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Juarez used the digits 1, 2, 3, 4, and 5 to write the five-digit number $\overline{a b c d e}$. Without revealing what this number is, he told Luciana that:
$\cdot$ the number $\overline{a b c}$ is divisible by 4;
- the number $\overline{b c d}$ is divisible by 5;
$\cdot$ the number $\overline{c d e}$ is divisible b... | Solution
Every number divisible by 5 has the unit digit equal to 5 or equal to 0. Since the digit 0 was not used by Juarez, for $\overline{b c d}$ to be divisible by 5, we must necessarily have $d=5$.
Every number divisible by 3 is such that the sum of its digits is also divisible by 3. Thus, since $\overline{c d e}$... | 12453 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A natural number $N$ greater than 10 is called a "super-square" if the number formed by each pair of consecutive digits of the number $N$ (considered in the same order) is always a perfect square. For example, 8164 is a "super-square" because the numbers 81, 16, and 64 are perfect squares. Other examples of super-squar... | Solution
a) It will be useful later to observe the list of all two-digit perfect squares:
$$
16, \quad 25, \quad 36, \quad 49, \quad 64, \quad 81
$$
The number formed by the first two digits of a super-square must be one of the numbers in list (.1). Thus, let's start counting the super-squares from the number 16.
-... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The figure below is composed of a square and a regular pentagon.
Calculate the sum of the angles $a^{\circ}$ and $b^{\circ}$.
Facts that can help. (You can use them!). The sum of the interi... | Solution
Consider the triangle $A P M$ in the following figure:
Since $A \hat{M} P=(180-a)^{\circ}$, and the sum of the internal angles of a triangle is $180^{\circ}$, then $A \hat{P} M=(a... | 324 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The figure below contains a square and two congruent right triangles.
Using these polygons, we form a rectangle and a trapezoid as shown in the following figure:
.
For example, if the numbers 2, 3, 4, an... | Solution
a) The number of numbers written by Yndira is equal to the number of pairs that can be formed with the 20 numbers written by Kiara.
In how many ways can we choose a pair of elements from a set containing 20 elements? For the choice of the first element, we have 20 possibilities. For each fixed choice of the ... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
If four distinct positive integers $m, n, p$ and $q$ satisfy the equation
$$
(7-m)(7-n)(7-p)(7-q)=4
$$
then the sum $m+n+p+q$ is equal to:
(a) 10
(b) 21
(c) 24
(d) 26
(e) 28 | As $m, n, p$ and $q$ are integers, then $7-m$, $7-n$, $7-p$ and $7-q$ are also integers. Now,
$$
4=(-1) \times(-2) \times 1 \times 2
$$
is the only decomposition of 4 into a product of distinct integers. It follows that
$$
(7-m)+(7-n)+(7-p)+(7-q)=(-1)+(-2)+1+2
$$
and from this we obtain $m+n+p+q=28$. The correct op... | 28 | Algebra | MCQ | Yes | Yes | olympiads | false |
To number the pages of a dictionary, the digit 1 was printed 1988 times. How many pages does this dictionary have? | Let's observe that:
- every 10 numbers, the digit 1 is printed once in the units place,
- every 100 numbers, the digit 1 is printed 10 times in the tens place,
- every 1000 numbers, the digit 1 is printed 100 times in the hundreds place.
Thus, from 1 to 999, the digit 1 is printed:
100 times in the units place + 100... | 3144 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine um valor de $n$ para o qual o numero $2^{8}+2^{11}+2^{n}$ seja um quadrado perfeito. | Observe que
$$
2^{8}+2^{11}+2^{n}=\left(2^{4}\right)^{2}+2 \times 2^{4} \times 2^{6}+\left(2^{\frac{n}{2}}\right)^{2}
$$
Logo, se $n=12$, temos
$$
2^{8}+2^{11}+2^{12}=\left(2^{4}+2^{6}\right)^{2}
$$
Logo $n=12$ é uma solução.
Solução Geral: $\operatorname{Se} 2^{8}+2^{11}+2^{n}=k^{2}$, então:
$$
\begin{aligned}
2... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The reverse of a two-digit integer is the number obtained by reversing the order of its digits. For example, 34 is the reverse of 43. How many numbers exist such that when added to their reverse, they give a perfect square? | Let's denote by $a b \mathrm{e}$ ba the number and its reverse. We have that
$$
a b+b a=10 a+b+10 b+a=11(a+b)
$$
On the other hand, $a \leq 9$ and $b \leq 9$, so, $a+b \leq 18$. Since 11 is a prime number and $a+b \leq 18$, for $11(a+b)$ to be a perfect square, we can only have $a+b=11$.
Thus, we have 8 numbers sati... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A game is played with four integers in the following way: add three of these numbers, divide this sum by 3, and add the result to the fourth number. There are four ways to do this game, yielding the following results: 17, 21, 23, and 29. What is the largest of the four numbers? | Let $a, b, c$ and $d$ be the numbers we are looking for. The given numbers are
$$
\frac{a+b+c}{3}+d, \frac{a+b+d}{3}+c, \frac{a+c+d}{3}+b \text{ and } \frac{b+c+d}{3}+a
$$
but we do not know their order. Since
$$
\begin{aligned}
\frac{a+b+c}{3}+d+\frac{a+b+d}{3}+c+\frac{a+c+d}{3}+b+\frac{b+c+d}{3}+a & =2(a+b+c+d) \\... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In a warehouse, a dozen eggs and 10 apples had the same price. After a week, the price of eggs fell by $2 \%$ and the price of apples rose by 10\%. How much more will be spent on the purchase of a dozen eggs and 10 apples?
(a) $2 \%$
(b) $4 \%$
(c) $10 \%$
(d) $12 \%$
(e) $12.2 \%$ | We can assume that the initial price of a dozen eggs is $R \$ 1.00$, so 10 apples also cost $R \$ 1.00$. Since the price of eggs has dropped by $2 \%$, the new price of a dozen eggs is $R \$ 0.98$. The price of apples has increased by $10 \%$, so the new price of 10 apples is $R \$ 1.10$. Thus, before it cost $R \$ 2.0... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
The number
$$
A=(\sqrt{6}+\sqrt{2})(\sqrt{3}-2) \sqrt{\sqrt{3}+2}
$$
is equal to:
(a) $-\sqrt{3}$
(b) $-\sqrt{2}$
(c) -2
(d) 1
(e) 2 | (c) Like
$$
\begin{aligned}
A^{2} & =[(\sqrt{6}+\sqrt{2})(\sqrt{3}-2) \sqrt{\sqrt{3}+2}]^{2} \\
& =(\sqrt{6}+\sqrt{2})^{2}(\sqrt{3}-2)^{2}(\sqrt{\sqrt{3}+2})^{2} \\
& =(\sqrt{6}+\sqrt{2})^{2}(\sqrt{3}-2)^{2}(\sqrt{3}+2) \\
& =(\sqrt{6}+\sqrt{2})^{2}(\sqrt{3}-2)[(\sqrt{3}-2)(\sqrt{3}+2)] \\
& =(6+2 \sqrt{12}+2)(\sqrt{3... | -2 | Algebra | MCQ | Yes | Yes | olympiads | false |
Five points lie on the same line. When we list the ten distances between any two of these points, from smallest to largest, we get $2,4,5,7,8, k, 13,15,17,19$. What is the value of $k$? | Solution 1: - This solution is a bit difficult to write because it is based on "trial and error." We start by drawing a number line and placing the points 0 and 19. Since the first distance is 2, we mark our first three points:
 1
(B) 2
(C) 4
(D) 6
(E) 8 | The correct answer is (B).
Of the total number of students in this class, $60 \%$ went to do community work, that is, $0.6 \times 40=24$. The minimum number of female students who participated in this work is obtained when the number of male students who participated is maximum, that is, when 22 male students are invo... | 2 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
By joining four identical trapezoids with bases $30 \mathrm{~cm}$ and $50 \mathrm{~cm}$ and equal non-parallel sides, as shown in the figure, we can form a square with an area of $2500 \mathrm{~cm}^{2}$, with a square "hole" in the middle. What is the area of each trapezoid, in $\mathrm{cm}^{2}$?
(A) 200
(B) 250
(C) 30... | The correct answer is (E).
By combining the four trapezoids, we form a square with a side length of $50 \mathrm{~cm}$, and thus an area of $2500 \mathrm{~cm}^{2}$. Since the "hole" square h... | 400 | Geometry | MCQ | Yes | Yes | olympiads | false |
Discover the rule used for the filled cells and complete the table. What is the value of A?
| $\mathbf{0}$ | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ |
| :---: | :---: | :---: | :---: | :---: |
| $\mathbf{1}$ | 2 | 5 | 10 | |
| $\mathbf{2}$ | | | | |
| $\mathbf{3}$ | | | | |
| $\mathbf{4}$ | ... | Observe that in each square formed by 4 small squares, the number in the bottom right is the sum of the other 3 numbers. Thus, we have:
| $\mathbf{0}$ | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ |
| :---: | :---: | :---: | :---: | :---: |
| $\mathbf{1}$ | 2 | 5 | 10 | $3+4+10=17$ |
| $\mathbf{2}$ | $1+... | 360 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A baker went to the market to buy eggs to make 43 cakes, all with the same recipe, which uses fewer than 9 eggs. The seller notices that if he tries to wrap the eggs the baker bought in groups of 2 or 3 or 4 or 5 or 6 eggs, there is always 1 egg left over. How many eggs does she use in each cake? What is the minimum nu... | Since the 43 cakes have the same recipe, the number of eggs the baker needs is a multiple of 43. On the other hand, this number is also a multiple of $2,3,4,5$ and 6 plus 1. The least common multiple (LCM) of $2,3,4,5$ and 6 is 60, but $60+1=61$ is not a multiple of 43! We need to find a number with these two propertie... | 301 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The digits $1,2,3,4$ and 5 were used, each one only once, to write a 5-digit number $a b c d e$, such that: $a b c$ is divisible by $4, b c d$ by 5, and $c d e$ by 3. Find this number. | For $a b c$ to be divisible by 4, its last two digits must form a number divisible by 4. Since the digits are 1, 2, 3, 4, and 5, the only possibilities are: $b c=12, b c=24, b c=32, b c=52$. On the other hand, numbers divisible by 5 end in 0 or 5. Since 0 is not included, it follows that $d=5$ because $b c d$ is divisi... | 12453 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a city with almost 30,000 inhabitants, two ninths of the men and two fifteenths of the women practice sports only on weekends, and the number of inhabitants who do not practice sports is five times the number of those who practice sports regularly. With this data, complete the table.
| Do not practice sports | | P... | From the data in the table, we have $8563+8322=16885$ people who do not practice sports. Therefore, the city has $16885 \div 5=3377$ people who practice sports regularly, and thus $3377-1252=2125$ women practice sports regularly.
Note that the number of people who practice sports only on weekends is divisible by 15 an... | 29970 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the luminous mechanism of the figure, each of the eight buttons can light up in green or blue. The mechanism works as follows: when turned on, all buttons light up blue, and if we press a button, that button and its neighbors switch colors. If we turn on the mechanism and successively press buttons 1, 3, and 5, how ... | The correct answer is (C).
The table shows the color of each button at each step.
| | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ | $\mathbf{5}$ | $\mathbf{6}$ | $\mathbf{7}$ | $\mathbf{8}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| start | blue | blue | blue | blue |... | 5 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
A 6-digit number starts with 1. If we move this digit 1 from the first position to the last on the right, we get a new 6-digit number that is three times the original number. What is this number? | The problem is to determine the digits $a, b, c, d$ and $e$ such that the number $a b c d e 1$ is three times $1 a b c d e$:
$$
\frac{\times 3}{a b c d e 1}
$$
Initially, we see that $e=7$, and from there we can start discovering each of the digits:
| $1 a b c d 7$ | $1 a b c 57$ | $1 a b 857$ | $1 a 2857$ |
| :---:... | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Luis wrote the sequence of natural numbers starting from 1:
$$
1,2,3,4,5,6,7,8,9,10,11,12, \cdots
$$
When did he write the digit 3 for the $25^{\text {th }}$ time? | Let's see each time Luis wrote the digit 3:
- $3 \rightarrow 1$;
- $\underbrace{13,23}_{2}, \underbrace{30,31,32,33, \ldots, 39}_{11}, \underbrace{43, \ldots, 93}_{6} \rightarrow 2+6+11=19$;
So far, he has written the digit 3 twenty times. From there, we have:
$$
\underbrace{103}_{21^{\text{st}}}, \underbrace{113}_{... | 131 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Is the number $3^{444}+4^{333}$ divisible by 5?
## List 5 | There is a pattern for the unit digit of a power of 3: it has a period of 4, as it repeats every 4 times.
$$
\begin{aligned}
& 3 \\
& 3^{2}=9 \\
& 3^{3}=27 \\
& 3^{4}=81 \\
& 3^{5}=243 \\
& 3^{6}=\ldots 9 \\
& 3^{7}=\ldots 7 \\
& 3^{8}=\ldots 1
\end{aligned}
$$
Since 444 is a multiple of 4, the unit digit of $3^{444}... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Fill in the squares with the numbers $1,2,3,5,6$.
$$
(\square+\square-\square) \times \square \div \square=4
$$ | The operation is equivalent to
$$
(\square+\square-\square) \times \square=4 \times
$$
Therefore, the left side of the equality is a multiple of 4, so the only possibilities are:
$$
(\square+\square-\square) \times \square=4 \times \square \quad \text { or } \quad(\square+\square-\square) \times \square=4 \times \sq... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Jandira must fill a $4 \times 4$ table that already comes with two cells pre-filled with the numbers 1 and 2 - see the side. Two cells are considered neighbors if they share a vertex or a side.
The candidates for the last 2 digits are the powers of 2 with 2 digits: 16, 32, and 64:
$$
25 \_16, \quad 25 ... | 25916 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Find an irreducible fraction such that the product of its numerator by its denominator is $2 \times 3 \times 4 \times 5 \times \ldots \times 10$. How many of these irreducible fractions exist?
## List 8 | For a fraction to be irreducible, the numerator and the denominator must not have a common factor.
Initially, let's see what the prime factors of \(N=2 \times 3 \times 4 \times 5 \times \ldots \times 10\) are:
\[
2 \times 3 \times \underbrace{4}_{2^{2}} \times 5 \times \underbrace{6}_{2 \times 3} \times 7 \times \und... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Writing successively the natural numbers, we obtain the sequence:
$$
12345678910111213141516171819202122 \ldots
$$
Which digit is in the $2009^{th}$ position of this sequence? | - the numbers 1 to 9 occupy 9 positions;
- the numbers 10 to 99 occupy $2 \times 90=180$ positions;
- the numbers 100 to 199 occupy $3 \times 100=300$ positions; the numbers 200 to 299 occupy $3 \times 100=300$ positions; the numbers 300 to 399 occupy $3 \times 100=300$ positions; etc.
$$
\underbrace{100, \ldots 199}_... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find $B \widehat{A} D$, knowing that $D \widehat{A C}=39^{\circ}, A B=A C$ and $A D=B D$.
## List 9 | Given $A B=A C$, the triangle $\triangle A B C$ is isosceles, hence $A \widehat{B} C=A \widehat{C} B$. Since $A D=B D$, the triangle $\triangle A B D$ is also isosceles, hence $A \widehat{B} D=B \widehat{A} D$. Therefore,
$$
A \widehat{B} C=A \widehat{C} B=B \widehat{A} D
$$
 / 3=140$ minutes to dig one hole. Similarly, 1 bird needs 40... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
At a circular table, 5 people are sitting: Arnaldo, Bernaldo, Cernaldo, Dernaldo, and Ernaldo, each in a chair. Analyzing in a clockwise direction, we have:
I. Between Arnaldo and Bernaldo there is 1 empty chair;
II. Between Bernaldo and Cernaldo are 5 chairs;
III. Between Dernaldo and Ernaldo are 4 chairs, almost a... | Solution
Let's position Arnaldo in the chair we will call 1, and by information $I$, Bernaldo should sit in chair 3, and consequently, by information $II$, Cernaldo should sit in chair 9. Since there are 6 chairs between Dernaldo and Ernaldo and 2 chairs between Dernaldo and Cernaldo, Cernaldo is between Dernaldo and ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A way to check if a number is divisible by 7 is to subtract, from the number formed by the remaining digits after removing the units digit, twice the units digit, and check if this number is divisible by 7. For example, 336 is divisible by 7, because $33 - 2 \cdot 6 = 21$ is divisible by 7, but 418 is not because $41 -... | Solution
a) $457-2 \cdot 8=441 \rightarrow 44-2 \cdot 1=42$, which is divisible by 7, so 4,578 is also divisible by 7.
b) Since $\overline{A B 5}$ has three digits, then $A \neq 0$. Moreover, $\overline{A B}-14$, by the rule of divisibility, is a multiple of 7, and consequently, $\overline{A B}$ must be a multiple of... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Mario bought some sweets at the market, where 3 chocolates cost the same as 2 popsicles and 2 lollipops cost the same as 5 chocolates.
a) Mario decided to return to the market with money to buy exactly 3 lollipops but decided to buy popsicles instead. How many popsicles did he manage to buy?
b) If he had used the mon... | Solution
a) 15 chocolates cost the same as 10 popsicles and the same as 6 lollipops. Therefore, 10 popsicles are worth the same as 6 lollipops, and consequently, 5 popsicles the same as 3 lollipops. Thus, with the money for 3 lollipops, Mario can buy 5 popsicles.
b) From the previous item, we saw that the largest num... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Marta chose a 3-digit number with different non-zero digits and multiplied it by 3. The result was a 3-digit number with all digits equal to the tens digit of the chosen number. What number did Marta choose?
# | Solution
Let the chosen number be $a b c$. Then, we have:
| $a b c$ |
| ---: |
| $a b c$ |
| $+\quad b \quad c$ |
| $b b b$ |
Let's analyze case by case: If $c=1$, then $b=3$, which is not possible because the tens digit of the result should be 9; if $c=2$, then $b=6$, which is not possible because the tens digit of... | 148 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Jonas wrote a sequence with the positive multiples of 13 in ascending order.
## $1326395265 \ldots$
a) What is the $2,019^{\text{th}}$ digit of Jonas's sequence?
b) Will the number 2,019 appear in this sequence? | Solution
a) There are 7 multiples of 13 with 2 digits (14 digits); with 3 digits, there are 69 multiples of $13(3 \cdot 69=207$ digits). Already, there are $14+207=221$ digits, so $2,019-221=1,798$ are still needed. Dividing 1,798 by 4, we get 449 and a remainder of 2, that is, the first multiple of 13 with 4 digits i... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a magic square, the sum of the numbers in each row, column, and diagonal is the same. In the magic square below, what is the value of $a+b+c$?
| 16 | 2 | $a$ |
| :---: | :---: | :---: |
| $c$ | 10 | $d$ |
| $b$ | $e$ | 4 | | Solution
The common sum for the rows, columns, and diagonals is $16+10+4=30$. Analyzing the sums of the rows and columns, we have:
I) From $2+10+e=30$, it follows that $e=18$.
II) From $b+e+4=30$, it follows that $b=8$.
III) From $16+c+b=30$, it follows that $c=6$.
IV) From $16+2+a=30$, it follows that $a=12$.
Th... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
To enter a park, a group with two men, four women, and two children paid 226 reais, while a group with three men, three women, and one child paid 207 reais.
a) How much would a group with 8 men, 10 women, and 4 children pay to enter the park?
b) If the ticket prices are all natural numbers, how many possible prices a... | Solution
a) Let $h$ be the ticket price for men, $m$ be the ticket price for women, and $c$ be the ticket price for children. Organizing the information, we have:
$$
\left\{\begin{aligned}
2 h+4 m+2 c & =226 \\
3 h+3 m+c & =207
\end{aligned}\right.
$$
A group with 8 men, 10 women, and 4 children is the same as a gro... | 640 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A numerical sequence is called a SEQUENLADA when, starting from the second number, the next element is formed by the rules:
I) if it has more than 2 digits, the last digit is moved to the $1 \underline{st}$ position and then the two last digits are added;
II) if it has two digits, these 2 are added until a single dig... | Solution
a) $246.831 \rightarrow 124.611 \rightarrow 11.247 \rightarrow 5.443 \rightarrow 358 \rightarrow 88 \rightarrow 16 \rightarrow 7$.
b) Let's analyze the sequence in reverse, with the first number being 1. The second number can only be 10 or 100. In the first case (10), the $3^{\text{rd}}$ term can be $19, 28,... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The owner of a clothing factory is a math enthusiast and organizes his products into packages whose number of items is a prime number. For example, the green-colored pieces, he organizes into packages of 2 pieces each; the blue-colored pieces, into packages of 3 pieces each; the pink-colored pieces, into packages of 5 ... | Solution
The colors green, blue, and pink have 2, 3, and 5 pieces per pack, respectively. We can take various quantities of packs of each color, including not taking any, as long as the total number of pieces is exactly 20. Let's create a table to list the possibilities:
| Green | Blue | Pink |
| :---: | :---: | :---... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The road connecting two villages on a mountain consists only of uphill or downhill sections. A bus always travels at $15 \mathrm{~km} / \mathrm{h}$ on uphill sections and at $30 \mathrm{~km} / \mathrm{h}$ on downhill sections. Find the distance between the villages if the bus takes exactly 4 hours to complete the round... | Observe that the uphill sections on the outbound journey are exactly the downhill sections for the return journey and vice versa. Thus, in a round trip, the distance covered on the ascents is equal to the distance covered on the descents.
Let's call the distance between the two villages $\mathrm{d}$. Since the total d... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In a circle, 15 white points and 1 black point were marked. Consider all possible polygons (convex) with their vertices at these points.
Let's separate them into two types:
- Type 1: those that have only white vertices.
- Type 2: those that have the black point as one of the vertices.
Are there more polygons of type... | Notice that for each type 1 polygon, we can construct a type 2 polygon by adding the black point.
On the other hand, if we have a type 2 polygon and remove the black point, the only way not to generate a polygon is if exactly two white points remain.
Therefore, there are more type 2 polygons than type 1 polygons.
To... | 105 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The sums of the three columns and the three rows of the table are equal.
| 4 | 9 | 2 |
| :--- | :--- | :--- |
| 8 | 1 | 6 |
| 3 | 5 | 7 |
What is the minimum number of cells in the table that need to be changed so that all the new six sums are different from each other? | Solution
If three or fewer cells are changed, there will either be two rows without any changed cells or one cell is the only one changed in its row and column. In the first case, these two rows without any changed cells have the same sum. In the second case, if only one cell is the only one changed in its row and col... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In a class with 35 students, a survey was conducted on preferences for mathematics and literature, and it was found that:
- 7 men like mathematics;
- 6 men like literature;
- 5 men and 8 women said they do not like either;
- there are 16 men in the class;
- 5 students like both; and
- 11 students like only mathematics... | Solution
Let $H$ be the set of men and $U$ the total set of people, so $U-H$ is the set of women. In addition, consider the sets Mat and $L$ of people who like mathematics and literature, respectively. If $x$ represents the number of men who like both mathematics and literature and $y$ the number of women who like onl... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The rectangle $A B C D$ with dimensions $A B=240 \mathrm{~cm}$ and $B C=288 \mathrm{~cm}$ represents a piece of paper that will be folded along the segment $E F$, where $E$ belongs to $A D$ and $F$ belongs to $B C$, such that point $C$ will be on the midpoint of $A B$.
 Determine the prime factorization of the number 344.
b) Find the sum of the first and the last number among those written on the t... | Solution
a) Since 44 is a multiple of 4, then $344=4 \cdot 86=2^{3} \cdot 43$.
b) If $y$ is the number of leaves torn out, and $2 y$ is the number of pages, and $x+1$ is the first number that appears on them, we have
$$
\begin{aligned}
(x+1)+(x+2)+(x+3)+\ldots+(x+2 y) & =344 \\
2 x y+(1+2+\ldots+2 y) & =2^{3} \cdot ... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A tile, in the shape of a regular polygon, was removed from the place it occupied on a panel. It was then observed that if this tile were to undergo a rotation of $40^{\circ}$ or $60^{\circ}$ around its center, it could be perfectly fitted into the space left vacant on the panel. What is the smallest number of sides th... | Solution
For such rotations to be possible, with the fitting of the polygon, it is necessary and sufficient that the central angle be a divisor of $40^{\circ}$ and $60^{\circ}$. If $n$ is the number of sides of the tile, the central angle is given by $\frac{360^{\circ}}{n}$. Thus, the ratios
$$
\frac{40^{\circ}}{\fra... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
When Paulo turned 15, he invited 43 friends to a party. The cake was in the shape of a regular 15-sided polygon and had 15 candles on it. The candles were placed in such a way that no three candles were in a straight line. Paulo divided the cake into triangular pieces where each cut connected two candles or connected a... | Solution
Let $n$ be the number of triangles into which the cake can be divided under the given conditions. We will sum the interior angles of these triangles in two ways:
- On the one hand, since each triangle has an internal angle sum of $180^{\circ}$, the sum of all the interior angles is $180^{\circ} \cdot n$.
- O... | 43 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The table below shows some of the results from the last Pirajuba Fishing Festival, displaying how many competitors $q$ caught $n$ fish for some values of $n$.
| $n$ | 0 | 1 | 2 | 3 | $\ldots$ | 13 | 14 | 15 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $q$ | 9 | 5 | 7 | 23 | $\ldots$ | ... | Solution
Let $N$ and $P$ be the number of competitors and fish caught in the event, respectively. Analyzing the competitors who caught 3 or more fish, we have:
$$
N-9-5-7=N-21 \text{ fishermen and } P-0 \cdot 9-1 \cdot 5-2 \cdot 7=P-19 \text{ fish, }
$$
with the average written as:
$$
\begin{aligned}
\frac{P-19}{N-... | 883 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a shooting contest, 8 targets are arranged in two columns with 3 targets and one column with 2 targets. The rules are:
- The shooter freely chooses which column to shoot at.
- He must aim at the lowest target not yet hit.
 If $x, y$ and $z$ are the positions of the targets, in principle, the shooter has 8 choices for $x$, $8-1=7$ for $y$, since we cannot repeat the chosen position, and $8-2=6$ choices for $z$, since we cannot repeat any of the two already selected positions. This gives $8 \cdot 7 \cdot 6$ choices. However, t... | 560 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
João's father has between 200 and 300 books in his library. One fifth of these books are in English, one seventh in French, one quarter in Italian, and the rest are in Spanish. What is the total number of Spanish books in this library?
# | Solution
The number of books is a multiple of 5, 7, and 4. Since these numbers do not have common prime factors, it must necessarily be a multiple of \(5 \cdot 7 \cdot 4 = 140\). In the given range, there is a unique multiple of 140, namely the number 280. Therefore, the number of books that are not in Spanish is
$$
... | 114 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A hotel has 5 distinct rooms, all with single beds for up to 2 people. The hotel has no other guests and 5 friends want to spend the night there. In how many ways can the 5 friends choose their rooms? | Solution
Analyzing only the number of people per room, without considering the order, the possible distributions can be associated with the lists:
$$
(1,1,1,1,1),(1,1,1,2) \text { or }(2,2,1)
$$
We will now analyze list by list the number of ways to distribute the friends.
i) In the list ( $1,1,1,1,1)$, all friends... | 2220 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a tournament with 5 teams, there are no ties. In how many ways can the $\frac{5 \cdot 4}{2}=10$ games of the tournament occur such that there is neither a team that won all games nor a team that lost all games? | Solution
Represent each team by a vertex of a pentagon and the result of each game by an arrow from the player who won to the player who lost.
Since each match has 2 possible outcomes, the... | 544 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
a) Which of the numbers is larger: $2^{100}+3^{100}$ or $4^{100}$?
b) Let $x$ and $y$ be natural numbers such that
$$
2^{x} \cdot 3^{y}=\left(24^{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{60}}\right) \cdot\left(24^{\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots+\frac{1}{60}}\right)^{2} \cdot\left(24^{\frac{1}{4... | Solution
a) Notice that
$$
4^{100}=4^{99}+4^{99}+4^{99}+4^{99}=3 \cdot 4^{99}+4^{99}
$$
Let's compare the terms separately:
$$
\begin{aligned}
4 & >3 \\
4^{99} & >3^{99} \\
3 \cdot 4^{99} & >3 \cdot 3^{99} \\
3 \cdot 4^{99} & >3^{100}
\end{aligned}
$$
and
$$
4^{99}=2^{198}>2^{100}
$$
Adding the inequalities
$$
... | 3540 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A circular wheel has 7 equal-sized sections, and each of them will be painted with one of two colors. Two colorings are considered equivalent if one can be rotated to produce the other. In how many non-equivalent ways can the wheel be painted?
Since $\triangle A D E$ is equilateral, it follows that $\angle A D E=60^{\circ}$. Therefore, $\angle B D E=360^{\circ}-150^{\c... | 39 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the triangle $ABC$ below, there is a point $D$ on its side $AB$, such that $AB=CD$, $\widehat{ABC}=100^{\circ}$, and $\widehat{DCB}=40^{\circ}$.
a) What is the measure of the angle $\wide... | Solution
a) Analyzing the sum of the internal angles of the triangle $\triangle B D C$, we have $100^{\circ}+40^{\circ}+\angle B D C=180^{\circ}$, that is, $\angle B D C=40^{\circ}$.
b) Due to the previous item, $\triangle B D C$ is isosceles with base $C D$ and thus $B C=B D$. Construct the auxiliary triangle $\tria... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
a) Consider a prime $p$ that divides $10^{n}+1$ for some positive integer $n$. For example, $p=7$ divides $10^{3}+1$. By analyzing the principal period of the decimal representation of $\frac{1}{p}$, verify that the number of times the digit $i$ appears is equal to the number of times the digit $9-i$ appears for each $... | Solution
a) We can write $10^{n}+1=p \cdot a$ where $a$ is a number with no more than $n$ digits in base 10, say $a=a_{1} a_{2} \ldots a_{n}$. We mean by this that each number $a_{i}$ is one of the digits of $a$. Even if it has strictly fewer than $n$ digits, we can place some $a_{i}$'s on the left as being 0. We have... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Consider the $9 \times 9$ board shown below. The rows are numbered from 1 to 9.
Figure 116.1
We color the cells of the odd-numbered rows of the board with the colors blue and white, alternat... | (a) Each odd row contains 5 blue houses and 4 white houses. Since the board has 5 odd rows, the number of blue houses is $5 \times 5 = 25$ and the number of white houses is $5 \times 4 = 20$.
Similarly, each even row has 5 gray houses and 4 red houses, and the board has 4 even rows. Thus, the number of gray houses is ... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$P$ is a point inside the square $A B C D$ such that $P A=1, P B=2$ and $P C=3$. What is the measure of the angle $A \hat{P} B$?
Figure 119.1 | Let $Q$ be a point such that the triangles CQB and APB are congruent, as shown in the figure. This is equivalent to rotating the triangle APB with center at B and angle $90^{\circ}$ clockwise. In particular, we have that $\mathrm{PBQ}=90^{\circ}$. Thus, $\mathrm{PQ}^{2}=$ $\mathrm{PB}^{2}+\mathrm{BQ}^{2}=2^{2}+2^{2}$, ... | 135 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Maria has just won a huge chocolate bar as an Easter gift. She decides to break it into pieces to eat it gradually. On the first day, she divides it into 10 pieces and eats only one of them. On the second day, she divides one of the remaining pieces from the previous day into 10 more pieces and eats only one of them. O... | Solution
a) At the end of the first day, she will have $10-1=9$ pieces. At the end of the next day, she will have $9-1+10-1=17$ pieces. From a practical standpoint, it's as if she added $10-1-1=8$ new pieces, because one piece is always lost to the division into $10$ and another is always eaten. At the end of the thir... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the figure below, the sides of rectangle $A B C D$ have been extended so that $E B=2 A B$, $A H=3 A D$, $D G=2 D C$, and $F C=3 B C$. Find the ratio between the areas of quadrilateral $E H G F$ and rectangle $A B C D$.
 / 2 \\
& =(A B \cdot 3 A D) / 2 \\
& =3 S / 2 \\
A_{D H G} & =(D H \cdot D G) / 2 \\
& =(2 A D \cdot 2 D C) / 2 \\
& =2 S \\
A_{F C G} & =(F... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
a) Given four distinct points $A, B, C$ and $D$, all on the same line, as indicated in the figure below, determine the number of distinct segments that can be formed with vertices at these points.
 There are 6 line segments with vertices at these 4 points: $A B, A C, A D, B C, B D$ and $C D$. Note that the answer would not be different if the points were not collinear.
b) Let's call the 10 points: $A, B, C, D, E, F, G, H, I$ and $J$. Observing the solution from part a), we can organize our count by i... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the figure below, Maria arranged 24 sticks to form a $3 \times 3$ square.
a) How many sticks would she need to form a $4 \times 4$ square?
b) What is the side length of the largest squa... | Solution
a) The figure below shows that she will need to use 40 sticks
b) If we want to form a square with a side composed of $n$ sticks, we will need $n+1$ rows of $n$ sticks and $n+1$ col... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
João is playing a game where the only allowed operation is to replace the natural number $n$ with the natural number $a \cdot b$ if $a+b=n$, with $a$ and $b$ being natural numbers. For example, if the last number obtained was 15, he can replace it with $56=7 \cdot 8$, since $7+8=15$ and both are natural numbers.
a) St... | Solution
a) Let's indicate the swap operations with an arrow $(\rightarrow)$. One way would be:
$$
7=2+5 \rightarrow 10=2+8 \rightarrow 16=12+4 \rightarrow 48
$$
b) Since $n=(n-1)+1$, it is possible to decrease $n$ by one unit, transforming it into $1 \cdot(n-1)=n-1$. Therefore, a good strategy would be to create a ... | 2014 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
During the FIFA World Cup, several mathematicians were called upon to speak about the teams' chances of qualification. In the group stage, each group is formed by four teams, and each team faces each of the other teams exactly once. In case of a win, a team earns 3 points; in case of a draw, 1 point; and in case of a l... | Solution
The minimum number of points is 7 points. First, let's see that with 6 points a team may not qualify. Let $A, B, C$ and $D$ be the teams in a certain group and consider the following table of results:
| Winner | Result | Loser |
| :---: | :---: | :---: |
| $A$ | $2 \times 0$ | $D$ |
| $B$ | $1 \times 0$ | $D... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
João has more than 30 and less than 100 chocolates. If he arranges the chocolates in rows of 7, one will be left over. If he arranges them in rows of 10, two will be left over. How many chocolates does he have?
# | Solution
In the first organization, with $x$ being the number of rows, the number of chocolates João has is of the form $7 x+1$. In the second organization, with $y$ being the number of rows, the number of chocolates João will have is $10 y+2$. That is, the number of chocolates João has leaves a remainder of 1 when di... | 92 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Around a circle, the natural numbers from 1 to $N$ with $N>2$ are written, each exactly once, in such a way that two neighbors have at least one digit in common. Find the smallest $N>2$ for which this is possible.
# | Solution
Since 3 will appear, we should use 13 and 23. Thus, as $9<23$, 9 must be in the circle. Its two neighbors must contain the digit 9, the smallest would be 19 and 29, consequently $N$ must be at least 29. Now, to prove that 29 is the minimum, it suffices to construct an example:
$$
1,12,2,22,20,21,23,3,13,14,4... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Pedrinho is playing with three triangular pieces with sides $(5,8,10),(5,10,12)$ and $(5,8,12)$ as shown in the drawing below. He can join two pieces by exactly gluing the sides of the same length. For example, he can join the side 10 of the first piece with the side 10 of the second, but he cannot join the side 10 of ... | Solution
By adding the perimeters of the three triangles, we have:
$$
(5+8+10)+(5+10+12)+(5+8+12)=23+27+25=75
$$
When we join two triangles using a specific side, the practical effect on the above sum is to reduce it by twice the length of that side, as it no longer contributes to two triangles. To maximize the sum ... | 49 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
To cook instant noodles, it is necessary to cook the noodles for exactly 3 minutes. Marking exactly 3 minutes is very complicated without a clock, but it is possible if you have certain hourglasses that mark exact times in minutes. For example, suppose you have two hourglasses, one that marks exactly 7 minutes and anot... | Solution
a) To mark 4 minutes, we should flip the 9-minute and 7-minute hourglasses several times so that the difference in time is 4 minutes. Since $2 \cdot 9 - 2 \cdot 7 = 4$, one procedure would be to flip the 9-minute hourglass twice and the 7-minute hourglass also twice. Initially, both should be flipped at the s... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A cricket can jump two distances: 9 and 8 meters. It competes in a 100-meter race to the edge of a cliff. How many jumps must the cricket make to reach the end of the race without going past the finish line and falling off the cliff?
# | Solution
First solution: Suppose the cricket only jumps 9 meters. On its twelfth jump, it would fall off the cliff, since $9 \cdot 12 = 108 \, \text{m}$. Since it can also jump 8 meters, it is enough to "exchange" 8 jumps of 9 meters for jumps of 8 meters. Thus, we have 4 jumps of 9 meters and 8 jumps of 8 meters, for... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the drawing below, three buildings have been constructed on a plot of land divided into rectangular lots. The perimeters of buildings $A$ and $B$ are $400 \mathrm{~m}$ and $240 \mathrm{~m}$, respectively. What is the perimeter of building $C$?
| | | | | | | | |
| :--- | :--- | :--- | :--- | :--- | :--- | :-... | Solution
In a plot, we have three important dimensions: the width $l$, the height $h$, and the diagonal $d$.
Let's call the perimeters of buildings $A, B$, and $C$ as $P_{A}, P_{B}$, and $... | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The king is a chess piece that can move only one square vertically, one horizontally, or one diagonally. We say that a king attacks a square if it can occupy it with a single move. For example, a king placed in the central squares of a $6 \times 6$ board attacks 8 squares, a king placed on the side squares attacks 5 sq... | Solution
a) Divide the $6 \times 6$ board into 4 $3 \times 3$ boards. If one of these four regions does not have a king, the central cell of that region will not be occupied or attacked by any king. Therefore, at least 4 kings are needed. If we place a king in the central cell of each $3 \times 3$ board, then all cell... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
On a $6 \times 6$ toy board, each cell represents a light button. When someone presses a button, it turns on if it is off and turns off if it is on. In addition, all buttons that share a side with a pressed button also change their state: from on to off or from off to on. Starting with all buttons off and pressing each... | Solution
Notice that a button will end up lit if it changes state an odd number of times and will end up off otherwise. Each button changes state when it or one of its neighbors is pressed, so the number of times it will change state will be equal to its number of neighbors plus one. We can thus analyze each cell of t... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
a) Show a way to separate all the numbers from 1 to 16 into four sets with four numbers each, so that each set has the same sum.
b) Show that there are at least 1024 ways to write the numbers from 1 to 16 in each of the cells of a $4 \times 4$ board so that the sum of the numbers in each row is equal. | Solution
a) First, form eight pairs of numbers by choosing opposite numbers around the "middle" of the sequence, that is, $(1,16),(2,15), \ldots,(7,10)$ and $(8,9)$. Notice that each pair has a sum of 17. Now, group the pairs into four groups, each with a sum of 34, for example: $(1,16,2,15)$, $(3,14,4,13)$, $(5,12,6,... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
a) In the figure below, two gears are meshed, one gear $A$ with 6 teeth and another gear $B$ with 8 teeth. All the teeth are the same size. If gear $A$ makes 12 revolutions, how many revolutions will gear $B$ make?
 Let $N$ be the number of turns made by gear $B$. The reader can verify that in the contact region between the two gears, each tooth of gear $A$ is touched above by a tooth of gear $B$. The number of such contacts is equal to $12 \times 6$, since gear $A$ makes a total of 12 turns and has 6 teeth. On the ot... | 5407 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Teacher Lorena taught her students the following notable product: for any real numbers $a$ and $b$,
$$
a^{2}-b^{2}=(a+b)(a-b)
$$
For example, $4^{2}-3^{2}=16-9=7$. On the other hand, $(4+3)(4-3)=7 \times 1=7$. Using this teaching from Teacher Lorena,
a) Calculate
$$
100^{2}-99^{2}+98^{2}-97^{2}+96^{2}-95^{2}+\cdots... | Solution
a) Using the notable product $a^{2}-b^{2}=(a+b)(a-b)$, we will rewrite the expression
$$
100^{2}-99^{2}+98^{2}-97^{2}+96^{2}-95^{2}+\cdots+2^{2}-1^{2}
$$
as
$$
(100+99)(100-99)+(98+97)(98-97)+(96+95)(96-95)+\ldots+(2+1)(2-1)
$$
Notice that many of the differences above are equal to one! For example, $100-... | 5050 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The following figure shows a cube.
a) Calculate the number of triangles whose three vertices are vertices of the cube.
b) How many of these triangles are not contained in a face of the cub... | Solution
a) Let's start by fixing a vertex, say $A$, and counting the number of triangles that use this vertex. Each of these triangles will be determined by indicating its other two vertices distinct from $A$.
 que é igual a 7.
(a) Quantos algarismos 9 escreveu Paul na lista?
(b) Se $S$ é a soma de todos os números na lista, qual a soma dos algarismos de ... | Solução
(a) Os números na lista de Paul são:
$$
97,997,9997,99997, \ldots, \underbrace{9999 \ldots 9}_{100 \text { vezes }} 7 .
$$
O número de algarismos 9 nessa lista é
$$
1+2+3+\cdots+100=\frac{(100)(101)}{2}=5050
$$
(b) Os números na lista de Paul podem ser escritos na forma:
$$
100-3,1000-3,10000-3,100000-3, ... | 106 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Pedro has a $6 \times 2$ board, with the two rightmost cells painted gray as illustrated below:
He must fill all the cells of his board with the numbers from 1 to 12 such that:
- in each r... | Solution
Let $A$ and $B$ be the numbers placed in the shaded cells as indicated in the following figure:
Observe that, by the rules, $B$ must be greater than all the other numbers in its r... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the following figure, the square $A B C D$ has been divided into three rectangles of the same area.
If the length of the segment $B M$ is equal to 4, calculate the area of the square $A B... | Solution
In the following figure, the area of rectangle $M N P C$ is $|M N| \times|N P|$, while the area of rectangle $N Q D P$ is $|N Q| \times|N P|$.
For the areas of these rectangles to ... | 144 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\mathrm{Na}$ expression
$$
* 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10
$$
Luís replaced each symbol $*$ with a + or a - sign, using five of each type. When calculating the value of the expression, the result obtained was a positive two-digit number $N$, which is a multiple of 7.
a) Show that $N$ is less than 25.
b) S... | Solution
a) Let's call the numbers that remained with a negative sign $a, b, c, d$ and $e$. Since
$$
-a-b-c-d-e=(a+b+c+d+e)-2(a+b+c+d+e)
$$
the number $N$ can also be calculated as
$N=(1+2+3+4+5+6+7+8+9+10)-2(a+b+c+d+e)=55-2(a+b+c+d+e)$.
Notice that the smallest possible value for the sum $a+b+c+d+e$ is $1+2+3+4+5... | 21 | Number Theory | proof | Yes | Yes | olympiads | false |
Pepi needs to place all the numbers from 1 to 8 on the following board in such a way that the sum of the two numbers placed in each column is always the same value $S$.
| | | | |
| :--- | :--- | :--- | :--- |
| | | | |
a) Show Pepi one way to place the numbers.
b) Convince Pepi that the only possible value f... | Solution
a) One way to place the numbers is as follows
| 1 | 2 | 3 | 4 |
| :--- | :--- | :--- | :--- |
| 8 | 7 | 6 | 5 |
b) The sum of all the numbers placed on the board can be calculated by first adding the two numbers in each column and then adding the sums obtained. Since the sum in each column is equal to $S$, ... | 384 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The square $A B C D$ is divided into 6 isosceles right triangles as indicated in the following figure:
If the area of the shaded triangle is 2, calculate the area of the square.
# | Solution
From the information given in the problem, the triangle $P Q R$ in the figure has an area of 2.
However, since it is a right triangle, the area of triangle $P Q R$ can be calculate... | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Francisco has 3 daughters: Alina, Valentina, and Civela. An interesting fact is that all three daughters were born on March 18. Today, March 18, 2014, is their birthday. Noting another curious fact, Francisco says:
- Alina, your age is now double the age of Valentina.
a) Show that this could never have happened befor... | Solution
a) It is clear that Alina is older than Valentina. Let's call the difference in age between Alina and Valentina $d$. Since they were born on the same day of the year, $d$ does not vary (note that this would not be true if they were born on different days, as in that case, the difference would change on the bi... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Ximena wishes to number the pages of a notebook. For this, she has a large quantity of stickers with the digits $0,1,3,4,5,6,7,8$ and 9, but she has only 100 stickers with the digit 2. Determine up to which page Ximena can number this notebook.
# | Solution
Let's start by counting how many times the digit 2 appears in the numbers from 1 to 199. As the units digit, the digit 2 appears on the pages
$$
2,12,22,32,42,52,62, \ldots, 152,162,172,182,192
$$
so far, we have counted 20 numbers.
As the tens digit, the digit 2 appears on the pages
$$
20,21,22,23, \ldot... | 244 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
a) How many zeros are at the end of the number $A=2^{5} \times 3^{7} \times 5^{7} \times 11^{3}$?
b) How many zeros are at the end of the number $B=1 \times 2 \times 3 \times 4 \times \cdots \times 137$? | Solution
a) To count the number of zeros at the end of a number, it is enough to find out which is the highest power of 10 that is a divisor of that number. On the other hand, since $10=2 \times 5$, it is enough to know the powers of 2 and 5 that appear in the factorization of the number. So we can write:
$$
A=2^{5} ... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all values of $x$ satisfying
$$
\frac{x+\sqrt{x+1}}{x-\sqrt{x+1}}=\frac{11}{5}
$$ | Solution
The equation can be rewritten as
$$
\begin{aligned}
\frac{x+\sqrt{x+1}}{x-\sqrt{x+1}} & =\frac{11}{5} \\
5 x+5 \sqrt{x+1} & =11 x-11 \sqrt{x+1} \\
16 \sqrt{x+1} & =6 x \\
8 \sqrt{x+1} & =3 x
\end{aligned}
$$
Squaring both sides of the last equation, we get
$$
\begin{aligned}
9 x^{2} & =64(x+1) \\
9 x^{2}-6... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $x$ and $y$ be real numbers such that $x+y=10$ and $x^{3}+y^{3}=400$, determine the value of $x^{2}+y^{2}$. | Solution
Given
$$
\begin{aligned}
x^{3}+y^{3} & =(x+y)\left(x^{2}-x y+y^{2}\right) \\
& =(x+y)\left((x+y)^{2}-3 x y\right) \\
& =10(100-3 x y)
\end{aligned}
$$
it follows that $40=100-3 x y$, that is, $x y=20$. Therefore,
$$
\begin{aligned}
x^{2}+y^{2} & =(x+y)^{2}-2 x y \\
& =100-40 \\
& =60
\end{aligned}
$$ | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Consider a square $ABCD$ with center $O$. Let $E, F, G$, and $H$ be points on the interiors of sides $AB, BC, CD$, and $DA$, respectively, such that $AE = BF = CG = DH$. It is known that $OA$ intersects $HE$ at point $X$, $OB$ intersects $EF$ at point $Y$, $OC$ intersects $FG$ at point $Z$, and $OD$ intersects $GH$ at ... | Solution
a) Let $x$ and $y$ be the lengths of $A E$ and $A H$, respectively. Given that $A H = E B$, $A E = B F$, and $\angle H A E = \angle E B F$, it follows that triangles $A E H$ and $E ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a school, $n$ clubs must be formed, with $n \geq 3$ and each with 3 members, such that for each pair of clubs there is exactly one student who is a member of both.
a) Give an example of a distribution of 7 clubs that satisfy the mentioned conditions.
b) Verify that if a student belongs to 4 clubs, then he must bel... | Solution
a) Naming the students as $A, B, C, D, E$ and $F$, the following sets represent 7 clubs that meet the conditions stated:
$$
\{A, B, C\},\{A, D, E\},\{A, F, G\},\{B, E, F\},\{B, D, G\},\{C, D, F\},\{C, E, G\}
$$
b) Suppose student $A$ belongs to the following 4 clubs:
$$
\{A, B, C\},\{A, D, E\},\{A, F, G\},... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The geometric progressions $a_{1}, a_{2}, a_{3}, \ldots$ and $b_{1}, b_{2}, b_{3}, \ldots$ have the same ratio, with $a_{1}=$ $27, b_{1}=99$ and $a_{15}=b_{11}$. Find the value of $a_{9}$.
# | Solution
Let $r$ be the common ratio of the two geometric progressions. We have
\[
\begin{aligned}
a_{15} & =a_{1} r^{14} \\
& =27 r^{14} \\
b_{11} & =b_{1} r^{10} \\
& =99 r^{10}
\end{aligned}
\]
Therefore, $27 r^{14}=99 r^{10}$ and thus $3 r^{4}=11$. Finally
\[
\begin{aligned}
a_{9} & =a_{1} r^{8} \\
& =27\left(r... | 363 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Starting with a positive integer $n$, a sequence is created satisfying the following rule: each term is obtained from the previous one by subtracting the largest perfect square that is less than or equal to the previous term, until reaching the number zero. For example, if $n=142$, we will have the following sequence o... | Solution
a) An example is the sequence
$$
a_{1}=23, a_{2}=7=23-16, a_{3}=3=7-4, a_{4}=2=3-1, a_{5}=1=2-1, a_{6}=0=1-1
$$
b) Since $a_{n+1}=a_{n}-x^{2}$, with $x^{2} \leq a_{n}\frac{a_{n+1}-1}{2}$. To obtain the minimum value, the sequence will be constructed backwards, and at each step, the minimum estimate of the i... | 167 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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