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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 $A B C D$ be a rectangle with $B C=2 \cdot A B$. Let $E$ be the midpoint of $B C$ and $P$ an arbitrary point on the side $A D$. Let $F$ and $G$ be the feet of the perpendiculars drawn from $A$ to $B P$ and from $D$ to $C P$. We know that $\angle B P C=85^{\circ}$.
 Due to the metric relations in right triangles applied to triangle $A B P$, we have
$$
B E^{2}=A B^{2}=B F \cdot B P
$$
Therefore,
$$
\frac{B E}{B P}=\frac{B F}{B E}
$$
Given the proportional relationship of the last equation and $\angle E B F=\angle E B P$, it follows that triangles $B E F$ and $B E P... | 85 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the figure below, the segments $A M, A K, B_{2} C_{2}$, and $B_{1} C_{1}$ are tangent to the circle with center $O$. If $\angle B_{1} O B_{2}=30^{\circ}$, determine the value of the angle $\angle B_{1} D B_{2}$.
=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}
$$
Therefo... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
For each positive integer $n$, an non-negative integer $f(n)$ is associated in such a way that the following three rules are satisfied:
i) $f(a b)=f(a)+f(b)$.
ii) $f(n)=0$ if $n$ is a prime greater than 10.
iii) $f(1)<f(243)<f(2)<11$.
Knowing that $f(2106)<11$, determine the value of $f(96)$. | Solution
By property $i$), we have $f(243)=f\left(3^{5}\right)=5 f(3)$. Given that
$$
0 \leq f(1)<5 f(3)<f(2)<11
$$
and that $5 f(3)$ is a multiple of 5, we have $5 f(3)=5$, that is, $f(3)=1$. Note that $2106=2 \cdot 3^{4} \cdot 13$. Thus, by property $i$),
$$
f(2106)=f(2)+4 f(3)+f(13)=f(2)+4
$$
From $f(2106)<11$,... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A school has 100 students and 100 lockers numbered from 1 to 100. Initially, all lockers are closed. The first student passes and opens all the lockers; the second passes and closes all the even-numbered ones; the third passes and changes the position of all multiples of 3, that is, he closes the ones that are open and... | Solution
The lockers that remain open are those with numbers that have an odd number of divisors, and this only happens with numbers that are perfect squares. From 1 to 100, we have 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100, which are perfect squares, that is, there are 10 perfect squares. Therefore, the number of lock... | 90 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
a) Verify that if we choose 3 or more integers from the set $\{6 k+1,6 k+2,6 k+3,6 k+4,6 k+5,6 k+6\}$, at least two will differ by 1, 4, or 5.
b) What is the largest number of positive integers less than or equal to 2022 that we can choose so that there are no two numbers whose difference is 1, 4, or 5? | Solution
a) Suppose, for the sake of contradiction, that we choose 3 or more integers and none of them differ by 1, 4, or 5. Since no difference is 1, we can assume that all differ by at least 2, and since no difference is 5, we cannot simultaneously choose $6k+1$ and $6k+6$. If $6k+1$ is not chosen, we must choose $6... | 674 | Combinatorics | 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}
$$
Si... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a competition, the competitors occupy all the seats in a rectangular hall where the seats are arranged in rows and columns such that there are more than two rows and in each row there are more than two seats. At a given moment, these competitors receive the order to shake hands only with their direct neighbors in th... | Solution
a) A competitor who sat in one of the 4 corners of the hall gave 3 handshakes. If a competitor sat on any of the 4 edges, but not positioned in the corners, he gave 5 handshakes. Finally, if a competitor sat in the interior of the hall, he gave 8 handshakes.
The following figure illustrates these possibiliti... | 280 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A company has a profit of $6 \%$ on the first $R \$ 1000$, 00 in daily sales, and $5 \%$ on all sales that exceed $R \$ 1000,00$ on the same day. What is the company's profit on a day when sales reach $R \$ 6000,00$?
(a) $R \$ 250$
(b) $R \$ 300$
(c) $\$ 310$
(d) $R \$ 320$
(e) $R \$ 360$ | (c) For the first $R \$ 1000$ reals, the company has a profit of $R \$ 60$ reals, and for the remaining $R \$ 5000$ reals, the profit is $5000 \times$ $5 \% = 250$ reals. Therefore, the company's profit for that day is $R \$ 310$ reals. | 310 | Algebra | MCQ | Yes | Yes | olympiads | false |
What is the $21^{\varrho}$ term of the sequence
$$
1 ; 2+3 ; 4+5+6 ; 7+8+9+10 ; 11+12+13+14+15 ; \ldots ?
$$ | Observe that the $21^{\text{st}}$ term is the sum of 21 consecutive numbers. Taking the first term of each term, that is, 1, 2, 4, 7, 11, $16, \ldots$, we have that
$$
\begin{aligned}
2 & =1+1 \\
4 & =2+1+1 \\
7 & =3+2+1+1 \\
11 & =4+3+2+1+1 \\
16 & =5+4+3+2+1+1
\end{aligned}
$$
Thus, the first term of the $21^{\text... | 4641 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On a sheet of paper, 100 characters fit in width and 100 in height. On this sheet, the numbers $1,2,3, \ldots$ are written successively with a space between each one. When at the end of a line there is not enough space to write a number, it is written on the next line. What is the last number written on the sheet? | On the $1^{\underline{a}}$ line, we write the numbers from 1 to 9, each followed by a space, occupying 18 spaces, and 82 spaces remain. Each number of 2 digits plus one space occupies 3 places on the line. Since $82=27 \times 3+1$, we complete the $1 \underline{\underline{a}}$ line with 27 numbers of two digits startin... | 2802 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Maria ordered a certain number of televisions at $R \$ 1994.00$ each. She noticed that in the total amount to pay, there are no 0, 7, 8, or 9. What is the smallest number of televisions she ordered?
## Solutions from List 4
Translate the above text into English, please retain the original text's line breaks and forma... | If Maria bought $n$ televisions, then she spent 1994n, which is a multiple of 1994 where the digits $0, 7, 8$, and 9 do not appear. Let's try to limit the value of $n$. First observe that
$$
1994 n=2000 n-6 n
$$
and also that if
$$
6 n<300
$$
then the number $2000 n-6 n$ has 7 or 8 or 9 in the hundreds digit (try s... | 56 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In 1972, filling the tank of a small car cost $R \$ 29.90$, and in 1992, it cost $\$ 149.70$ to fill the same tank. Which of the following values best approximates the percentage increase in the price of gasoline over this 20-year period?
(a) $20 \%$
(b) $125 \%$
(d) $300 \%$
(d) $400 \%$
(e) $500 \%$ | The increase in value was
$$
149.70-29.90=119.80 \text { reais }
$$
which corresponds to:
$$
\frac{119.80}{29.90} \times 100 \% = 400.66 \%
$$
The correct option is (d). | 400 | Algebra | MCQ | Yes | Yes | olympiads | false |
A boy tried to align 480 cans in the shape of a triangle with one can in the $1^{\text{st}}$ row, 2 cans in the $2^{\text{nd}}$ row, and so on. In the end, 15 cans were left over. How many rows does this triangle have? | Suppose the triangle is composed of $n$ rows, so $1+2+3+\cdots+n$ cans were used, thus
$$
480-15=1+2+\cdots+n=\frac{n(n+1)}{2} \Longrightarrow n^{2}+n-930=0
$$
Solving the equation $n^{2}+n-930=0$, we get:
$$
n=\frac{-1 \pm \sqrt{1+4 \times 930}}{2}=\frac{-1 \pm 61}{2}
$$
Thus, $n=30$ is the only positive solution ... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A circle is inscribed in a right triangle. The point of tangency $\mathrm{O}$ divides the hypotenuse into two segments of lengths $6 \mathrm{~cm}$ and $7 \mathrm{~cm}$. Calculate the area of the triangle. | Let $r$ be the radius of the inscribed circle. Using the Pythagorean theorem, we have $(6+7)^{2}=(r+6)^{2}+(r+7)^{2}=r^{2}+12 r+36+r^{2}+14 r+49=2\left(r^{2}+13 r\right)+85$, thus we have $r^{2}+13 r=\frac{169-85}{2}=42$.
On the other hand, the area of the triangle is
$\frac{(r+6)(r+7)}{2}=\frac{r^{2}+13 r+42}{2}=\fr... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Mr. and Mrs. Seventh have 7 children, all born on April 1st, actually over six consecutive April 10ths. This year, for their birthdays, Mrs. Seventh made a cake with candles for each one - the number of candles equal to the number of years of each one. João Seventh, the son who loves Math the most, noticed that this ye... | The births occurred on six 1st of April, so there are twin siblings. Since this year we have 2 more cakes than 2 years ago, it means that 2 years ago the youngest had not been born yet, the second youngest had just been born, and the twins had already been born. Currently, the youngest is 1 year old and the twins are $... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
What is the greatest result we can find when we divide a two-digit number by the sum of its digits? | We are looking for the maximum value of $\frac{10 a+b}{a+b}$, where $a$ and $b$ represent digits, with at least one different from 0. We have
$$
\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
$$
Therefore, if we can find $a$ and $b$ such that $\frac{10 a+... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Starting at noon, João makes, every 80 minutes, a mark at the position of the hour hand of his clock.
a) After how much time will it no longer be necessary to make new marks on the clock?
b) What is the sum of the interior angles of the polygon formed by the marks? | Solution
a) The hour hand will complete one full revolution after $12 \cdot 60=720$ minutes, and during this time no mark will be repeated. Since 720 is a multiple of 80, during this period exactly $\frac{12 \cdot 60}{80}=9$ marks are made on the clock, and in addition, both hands return to their initial positions. Th... | 720 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
a) José learned a method for calculating the product of two numbers more quickly based on factorization:
$$
(n-k)(n+k)=n^{2}-k^{2}
$$
To calculate 23 $\cdot 17$, he chooses $n=20, k=3$ and calculates:
$$
23 \cdot 17=20^{2}-3^{2}=400-9=391
$$
Determine, without using a calculator, the value of $\sqrt{1001 \cdot 1003... | Solution
a) It suffices to choose $n=1002$ and $k=1$, since
$$
\begin{aligned}
\sqrt{1001 \cdot 1003+1} & =\sqrt{1002^{2}-1^{2}+1} \\
& =\sqrt{1002^{2}} \\
& =1002
\end{aligned}
$$
b)
$$
\begin{aligned}
(n(n+3)+1)^{2} & =n^{2}(n+3)^{2}+2 n(n+3)+1 \\
& =n(n+3)[n(n+3)+2]+1 \\
& =n(n+3)\left[n^{2}+3 n+2\right]+1 \\
& ... | 4062239 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
An arithmetic progression, commonly referred to as an A.P., is a sequence in which each term, starting from the second, is equal to the sum of the previous term and a fixed value \( r \) called the common difference or the ratio of the progression. For example, the sequence below is an arithmetic progression with the f... | Solution
a) If $a_{1}=2$ and $r=3$, we have
$$
\begin{aligned}
& a_{2}=a_{1}+3=5 \\
& a_{3}=a_{2}+3=8 \\
& a_{4}=a_{3}+3=11
\end{aligned}
$$
b) Let $a_{1}=d$ and $r$ be the common difference. Then, we have:
$$
\begin{aligned}
& a_{1}=d, \quad a_{2}=d+r, \quad a_{3}=d+2 r, \quad a_{4}=d+3 r, \quad a_{5}=d+4 r, \quad... | 109 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Pedro decided to take all his children, boys and girls, for ice cream at the Ice Cream Math parlor. At the parlor, there are 12 different ice cream flavors, and each child ordered a combo with 3 scoops of ice cream. After leaving the parlor, Pedro realized that exactly two scoops of each available flavor were ordered i... | Solution
a) Let $n$ be the number of Pedro's children. In total, $3 n$ scoops of ice cream were ordered. Since each of the 12 flavors was ordered twice, we have $3 n=2 \cdot 12$, that is, $n=8$. Therefore, Pedro has 8 children.
b) Let $x$ be the number of boys and $y$ be the number of girls. From the previous item, w... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the drawing below, $ABCD$ is a trapezoid and its diagonals $AC$ and $BD$ are perpendicular. Additionally, $BC=10$ and $AD=30$.
a) Determine the ratio between the segments $BE$ and $ED$.
... | Solution
a) Since $B E$ and $A D$ are parallel, $\angle E B C=\angle E D A$ and $\angle B C E=\angle C A D$. Consequently, triangles $\triangle B E C$ and $\triangle E A D$ are similar, and $B E / E D=B C / A D=10 / 30=1 / 3$. Similarly, we can show that $E C / A E=1 / 3$.
b) From the previous item, we know that $E D... | 192 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A technique widely used to calculate summations is the Telescoping Sum. It consists of "decomposing" the terms of a sum into parts that cancel each other out. For example,
$$
\begin{aligned}
& \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}= \\
& \left(\frac{1}{1}-\frac{1}{2}\right)+\le... | Solution
a) See that the first odd number is $2 \cdot 1-1$ and, knowing that odd numbers increase by 2, we can conclude that the odd number at position $m$ in our count is
$$
\begin{aligned}
2 \cdot 1-1+\underbrace{2+2+\ldots+2}_{m-1 \text { times }} & =2 \cdot 1-1+2(m-1) \\
& =2 m-1
\end{aligned}
$$
To verify that ... | 764049 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
How many digits does the number $3^{100}$ have? Well, we can give an approximate answer to this question without using a calculator, simply by comparing it with powers of 10. Notice that $3^{2}<10$ allows us to conclude that $\left(3^{2}\right)^{50}=3^{100}<10^{50}$. Therefore, $3^{100}$ has at most 50 digits since $10... | Solution
See that $2^{3}=8$ is less than 10, so
$$
2^{100}2^{98}=\left(2^{7}\right)^{14}>\left(10^{2}\right)^{14}=10^{28}
$$
Since $10^{28}$ is the smallest number with 29 digits, $2^{100}$ has at least 29 digits. It is thus proven that $k=29$ satisfies the condition given that $29 \leq N \leq 34$.
# | 29 | Number Theory | 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
$$
\begin{aligned}
\frac{\sqrt{\frac... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Observe the equation:
$$
\begin{aligned}
(1+2+3+4)^{2} & =(1+2+3+4)(1+2+3+4) \\
& =1 \cdot 1+1 \cdot 2+1 \cdot 3+1 \cdot 4+2 \cdot 1+2 \cdot 2+2 \cdot 3+2 \cdot 4+ \\
& +3 \cdot 1+3 \cdot 2+3 \cdot 3+3 \cdot 4+4 \cdot 1+4 \cdot 2+4 \cdot 3+4 \cdot 4
\end{aligned}
$$
Note that $4 \times 4=16$ products are formed when ... | Solution
a) Each product that appears in the final sum is an expression of the form $x \cdot y \cdot z$ where $x$ is a number from the first parenthesis, $y$ is a number from the second, and $z$ is a number from the third. Since there are 4 possible options for each of these numbers, by the multiplicative principle, w... | 10000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
a) In the figure below, $A D=D C, A E=B D, \angle A E C=90^{\circ}$. Determine the value of the angle $\angle C B D$.
b) In the triangle $\triangle A B C$ below, $B D$ is the bisector of the... | Solution
a) From point $D$, draw the segment perpendicular to side $B C$ as indicated in the figure below. Since $D$ is the midpoint of $A C$ and $D F \| A E$, we can conclude that $D F$ is the midline of triangle $\triangle A E C$ with respect to base $A E$. Therefore, if $A E=2 x$, then $D F=A E / 2=x$. Thus, $\sin ... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
a) Verify that $(1+\operatorname{tg} k)\left(1+\operatorname{tg}\left(45^{\circ}-k\right)\right)=2$.
b) Given that
$$
\left(1+\operatorname{tg} 1^{\circ}\right)\left(1+\operatorname{tg} 2^{\circ}\right) \cdot \ldots \cdot\left(1+\operatorname{tg} 45^{\circ}\right)=2^{n}
$$
find $n$. | Solution
a)
$$
\begin{aligned}
\operatorname{tg}\left(45^{\circ}-k\right)+1 & =\frac{\operatorname{sen}\left(45^{\circ}-k\right)}{\cos \left(45^{\circ}-k\right)}+1 \\
& =\frac{\operatorname{sen} 45^{\circ} \cos k-\cos 45^{\circ} \operatorname{sen} k}{\cos 45^{\circ} \cos k+\operatorname{sen} 45^{\circ} \operatorname{... | 23 | 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 |
On a blackboard, the 2014 positive integers from 1 to 2014 are written. The allowed operation is to choose two numbers $a$ and $b$, erase them, and write in their place the numbers $gcd(a, b)$ (Greatest Common Divisor) and $lcm(a, b)$ (Least Common Multiple). This operation can be performed on any two numbers on the bl... | Solution
The maximum number of 1s we can leave is 1007. First, we will show how to obtain them. For this, it is enough to take the pairs of consecutive numbers, $(1,2),(3,4),(5,6)$, ..., $(2013,2014)$ and perform the operation on each pair. Knowing that consecutive numbers do not have a common factor, each of the grea... | 1007 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a chess tournament, all players faced each other exactly once. In each match, a player earns 1 point for a win, $1 / 2$ for a draw, and 0 points for a loss. At the end of the tournament, a reporter summed up the scores of all the players and obtained 190 points. In this type of tournament, the winner is the one who ... | Solution
a) Let $J$ be the number of players. Each match is worth a total of 1 point, whether it's $1+0=1$ or $1 / 2+1 / 2=1$. Therefore, the total score is equal to the number of matches. Since each of the $J$ players faces each of the other $J-1$ players, we might think that the total number of games would be $J(J-1... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
a) A worm always moves along a straight line. Every day, it advances 5 m and retreats 3 m. At the end of 15 days, how far from the starting point will the worm be?
b) At the end of these 15 days, how many meters in total will this worm have walked?
c) Another worm also moves along a straight line, but in a different ... | Solution
a) Each day, the worm advances 5 meters and retreats 3 meters. Therefore, at the end of the day, the worm advances $5-3=2$ meters. After 15 days, the worm will be $15 \times 2=30$ meters from the starting point.
b) Each day, the worm travels $5+3=8$ meters. Therefore, after these 15 days, the worm will have ... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
On the board below, it is allowed to move any object from its square to any adjacent empty square above, below, to the side, or diagonally.
a) Show how to swap the positions of all the hats ... | Solution
a) Below, we show a sequence of five moves to swap the hats with the trophies (there is another one too)!
The argument to show that it is not possible to swap them with fewer than... | 2001 | Logic and Puzzles | 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 |
In the following $4 \times 4$ board, 4 towers, 4 knights, 4 bishops, and 4 pawns must be placed such that in each row and each column the pieces placed are distinct, as in the example:
| $B$ | $T$ | $P$ | $C$ |
| :--- | :--- | :--- | :--- |
| $P$ | $B$ | $C$ | $T$ |
| $T$ | $C$ | $B$ | $P$ |
| $C$ | $P$ | $T$ | $B$ |
... | Solution
Three pieces must be placed in the first row, let's call them $\alpha$, $\beta$, and $\theta$ as in the following figure:
In the cell $(b, 2)$, a piece different from $\beta$ must ... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The following figure shows a square $A B C D$.
If $\measuredangle A M B=60^{\circ}$ and $\measuredangle D M N=60^{\circ}$, calculate $\measuredangle M B N$.
# | Solution
Observe that $\measuredangle B M N=(180-60-60)^{\circ}=60^{\circ}$. In particular, $M B$ is the bisector of the angle $A M N$. If we draw the segment $\overline{B D}$, we see that this segment is the bisector of the right angle $A D C$.
 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 |
Santa Claus arrived at Arnaldo and Bernaldo's house carrying ten distinct toys numbered from 1 to 10 and said to them: "the toy number 1 is for you, Arnaldo and the toy number 2 is for you, Bernaldo. But this year, you can choose to keep more toys as long as you leave at least one for me". In how many ways can Arnaldo ... | Solution
For each of the 8 toys, from number 3 to number 10, we must decide whether it belongs to Arnaldo, Bernaldo, or should be left for Santa Claus. If we multiply then
$$
\underbrace{3 \times 3 \times \cdots \times 3}_{8 \text { times }}
$$
we will count the ways to distribute the toys among Arnaldo, Bernaldo, a... | 6305 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The pirate law stipulates that, to divide the coins of a treasure, the captain must choose a group of pirates (excluding himself). Then, the captain must distribute the same number of coins to each of the pirates in this group, in such a way that it is not possible to give each of them any of the remaining coins (respe... | Solution
a) Let $N$ be the number of coins in the treasure. When Barbaroxa chooses 99 pirates to divide the $N$ coins, 51 coins are left for him, that is, he can divide $N-51$ coins among 99 pirates, or equivalently, the number $N-51$ is divisible by 99. In particular, the number $N-51+99=N+48$ is also divisible by 99... | 93 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the following sheet, a $4 \times 6$ grid was drawn and then the diagonal from $A$ to $B$ was traced.
Notice that the diagonal $A B$ intersects the grid at 9 points:
The diagonal $AB$ intersects each vertical line at exactly one point. This gives us 18 points of intersec... | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Knowing that $144 \times 177=25488$ we can conclude that $254.88 \div 0.177$ is equal to
(a) 1440
(b) 14.4
(c) 1.44
(d) 0.144
(e) 144 | Performing the division we have:
$$
\frac{254.88}{0.177}=\frac{254880}{177}=\frac{144 \times 177 \times 10}{177}=1440
$$ | 1440 | Algebra | MCQ | Yes | Yes | olympiads | false |
In a test with 84 questions, if you answer 58/84 correctly, then what is your percentage of correct answers? | The division of 58 by 84 is: $58 \div 84=0.69047 \ldots$ Multiplying by 100 gives us the percentage of correct answers as $0.69047 \times 100=$ $69.047\%$, which is approximately $69\%$. | 69 | Other | math-word-problem | Yes | Yes | olympiads | false |
Jorge walks along a rectangular path, where twelve trees are placed with a $5 \mathrm{~m}$ distance between two consecutive ones. Jorge plays a game of touching each tree during his walk.
F... | The figures illustrate the route that Jorge took:
- walking $32 m$ at the beginning, he touches 7 trees and stops $2 m$ from the last one he touched;
- going back $18 m$, he touches 4 trees and stops $1 m$ from the last one he touched;
- upon returning $22 m$, he touches 5 trees and stops $1 m$ from the last one he to... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The squares of the natural numbers from 1 to 99 were written one after another, forming the number 14916253649 ... What is the digit that occupies the $100^{th}$ position? (The positions are counted from left to right: the $1^{st}$ position is the 1, the $2^{nd}$ is the 4, etc.) | Separating the numbers whose squares have 1, 2, and 3 digits, we have:
1 digit: $1, 2, 3$
2 digits: $4, 5, 6, 7, 8, 9$
3 digits: $10, 11, 12, \ldots, 31$
Up to $31^{2}$, the sequence has $3+12+66=81$ digits.
$$
\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 |
A school library bought 140 new books, resulting in $\frac{27}{25}$ of books. The number of books before the purchase, is:
(a) 1750
(b) 2500
(c) 2780
(d) 2140
(e) 1140 | When buying 140 books, the library ended up with $\frac{27}{25}$ of the number of books, so 140 corresponds to $\frac{2}{25}$ of the books in the library. Therefore, we have:
$$
\begin{aligned}
& \frac{2}{25} \longrightarrow 140 \\
& \frac{1}{25} \longrightarrow 140 \div 2=70 \\
& \frac{25}{25} \longrightarrow 70 \tim... | 1750 | Algebra | MCQ | Yes | Yes | olympiads | false |
How many fractions less than 1 exist, such that the numerator and denominator are one-digit natural numbers? | For a fraction to be less than 1, the numerator must be less than the denominator. The fractions are:
- with denominator 2: $\frac{1}{2}$
- with denominator 3: $\frac{1}{3}$ and $\frac{2}{3}$
- with denominator 4: $\frac{1}{4}, \underbrace{\frac{2}{4}}_{1 / 2}$ and $\frac{3}{4}$
- with denominator 5: $\frac{1}{5}, \fr... | 27 | 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.
 Who is Paulo's wife?
b) Who is Valdir's wi... | Solution
a) Let's build a table with the initials of the people, where the husbands will be in the first column and the wives in the first row. Each box associates a man with a woman. We will also use the notation $x$ for "not a couple" and $\checkmark$ for "is a couple".
| | $C$ | $L$ | $B$ |
| :--- | :--- | :--- |... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the figure, each of the 5 letters has a certain number of circles. We have 3 colors (blue, red, orange) 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 |
The numbers $1,2,3, \ldots, 14$ must be written at the 14 vertices of the polygonal line below so that the sums of the 4 numbers written on each of the 7 segments of the polygonal line are the same.
 The sum of all the numbers that must be written is
$$
1+2+3+\ldots+14=105
$$
b) When we sum the numbers written in each of the 7 segments, each of them will be added twice. Therefore, the common sum for each of these segments is $\frac{105 \cdot 2}{7}=$ 30.
c) The figure below shows a possible distribut... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
How many positive integers $n$ exist such that $\frac{2 n^{2}+4 n+18}{3 n+3}$ is an integer? | As
$$
\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)
$$
it follows that $n+1$ must divide 16. Thus, $n$ must belong to the set $\{1,3,7,15\}$. In each of these cases, we have
| $n$ | $\frac{2 n^{2}+4 n+18}{3 n+3}$ |
| :---: | :---: |
| 1 |... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
If $n$ is a natural number, we denote by $n!$ the product of all integers from 1 to $n$. For example: $5! = 1 \times 2 \times 3 \times 4 \times 5$ and 13! $=$ $1 \times 2 \times 3 \times 4 \times 5 \times \ldots \times 12 \times 13$. By convention, $0! = 1$. Find three different integers $a, b$, and $c$, between 0 and ... | Firstly, note that since the number has 3 digits, the largest digit must be less than or equal to 6, as $7!>1000$. Since the number must have 3 digits and $4!=1 \times 2 \times 3 \times 4=24$, one of the digits must be 5 or 6, but $6!=720$ would imply that the sum has a digit greater than or equal to 7, so the largest ... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Riquinho distributed $R \$ 1000.00$ reais among his friends: Antônio, Bernardo, and Carlos in the following manner: he gave, successively, 1 real to Antônio, 2 reais to Bernardo, 3 reais to Carlos, 4 reais to Antônio, 5 reais to Bernardo, etc. How much did Bernardo receive? | The money was divided into installments in the form
$$
1+2+3+\cdots+n \leq 1000
$$
Since $1+2+3+\cdots+n$ is the sum $S_{n}$ of the first $n$ natural numbers starting from $a_{1}=1$, we have:
$$
S_{n}=\frac{\left(a_{1}+a_{n}\right) n}{2}=\frac{(1+n) n}{2} \leq 1000 \Longrightarrow n^{2}+n-2000 \leq 0
$$
We have tha... | 345 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The five cards below are on a table, and each one has a number on one side and a letter on the other. Simone must decide whether the following statement is true: "If a card has a vowel on one side, then it has an even number on the other." What is the minimum number of cards she needs to turn over to decide correctly?
... | 
She does not need to turn over the card with the number 2, because whether it is a vowel or a consonant, it meets the condition in the same way. She also does not need to turn over the card w... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The sum $S_{n}=9+19+29+39+\cdots+a_{n}$ denotes the sum of the first $n$ natural numbers ending in 9. What is the smallest value of $n$ for which $S_{n}$ is greater than $10^{5}$? | Note that $S_{n}$ is the sum of the first $n$ terms of an arithmetic progression whose first term is $a_{1}=9$ and the common difference is $r=10$. Substituting these data into the formula $a_{n}=a_{1}+(n-1) r$ we get $a_{n}=9+10(n-1)$. On the other hand, note that:
$$
\begin{aligned}
& 9=9+0 \cdot 10 \\
& 19=9+1 \cdo... | 142 | Algebra | 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 the other in the afternoon, not on the same day nor on consecutive days. In the morning, there are swimming classes from Monday to Saturday, at $9 \text{ am}$, $10 \text{ am}$, and $11 \text{ am}$, and in ... | 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 3 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, then the af... | 96 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Cláudia likes to play with numbers of two or more digits. She chooses one of these numbers, multiplies its digits, and if the product has more than one digit, she adds them. She calls the final result the transformed number of the chosen number. For example, the transformed number of 187 is 11, because $1 \times 8 \tim... | Solution
a) First, we multiply the digits of 79, obtaining $7 \times 9=63$, and then we sum the digits of this product, obtaining $6+3=9$. Therefore, the transformed number of 79 is 9.
b) Claudia's game has two stages: the first, in which she multiplies the digits, and the second, in which she sums the digits of the ... | 171 | Number Theory | 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 |
A "mathemagician" performs magic with green, yellow, blue, and red cards, numbered from 1 to 13 for each color. He shuffles the cards and says to a child: "Without me seeing, choose a card, calculate double the number on the card, add 3, and multiply the result by 5. Then
add 1, if the card is green;
add 2, if the ca... | Solution
a) To know the number that Joãozinho should tell the mathemagician, he must do four calculations:
$1^{st}$ calculation: multiply the number on the chosen card by 2;
$2^{nd}$ calculation: add 3 to the result of the first calculation;
$3^{rd}$ calculation: multiply by 5 the result of the second calculation;
... | 49 | Logic and Puzzles | math-word-problem | 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 |
To obtain the summary of a number with up to 9 digits, you should write how many digits it has, then how many of those digits are odd, and finally how many are even. For example, the number 9103405 has 7 digits, 4 of which are odd and 3 are even, so its summary is 743.
a) Find a number whose summary is 523.
b) Find a... | Solution
a) Consider a number whose summary is 523. Then it has five digits (523), of which two are odd (523) and three are even (523). We can form many numbers satisfying these conditions; some examples are 11222, 23456, and 36854.
b) Since the summary of any number has three digits, we see that for a number to be e... | 321 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Two natural numbers form a pair when they have the same number of digits and their sum only contains the digit 9. For example, 225 and 774 form a pair, because both have three digits and \(225 + 774 = 999\).
a) What is the number that forms a pair with 2010?
b) How many pairs are formed by two-digit numbers?
Special... | Solution
a) The number that forms a pair with 2010 is 7989, because both have 4 digits and their sum is $2010+7989=$ 9999.
b) There are ninety two-digit numbers, namely, the numbers from 10 to 99. Of these numbers, only those that start with 9 do not have a pair, that is, the ten numbers from 90 to 99. Therefore, eig... | 7989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A two-digit number $A$ is a supernumber if it is possible to find two numbers $B$ and $C$, both also two-digit numbers, such that:
- $A = B + C$;
- the sum of the digits of $A$ = (sum of the digits of $B$) + (sum of the digits of $C$).
For example, 35 is a supernumber. Two different ways to show this are $35 = 11 + 2... | Solution
a) Two ways to show that 22 is a super number are $22=10+12$ and $22=11+11$, since $2+2=$ $(1+0)+(1+2)$ and $2+2=(1+1)+(1+1)$.
b) We present below all the ways to write 49 as the sum of two two-digit numbers, always placing the smaller of the two on the left:
$$
\begin{aligned}
49 & =10+39 \\
49 & =11+38 \\... | 80 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Gabriel draws squares divided into nine cells and writes the natural numbers from 1 to 9, one in each cell. He then calculates the sum of the numbers in each row and each column. The figure shows one of Gabriel's squares; note that the sum of the numbers in the third row is \(5+8+2=15\) and the sum of the numbers in th... | Solution
a) Adding the sums of the rows is the same as adding all the numbers in the square; thus, the sum of the sums of the rows is $1+2+3+4+5+6+7+8+9=45$. The same can be said about the sum of the sums of the columns, and we conclude that the sum of all the sums is $2 \times 45=90$. Therefore, the missing sum is $9... | 19 | Logic and Puzzles | math-word-problem | 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.
 All faces have the same color?
b) Five faces have the same color and one face has a different color from the rest?
c) All faces have different colors?
Note: remember, for example, that the two paintings below are the same, s... | Solution
a) Since all faces must be the same color, we can only paint the die with one color. With six colors to choose from, there are six distinct ways to paint the cube.
b) Paint one face with one of the six possible colors. There are six possibilities for this. For each of these possibilities, the remaining faces... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Emílio likes to propose mathematical and animal challenges. He wrote on a piece of paper the following sum:
$$
\begin{array}{rcccc}
& \mathbf{G} & \mathbf{A} \mathbf{O} \\
+ & \mathbf{P} & \mathbf{U} & \mathbf{A} \\
\hline \mathbf{P} & \mathbf{U} & \mathbf{M} & \mathbf{S}
\end{array}
$$
Emílio said that the sum abov... | Solution
a) Since the number PUMAS has five digits, and the numbers GATO and PUMA have only four digits, it is necessarily true that the first digit of the number PUMAS is equal to 1, that is, $\mathbf{P}=1$.
b) To find the value of $\mathbf{U}$, we need to note that there are two possibilities:
$$
\mathbf{G}+1=10+\... | 10652 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Nine points are drawn on a sheet of paper, as shown in the following figure:
a) In how many ways is it possible to choose three collinear points?
b) In how many ways is it possible to choose four points such that three of them are collinear? | Solution
a) There are eight ways to choose three collinear points as illustrated in the figure below:
b) To choose a set of four points containing three collinear points, we can adopt a pro... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A number is called special if it does not contain the digit zero and, in addition, the sum of its digits is equal to twice the first digit. For example, the number 8161 is special because:
- none of its digits is zero;
- the sum of all its digits is $8+1+6+1=16$;
- the double of its first digit is $8 \times 2=16$.
a)... | Solution
a) They exist. For example, the number 51112 is special, as zero does not appear among its digits, and, moreover, the double of its first digit is $2 \times 5=10$, equal to the sum of its digits, given by $5+1+1+1+2=10$.
b) The largest special number is 9111111111. No other number larger than it can be speci... | 9621 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A Professor Fernando's tree grows according to the following rule:
- in the first week, the tree starts growing with just one branch;
- after growing for two weeks, this branch gives rise to a new branch per week;
- each new branch generated continues to grow, and after growing for two weeks, it gives rise to a new br... | Solution
a) Following the growth rules of Professor Fernando's tree, we can continue the drawing shown in the problem statement to obtain an illustration of this tree after the sixth week of growth:
Then she declared that some points of the hexagon would be called cool. A point of Clariss... | Solution
a) Consider the figure below that appears in the problem statement:
In this figure, there are four legal triangles. If we add the diagonal $\overline{B D}$, it will create one mor... | 2014 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Aline received as a gift a cube composed of $4 \times 4 \times 4$ small white cubes, as shown in the following figure.
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 natural number is called a square if it can be written as the product of two equal numbers. For example, 9 is a square, since $9=3 \times 3$. The first squares are 1, 4, 9, 16, 25, ... A natural number is called a cube if it can be written as the product of three equal numbers. For example, 8 is a cube, since $8=2 \t... | Solution
a) Let's underline the numbers that are squares or cubes:
$$
\underline{\underline{1}}, 2,3, \underline{\underline{4}}, 5,6,7, \underline{\underline{8}}, \underline{\underline{9}}, 10,11,12,13,14,15, \underline{\underline{16}}, 17, \ldots
$$
Counting the positions of the numbers that were not underlined, we... | 2067 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
If $(x, y)$ is a solution to the system
$$
\left\{\begin{array}{l}
x y=6 \\
x^{2} y+x y^{2}+x+y=63
\end{array}\right.
$$
determine the value of $x^{2}+y^{2}$. | Solution
We have
$$
\begin{aligned}
63 & =x^{2} y+x y^{2}+x+y \\
& =x y(x+y)+(x+y) \\
& =6(x+y)+(x+y) \\
& =7(x+y)
\end{aligned}
$$
Therefore, $x+y=9$. Thus,
$$
\begin{aligned}
x^{2}+y^{2} & =(x+y)^{2}-2 x y \\
& =81-12 \\
& =69
\end{aligned}
$$ | 69 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $S(n)$ be the sum of the digits of an integer $n$. For example, $S(327)=3+2+7=12$. Find the value of
$$
A=S(1)-S(2)+S(3)-S(4)+\ldots-S(2016)+S(2017)
$$ | Solution
If $m$ is even, the number $m+1$ has the same digits as $m$ except for the units digit, which is one unit larger. Therefore, $S(m+1)-S(m)=1$. This allows us to group the terms of the sequence into pairs with a difference of 1:
$$
\begin{array}{r}
S(1)-S(2)+S(3)-S(4)+\ldots-S(2016)+S(2017)= \\
S(1)+(S(3)-S(2)... | 1009 | Number Theory | 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 |
A quadrilateral $A B C D$ is inscribed in a circle with center $O$. It is known that the diagonals $A C$ and $B D$ are perpendicular. On each side, we construct semicircles externally, as shown in the figure below.
 Initially, observe that
$$
\begin{aligned}
A \hat{O} B + C \hat{O} D & = \\
2 \cdot (A \hat{C} B + C \hat{B} D) & = \\
2 \cdot 90^{\circ} & = 180^{\circ}
\end{aligned}
$$
Let $M$ and $N$ be the midpoints of the sides $AB$ and $CD$, respectively. Then $B O \hat{O} M = \frac{A \hat{O} B}{2} = \frac{180^{\c... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $ABC$ be any triangle, and construct equilateral triangles $ABD$ and $ACE$ externally to $ABC$.
a) Verify that $DC = BE$.
b) Let $F$ be the intersection point of $DC$ and $BE$. Find the angle $\angle AFB$. | Solution
a) Observe that $A D=A B, A C=A E$ and $\angle D A C=\angle B A E=\angle B A C+60^{\circ}$. Therefore, $\triangle A D C \equiv$ $\triangle A B E$, by the $S A S$ case, and thus $D C=B E$.
b) From the previous congruence,
$$
\angle A D F=\angle A D C=\angle A B E=\angle A B F
$$
If $C D$ intersects the circ... | 120 | Geometry | proof | 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 |
In a certain country, there are 21 cities and the government plans to build $n$ roads (all two-way), each connecting exactly 2 of the country's cities. What is the smallest value of $n$ such that, regardless of how the roads are built, it is possible to travel between any 2 cities (possibly passing through intermediate... | Solution
We will say that a group of cities is connected if it is possible to travel from one of them to any other by means of roads. Naturally, distinct connected groups of cities cannot have cities in common. If the government builds all the roads of a group of 20 cities and leaves one city isolated without roads, t... | 191 | Combinatorics | 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 |
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