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In the star $A B C D E$ of the given figure, it is known that $G \widehat{B} F=20^{\circ}$ and $G \widehat{H} I=$ $130^{\circ}$. What is the value of the angle $J \widehat{E} I$ ?
 | Observe that $J \widehat{E} I=B \widehat{E} H$.
In the triangle $\triangle B E H$ we have
$$
20^{\circ}+130^{\circ}+B \widehat{E} H=180^{\circ}
$$
therefore,
$$
J \widehat{E} I=B \widehat{E} H=30^{\circ}
$$
 never to seat husband and wife as neighbors at the table;
(b) the arrangement of the six at the table is different each Saturday.
For how many Saturdays can the couples go to the restaurant without repea... | To simplify, let's denote each couple by a pair of numbers: $(1,2),(3,4),(5,6)$, where in each pair, one number represents the husband and the other the wife. Three pairs cannot be neighbors: $(1,2),(3,4)$, $(5,6)$.
See two possible arrangements; starting from 1 in a clockwise direction: 1-3-2-5-4-6 and 1-6-4-5-2-3.
... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a group of 40 students, 20 play football, 19 play volleyball, and 15 play exactly one of these two sports. How many students do not practice football and volleyball?
(a) 7
(b) 5
(c) 13
(d) 9
(e) 10 | Let $x$ be the number of students who practice both sports simultaneously. Therefore, the number of students who practice only football is $20-x$ and those who practice only volleyball is $19-x$. Thus, the students who practice exactly one sport are
$$
(20-x)+(19-x)=15
$$
It follows that $x=12$ and we will have that ... | 13 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
Professor Newton divided his students into groups of $4$ and $2$ were left over. He divided his students into groups of 5 and one student was left out. If 15 students are women and there are more women than men, the number of male students is:
(a) 7
(b) 8
(c) 9
(d) 10
(e) 11 | Solution 1: Since the number of male students is less than 15 and the number of female students is 15, we have
$$
15<\text { male students }+ \text { female students }<15+15=30
$$
that is: the number of students is between 15 and 30.
On the other hand, when we divide by 4, there are 2 students left, so the number of... | 26 | Number Theory | MCQ | Yes | Yes | olympiads | false |
40 - 120
In triangle $ABC$, the angle $A \hat{B C}$ measures $20^{\circ}$ and the angle $A \hat{C} B$ measures $40^{\circ}$. Let $E$ be a point on $BC$ such that $BE = BA$.
(a) Show that triangle CEA is isosceles.
(b) Knowing that the length of the angle bisector of angle BÂC is 2, determine $\mathrm{BC}-\mathrm{AB}... | (a) We have $C \hat{A B}=180^{\circ}-20^{\circ}-40^{\circ}=120^{\circ}$. Since triangle $A B E$ is isosceles, it follows that
$$
A \hat{E} B=E \hat{A} B=\frac{180^{\circ}-20^{\circ}}{2}=80^{\circ}
$$
Thus, $\mathrm{CA} E=120^{\circ}-80^{\circ}=40^{\circ}$ and triangle ACE has two angles of $40^{\circ}$, and therefore... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $A B C$ be an isosceles triangle with $A B=A C$ and $\widehat{A}=30^{\circ}$. Let $D$ be the midpoint of the base $B C$. On $A D$ and $A B$, take two points $P$ and $Q$, respectively, such that $\mathrm{PB}=\mathrm{PQ}$. Determine the measure of the angle $\angle PQC$. | Let us observe that
$$
A \hat{B C}=A \hat{C} B=\frac{180^{\circ}-30^{\circ}}{2}=75^{\circ}
$$
Since all points on the altitude AP are equidistant from B and C, in particular, triangle BPC is isosceles with $B P=P C$. By the problem's hypothesis, triangle BPQ is also isosceles. We denote by $\alpha$ the measure of ang... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $ABC$ be a triangle such that $AB=55, AC=35$ and $BC=72$. Consider a line $\ell$ that intersects side $BC$ at $D$ and side $AC$ at $E$ and that divides the triangle into two figures with equal perimeters and equal areas. Determine the measure of segment $CD$.
## | Let $CD = x$, $CE = y$, and $DE = z$.
(1) Since the triangle CED has the same perimeter as the quadrilateral ABDE, we have
$$
x + y + z = (35 - y) + z + (72 - x) + 55 \Longleftrightarrow y = 81 - x
$$
(2) Since they also have the same area, the area of triangle DCE must be equal to half the area of triangle $ABC$. T... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$O$ triangle $A B C$ with orthocenter $H$ is such that $A B=48$ and $\mathrm{HC}=14$. The midpoint of side $A B$ is $M$ and the midpoint of segment $HC$ is $N$.
(a) Show that the angle $MÊN$ is a right angle.
(b) Determine the length of the segment $\mathrm{MN}$.
Figure 108.3 tenuse of triangle $AEB$. Therefore, $ME=AM=MB=24$. From this fact it follows that triangle $BME$ is isosceles. Thu... | 25 | Geometry | proof | Yes | Yes | olympiads | false |
Suggestion: Let $a=\sqrt{2 n+1}$ and $b=$ $\sqrt{2 n-1}$.
Facts that Help: Use the identity
$\left(a^{2}+a b+b^{2}\right)(a-b)=a^{3}-b^{3}$.
For a positive integer $n$ consider the function
$$
f(n)=\frac{4 n+\sqrt{4 n^{2}-1}}{\sqrt{2 n+1}+\sqrt{2 n-1}}
$$
Calculate the value of
$$
f(1)+f(2)+f(3)+\cdots+f(40)
$$ | Let $a=\sqrt{2 n+1}$ and $b=\sqrt{2 n-1}$. Then $a b=\sqrt{4 n^{2}-1}$, $a^{2}+b^{2}=4 n$ and $a^{2}-b^{2}=2$. Therefore,
$$
f(n)=\frac{a^{2}+b^{2}+a b}{a+b}
$$
Since $a-b \neq 0$, we can write
$f(n)=\frac{a^{2}+b^{2}+a b}{a+b} \cdot \frac{a-b}{a-b}=\frac{a^{3}-b^{3}}{a^{2}-b^{2}}=\frac{(\sqrt{2 n+1})^{3}-(\sqrt{2 n... | 364 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(a) Show that the identity below is always true:
$$
a^{n+1}+b^{n+1}=(a+b)\left(a^{n}+b^{n}\right)-a b\left(a^{n-1}+b^{n-1}\right)
$$
(b) Let a and b be real numbers such that $\mathrm{a}+\mathrm{b}=1$ and $\mathrm{ab}=-1$. Show that the number $a^{10}+b^{10}$ is an integer, calculating its value. | (a) Let's observe that
$$
\begin{aligned}
(a+b)\left(a^{n}+b^{n}\right) & =a^{n+1}+a b^{n}+b a^{n}+b^{n+1}= \\
& =a^{n+1}+b^{n+1}+a b\left(a^{n-1}+b^{n-1}\right)
\end{aligned}
$$
and the identity follows.
(b) Let's call $f_{n}=a^{n}+b^{n}$. Observe that $f_{1}=a+b=1$. Let's calculate $\mathrm{f}_{2}$:
$$
f_{2}=a^{2... | 123 | Algebra | proof | Yes | Yes | olympiads | false |
(a) Determine $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ such that the equality
$$
(n+2)^{2}=a(n+1)^{2}+b n^{2}+c(n-1)^{2}
$$
is true for any number $n$.
(b) Suppose that $x_{1}, x_{2}, \ldots, x_{7}$ satisfy the system
$$
\left\{\begin{array}{l}
x_{1}+4 x_{2}+9 x_{3}+16 x_{4}+25 x_{5}+36 x_{6}+49 x_{7}=1 \\
4 x_{1}... | (a) If a polynomial vanishes for infinitely many values, then all its coefficients are zero.
Expanding the equality we have
$$
n^{2}+4 n+4=a\left(n^{2}+2 n+1\right)+b n^{2}+c\left(n^{2}-2 n+1\right)
$$
Thus,
$$
(a+b+c-1) n^{2}+(2 a-2 c-4) n+(a+c-4)=0
$$
for any number $n$. Therefore,
$$
\left\{\begin{array}{l}
a+... | 334 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The square of 13 is 169, which has the digit 6 in the tens place. The square of another number has the digit 7 in the tens place. What are the possible values for the digit in the units place of this square? | Suppose the number is $10a + b$, with $b$ being a digit. When we square it, we get
$$
(10a + b)^2 = 100a^2 + 20ab + b^2
$$
which has three terms: $100a^2$, $20ab$, and $b^2$.
The first term ends in 00, while the second term ends in an even number followed by zero. For the tens digit to be 7, which is odd, it is nece... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The fractions of the form $\frac{n}{n+1}$, with $n$ a positive integer are:
$$
\underbrace{\frac{1}{2}}_{n=1} ; \quad \underbrace{\frac{2}{3}}_{n=2} ; \quad \underbrace{\frac{3}{4}}_{n=3} ; \quad \underbrace{\frac{4}{5}}_{n=4} ; \quad \underbrace{\frac{5}{6}}_{n=5} \cdots
$$
Observe that this sequence of fractions is... | 2 - Converting to decimal numbers we have: $7 / 9=0.777 \ldots$ and $1 / 2=0.5$; $2 / 3=0.666 \ldots ; 3 / 4=0.75 ; 4 / 5=0.8 ; 5 / 6=0.8333 \ldots$
Therefore, the sequence is increasing and only $1 / 2=0.5 ; 2 / 3=0.666 \ldots ; 3 / 4=0.75$ are less than $7 / 9=0.777 \ldots$
Solution 3 - If $\frac{n}{n+1}<\frac{7}{9... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
From the statement, we have that the number of women who wear only one earring is $0.03 \times 800=24$. There remain $800-24=776$, of which 388 wear two earrings and 388 do not wear earrings. Therefore, the total number of earrings worn by all the women is: $24+388 \times 2=800$. | 2 - If each woman with two earrings gives one of hers to one of those who don't have any, all 800 women will end up with a single earring. Therefore, the number of earrings is equal to the number of women, which is 800. | 800 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Comment: Note that:
the units digit of $\left(9867^{3}-9867^{2}\right)=$ the units digit of $\left(7^{3}-7^{2}\right)$ | 2: $n^{3}-n^{2}=n^{2}(n-1)$. Thus, $n^{2}=(9867)^{2}$ ends in 9 and $n-1=9866$ ends in 6. Since, $9 \times 6=54$, the last digit of the result is 4. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
As $96 \div 8=12$, we have $8 \times 12=96$.
Notice that the solution is equivalent to solving the equation $8 x=96$, whose root is $x=\frac{96}{8}=12$. | 2 - We must find among the list of five options which number, when multiplied by 8, gives 96. The unit digit of this number can only be 2 or 7.
Therefore, it can only be the number 12. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The person will pay 120 reais minus the discount which is $30\%$ of 120. That is: $120-0.3 \times 120=120-36=84$ reais. | 2- We can also solve this problem by noting that if the discount is $30 \%$, then the price the person will pay is $70 \%$ of 120, that is: $0.7 \times 120=84$ reais. | 84 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
We have $C D=80-50=30$ and $A B=80-45=35$. Therefore, $B C=80-35-30=15$.
## | 2 -
From the statement, we have: \( AC = 50 \), \( BD = 45 \), and \( AD = 80 \). From the figure, it follows that \( BC = AC - AB \), so \( BC = 50 - AB \).
Therefore, it is enough to cal... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The figures show that the volumes occupied by the liquids correspond, approximately to more than half in bottle A, half in bottle B, and less than half in bottle C.
The only group of fractions that corresponds to these estimates is: $\frac{2}{3}$ (more than half); $\frac{1}{2}$ (half); $\frac{1}{4}$ (less than half). | 2 - The figures show that the volumes occupied by the liquids are decreasing numbers. The only possible options are B and E. Since $\frac{3}{3}=1$ and none of the flasks are full, the answer is B. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
At Zé's office, there are six pieces of chains with the following number of links: $10, 10, 8, 8, 5$ and 2. He needs to join these pieces to form a circular chain. He spends 1 minute to cut a link and 2 minutes to join it, totaling 3 minutes per link. If he cuts a link at the end of each separate piece, joining the pie... | Solution
a) One way for him to form the chain in 15 minutes is to initially open all the links of the 5-link piece. In this process, he will spend $5 \cdot 1=5$ minutes. Next, he should use each of these open links between the remaining 5 pieces of chain, using exactly one open link to join two distinct pieces. For th... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the figure below, segments $A B$ and $C D$ are parallel. If $\angle C I J=10 \beta, \angle A G J=10 \alpha$, $\angle C E J=6 \alpha$ and $\angle J F G=6 \beta$, determine the value of the angle $\angle I J K$.
Since $\angle C E J$ and $\angle J F A$ are consecutive interior angles, $6 \alpha + 6 \beta = 180^{\circ}$, which means $\alpha + \beta = 30^{\circ}$. From point $J$, consider the... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Two tangents are drawn from a point $A$ to a circle with center $O$, touching it at $B$ and $C$. Let $H$ be the orthocenter of triangle $A B C$, given that $\angle B A C=40^{\circ}$, find the value of the angle $\angle H C O$.
 The first two three-digit numbers are $100=2 \cdot 2 \cdot 5 \cdot 5$ and $101=101$ (which is prime). When testing 102, we have $102=2 \cdot 3 \cdot 17$, which is the smallest three-star number.
(b) It suffices to show that every three-star number has at least one of the prime factors in the set $\{2,3,5... | 102 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the following figure, the circle with center $B$ is tangent to the circle with center $A$ at $X$. The circle with center $C$ is tangent to the circle with center $A$ at $Y$. Additionally, the circles with centers $B$ and $C$ are also tangent. If $A B=6, A C=5$ and $B C=9$, what is the measure of $A X$?
# | Solution
Since the sum of the interior angles of a quadrilateral is $360^{\circ}$, we have:
$$
\begin{aligned}
\alpha+\beta & =360^{\circ}-\angle I H E-\angle I J E \\
& =360^{\circ}-\ang... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
a) João arranged 13 sticks in the shape of a rectangular fence $1 \times 4$ as shown in the figure below. Each stick is the side of a $1 \times 1$ square, and inside each of these squares, he placed an ant. What is the minimum number of sticks we need to remove to ensure that all 4 ants can escape and return to their a... | Solution
a) It is possible to free all the ants by removing 4 matchsticks as indicated in the following figure.
Since each matchstick is shared by at most two squares and each square must ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
All vertices of the pentagon $A B C D E$ lie on the same circle. If $\angle D A C=50^{\circ}$, determine $\angle A B C+\angle A E D$.
# | Solution
Since inscribed angles associated with the same arc are equal, we have $\angle D A C=\angle D B C$. Furthermore, knowing that the sum of opposite angles of a cyclic quadrilateral is $180^{\circ}$, it follows that
$$
\begin{aligned}
\angle A B C+\angle A E D & =(\angle A B D+\angle A E D)+\angle D B C \\
& =1... | 230 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The points $E$ and $F$ are on the sides $A D$ and $B C$, respectively, of the square $A B C D$. Knowing that $B E=E F=F D=30$, find the area of the square.
# | Solution
Let $G$ and $H$ be the feet of the perpendiculars drawn from $E$ and $F$ to the sides $B C$ and $A D$, respectively.
Since $A B=C D$ and $B E=F D$, applying the Pythagorean Theorem... | 810 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the figure, line $t$ is parallel to segment $E F$ and tangent to the circle. If $A E=12, A F=10$ and $F C=14$, determine the length of $E B$.
 | Solution
Since the line $t$ is tangent to the circle, we have $\angle X A E=\angle A C B$. Furthermore, since $t$ and $E F$ are parallel, we have $\angle X A E=\angle A E F$. Similarly, we have $\angle Y A F=\angle A F E=\angle A B C$. Therefore, $\triangle A F E \simeq \triangle A B C$. Thus,
$$
\begin{aligned}
\fra... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On a blackboard, the numbers 1, 2, 3, 4, 5, and 6 are written. In each move, João can replace two numbers $a$ and $b$ with $a \cdot b + a + b$. Find all the possibilities for the last number on the blackboard. | Solution
Let's perform a sequence of moves to analyze the behavior of the list of numbers over the moves:
$$
\begin{aligned}
1,2,3,4,5,6 & \rightarrow \\
1,11,4,5,6 & \rightarrow \\
1,11,5,34 & \rightarrow \\
\mathbf{6 9}, 11,5 & \rightarrow \\
\mathbf{1 1 , 4 1 9} & \rightarrow \\
5039 . &
\end{aligned}
$$
The expr... | 5039 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Three cars depart from city $A$ at the same time and travel a closed path composed of three straight segments $A B, B C$, and $C A$. The speeds of the first car on these segments are 12, 10, and 15 kilometers per hour, respectively. The speeds of the second car are 15, 15, and 10 kilometers per hour, respectively. Fina... | Solution
Let $x, y$ and $z$ be the lengths of $AB, BC$ and $AC$, respectively. The arrival time $t$, common to the three cars, can be found through the equations:
$$
\left\{\begin{array}{l}
\frac{x}{12}+\frac{y}{10}+\frac{z}{15}=t \\
\frac{x}{15}+\frac{y}{15}+\frac{z}{10}=t \\
\frac{x}{10}+\frac{y}{20}+\frac{z}{12}=t... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the figure below, triangle $A B C$ is a right triangle at $C$ and both $B C D E$ and $C A F G$ are squares. If the product of the areas of triangles $E A B$ and $B F A$ is 64, determine the area of triangle $A B C$.
}{2}
$$
Since this story is from a long time a... | Solution
(a) Let $p$ be a prime greater than 3. This prime, like any positive integer, can only leave remainders from 0 to 5 when divided by 6. Notice that if it left remainders of $0, 2, 3$, or 4, we could factor it:
$$
\begin{aligned}
& p=6 q+0=2 \cdot 3 q \\
& p=6 q+2=2 \cdot(3 q+1) \\
& p=6 q+3=3 \cdot(2 q+1) \\
... | 166000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the city of Oropis, there is a triangular lake with each of its three sides being part of the perimeter of a square-shaped land with areas of $370 \mathrm{~m}^{2}$, $116 \mathrm{~m}^{2}$, and $74 \mathrm{~m}^{2}$, as shown in the first figure below. The mayor of Oropis, Arnaldo, wants to calculate the area of the la... | Solution
(a) See that the sides of the lake squared result in 74, 116, and 370. The conditions that must be met are the equations derived from three applications of the Pythagorean Theorem:
$$
\begin{aligned}
a^{2}+c^{2} & =74 \\
b^{2}+d^{2} & =116 \\
(a+b)^{2}+(c+d)^{2} & =370
\end{aligned}
$$
We can limit the test... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
We know that
$$
\frac{8^{x}}{2^{x+y}}=64 \text{ and } \frac{9^{x+y}}{3^{4 y}}=243
$$
Determine the value of $2 x y$.
# | Solution
Since $8=2^{3}$ and $9=3^{2}$, we have
$$
\begin{aligned}
64 & =\frac{8^{x}}{2^{x+y}} \\
2^{6} & =2^{3 x-(x+y)} \\
& =2^{2 x-y}
\end{aligned}
$$
$$
\begin{aligned}
243 & =\frac{9^{x+y}}{3^{4 y}} \\
3^{5} & =3^{(2 x+2 y)-4 y} \\
& =3^{2 x-2 y}
\end{aligned}
$$
Thus, we have the following system:
$$
\left\{... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
João wrote all the powers of 2, 3, and 5 greater than 1 and less than 2017 on a sheet of paper. Then, he performed all possible products of two distinct numbers from this sheet and wrote them on another sheet of paper. What is the number of integers that João recorded on the second sheet?
# | Solution
Initially, we need to find the powers of 2, 3, and 5 recorded on the first sheet. Since $2^{10}<2017<2^{11}, 3^{6}<2017<3^{7}$, and $5^{4}<2017<5^{5}$, the powers written on the first sheet can be divided into three sets:
$$
P_{2}=\left\{2^{1}, 2^{2}, \ldots, 2^{10}\right\}, P_{3}=\left\{3^{1}, 3^{2}, \ldots... | 155 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A fraction is said to be irreducible when its numerator and denominator do not have common factors, that is, when the greatest common divisor between the two numbers is 1. For example, the fraction $\frac{3}{7}$ is irreducible, but the fraction $\frac{10}{14}$ is not, since 2 is a common factor of 10 and 14. For which ... | Solution
a) Since $6 n+5$ and $5 n+6$ are multiples of $d$, there exist integers $x$ and $y$ such that $6 n+5=d x$ and $5 n+6=d y$. Therefore,
$$
\begin{aligned}
n-1 & =(6 n+5)-(5 n+6) \\
& =d x-d y \\
& =d(x-y)
\end{aligned}
$$
implying that $d$ is a divisor of $n-1$.
b) Using the fact that $6 n+5$ and $n-1$ are m... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the drawing below, $ABCD$ and $EFGC$ are squares. The lines $BG$ and $DE$ intersect at point $H$.
a) Verify that $\angle BHD=90^{\circ}$ and conclude that point $H$ is simultaneously on the circumferences of diameters $BD$ and $EG$.
b) Find the value of $\angle AHD + \angle DHG + \angle GHF$.
 The figure below shows a way to cover the $3 \times 4$ board using only L-triminós.
b) Consider the $3 \times 5$ board below with the 6 shaded squares.
... | 7 | Combinatorics | proof | Yes | Yes | olympiads | false |
To determine the number of positive divisors of a number, it is enough to factor it as powers of distinct primes and multiply the successors of the exponents. For example, $2016=2^{5} \cdot 3^{2} \cdot 5^{1}$ has $(5+1)(2+1)(1+1)=36$ positive divisors. Consider the number $n=2^{7} \cdot 3^{4}$.
a) Determine the number... | Solution
a) From the factorization of $n$, we can determine the factorization of $n^{2}$:
$$
\begin{aligned}
n^{2} & =\left(2^{7} \cdot 3^{4}\right)^{2} \\
& =2^{14} \cdot 3^{8}
\end{aligned}
$$
Thus, the number $n^{2}$ has $(14+1)(8+1)=15 \cdot 9=135$ positive divisors.
b) Note that all positive divisors of $n^{2}... | 28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Consider a triangle $A B C$ and squares $A B D E$ and $A C F G$ constructed externally on its sides. The segments $B G$ and $E C$ intersect at $P$.
a) Prove that the triangles $B A G$ and $... | Solution
Consider the figure below where we have marked the equal sides of the two squares.
a) We can state that triangles $B A G$ and $E A C$ are congruent by the $S A S$ case, since $B A... | 90 | Geometry | proof | Yes | Yes | olympiads | false |
There are 2017 unoccupied chairs in a row. Every minute, a person arrives and sits in one of them that is empty, and at the same instant, if it is occupied, a person in an adjacent chair stands up and leaves. What is the maximum number of people that can be sitting simultaneously in the row of chairs? | Solution
It is not possible for all chairs to be simultaneously occupied, as the last person to sit down would inevitably sit next to an occupied chair and force, according to the rule in the problem statement, someone to leave. Our goal now is to show a sequence of moves where it is possible for 2016 people to be sea... | 2016 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
a) Consider the $4 \times 4$ board shown in the following figure. The letters and numbers have been placed to help locate the small squares. For example, the square $A 1$ is in the top-left corner and the $D 4$ is in the bottom-right corner.
 There are several ways to do this (we will see exactly how many in the next item). In the example on the left below, only the squares $A 1, B 2, C 3$ and $D 4$ are not chosen.
b... | 576 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(a) What is the smallest (positive) multiple of 9 that is written only with the digits 0 and 1?
(b) What is the smallest (positive) multiple of 9 that is written only with the digits 1 and 2? | (a) A number is divisible by 9 if the sum of its digits is a multiple of 9. Therefore, the number must have 9 digits equal to 1. Thus, the smallest number is: 111111111.
(b) We should use the largest possible number of digits equal to 2, which should be placed in the rightmost positions. Thus, the smallest number is: ... | 111111111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A city still does not have electric lighting, so candles are used in houses at night. In João's house, one candle is used per night, without burning it completely, and with four of these candle stubs, João makes a new candle. For how many nights can João light his house with 43 candles?
(a) 43
(b) 53
(c) 56
(d) 57
(e) ... | The correct option is (d).
With 43 candles, João's house can be lit for 43 nights, leaving 43 candle stubs. Since $43=4 \times 10+3$, with these 43 stubs, 3 stubs can be saved and 10 new candles can be made to light 10 nights. From these 10 candles, we get 10 stubs, which, together with the 3 that were left, give 13 s... | 57 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
A factory packages cans of palm heart in cardboard boxes of cubic shape with a side of $20 \mathrm{~cm}$. In each box, 8 cans are placed, and the boxes are placed, without leaving any empty spaces, in wooden crates measuring $80 \mathrm{~cm}$ in width by $120 \mathrm{~cm}$ in length and $60 \mathrm{~cm}$ in height. Wha... | The correct option is (a).
In each wooden crate of dimensions $a \times b \times c$, $(a \times b \times c) / l^{3}$ cubes with side $l$ can fit when stacked regularly. In the case of the palm hearts, we have, in centimeters, $a=60, b=80, c=120$ and $l=20$. Since 60, 80, and 120 are multiples of 20, we can fill the cr... | 576 | Geometry | MCQ | Yes | Yes | olympiads | false |
How many fractions of the form $\frac{n}{n+1}$ are less than $7 / 9$, knowing that $n$ is a positive integer?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5 | The correct option is (c).
Solution 1: The fractions of the form $\frac{n}{n+1}$, with $n$ a positive integer, are
Observe that we have $\frac{1}{2} \cdot \frac{35}{45} = \frac{7}{9}$. Th... | 3 | Number Theory | MCQ | Yes | Yes | olympiads | false |
In a certain African village, 800 women live, $3 \%$ of whom wear only one earring. Of the rest, half wear two earrings and the other half wear none. What is the total number of earrings worn by all the women in this village?
(a) 776
(b) 788
(c) 800
(d) 812
(e) 824 | The correct option is (c).
Solution 1: We know that the number of women who wear only one earring is $0.03 \times 800=24$. There remain $800-24=776$ women, of which 388 wear two earrings and 388 do not wear earrings. Therefore, the total number of earrings worn by all the women is $24+388 \times 2=800$.
Solution 2: I... | 800 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
The weight limit that a truck can transport corresponds to 50 bags of sand or 400 bricks. If this truck is already carrying 32 bags of sand, how many bricks, at most, can it still carry?
(a) 132
(b) 144
(c) 146
(d) 148
(e) 152 | The correct option is (b).
Since the weight of a bag of sand is equal to that of eight bricks and there are already 32 bags of sand in the truck, it can still carry 18 more bags of sand, which is equivalent to $18 \times 8=144$ bricks. | 144 | Algebra | MCQ | Yes | Yes | olympiads | false |
The absolute value $|a|$ of any number $a$ is defined by
$$
|a|=\left\{\begin{array}{cl}
a & \text { if } a>0 \\
0 & \text { if } a=0 \\
-a & \text { if } a<0
\end{array}\right.
$$
For example, $|6|=6,|-4|=4$ and $|0|=0$. What is the value of $N=|5|+|3-8|-|-4|$?
(a) 4
(b) -4
(c) 14
(d) -14
(e) 6 | The correct option is (e).
We have: $|5|=5,|3-8|=|-5|=5$ and $|-4|=4$. Therefore, $N=5+5-4=6$. | 6 | Algebra | MCQ | Yes | Yes | olympiads | false |
What is the smallest positive integer $N$ such that $N / 3, N / 4, N / 5, N / 6$ and $N / 7$ are all integers?
(a) 420
(b) 350
(c) 210
(d) 300
(e) 280 | The correct option is (a).
For $\frac{N}{3}, \frac{N}{4}, \frac{N}{5}, \frac{N}{6}$ and $\frac{N}{7}$ to be integers, $N$ must be a common multiple of $3, 4, 5, 6$ and 7. Since we want the smallest possible $N$, it must be the least common multiple (LCM) of 3, 4, 5, 6 and 7, that is,
$$
N=3 \times 4 \times 5 \times 7... | 420 | Number Theory | MCQ | Yes | Yes | olympiads | false |
The vertices of a cube are numbered from 1 to 8, such that one of the faces has the vertices $\{1,2,6,7\}$ and the other five have the vertices $\{1,4,6,8\},\{1,2,5,8\}$, $\{2,3,5,7\},\{3,4,6,7\}$ and $\{3,4,5,8\}$. Which is the number of the vertex that is farthest from the vertex numbered 6?
(a) 1
(b) 3
(c) 4
(d) 5
(... | The correct option is (d).
Solution 1: By drawing the cube and numbering its vertices according to the question statement, we obtain a figure in which we can see that vertex 5, being diametrically opposite, is the farthest from vertex 6.
 54
(b) 8
(c) 12
(d) 58
(e) 46
.
Just read the graph to get the number of points for each player. The sum of these points gives a total of $7+8+2+11+6+12+1+7=54$ points scored by the team. | 54 | Other | MCQ | Yes | Yes | olympiads | false |
Given two real numbers $a$ and $b$, consider $ab = a^2 - ab + b^2$. What is the value of 1?
(a) 1
(b) 0
(c) 2
(d) -2
(e) -1 | The correct option is (a).
By setting $a=1$ and $b=0$ in $a=a^{2}-a b+b^{2}$, we obtain $1 \times 1^{2}-1 \times 0+0^{2}=1$. | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
The bar chart shows the distribution of students in a school according to the time they spend traveling from home to school. Fractions of a minute were disregarded; for example, if a student spends 40 minutes and 15 seconds on this journey, it is considered that the time spent is 40 minutes.
 The students who spend less than 20 mi... | 180 | Other | math-word-problem | Yes | Yes | olympiads | false |
Manoel tests his aim by launching five arrows that hit the target at points $A, B, C, D$ and $E$, with coordinates $A=(1,-1), B=(2.5, 1), C=(-1,4)$, $D=(-4,-4)$ and $E=(6,5)$.
The table shows how many points are earned when an arrow hits a point within each of the three regions, as shown in the figure.
(a) Mark the p... | (a) The five given points are marked in the figure.
(b) In the smaller circle, we have only point $A$, so Manoel hit this circle once, which gives him 300 points.
(c) To calculate the tota... | 500 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
André's birthday party has fewer than 120 guests. For dinner, he can divide the guests into complete tables of six people or into complete tables of seven people. In both cases, more than 10 tables are needed and all guests are seated at some table. How many guests are there? | We can distribute the total number of guests at tables of 6 or 7, and the number of guests is a multiple of 6 and 7. Since the least common multiple of 6 and 7 is 42, we can have $42, 84, 126, \ldots$ guests. As there are fewer than 120 guests, we can only have 42 or 84 guests. On the other hand, since more than 10 tab... | 84 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the given figure, $ABCD$ is a rectangle and $\triangle ABE$ and $\triangle CDF$ are right triangles. The area of triangle $\triangle ABE$ is $150 \mathrm{~cm}^{2}$ and the segments $AE$ and $DF$ measure, respectively, 15 and $24 \mathrm{~cm}$. What is the length of the segment $CF$?
^{2}=(C F)^{2}+(F D)^{2}=(C F)^{2}+24^{2}$, and from this, we get $(C F)^{2}=(C D)^{2}-24^{2}$. In other words, to find $C F$, it is enough to know... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Using only the digits 1, 2, 3, 4, and 5, Peri constructed the sequence
$$
1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,1,1,1,1,1,1,2,2,2,2,2,2,2, \ldots
$$
starting with one 1, followed by two 2s, three 3s, four 4s, five 5s, six 1s, seven 2s, and so on. What is the hundredth term of this sequence? | We group the sequence into consecutively numbered blocks, each block formed by consecutive equal terms, as shown below.
$$
\begin{aligned}
& \underbrace{1}_{\text {block } 1}, \underbrace{2,2}_{\text {block } 2}, \underbrace{3,3,3}_{\text {block } 3}, \underbrace{4,4,4,4}_{\text {block } 4}, \underbrace{5,5,5,5,5}_{\t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Capitu has a hundred cards numbered from 1 to 100. All cards have one yellow side and the other red, and the number of each card is written on both sides. The cards were placed on a table, all with the red side facing up. Capitu flipped all the cards with even numbers and then all the cards with numbers that are multip... | Solution 1: First, Capitu turned over the 50 even-numbered cards. After that, the 50 even-numbered cards were face up with the yellow side, and the 50 odd-numbered cards were face up with the red side. When she then turned over the multiples of 3, she only turned over the odd multiples of 3, which are 3, 9, 15, 21, 27,... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
To fill a rectangular tank with water, which has a length of $3 \mathrm{~m}$, a width of $50 \mathrm{~cm}$, and a height of $0.36 \mathrm{~m}$, a man uses a cylindrical bucket with a base diameter of $30 \mathrm{~cm}$ and a height of $48 \mathrm{~cm}$ to fetch water from a spring. Each time he goes to the spring, he fi... | In what follows, all volume measures are given in $\mathrm{cm}^{3}$.
The volume $V$ of the bucket is given by the usual formula for the volume of a cylinder, that is, $V=$ base area $\times$ height. The base of the bucket is a circle with a diameter of $30 \mathrm{~cm}$; its radius, then, measures $r=15 \mathrm{~cm}$ ... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
What is the largest prime factor of 2006? | The prime factorization of 2006 is $2006=2 \times 17 \times 59$. Therefore, the largest prime factor of 2006 is 59. | 59 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Between 1986 and 1989, the currency of our country was the cruzado ( $\mathrm{Cz} \$$ ). Since then, we have had the new cruzado, the cruzeiro, the new cruzeiro, and today, we have the real. To compare values from the time of the cruzado and today, economists calculated that 1 real is equivalent to 2750000000 cruzados.... | The correct option is (d).
The statement says that 1 real $=275 \times 10^{7}$ cruzados. João's salary is 640 reais, which is equivalent to $640 \times 275 \times 10^{7}=176000 \times 10^{7}=176 \times 10^{10}$ cruzados. The number of stacks of one hundred notes that can be made with this amount of 1 cruzado notes is ... | 264000 | Algebra | MCQ | Yes | Yes | olympiads | false |
A point $P$ is at the center of a square with a side length of $10 \mathrm{~cm}$. How many points on the edge of the square are $6 \mathrm{~cm}$ away from $P$?
(a) 1
(b) 2
(c) 4
(d) 6
(e) 8 | The correct option is (e).
The points that are $6 \mathrm{~cm}$ away from point $P$ form a circle with center $P$ and radius $R=6$ $\mathrm{cm}$. If $d$ denotes the diagonal of the square, by the Pythagorean Theorem we have
$$
d=\sqrt{10^{2}+10^{2}}=\sqrt{2 \times 10^{2}}=10 \sqrt{2}
$$
The circle with radius $L / 2... | 8 | Geometry | MCQ | Yes | Yes | olympiads | false |
If $2\left(2^{2 x}\right)=4^{x}+64$, what is the value of $x$?
(a) -2
(b) -1
(c) 1
(d) 2
(e) 3 | The correct option is (e).
Solution 1: We observe that the terms on the right side of the given equation can be written as powers of 2. Indeed, $4^{x}=\left(2^{2}\right)^{x}=2^{2 x}$ and $64=2^{6}$. Thus, the equation becomes $2\left(2^{x}\right)=2^{2 x}+2^{3}$. We then have $2\left(2^{2 x}\right)-2^{2 x}=2^{6}$, whic... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
Determine the value of $(666666666)^{2}-(333333333)^{2}$. | Using the factorization $x^{2}-y^{2}=(x-y)(x+y)$ with $x=666666666$ and $y=333333333$, we see that $x-y=y$ and $x+y=999999999$, therefore,
$$
\begin{aligned}
666666666^{2}-333333333^{2} & =333333333 \times 999999999 \\
& =333333333 \times(1000000000-1) \\
& =333333333000000000-333333333 \\
& =333333332666666667
\end{a... | 333333332666666667 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the figure, the number 8 was obtained by adding the two numbers directly below its position. By doing the same to fill in the blank spaces, one gets 42 in the indicated position. What is the value of $x$?
(a) 7
(b) 3
(c) 5
(d) 4
(e) 6
.
Using the given rule, we fill in the empty cells starting from the second row from the bottom and obtain the figure. Therefore,
$$
(13+x)+(11+2 x)=42
$$
and thus, $24+3 x=42$, which means $x=6$.
 0
(b) 2
(c) 4
(d) 6
(e) 8 | The correct option is (c).
Solution 1: The last digit of $9867^{3}$ is the same as that of $7^{3}=343$, which is 3. The last digit of $9867^{2}$ is the same as that of $7^{2}=49$, which is 9. If we subtract a number ending in 9 from another ending in 3, the last digit of the result is 4.
Note: Observe that the last d... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
A baker wants to use all his flour to make bread. Working alone, he could finish the flour in 6 hours. With an assistant, the same could be done in 2 hours. The baker started working alone and, after some time, tired, he called his assistant and thus, after 150 minutes the flour ran out. For how many minutes did the ba... | The correct option is (d).
Let $x$ be the amount of flour, in kilograms, that the baker has. Working alone, he would use $x / 6$ kilograms of flour per hour. Working with his assistant, they would use $x / 2$ kilograms of flour per hour. Let $t$ be the time, in hours, that the baker worked alone. Since the flour runs ... | 45 | Algebra | MCQ | Yes | Yes | olympiads | false |
In the figure, the two triangles $\triangle A B C$ and $\triangle D E F$ are equilateral. What is the value of the angle $x$?
(a) $30^{\circ}$
(b) $40^{\circ}$
(c) $50^{\circ}$
(d) $60^{\circ}$
(e) $70^{\circ}$
.
Since $\triangle A B C$ and $\triangle D E F$ are equilateral triangles, each of their internal angles measures $60^{\circ}$. In the triangle $\triangle A G D$ we have
$$
G \widehat{A} D=180^{\circ}-75^{\circ}-60^{\circ}=45^{\circ} \text { and } \mathrm{G} \widehat{D A}=180^{\circ}-65^{\cir... | 40 | Geometry | MCQ | Yes | Yes | olympiads | false |
The figure shows a piece of cardboard that will be folded and glued along the edges to form a rectangular box. The angles at the corners of the cardboard are all right angles. What will be the volume, in $\mathrm{cm}^3$, of the box?
(a) 1500
(b) 3000
(c) 4500
(d) 6000
(e) 12000
.
The figure shows the folds that will be made to assemble the box, which will have the following dimensions: $20 \mathrm{~cm}$ in width, $15 \mathrm{~cm}$ in leng... | 3000 | Geometry | MCQ | Yes | Yes | olympiads | false |
In a sequence, each term, starting from the third, is the sum of the two immediately preceding terms. If the second term is 1 and the fifth term is 2005, what is the sixth term?
(a) 3002
(b) 3008
(c) 3010
(d) 4002
(e) 5004 | The correct option is (b).
Let $x$ be the first term of the sequence. Since the second term is $1$, and from the third term onwards, each term is the sum of the two preceding ones, we have:
- the third term is $1+x$;
- the fourth term is $1+(1+x)=2+x$;
- the fifth term is $(1+x)+(2+x)=3+2x$;
- the sixth term is $(2+x... | 3008 | Algebra | MCQ | Yes | Yes | olympiads | false |
How many numbers between 10 and 13000, when read from left to right, are formed by consecutive digits in ascending order? For example, 456 is one of these numbers, but 7890 is not.
(a) 10
(b) 13
(c) 18
(d) 22
(e) 25 | The correct option is (d).
The numbers in question, with
- two digits, are 12, 23, 34, 45, 56, 67, 78, and 89 (8 numbers);
- three digits, are 123, 234, 345, 456, 567, 678, and 789 (7 numbers);
- four digits, are 1234, 2345, 3456, ..., 6789 (6 numbers);
- finally, with five digits, only 12345,
for a total of $8+7+6+... | 22 | Number Theory | MCQ | Yes | Yes | olympiads | false |
In a block of $1 \times 2 \times 3$ centimeters, we mark three faces with the letters X, Y, and Z, as shown in the figure. The block is placed on an $8 \times 8$ cm board with the face X facing down, in contact with the board, as shown in the figure. We rotate the block $90^{\circ}$ around one of its edges so that the ... | The correct option is (b).
According to the figure, we can conclude that the dimensions of the faces X, Y, and Z are 2, 3, and $6 \mathrm{~cm}^{2}$, respectively. Next, we will indicate th... | 19 | Geometry | MCQ | Yes | Yes | olympiads | false |
The function $f$ is given by the table
| $x$ | 1 | 2 | 3 | 4 | 5 |
| :---: | :--- | :--- | :--- | :--- | :--- |
| $f(x)$ | 4 | 1 | 3 | 5 | 2 |
For example, $f(2)=1$ and $f(4)=5$. What is the value of $\underbrace{f(f(f(f(\ldots f}_{2004 \text { times }}(4) \ldots)))$ ?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5 | The correct option is (d).
From the table
| $x$ | 1 | 2 | 3 | 4 | 5 |
| :---: | :--- | :--- | :--- | :--- | :--- |
| $f(x)$ | 4 | 1 | 3 | 5 | 2 |
we obtain
$$
\begin{aligned}
& f(4)=5, f(\underbrace{f(4)}_{5})=f(5)=2, f(f(\underbrace{f(4)}_{5}))=f(\underbrace{f(5)}_{2})=f(2)=1 \text{ and} \\
& \underbrace{f(f(f(f(4... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
How many degrees does the angle $\alpha$ in the figure measure?
(a) 20
(b) 25
(c) 30
(d) 35
(e) 40
 | The correct option is (a).
The internal angles of the given quadrilateral are $50^{\circ}, 180^{\circ}-30^{\circ}=150^{\circ}, \alpha$ and $180^{\circ}-40^{\circ}=140^{\circ}$. Since the sum of the internal angles of a quadrilateral is $360^{\circ}$, we have $50^{\circ}+150^{\circ}+\alpha+140^{\circ}=360^{\circ}$, fro... | 20 | Geometry | MCQ | Yes | Yes | olympiads | false |
Artur wants to draw a "spiral" 4 meters long, formed of straight segments. He has already traced seven segments, as shown in the figure. How many segments are still missing?
(a) 28
(b) 30
(c) 24
(d) 32
(e) 36 | The correct option is (d).
The figure shows that the "spiral" is made up of segments whose lengths form a finite sequence of the form $1,1,2,2,3,3,4,4, \ldots, n, n$ (if the last two segments of the spiral have the same length) or, alternatively, of the form $1,1,2,2,3,3,4,4, \ldots$, $n, n, n+1$ (if the last two segm... | 32 | Geometry | MCQ | Yes | Yes | olympiads | false |
The extensions of a telephone switchboard have only 2 digits, from 00 to 99. Not all extensions are in use. By swapping the order of two digits of an extension in use, one either gets the same number or a number of an extension that is not in use. What is the maximum possible number of extensions in use?
(a) Less than ... | The correct option is (e).
There are two types of extensions that may be in use. We have extensions with
- the two digits being the same $(00,11,22,33,44,55,66,77,88$ and 99 ), for a total of 10, and those with
- the two digits being different. In this case, we have $10 \times 9=90$ numbers, and half of them can be u... | 55 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
A bus, a train, and a plane depart from city A to city B at the same time. If I take the bus, which has an average speed of 100 $\mathrm{km} / \mathrm{h}$, I will arrive in city B at 8 PM. If I take the train, which has an average speed of $300 \mathrm{~km} / \mathrm{h}$, I will arrive in city B at 2 PM. What will be t... | Let $d$ be the distance between the two cities and $h$ be the common departure time of the bus, the train, and the airplane. Since distance $=$ speed $\times$ time, we have $d=100 \times(20-h)$ and $d=300 \times(14-h)$. Therefore, $100 \times(20-h)=300 \times(14-h)$, from which $h=11$. Thus, the distance between the tw... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
If $\frac{\sqrt{x}}{\sqrt{y}}=5$, what is $\frac{x+y}{2 y}$?
(a) $\frac{5}{2}$
(b) $3 \sqrt{2}$
(c) $13 y$
(d) $\frac{25 y}{2}$
(e) 13 | The correct option is (e).
Squaring both sides of $\frac{\sqrt{x}}{\sqrt{y}}=5$, we get $\frac{x}{y}=25$. Thus,
$$
\frac{x+y}{2 y}=\frac{1}{2} \times \frac{x+y}{x}=\frac{1}{2} \times\left(\frac{x}{y}+\frac{y}{y}\right)=\frac{1}{2} \times\left(\frac{x}{y}+1\right)=\frac{1}{2} \times(25+1)=13
$$ | 13 | Algebra | MCQ | Yes | Yes | olympiads | false |
How many different pairs of positive integers $(a, b)$ are there such that $a+b \leq 100$ and $\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=13$?
(a) 1
(b) 5
(c) 7
(d) 9
(e) 13 | The correct option is (c).
We have
$$
13=\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=\frac{\frac{a b+1}{b}}{\frac{1+a b}{a}}=\frac{(a b+1) \times a}{(1+a b) \times b}=\frac{a}{b}
$$
Thus, $a=13 b$. Since $a+b \leq 100$, it follows that $14 b \leq 100$ and, therefore, $b \leq 7.14$. Since $b$ is an integer, we must have $b \... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
In the figure, the three circles are concentric, and the area of the smallest circle coincides with the area of the largest ring, highlighted in gray. The radius of the smallest circle is $5 \mathrm{~cm}$ and of the largest $13 \mathrm{~cm}$. What is the radius (in cm) of the intermediate circle?
(a) 12
(c) $10 \sqrt{6... | The correct option is (a).
The area of the largest circle is $13^{2} \pi=169 \pi$ and that of the smallest is $5^{2} \pi=25 \pi$, which is also the area of the largest ring. Let $r$ be the radius of the intermediate circle. Then, the area of the largest ring is $169 \pi-\pi r^{2}$. Therefore, $169 \pi-\pi r^{2}=25 \pi... | 12 | Geometry | MCQ | Yes | Yes | olympiads | false |
How many of the numbers $-5,-4,-3,-2,-1,0$, $1,2,3$ satisfy the inequality $-3 x^{2}<-14$?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5 | The correct option is (d).
If $-3 x^{2}14$, or $x^{2}>\frac{1}{3} 14=4 \frac{2}{3}$. Since we are only looking at integer values of $x$, then $x^{2}$ is also an integer. Given that $x^{2}>4 \frac{2}{3}$, we conclude that $x^{2}$ is at least 5. Among the numbers $-5,-4,-3,-2,-1,0,1,2,3$ only four, namely, $-5,-4,-3$ an... | 4 | Inequalities | MCQ | Yes | Yes | olympiads | false |
If $S_{n}=1-2+3-4+5-6+\cdots+(-1)^{n+1} n$ for each positive integer $n$, then $S_{1992}+S_{1993}$ is equal to
(a) -2 ;
(b) -1 ;
(c) 0 ;
(d) 1 ;
(e) 2 . | The correct option is (d).
The expression $(-1)^{n+1}$ in the definition of $S_{n}$ has a value of 1 if $n$ is even and a value of -1 if $n$ is odd.
Solution 1: By associating consecutive terms in pairs, we obtain a sum of several terms equal to -1: $(1-2)+(3-4)+(5-6)+\cdots$. Therefore,
$$
S_{1992}=\underbrace{(1-2... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
An arc of a circle measures $300^{\circ}$ and its length is $2 \mathrm{~km}$. What is the nearest integer to the measure of the radius of the circle, in meters?
(a) 157
(b) 284
(c) 382
(d) 628
(e) 764 | The correct option is (c).
Solution 1: If the radius is $r$, then the length of an arc of $\theta$ degrees is $\frac{2 \pi}{360} \theta r$. Thus, in the given problem, we have that
$$
2000 \mathrm{~m}=\frac{2 \pi}{360} 300 r=\frac{5 \pi}{3} r
$$
therefore $r=2000 \times(3 / 5 \pi) \approx 382.17 \mathrm{~m}$.
Solut... | 382 | Geometry | MCQ | Yes | Yes | olympiads | false |
In a taxi, a passenger can sit in the front and three passengers can sit in the back. In how many ways can four passengers sit in a taxi if one of these passengers wants to sit by the window? | The passenger who wants to sit by the window has three possible seats, the next can sit in any free seat, thus having three possible seats; the next has two possible seats, and the last one has no choice. We conclude that the number of these ways of sitting is $3 \times 3 \times 2=18$. | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Eliane wants to choose her schedule for swimming. She wants to attend two classes per week, one in the morning and one in the afternoon, not on the same day, nor on consecutive days. In the morning, there are swimming classes from Monday to Saturday, at $9 \mathrm{~h}$, $10 \mathrm{~h}$, and $11 \mathrm{~h}$, and in th... | If the morning class is on Monday or Friday (in any of the three time slots), then the afternoon class day can be chosen in three different ways (in any of the two time slots), so we have $2 \times 3 \times 3 \times 2=36$ different ways to choose the schedule. In the case where the morning class is on Saturday, the aft... | 96 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
How many positive integers $n$ exist such that the quotient $\frac{2 n^{2}+4 n+18}{3 n+3}$ is an integer? | As
$$
\frac{2 n^{2}+4 n+18}{3 n+3}=\frac{2}{3}\left[\frac{\left(n^{2}+2 n+1\right)+8}{n+1}\right]=\frac{1}{3}\left(2 n+2+\frac{16}{n+1}\right)
$$
it follows that the expression in parentheses must be a multiple of 3 and, in particular, $n+1$ must divide 16. Thus, $n$ can be $1, 3, 7$ or 15. From the table below, in e... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Four mayors decide to build a circular highway that passes within the boundaries of their cities. Since the four cities are not on the same circle, the mayors hire a company to develop a project for the construction of a circular highway equidistant from the four cities. What is the maximum number of geographically dis... | The number of highways is equal to the number of points that can be the center of a circle (highway) that is equidistant from four given points (cities). Since no circle passes through all four points, if any circle is equidistant from the four points, this circle cannot leave all four points on the inside or all on th... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Riquinho distributed 1000.00 reais among his friends Antônio, Bernardo, and Carlos in the following manner: he gave, successively, 1 real to Antônio, 2 reais to Bernardo, 3 reais to Carlos, 4 reais to Antônio, 5 reais to Bernardo, etc. What was the amount received by Bernardo? | The 1000 reais of Riquinho were divided into increasing installments starting from 1, such that $1+2+3+\cdots+n \leq 1000$. Since $1+2+3+\cdots+n$ is the sum of the first $n$ integers starting from 1, we have $1+2+3+\cdots+n=\frac{1}{2}(1+n) n$. Thus, we want to find the largest $n$ such that $\frac{1}{2}(1+n) n=1+2+3+... | 345 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The diagonals of a rectangle measure $\sqrt{1993} \mathrm{~cm}$. What are the dimensions of the rectangle, knowing that they are integers? | Let $a$ and $b$ be the lengths of the sides of the rectangle. Assuming $a \leq b$, we have $b^{2}<a^{2}+b^{2} \leq 2 b^{2}$, since $a^{2} \leq b^{2}$. The diagonals of the rectangle measure $\sqrt{1993}$, so by the Pythagorean Theorem, $a^{2}+b^{2}=1993$. Thus, $b^{2}<1993 \leq 2 b^{2}$, or $996.5 \leq b^{2}<1993$. The... | 43 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Write in ascending order the multiples of 3 that, when added to 1, are perfect squares, that is, $3,15,24,48, \ldots$ What is the multiple of 3 in the $2006^{\mathrm{th}}$ position? | Writing $a$ for any number in the list, we know that $a$ is a multiple of 3 and that $a+1$ is a perfect square, that is, $a+1=k^{2}$, for some positive integer $k$. Thus, $a=k^{2}-1=(k-1)(k+1)$.
Since $a$ is divisible by 3, then either $k+1$ or $k-1$ is divisible by 3. Therefore, $k$ cannot be divisible by 3, and thus... | 9060099 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Five cards are on a table, and each 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 make a correct decision?
![](ht... | Simone doesn't need to turn over the card that has the number $\mathbf{2}$ because the other side, whether a vowel or a consonant, will satisfy the condition "If a card has a vowel on one side, then it has an even number on the other."
 250
(b) 300
(c) 310
(d) 320
(e) 360 | The correct option is (c).
For the first $\mathrm{R} \$ 1000.00$, the company makes a profit of $1000 \times 6\% = 60$ reais, and for the remaining $\mathrm{R} \$ 5000.00$, the profit is $5000 \times 5\% = 250$ reais. Therefore, the company's profit for that day is $\mathrm{R} \$ 310.00$. | 310 | Algebra | MCQ | Yes | Yes | olympiads | false |
Find the 21st term of the sequence that starts like this:
$$
1 ; 2+3 ; 4+5+6 ; 7+8+9+10 ; 11+12+13+14+15 ; \ldots
$$ | Observe that the 21st term is the sum of 21 consecutive numbers. To find out what these numbers are, we just need to find the first term of the 21st term, which is the sum of 21 consecutive numbers. We observe that, starting from the second term, the first term of
$$
\begin{aligned}
2^{\text{nd}} \text{ term is } 2 & ... | 4641 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On a sheet of paper, 100 characters fit in width and 100 in height. On this sheet, the numbers $1, 2, 3$, and so on, are written successively, with a space between each one and the next. If at the end of a line there is not enough space to write the next number, it is written at the beginning of the next line. What is ... | On the 1$^{\text {st }}$ line, we write the numbers from 1 to 9, each followed by a space, which occupies 18 spaces; 82 spaces remain. Each two-digit number plus one space occupies three places on the line. Since $82=27 \times 3+1$, we complete the 1st line with 27 two-digit numbers, starting from the number 10. Thus, ... | 2220 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Emília's tree grows according to the following rule: two weeks after a branch appears, that branch produces a new branch every week, and the original branch continues to grow. After five weeks, the tree has five branches, as shown in the figure. How many branches, including the main branch, will the tree have at the en... | Let $g_{n}$ denote the number of branches on the tree after $n$ weeks. Since only after two weeks does a branch appear, $g_{2}=1$. In the third week, this branch produces a new branch, that is, $g_{3}=2$. The number of branches in a week is equal to the number of branches that existed in the previous week, plus the new... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Maria ordered a certain number of televisions for the stock of a large store, paying $\mathrm{R} \$ 1994.00$ per television. She noticed that, in the total amount to pay, the digits 0, 7, 8, and 9 do not appear. What is the smallest number of televisions she could have ordered? | Let's say Maria ordered $n$ televisions, paying R \$ 1994.00 per television. The total paid is $1994 n$, and we know that in this multiple of 1994, the digits 0, 7, 8, and 9 do not appear. To limit the value of $n$, we observe that
$$
1994 n=2000 n-6 n
$$
If $6 n<300$, then the number $2000 n-6 n$ has 7 or 8 or 9 in ... | 56 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Filling up the tank of a small car cost, in current values, R$ 29.90 in 1972 and R$ 149.70 in 1992. Which of the values below best approximates the percentage increase in the price of gasoline over this 20-year period?
(a) $20 \%$
(b) $125 \%$
(c) $300 \%$
(d) $400 \%$
(e) $500 \%$ | The correct option is (d).
Solution 1: The price increase was 149.70 - 29.90 = 119.80 reais, which corresponds to
$$
\frac{119.80}{29.90} \times 100 \% = 400.66 \%
$$
Solution 2: The price practically went from 30 to 150, that is, it was multiplied by 5, which is equivalent to the price going from 100 to 500, repres... | 400 | Algebra | MCQ | Yes | Yes | olympiads | false |
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