$ and the sum of its digits is 3. We will now verify that the sum of the digits of $3 n^{2}+n+1$ cannot be 1 or 2 and conclude that the smallest possible value is 3. Since $n(n+1)$ is the product of two consecutive numbers, it is even, and thus $3 n^{2}+n+1=2 n^{2}+n(n+1... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the following drawing, the chords $D E$ and $B C$ are perpendicular, with $B C$ being a diameter of the circle with center at $A$. Additionally, $\angle C G F=40^{\circ}$ and $G H=2 \text{~cm}$.
a) Determine the value of the angle $\angle C H F$.
b) Find the length of $H J$.
 Since $BC$ is a diameter, it follows that $\angle BFC=90^{\circ}$. Thus, as we also have $\angle CHG=90^{\circ}$, the circle $\Gamma$ with diameter $CG$ passes through $F$ and $H$. In this circle, the angles $\angle CGF$ and $\angle CHF$ are inscribed in the same arc $CF$, so $\angle CHF = \angle CGF = 40^... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Two precocious students from Level 3 participated in a university chess tournament. Each participant plays against all the others exactly once. A win is worth 1 point, a draw is worth 0.5 points, and a loss is worth 0 points. The sum of the scores of the two Level 3 students is 6.5. All university students scored the s... | Solution
Let $x$ be the number of college students and $p$ the common score of all of them. Since exactly 1 point is contested in each game, it follows that the total score of the tournament, which is $6.5 + p x$, coincides with the number of games, which is $\frac{(x+2)(x+1)}{2}$. In addition, the score of each parti... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
a) Verify that for any positive integer $a, \operatorname{with} a>1$, the equation
$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{a}
$$
has at least three solutions of the form $(x, y)$, with $x$ and $y$ positive integers. For example, for $a=3$, the pairs $(6,6),(4,12)$ and $(12,4)$ are solutions.
b) Find the number of pairs ... | Solution
a) We can find an equivalent equation:
$$
\begin{aligned}
\frac{1}{x}+\frac{1}{y} & =\frac{1}{a} \Leftrightarrow \\
(x-a)(y-a) & =a^{2}
\end{aligned}
$$
Since $1 / x$ and $1 / y$ are less than $1 / a$, it follows that $x-a$ and $y-a$ are positive. To find solutions to the last equation, consider the followi... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
An empty swimming pool was filled with water by two faucets $A$ and $B$, both with constant flow rates. For 4 hours, both faucets were open and filled $50 \%$ of the pool. Then, faucet B was turned off and for 2 hours faucet A filled $15 \%$ of the pool's volume. After this period, faucet A was turned off and faucet $\... | Since 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's volume in 2 hours, then in 4 hours it will fill $30 \%$ of the pool's volume.
However, when taps A and B are both open for ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Davi has a very original calculator; it performs only two operations: the usual addition $(+)$ and another operation, denoted by $*$, which satisfies:
(i) $a * a=a$
(ii) $a * 0=2a$
(iii) $(a * b) + (c * d) = (a * c) + (b * d)$
What are the results of the operations $(2+3) * (0+3)$ and $1024 * 48$? | To calculate $(2 * 3)+(0 * 3)$, we use properties (i), (ii), and (iii). Then
$$
\begin{aligned}
& (2 * 3)+(0 * 3) \quad \stackrel{(\mathrm{iii})}{=} \quad(2 * 0)+(3 * 3) \\
& \text { (i) (ii) } 2 \times 2+3=7 \text {. }
\end{aligned}
$$
To calculate $1024 * 48$, observe that $1024=976+48$. We have:
$$
\begin{aligned... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The rectangular grid in the figure is made of 31 segments of $0.5 \mathrm{~cm}$ and comprises 12 squares. Rosa drew on a rectangular sheet of $21 \mathrm{~cm}$ by $29.7 \mathrm{~cm}$, grid-lined with squares of side $0.5 \mathrm{~cm}$, a large rectangular grid made with 1997 segments. How many squares does this rectang... | Let $m$ and $n$ be, respectively, the number of segments of $0.5 \, \text{cm}$ on two consecutive sides of the rectangle. We know that the total number of segments of $0.5 \, \text{cm}$ in the division of the rectangle into $m \times n$ squares of side $0.5 \, \text{cm}$ is: $m(n+1) + n(m+1)$ (prove this). Thus,
$$
m(... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A triangle has vertex $A=(3,0), B=(0,3)$ and $C$, where $C$ is on the line $x+y=7$. What is the area of the triangle? | Observe that the height $h$, relative to the side $A B$, of all triangles $A B C$ that have the vertex $C$ on the line $x+y=7$, is the same, since the latter line is parallel to the line passing through $A$ and $B$. Therefore, these triangles all have the same area, namely:
$$
\frac{A B \times h}{2}
$$
We need to det... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Three circles with radii $1 \mathrm{~cm}, 2 \mathrm{~cm}$, and $3 \mathrm{~cm}$ are pairwise externally tangent, as shown in the figure below.
Determine the radius of the circle that is externally tangent to the three circles.
$ the sum of the digits of the number $n$. For example, $s(2345) = 2 + 3 + 4 + 5 = 14$. Observe that:
$40 - s(40) = 36 = 9 \times 4; 500 - s(500) = 495 = 9 \times 55; 2345 - s(2345) = 2331 = 9 \times 259$.
(a) What can we say about the number $n - s(n)$?
(b) Using the previous item, calculate $s... | (a) Observe these two examples:
$$
\underbrace{2000}_{2 \cdot 10^3}-\underbrace{s(2000)}_{2}=1998, \underbrace{60000}_{6 \cdot 10^4}-\underbrace{s(60000)}_{6}=59994
$$
From these, it is easy to understand that if $a$ is a digit between 1 and 9, then $s\left(a \cdot 10^{k}\right)=a$.
Thus, we have:
$$
a \cdot 10^{k}... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A number is framed when, by adding it to the number obtained by reversing the order of its digits, the result is a perfect square. For example, 164 and 461 are framed, since 164+461 = $625=25^{2}$. How many framed numbers are there between 10 and 100?
A) 5
B) 6
C) 8
D) 9
E) 10 | Solution
## ALTERNATIVE C
Let $n$ be a number between 10 and 100, $a$ its tens digit, and $b$ its units digit; note that $1 \leq a \leq 9$ and $0 \leq b \leq 9$. Then $n=10a+b$ and the number obtained by reversing the digits of $n$ is $10b+a$. Since $n$ is a framed number, we have that $(10a+b) + (10b+a) = 11a + 11b ... | 8 | Number Theory | MCQ | Yes | Yes | olympiads | false |
In the figure, triangles $A B C$ and $D E F$ are equilateral with sides $14 \mathrm{~cm}$ and $13 \mathrm{~cm}$, respectively, and sides $B C$ and $E F$ are parallel.
a) Calculate the measu... | Solution
a) Since $BC$ and $EF$ are parallel, the angles $EUT$ and $ACB$ are alternate interior angles, hence $E \hat{U} T = A \hat{C} B = 60^{\circ}$. b) From item a) we can conclude that all triangles in the figure are equilateral. Thus, we have $QP = FP$, $UT = UE$, $TS = CS$, and $RQ = RB$. Therefore, the perimete... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A board of size $2013 \times 5$ (i.e., with 2013 rows and 5 columns) must be painted with the colors $A, B, C, D$. Some squares in the first row have already been painted, as shown in the figure below (the squares not represented in the figure have not been painted yet). To continue painting the board, we must follow t... | Solution
a) The house that is missing painting in the first row has neighbors painted with $B$ and $C$. Therefore, it can only be painted with $A$ or $D$. Let's analyze the case where it is painted with $A$, as shown in the figure below.
Using these polygons, we form a rectangle and a trapezoid as shown in the following figure:
^{2}+2 \times 2^{4} \times 2^{6}+\left(2^{\frac{n}{2}}\right)^{2}
$$
Logo, se $n=12$, temos
$$
2^{8}+2^{11}+2^{12}=\left(2^{4}+2^{6}\right)^{2}
$$
Logo $n=12$ é uma solução.
Solução Geral: $\operatorname{Se} 2^{8}+2^{11}+2^{n}=k^{2}$, então:
$$
\begin{aligned}
2... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The reverse of a two-digit integer is the number obtained by reversing the order of its digits. For example, 34 is the reverse of 43. How many numbers exist such that when added to their reverse, they give a perfect square? | Let's denote by $a b \mathrm{e}$ ba the number and its reverse. We have that
$$
a b+b a=10 a+b+10 b+a=11(a+b)
$$
On the other hand, $a \leq 9$ and $b \leq 9$, so, $a+b \leq 18$. Since 11 is a prime number and $a+b \leq 18$, for $11(a+b)$ to be a perfect square, we can only have $a+b=11$.
Thus, we have 8 numbers sati... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a warehouse, a dozen eggs and 10 apples had the same price. After a week, the price of eggs fell by $2 \%$ and the price of apples rose by 10\%. How much more will be spent on the purchase of a dozen eggs and 10 apples?
(a) $2 \%$
(b) $4 \%$
(c) $10 \%$
(d) $12 \%$
(e) $12.2 \%$ | We can assume that the initial price of a dozen eggs is $R \$ 1.00$, so 10 apples also cost $R \$ 1.00$. Since the price of eggs has dropped by $2 \%$, the new price of a dozen eggs is $R \$ 0.98$. The price of apples has increased by $10 \%$, so the new price of 10 apples is $R \$ 1.10$. Thus, before it cost $R \$ 2.0... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
The number
$$
A=(\sqrt{6}+\sqrt{2})(\sqrt{3}-2) \sqrt{\sqrt{3}+2}
$$
is equal to:
(a) $-\sqrt{3}$
(b) $-\sqrt{2}$
(c) -2
(d) 1
(e) 2 | (c) Like
$$
\begin{aligned}
A^{2} & =[(\sqrt{6}+\sqrt{2})(\sqrt{3}-2) \sqrt{\sqrt{3}+2}]^{2} \\
& =(\sqrt{6}+\sqrt{2})^{2}(\sqrt{3}-2)^{2}(\sqrt{\sqrt{3}+2})^{2} \\
& =(\sqrt{6}+\sqrt{2})^{2}(\sqrt{3}-2)^{2}(\sqrt{3}+2) \\
& =(\sqrt{6}+\sqrt{2})^{2}(\sqrt{3}-2)[(\sqrt{3}-2)(\sqrt{3}+2)] \\
& =(6+2 \sqrt{12}+2)(\sqrt{3... | -2 | Algebra | MCQ | Yes | Yes | olympiads | false |
Five points lie on the same line. When we list the ten distances between any two of these points, from smallest to largest, we get $2,4,5,7,8, k, 13,15,17,19$. What is the value of $k$? | Solution 1: - This solution is a bit difficult to write because it is based on "trial and error." We start by drawing a number line and placing the points 0 and 19. Since the first distance is 2, we mark our first three points:
 1
(B) 2
(C) 4
(D) 6
(E) 8 | The correct answer is (B).
Of the total number of students in this class, $60 \%$ went to do community work, that is, $0.6 \times 40=24$. The minimum number of female students who participated in this work is obtained when the number of male students who participated is maximum, that is, when 22 male students are invo... | 2 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
In the luminous mechanism of the figure, each of the eight buttons can light up in green or blue. The mechanism works as follows: when turned on, all buttons light up blue, and if we press a button, that button and its neighbors switch colors. If we turn on the mechanism and successively press buttons 1, 3, and 5, how ... | The correct answer is (C).
The table shows the color of each button at each step.
| | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ | $\mathbf{5}$ | $\mathbf{6}$ | $\mathbf{7}$ | $\mathbf{8}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| start | blue | blue | blue | blue |... | 5 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
Is the number $3^{444}+4^{333}$ divisible by 5?
## List 5 | There is a pattern for the unit digit of a power of 3: it has a period of 4, as it repeats every 4 times.
$$
\begin{aligned}
& 3 \\
& 3^{2}=9 \\
& 3^{3}=27 \\
& 3^{4}=81 \\
& 3^{5}=243 \\
& 3^{6}=\ldots 9 \\
& 3^{7}=\ldots 7 \\
& 3^{8}=\ldots 1
\end{aligned}
$$
Since 444 is a multiple of 4, the unit digit of $3^{444}... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Fill in the squares with the numbers $1,2,3,5,6$.
$$
(\square+\square-\square) \times \square \div \square=4
$$ | The operation is equivalent to
$$
(\square+\square-\square) \times \square=4 \times
$$
Therefore, the left side of the equality is a multiple of 4, so the only possibilities are:
$$
(\square+\square-\square) \times \square=4 \times \square \quad \text { or } \quad(\square+\square-\square) \times \square=4 \times \sq... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Writing successively the natural numbers, we obtain the sequence:
$$
12345678910111213141516171819202122 \ldots
$$
Which digit is in the $2009^{th}$ position of this sequence? | - the numbers 1 to 9 occupy 9 positions;
- the numbers 10 to 99 occupy $2 \times 90=180$ positions;
- the numbers 100 to 199 occupy $3 \times 100=300$ positions; the numbers 200 to 299 occupy $3 \times 100=300$ positions; the numbers 300 to 399 occupy $3 \times 100=300$ positions; etc.
$$
\underbrace{100, \ldots 199}_... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Elisa has 24 science books and others of mathematics and literature. If Elisa had one more mathematics book, then $\frac{1}{9}$ of her books would be mathematics and one quarter literature. If Elisa has fewer than 100 books, how many mathematics books does she have?
## List 10 | Let $N$ be the total number of books Elisa has. Since $N+1$ is a multiple of 9 and 4, it follows that it is a multiple of 36. Therefore, $N+1$ is 36 or 72, as Elisa has fewer than 100 books. If $N=35$, then the number of math books is $36 \div 9-1=3$ and the number of literature books is $36 \div 4=9$. Thus, Elisa woul... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$O$ double of a number divided by 5 leaves a remainder of 1. What is the remainder of the division of this number by 5?
Translating the text as requested, while preserving the original line breaks and format. | According to the data from the problem, double the number is a multiple of 5 plus 1. Since multiples of 5 end in 0 or 5, the double ends in 1 or 6. But the double is an even number, so it ends in 6. Thus, the number ends in 3 or 8, therefore when divided by 5, it leaves a remainder of 3. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
At a circular table, 5 people are sitting: Arnaldo, Bernaldo, Cernaldo, Dernaldo, and Ernaldo, each in a chair. Analyzing in a clockwise direction, we have:
I. Between Arnaldo and Bernaldo there is 1 empty chair;
II. Between Bernaldo and Cernaldo are 5 chairs;
III. Between Dernaldo and Ernaldo are 4 chairs, almost a... | Solution
Let's position Arnaldo in the chair we will call 1, and by information $I$, Bernaldo should sit in chair 3, and consequently, by information $II$, Cernaldo should sit in chair 9. Since there are 6 chairs between Dernaldo and Ernaldo and 2 chairs between Dernaldo and Cernaldo, Cernaldo is between Dernaldo and ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Mario bought some sweets at the market, where 3 chocolates cost the same as 2 popsicles and 2 lollipops cost the same as 5 chocolates.
a) Mario decided to return to the market with money to buy exactly 3 lollipops but decided to buy popsicles instead. How many popsicles did he manage to buy?
b) If he had used the mon... | Solution
a) 15 chocolates cost the same as 10 popsicles and the same as 6 lollipops. Therefore, 10 popsicles are worth the same as 6 lollipops, and consequently, 5 popsicles the same as 3 lollipops. Thus, with the money for 3 lollipops, Mario can buy 5 popsicles.
b) From the previous item, we saw that the largest num... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Jonas wrote a sequence with the positive multiples of 13 in ascending order.
## $1326395265 \ldots$
a) What is the $2,019^{\text{th}}$ digit of Jonas's sequence?
b) Will the number 2,019 appear in this sequence? | Solution
a) There are 7 multiples of 13 with 2 digits (14 digits); with 3 digits, there are 69 multiples of $13(3 \cdot 69=207$ digits). Already, there are $14+207=221$ digits, so $2,019-221=1,798$ are still needed. Dividing 1,798 by 4, we get 449 and a remainder of 2, that is, the first multiple of 13 with 4 digits i... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The owner of a clothing factory is a math enthusiast and organizes his products into packages whose number of items is a prime number. For example, the green-colored pieces, he organizes into packages of 2 pieces each; the blue-colored pieces, into packages of 3 pieces each; the pink-colored pieces, into packages of 5 ... | Solution
The colors green, blue, and pink have 2, 3, and 5 pieces per pack, respectively. We can take various quantities of packs of each color, including not taking any, as long as the total number of pieces is exactly 20. Let's create a table to list the possibilities:
| Green | Blue | Pink |
| :---: | :---: | :---... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The sums of the three columns and the three rows of the table are equal.
| 4 | 9 | 2 |
| :--- | :--- | :--- |
| 8 | 1 | 6 |
| 3 | 5 | 7 |
What is the minimum number of cells in the table that need to be changed so that all the new six sums are different from each other? | Solution
If three or fewer cells are changed, there will either be two rows without any changed cells or one cell is the only one changed in its row and column. In the first case, these two rows without any changed cells have the same sum. In the second case, if only one cell is the only one changed in its row and col... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In a class with 35 students, a survey was conducted on preferences for mathematics and literature, and it was found that:
- 7 men like mathematics;
- 6 men like literature;
- 5 men and 8 women said they do not like either;
- there are 16 men in the class;
- 5 students like both; and
- 11 students like only mathematics... | Solution
Let $H$ be the set of men and $U$ the total set of people, so $U-H$ is the set of women. In addition, consider the sets Mat and $L$ of people who like mathematics and literature, respectively. If $x$ represents the number of men who like both mathematics and literature and $y$ the number of women who like onl... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the triangle $ABC$ below, there is a point $D$ on its side $AB$, such that $AB=CD$, $\widehat{ABC}=100^{\circ}$, and $\widehat{DCB}=40^{\circ}$.
a) What is the measure of the angle $\wide... | Solution
a) Analyzing the sum of the internal angles of the triangle $\triangle B D C$, we have $100^{\circ}+40^{\circ}+\angle B D C=180^{\circ}$, that is, $\angle B D C=40^{\circ}$.
b) Due to the previous item, $\triangle B D C$ is isosceles with base $C D$ and thus $B C=B D$. Construct the auxiliary triangle $\tria... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
a) Consider a prime $p$ that divides $10^{n}+1$ for some positive integer $n$. For example, $p=7$ divides $10^{3}+1$. By analyzing the principal period of the decimal representation of $\frac{1}{p}$, verify that the number of times the digit $i$ appears is equal to the number of times the digit $9-i$ appears for each $... | Solution
a) We can write $10^{n}+1=p \cdot a$ where $a$ is a number with no more than $n$ digits in base 10, say $a=a_{1} a_{2} \ldots a_{n}$. We mean by this that each number $a_{i}$ is one of the digits of $a$. Even if it has strictly fewer than $n$ digits, we can place some $a_{i}$'s on the left as being 0. We have... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the figure below, the sides of rectangle $A B C D$ have been extended so that $E B=2 A B$, $A H=3 A D$, $D G=2 D C$, and $F C=3 B C$. Find the ratio between the areas of quadrilateral $E H G F$ and rectangle $A B C D$.
 / 2 \\
& =(A B \cdot 3 A D) / 2 \\
& =3 S / 2 \\
A_{D H G} & =(D H \cdot D G) / 2 \\
& =(2 A D \cdot 2 D C) / 2 \\
& =2 S \\
A_{F C G} & =(F... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
During the FIFA World Cup, several mathematicians were called upon to speak about the teams' chances of qualification. In the group stage, each group is formed by four teams, and each team faces each of the other teams exactly once. In case of a win, a team earns 3 points; in case of a draw, 1 point; and in case of a l... | Solution
The minimum number of points is 7 points. First, let's see that with 6 points a team may not qualify. Let $A, B, C$ and $D$ be the teams in a certain group and consider the following table of results:
| Winner | Result | Loser |
| :---: | :---: | :---: |
| $A$ | $2 \times 0$ | $D$ |
| $B$ | $1 \times 0$ | $D... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A cricket can jump two distances: 9 and 8 meters. It competes in a 100-meter race to the edge of a cliff. How many jumps must the cricket make to reach the end of the race without going past the finish line and falling off the cliff?
# | Solution
First solution: Suppose the cricket only jumps 9 meters. On its twelfth jump, it would fall off the cliff, since $9 \cdot 12 = 108 \, \text{m}$. Since it can also jump 8 meters, it is enough to "exchange" 8 jumps of 9 meters for jumps of 8 meters. Thus, we have 4 jumps of 9 meters and 8 jumps of 8 meters, for... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The king is a chess piece that can move only one square vertically, one horizontally, or one diagonally. We say that a king attacks a square if it can occupy it with a single move. For example, a king placed in the central squares of a $6 \times 6$ board attacks 8 squares, a king placed on the side squares attacks 5 sq... | Solution
a) Divide the $6 \times 6$ board into 4 $3 \times 3$ boards. If one of these four regions does not have a king, the central cell of that region will not be occupied or attacked by any king. Therefore, at least 4 kings are needed. If we place a king in the central cell of each $3 \times 3$ board, then all cell... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find all values of $x$ satisfying
$$
\frac{x+\sqrt{x+1}}{x-\sqrt{x+1}}=\frac{11}{5}
$$ | Solution
The equation can be rewritten as
$$
\begin{aligned}
\frac{x+\sqrt{x+1}}{x-\sqrt{x+1}} & =\frac{11}{5} \\
5 x+5 \sqrt{x+1} & =11 x-11 \sqrt{x+1} \\
16 \sqrt{x+1} & =6 x \\
8 \sqrt{x+1} & =3 x
\end{aligned}
$$
Squaring both sides of the last equation, we get
$$
\begin{aligned}
9 x^{2} & =64(x+1) \\
9 x^{2}-6... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Consider a square $ABCD$ with center $O$. Let $E, F, G$, and $H$ be points on the interiors of sides $AB, BC, CD$, and $DA$, respectively, such that $AE = BF = CG = DH$. It is known that $OA$ intersects $HE$ at point $X$, $OB$ intersects $EF$ at point $Y$, $OC$ intersects $FG$ at point $Z$, and $OD$ intersects $GH$ at ... | Solution
a) Let $x$ and $y$ be the lengths of $A E$ and $A H$, respectively. Given that $A H = E B$, $A E = B F$, and $\angle H A E = \angle E B F$, it follows that triangles $A E H$ and $E ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a school, $n$ clubs must be formed, with $n \geq 3$ and each with 3 members, such that for each pair of clubs there is exactly one student who is a member of both.
a) Give an example of a distribution of 7 clubs that satisfy the mentioned conditions.
b) Verify that if a student belongs to 4 clubs, then he must bel... | Solution
a) Naming the students as $A, B, C, D, E$ and $F$, the following sets represent 7 clubs that meet the conditions stated:
$$
\{A, B, C\},\{A, D, E\},\{A, F, G\},\{B, E, F\},\{B, D, G\},\{C, D, F\},\{C, E, G\}
$$
b) Suppose student $A$ belongs to the following 4 clubs:
$$
\{A, B, C\},\{A, D, E\},\{A, F, G\},... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $A B C$ be a right triangle with $\angle B A C=90^{\circ}$ and $I$ the point of intersection of its angle bisectors. A line through $I$ intersects the sides $A B$ and $A C$ at $P$ and $Q$, respectively. The distance from $I$ to the side $B C$ is $1 \mathrm{~cm}$.
a) Find the value of $P M \cdot N Q$.
b) Determine... | Solution
a) If $\angle A P Q=\alpha$, it follows that $\angle M I P=90^{\circ}-\alpha$ and
$$
\angle N I Q=180^{\circ}-\angle M I N-\angle M I P=\alpha
$$
Therefore, triangles $I M P$ and $N I Q$ are similar, and thus
$$
\frac{I M}{N Q}=\frac{P M}{I N} \Rightarrow P M \cdot N Q=I M \cdot I N=1
$$
b) Let $x=M P$ an... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be an odd function, that is, a function that satisfies $-f(x)=f(-x)$ for all $x \in \mathbb{R}$. Suppose that $f(x+5)=f(x)$ for all $x \in \mathbb{R}$ and that $f(1 / 3)=1$. Determine the value of the sum:
$$
f(16 / 3)+f(29 / 3)+f(12)+f(-7)
$$
# | Solution
We have
$$
\begin{aligned}
f\left(\frac{1}{3}\right) & =1 \\
f\left(\frac{1}{3}+5\right) & =1 \\
f\left(\frac{16}{3}\right) & =1
\end{aligned}
$$
Since $f$ is an odd function, i.e., $f(-x)=-f(x)$, it follows that
$$
\begin{aligned}
f\left(-\frac{1}{3}\right) & =-1 \\
f\left(-\frac{1}{3}+5+5\right) & =-1 \\... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the following figure, $\angle C A B=2 \cdot \angle C B A, A D$ is an altitude and $M$ is the midpoint of $A B$. If $A C=2 \text{ cm}$, find the length of the segment $D M$.
# | Solution
Let $K$ be the midpoint of $A C$. Since the circle with center $K$ and diameter $A C$ passes through $D$, it follows that $C K=A K=D K=1 \mathrm{~cm}$. From this last equality, it follows that triangle $A D K$ is isosceles and thus $\angle K D A=\angle K A D$. The segment $K M$ is the midline of triangle $A B... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $A B C D$ be a rectangle such that $A B=\sqrt{2} B C$. Let $E$ be a point on the semicircle with diameter $A B$, as indicated in the following figure. Let $K$ and $L$ be the intersections of $A B$ with $E D$ and $E C$, respectively. If $A K=2 \mathrm{~cm}$ and $B L=9 \mathrm{~cm}$, calculate, in $\mathrm{cm}$, the ... | Solution
Let $x$ and $y$ be the lengths of the orthogonal projections of segments $E K$ and $E L$ onto segment $A B$, and $P$ be the orthogonal projection of $E$ onto $A B$. Also, let $h = E P$. By the similarity of triangles, we have
$$
\frac{2}{x} = \frac{B C}{h} \text{ and } \frac{9}{y} = \frac{B C}{h}
$$
Therefo... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On planet $X$, there are 100 alien countries in conflict with each other. To prevent a world war, these countries organize themselves into military alliance groups for mutual protection. We know that the alliances follow these rules:
1) No alliance contains more than 50 countries.
2) Any two countries belong to at lea... | Solution
a) It is not possible. Suppose that country $A$ belongs to at most two alliances. In this case, since each alliance has at most 50 countries, country $A$ is a member of the same alliance with at most $49+49=98$ countries. Since there are 99 countries distinct from $A$, at least one of them will not be in an a... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the drawing to the side, triangle $ABC$ is equilateral and $BD = CE = AF = \frac{AB}{3}$. The ratio $\frac{EG}{GD}$ can be written in the form $\frac{m}{n}, \operatorname{gcd}(m, n)=1$. What is the value of $m+n$?
}{\frac{1}{2} B C \cdot B F \cdot \sin(\angle C B F)}$. Since $\angle E B G=\angle C B F$, we have
$$
\frac{[E B G]}{[B C F]}=\frac{E B \cdot B G}{B C \cdot B F}
$$
Si... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
What is the greatest result we can find when we divide a two-digit number by the sum of its digits? | We are looking for the maximum value of $\frac{10 a+b}{a+b}$, where $a$ and $b$ represent digits, with at least one different from 0. We have
$$
\frac{10 a+b}{a+b}=\frac{10 a+10 b-9 b}{a+b}=\frac{10 a+10 b}{a+b}-\frac{9 b}{a+b}=10-\frac{9 b}{a+b} \leq 10
$$
Therefore, if we can find $a$ and $b$ such that $\frac{10 a+... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Pedro decided to take all his children, boys and girls, for ice cream at the Ice Cream Math parlor. At the parlor, there are 12 different ice cream flavors, and each child ordered a combo with 3 scoops of ice cream. After leaving the parlor, Pedro realized that exactly two scoops of each available flavor were ordered i... | Solution
a) Let $n$ be the number of Pedro's children. In total, $3 n$ scoops of ice cream were ordered. Since each of the 12 flavors was ordered twice, we have $3 n=2 \cdot 12$, that is, $n=8$. Therefore, Pedro has 8 children.
b) Let $x$ be the number of boys and $y$ be the number of girls. From the previous item, w... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $a$ and $b$ be any positive real numbers. Determine the value of the expression
$$
\frac{\sqrt{\frac{a b}{2}}+\sqrt{8}}{\sqrt{\frac{a b+16}{8}+\sqrt{a b}}}
$$ | Solution
Let $x=\sqrt{\frac{a b}{2}}+\sqrt{8}$. Then:
$$
\begin{aligned}
x^{2} & =\frac{a b}{2}+4 \sqrt{a b}+8 \\
& =4\left(\frac{a b+16}{8}+\sqrt{a b}\right) \\
& =4\left(\sqrt{\frac{a b+16}{8}+\sqrt{a b}}\right)^{2}
\end{aligned}
$$
Thus, the value of the desired expression is
$$
\begin{aligned}
\frac{\sqrt{\frac... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $D$ be a point on side $A B$ of triangle $\triangle A B C$ and $F$ the intersection of $C D$ and the median $A M$. If $A F=A D$, find the ratio between $B D$ and $F M$.
 | Solution
Let $E$ be the intersection point of the line parallel to side $AB$, passing through point $M$, with segment $CD$. Since $ME \| AB$, it follows that $\angle MED = \angle EDA$. Moreover, since $\angle AFD = \angle EFM$ and triangle $\triangle AFD$ is isosceles, we can conclude that $\triangle EFM$ is also isos... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Augusto has a wire that is $10 \mathrm{~m}$ long. He makes a cut at a point on the wire, obtaining two pieces. One piece has a length of $x$ and the other has a length of $10-x$ as shown in the figure below:
 A piece of rope has length $x$ and another piece of rope has length $10-x$. Since a square has four sides of equal length, one square will have a side length of $\frac{x}{4}$ and the other square will have a side length of $\frac... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Julian trains on a $3 \mathrm{~km}$ track. He walks the first kilometer, runs the second, and cycles the third. If he had cycled the entire track, it would have taken him 10 minutes less. Julian runs at twice the speed he walks, and cycles at three times the speed he walks. How long does Julian take to run $1 \mathrm{~... | Solution
Let $t$ be the time, in minutes, that Julian takes to cycle one kilometer. Since he cycles at three times the speed he walks, he walks one kilometer in $3 t$ minutes, and since he runs at twice the speed he walks, he runs one kilometer in $3 t / 2$ minutes. Thus, he took
$$
t+3 t+\frac{3 t}{2}=\frac{11 t}{2}... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Carla wrote on the blackboard the integers from 1 to 21. Diana wants to erase some of them so that the product of the remaining numbers is a perfect square.
a) Show that Diana must necessarily erase the numbers 11, 13, 17, and 19 to achieve her goal.
b) What is the minimum number of numbers that Diana must erase to a... | Solution
a) If Diana decides not to erase the number 11, then the product of the remaining numbers will be of the form $P=11 \times A$. Since 11 is the only multiple of 11 among the numbers written by Carla, $A$ is the product of numbers not divisible by 11, and thus $A$ is not a multiple of 11. Therefore, $P$ would b... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The squares of the natural numbers from 1 to 99 were written one after another, forming the number 14916253649 ... What is the digit that occupies the $100^{th}$ position? (The positions are counted from left to right: the $1^{st}$ position is the 1, the $2^{nd}$ is the 4, etc.) | Separating the numbers whose squares have 1, 2, and 3 digits, we have:
1 digit: $1, 2, 3$
2 digits: $4, 5, 6, 7, 8, 9$
3 digits: $10, 11, 12, \ldots, 31$
Up to $31^{2}$, the sequence has $3+12+66=81$ digits.
$$
\underbrace{1^{2}, 2^{2}, 3^{2}}_{1 \times 3 \text { digits }}, \underbrace{4^{2}, \ldots, 9^{2}}_{2 \ti... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the figure, we have a target at which darts are thrown. The inner circle is worth 10 points, the intermediate ring is worth 5 points, the outer ring is worth 3 points, and the external area is worth 0 points. If a dart hits a line, the score will be the average of the points of the regions divided by the line. In ea... | Solution
a) Leo scored $10+5+\frac{5+3}{2}+\frac{3+0}{2}=20.5$ points.
b) The minimum score is 0 and the maximum is 40. Therefore, it is enough to verify for each of the natural numbers in this interval the possibility of being a possible sum. Let's construct a table, showing an example for each possible sum.
| $1^{... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a fruit shop, Jaime noticed that an orange costs the same as half an apple plus half a real, and he also noticed that a third of an apple costs the same as a quarter of an orange plus half a real. With the value of 5 oranges plus 5 reals, how many apples can Jaime buy?
# | Solution
If an orange costs the same as half an apple plus half a real, then, 2 oranges cost the same as 1 apple plus 1 real; we also have that if a third of an apple costs the same as a quarter of an orange plus half a real, then, multiplying each part by 12, we have that 4 apples cost the same as 3 oranges plus 6 re... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Joseane's calculator has gone crazy: for each digit she presses, its double appears on the screen. The addition, subtraction, multiplication, and division operation keys work normally and cannot be pressed twice in a row. For example, a permitted sequence of operations is to write $2 \rightarrow \times \rightarrow 3$, ... | Solũ̧̧̃o
a) By pressing the digits, the numbers that can appear on the screen are $0, 2, 4, 6, 8$, 10, 12, 14, 16, and 18. One way to make 80 appear on the display is to press the sequence $4 \rightarrow \times \rightarrow 5$, which results in $8 \times 10=80$.
b) The sequence should be: digit - operation - digit - o... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A school with 862 students will participate in a scavenger hunt, whose rules are:
I) The number of registrants must be a number between $\frac{2}{3}$ and $\frac{7}{9}$ of the total number of students in the school.
II) Since the students will be divided into groups of 11, the number of registrants must be a multiple ... | Solution
Since $\frac{2}{3} \cdot 862=574, \overline{6}$ and $\frac{7}{9} \cdot 862=670, \overline{4}$, the number of students must be a multiple of 11 in the interval $[575,670]$. The smallest multiple in the interval is $53 \cdot 11=583$ and the largest is $60 \cdot 11=660$. Therefore, there are $60-52=8$ possible d... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On a table are 10 coins, all with "heads" facing up. A move consists of flipping exactly 4 coins.
a) What is the minimum number of moves required for all to be "tails" facing up?
b) If there were 11 coins, would it be possible to have all of them with tails facing up? | Solution
a) With 2 moves it is not possible, as we can only change $2 \cdot 4=8$ coins and we need to change 10. We will now show that it is possible with 3 moves, analyzing the sequence in the figure, where $k$ represents heads and $C$, tails.
 to paint the circles (each one in a single color).
a) In how many different ways c... | Solution
a) There are 3 color possibilities for each of the circles. Therefore, we can paint it in $3 \cdot 3 \cdot 3 \cdot 3=81$ ways.
b) Without considering the restriction (using all 3 colors necessarily), the total number of possibilities, according to the reasoning from the previous item, is $3^{6}=729$. Now let... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
How many positive integers $n$ exist such that $\frac{2 n^{2}+4 n+18}{3 n+3}$ is an integer? | As
$$
\frac{2 n^{2}+4 n+18}{3 n+3}=\frac{2}{3} \frac{\left(n^{2}+2 n+1\right)+8}{n+1}=\frac{1}{3}\left(2 n+2+\frac{16}{n+1}\right)
$$
it follows that $n+1$ must divide 16. Thus, $n$ must belong to the set $\{1,3,7,15\}$. In each of these cases, we have
| $n$ | $\frac{2 n^{2}+4 n+18}{3 n+3}$ |
| :---: | :---: |
| 1 |... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The five cards below are on a table, and each one has a number on one side and a letter on the other. Simone must decide whether the following statement is true: "If a card has a vowel on one side, then it has an even number on the other." What is the minimum number of cards she needs to turn over to decide correctly?
... | 
She does not need to turn over the card with the number 2, because whether it is a vowel or a consonant, it meets the condition in the same way. She also does not need to turn over the card w... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A number is even and has 10 digits, and the sum of these digits is 89. What is the unit digit of this number?
A) 0
B) 2
C) 4
D) 6
E) 8 | Solution
## Alternative E
The greatest possible sum of ten digits is $10 \times 9=90$, which occurs when we have ten digits 9. To get a sum of 89, we just need to reduce one unit from one of the digits, that is, replace a 9 with an 8. Therefore, the number has nine digits 9 and one digit 8. Since it is even, its unit... | 8 | Number Theory | MCQ | Yes | Yes | olympiads | false |
By placing addition signs between some of the digits of the number 123456789, we can obtain various sums. For example, we can obtain 279 with four addition signs: $123+4+56+7+89=279$. How many addition signs are needed to obtain the number 54 in this way?
A) 4
B) 5
C) 6
D) 7
E) 8 | Solution
## ALTERNATIVE D
Since we want to obtain the sum of 54, we must place addition signs between all the digits starting from 5, that is, $1 ? 2 ? 3 ? 4 ? 5 + \underbrace{6+7+8+9}_{30} = 54$. Therefore, we need $1 ? 2 ? 3 ? 4 ? 5 = 24$.
Using the same argument as before, we see that this can only be done as $12... | 7 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
Using matchsticks, we form digits as shown in the figure. In this way, to write the number 188, we use 16 matchsticks.
César wrote the largest number possible using exactly 13 matchsticks. ... | Solution
## ALTERNATIVE B
A number with a certain number of digits, where the first digit from the left is not zero, is always greater than any number that has one less digit. For example, 1000 (with 4 digits) is greater than 999 (which has only 3 digits).
Thus, with exactly 13 sticks, we should form a number that h... | 9 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
A farmer asked his son: How many feet can I count when I am milking a cow? The boy replied: There are 6, 4 from the cow and 2 from you. The father then said: Actually, there are 9, because you forgot to count the 3 from the stool I sit on. Following this, the father proposed another problem to his son: In a corral, the... | Solution
## ALTERNATIVE C
The table below represents all possibilities for the number of heads to be 5 (we remember that stools do not have heads and there is at least one person and one cow).
| Heads | | Feet | Feet of stools |
| :---: | :---: | :---: | :---: |
| Cows | People | | (22 - feet, cows and people) |
|... | 3 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
Pedrinho wrote all the integers between 100 and 999 whose sum of digits is 12. For example, the numbers 129 and 750 appear among the numbers written.
a) How many of the written numbers have only two equal digits?
b) How many of the written numbers are formed only by odd digits? | Solution
a) The digit 1 cannot be repeated because it is not possible to write 12 as a sum of the form $1+1+x$ where $x$ is a digit; indeed, since $x$ is at most 9, this sum will be at most 11. The digit 4 also cannot be repeated, as in this case the number would have to be 444, which has three identical digits and do... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Four teams competed in a football tournament where each played once against each of the others. When a match ended in a draw, each team received one point; otherwise, the winner received three points and the loser, zero. The table shows the final score of the tournament. How many were the draws?
| Team | Points |
| :-... | Solution
## ALTERNATIVE D
$1^{st}$ solution: Each team played three times. With 5 points, Cruzinthians could only have won one match and drawn two, because if they had won two matches, they would have at least 6 points, and if they hadn't won any, they would have a maximum of 3 points. Greminense did not win any matc... | 5 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
Dona Benta divided the Sítio do Picapau Amarelo among six characters, keeping a part of the Sítio as a forest reserve. The division is indicated in the figure, where the area of each character is given in hectares and the shaded area is the forest reserve. The Sítio has a rectangular shape and $AB$ is a diagonal.
 A rectangle is divided into two regions of the same area by its diagonal. Therefore, the lands of Quindim, Visconde de Sabugosa, and Cuca together have an area equal to half the area of the Sítio. The area of these lands, in hectares, totals $4+7+12=23$. The other half of the Sítio has the same area and is... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Mário built a cube with twelve equal sticks and wants to paint them so that no vertex has sticks of the same color meeting there. What is the minimum number of colors he needs to use?
A) 2
B) 3
C) 4
D) 6
E) 8 | Solution
## ALTERNATIVE B
Each vertex is the endpoint of three edges and, therefore, at least three different colors are needed. On the other hand, three different colors are enough; we can see this in the figure, where three different colors are indicated by solid, dashed, and dotted lines.
Without separating the small cubes, Aline decided to paint all the fac... | Solution
a) On each face of the cube, the small cubes that end up with one face painted red are the four that are in the center, that is, those that do not have any edge contained in an edge of the large cube. The figure below illustrates which these cubes are.
From the drawing, we see that $\measuredangle B A O=20^{\circ}$, since the bounces are perfect.
Let's recall that ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A symmetric ring with $m$ regular polygons, each with $n$ sides, is formed according to the rules:
i) each polygon in the ring meets two others;
ii) two adjacent polygons share only one side;
iii) the perimeter of the region inside the ring consists of exactly two sides of each polygon.
The example in the following... | Solution
Let $\alpha=\frac{360^{\circ}}{n}$ be the measure of each exterior angle of a regular polygon $A B C D E \ldots$. Suppose that the path $B C D$ is part of the perimeter of some internal region of a ring. If $O$ is the intersection of the extensions of $A B$ and $D E$, by symmetry, it is the center of the inte... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The figure below shows a segment $AB$, its midpoint $C$, and the semicircles with diameters $AB$ and $AC$. A circle with center $P$ is tangent to both semicircles and also to the segment $AB$. Given $AB=8 \text{ cm}$, and $O$, the midpoint of $AC$, the questions are:
 Let $x$ be the radius of the circle centered at $P$. We draw $OP$, which passes through the point of tangency $D$; $CP$, which passes through the point of tangency $F$; and $PE$, perpendicular to $AB$ (and a radius of the circle highlighted in the figure). We have $CP=4-x$ and $OP=2+x$. The perimeter of tr... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Determine the minimum term of the sequence
$$
\sqrt{\frac{7}{6}}+\sqrt{\frac{96}{7}}, \sqrt{\frac{8}{6}}+\sqrt{\frac{96}{8}}, \sqrt{\frac{9}{6}}+\sqrt{\frac{96}{9}}, \ldots, \sqrt{\frac{95}{6}}+\sqrt{\frac{96}{95}}
$$ | Solution
Remember that $(x-y)^{2} \geq 0$ for all real numbers $x$ and $y$. Thus, we can rewrite the inequality as $\frac{x^{2}+y^{2}}{2} \geq x y$, and by substituting $x=\sqrt{a}$ and $y=\sqrt{b}$, with $a$ and $b$ being non-negative real numbers, we have $\frac{a+b}{2} \geq \sqrt{a b}$. Now, observe that all terms ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Some students from a school were divided into teams satisfying the following conditions:
i) Any 2 different teams have exactly 2 members in common.
ii) Every team has exactly 4 members.
iii) For any 2 students, there is a team of which neither is a member.
a) Explain why any pair of students can participate in at m... | Solution
a) Suppose there are students $A$ and $B$ participating in 4 teams called $E_{1}, E_{2}$, $E_{3}$, and $E_{4}$:
$$
\begin{aligned}
E_{1} & =\{A, B, C, D\} \\
E_{2} & =\{A, B, E, F\} \\
E_{3} & =\{A, B, G, H\} \\
E_{4} & =\{A, B, I, J\}
\end{aligned}
$$
By condition (iii), there exists a team $E_{5}$ that do... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the sum below, the same letters represent the same digits and different letters represent different digits.
$$
\begin{array}{ccc}
& & X \\
+ & & X \\
& Y & Y \\
\hline Z & Z & Z
\end{array}
$$
What digit does the letter $X$ represent?
# | Solution
Initially note that $X+X+Z \leq 2 \cdot 9+8=26$. Therefore, at most 2 units will be added to the tens place. Since $Y \neq Z$, when we add 1 or 2 units to $Y$, we should obtain the two-digit number $\overline{Z Z}$. As $Y+2 \leq 9+2=11$, the only possibility is to have $\overline{Z Z}=11$, thus, $Y=9$. For th... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A regular icosahedron is a geometric solid with 20 faces, which are equilateral triangles. By drawing the numbers from 1 to 20 on the faces of a regular icosahedron, it is transformed into a die. Luísa and Mathias invented a game in which each of them rolls the icosahedral die 5 times and records the numbers drawn in s... | Solution
a) 2,020,190,909. For this to happen, Luísa scored 20 in her 5 throws and Mathias scored 1 in his 5 throws, making the difference equal to 2,020,202,020 - 11,111 = 2,020,190,909.
b) No. The highest possible score would be by scoring 20 on the last throw, resulting in 141,181,920, which is less than 162,012,5... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
King Arthur had to fight the Three-Headed and Three-Tailed Dragon (known as $D T C T C$). His task was made easier when he managed to obtain a magical sword that could deliver the following strikes (one at a time):
A) cut off one head;
C) cut off one tail;
B) cut off two heads;
D) cut off two tails.
Moreover, the Fair... | Solution
a) Yes, it is enough for the King to use, for example, the sequence of strikes $D \rightarrow C \rightarrow C \rightarrow C$.
b) From a practical standpoint, the only strike that reduces the number of heads is $B$, but we need two of them to perform it (the same applies to tails and the strike $D$), unless s... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A tournament will take place with 100 competitors, all with different skill levels. The most skilled competitor always wins against the least skilled competitor. Each participant plays exactly twice, with two randomly drawn opponents (once against each). A competitor who wins two matches receives a medal. Determine the... | Solution
Let's denote that a player $A$ is more skilled than a player $B$ by $A \rightarrow B$. Suppose that
$$
L_{1} \rightarrow L_{2} \rightarrow L_{3} \rightarrow \ldots \rightarrow L_{100}
$$
are the 100 players in the tournament. Since $L_{1}$ is the most skilled of all, he will certainly win a medal by winning... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Consider the set $A=\{1,2,3, \ldots, 2011\}$. How many subsets of $A$ exist such that the sum of their elements is 2023060? | Observe that the sum $1+2+\cdots+2011=\frac{2011 \times 2012}{2}=$ 2023066. Therefore, to obtain a subset of $A$ that has a sum of its elements equal to 2023060, it suffices to remove from $A$ the elements whose sum is 6. The possible cases are:
- Subsets with one element: $\{6\}$.
- Subsets with two elements: $\{2,4\... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The 2,019 lockers of 2,019 students in a school are numbered with the squares of the first 2,019 positive natural numbers, that is, the first locker has the number $1^{2}=1$, the second locker has the number $2^{2}=4$, the third locker has the number $3^{2}=9$, and so on until the last locker which has the number $2,01... | Solution
a) Let's divide into groups by the number of digits:
I) 1 digit: 3 lockers $\left(1^{2}, 2^{2}, 3^{2}\right)$;
II) 2 digits: $9-3=6$ lockers ( $4^{2}$ to $9^{2}$ );
III) 3 digits: $31-9=22$ lockers $\left(10^{2}\right.$ to $\left.31^{2}\right)$;
IV) 4 digits: $99-31=68$ lockers $\left(32^{2}\right.$ to $\l... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
At the edge of a circular lake, there are stones numbered from 1 to 10, in a clockwise direction. Frog starts from stone 1 and jumps only on these 10 stones in a clockwise direction.
a) If Frog jumps 2 stones at a time, that is, from stone 1 to stone 3, from stone 3 to stone 5, and so on, on which stone will Frog be a... | Solution
a) After 5 jumps, Frog returns to stone 1 and starts the same sequence. Since 100 is a multiple of 5, on the $100^{\text{th}}$ jump, he goes to stone 1.
b) On the $1^{\text{st}}$ jump, he moves 1 stone; on the $2^{\text{nd}}$, 2 stones; on the $3^{\text{rd}}$, 3 stones, and so on until the last jump when he ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On a line $r$, points $A$ and $B$ are marked, and on a line $s$, parallel to $r$, points $C$ and $D$ are marked, such that $A B C D$ is a square. Point $E$ is also marked on the segment $C D$.
 Let $2 k$ be the area of square $A B C D$, then the area of triangle $A B E$ is equal to $k$ and the area of triangle $B C D$ is also equal to $k$, therefore, the ratio between the areas is 1.
b) If $\frac{A_{B F E}}{A_{D F E}}=2$, then $\frac{B F}{F D}=2$. If $r / / s$, then $\triangle A B F \sim \triang... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The table shows the final standings of the Quixajuba football championship. In this championship, each team played against each of the other teams four times. Each team earns 3 points for a win, 1 point for a draw, and no points for a loss.
| Team | Points |
| :---: | :---: |
| Bissetriz | 22 |
| Primo | 19 |
| Potênc... | (a) There are 6 possible matches between the four teams: (Bissetriz vs Primo), (Bissetriz vs Potência), (Bissetriz vs MDC), (Primo vs Potência), (Primo vs MDC), and (Potência vs MDC). Each of these matches occurred 4 times, so the number of matches is equal to $4 \times 6 = 24$.
(b) The maximum number of points in the... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Guilherme wrote 0 or 1 in each cell of a $4 \times 4$ board. He placed the numbers in such a way that the sum of the numbers in the neighboring cells of each cell on the board was equal to 1.
For example, in Figure 40.1, considering the cell marked with $\bullet$, the sum of the numbers in the shaded cells is equal to... | Each house can only have one neighbor with a number 1, and the other neighbors must be zeros, since the sum of the neighbors is 1.
Starting from the top left corner, we can assume without loss of generality that we fill the board as shown in Figure 40.2.
In the following steps, the filled cells are the neighbors of t... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A game starts with 7 coins aligned on a table, all with the head side facing up. To win the game, you need to flip some coins so that in the end, two adjacent coins are always with different sides facing up. The rule of the game is: in each move, you have to flip two adjacent coins. What is the minimum number of moves ... | If we assign the value of 1 to heads and -1 to tails and sum the results after each flip, the game starts with a sum of 7 and we need to reach alternating heads and tails, so the game ends at 1 or -1. We observe that at each step of the game, we have the following possibilities: we swap two heads for two tails and the ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The price of a kilogram of chicken was $R \$ 1.00$ in January 2000 and began to triple every 6 months. When will it reach $R \$ 81.00$?
(a) 1 year
(b) 2 years
(c) $21 / 2$ years
(d) 13 years
(e) $131 / 2$ years | As $81=3^{4}$, then the value of the franc has tripled 4 times, the number of months elapsed is $4 \times 6=24$ months, that is, 2 years, meaning that in January 2002 the chicken will reach the proposed price. The correct option is (b). | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
What is the unit digit of the product
$$
(5+1)\left(5^{3}+1\right)\left(5^{6}+1\right)\left(5^{12}+1\right) ?
$$
(a) 0
(b) 1
(c) 2
(d) 5
(e) 6 | The unit digit of any power of 5 is 5, so the unit digit of each factor of the product is $5+1=6$. But, $6 \times 6=36$, that is, the product of two numbers ending in 6 is also a number that ends in 6. Therefore, the unit digit of this product is 6. The correct option is (e). | 6 | Number Theory | MCQ | Yes | Yes | olympiads | false |
In a warehouse, a dozen eggs and 10 apples had the same price. After a week, the price of eggs dropped by $10 \%$ and the price of apples increased by $2 \%$. How much more will be spent on the purchase of a dozen eggs and 10 apples?
(a) $2 \%$
(b) $4 \%$
(c) $10 \%$
(d) $12 \%$
(e) $12.2 \%$
## Solutions to List 3 | Suppose, initially, that a dozen eggs cost $R \$ 1.00$. Thus, 10 apples also cost $R \$ 1.00$. Since the price of eggs increased by $10 \%$, the new price of eggs is $R \$ 1.10$. The price of apples decreased by $2 \%$, so the new price of apples is $R \$ 0.98$.
Thus, before it cost 2 reais to buy 1 dozen eggs and 10 ... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
What is the digit $a \mathrm{in}$
$$
a 000+a 998+a 999=22997 ?
$$ | Performing the addition
$$
\begin{array}{r}
a \\
a \\
a 998 \\
+\quad a 999 \\
\hline \square 997
\end{array}
$$
we find $\square 997=22997$, where $\square=a+a+a+1$.
Therefore, $22=a+a+a+1$. Thus, $a=7$. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A magic square is a square table in which the sum of the numbers in any row or column is constant. For example,
| 1 | 5 | 9 |
| :--- | :--- | :--- |
| 8 | 3 | 4 |
| 6 | 7 | 2 |
is a magic square, which uses the numbers from 1 to 9. As the reader can verify, the sum in any row or column is always equal to 15.
a) The ... | Solution
a) The sum of all odd numbers from 1 to 17 is 81. Since there are three columns in the square, and all columns (and rows) have the same sum, the sum in each column must be $81 / 3 = 27$. From this, we deduce that the missing number in the third column (the one with the numbers 13 and 3) is the number 11, beca... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Arnaldo, Bernaldo, Cernaldo, Dernaldo, and Ernaldo are students from different parts of Brazil who were chosen to represent their country in international olympiads. After several weeks of training, some friendships were formed. We then asked each of them how many friends they had made in the group. Arnaldo, Bernaldo, ... | Solution
Suppose Ernaldo has $x$ friends within the group. Since Dernaldo has 4 friends and the group has 5 members, everyone is a friend of Dernaldo. Let's remove Dernaldo from the group. Thus, Arnaldo, Bernaldo, Cernaldo, and Ernaldo now have $0, 1, 2$, and $x-1$ friends within the subgroup, respectively. Since Arna... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Show that $M=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$ is an integer. | Let $a=\sqrt[3]{\sqrt{5}+2}$ and $b=\sqrt[3]{\sqrt{5}-2}$. Thus, $M=a-b$ and we have:
$$
M^{3}=(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)
$$
We know that $a^{3}-b^{3}=4$ and $a b=1$. Therefore, $M^{3}+3 M-4=0$, which means the number $M$ is a root of the polynomial $x^{3}+3 x-4$.
In turn, the number 1 is a root of the polynom... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
The squared rectangle in the figure is made of 31 segments of $0.5 \mathrm{~cm}$, and comprises 12 squares. Rosa drew on a rectangular sheet of $21 \mathrm{~cm}$ by $29.7 \mathrm{~cm}$, quadriculated with squares of side $0.5 \mathrm{~cm}$, a large squared rectangle made with 1997 segments. How many squares does this r... | Let $m$ and $n$ be, respectively, the number of segments of $0.5 \, \text{cm}$ on two consecutive sides of the rectangle. We know that the total number of segments of $0.5 \, \text{cm}$ in the division of the rectangle into $m \times n$ squares of side $0.5 \, \text{cm}$ is: $m(n+1) + n(m+1)$ (prove this). Thus,
$$
m(... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Draw two circles with the same center, one with a radius of $1 \mathrm{~cm}$ and the other with a radius of $3 \mathrm{~cm}$. In the region outside the circle with a radius of $1 \mathrm{~cm}$ and inside the circle with a radius of $3 \mathrm{~cm}$, draw circles that are simultaneously tangent to both circles, as shown... | (a) Since the circles with radii $1 \mathrm{~cm}$ and $3 \mathrm{~cm}$ are concentric, the other circles shown in the figure must have a radius equal to $1 \mathrm{~cm}$.
(b) The centers of the 3 circles with radius $1 \mathrm{~cm}$ shown in the figure form an equilateral triangle with side length $2 \mathrm{~cm}$. Th... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.