problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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A positive integer is written on each of the six faces of a cube. For each vertex of the cube we compute the product of the numbers on the three adjacent faces. The sum of these products is 1001. What is the sum of the six numbers on the faces? | Let the numbers on the faces be $a_{1}, a_{2}, b_{1}, b_{2}, c_{1}, c_{2}$, placed so that $a_{1}$ and $a_{2}$ are on opposite faces etc. Then the sum of the eight products is equal to
$$
\left(a_{1}+a_{2}\right)\left(b_{1}+b_{2}\right)\left(c_{1}+c_{2}\right)=1001=7 \cdot 11 \cdot 13 .
$$
Hence the sum of the number... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The 25 member states of the European Union set up a committee with the following rules: (1) the committee should meet daily; (2) at each meeting, at least one member state should be represented; (3) at any two different meetings, a different set of member states should be represented; and (4) at the $n$ 'th meeting, fo... | If one member is always represented, rules 2 and 4 will be fulfilled. There are $2^{24}$ different subsets of the remaining 24 members, so there can be at least $2^{24}$ meetings. Rule 3 forbids complementary sets at two different meetings, so the maximal number of meetings cannot exceed $\frac{1}{2} \cdot 2^{25}=2^{24... | 16777216 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $A B C D$ be a parallelogram such that $\angle B A D=60^{\circ}$. Let $K$ and $L$ be the midpoints of $B C$ and $C D$, respectively. Assuming that $A B K L$ is a cyclic quadrilateral, find $\angle A B D$. | Let $\angle B A L=\alpha$. Since $A B K L$ is cyclic, $\angle K K C=\alpha$. Because $L K \| D B$ and $A B \| D C$, we further have $\angle D B C=\alpha$ and $\angle A D B=\alpha$. Let $B D$ and $A L$ intersect at $P$. The triangles $A B P$ and $D B A$ have two equal angles, and hence $A B P \sim D B A$. So
$$
\frac{A... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
. Positive integers $x_{1}, \ldots, x_{m}$ (not necessarily distinct) are written on a blackboard. It is known that each of the numbers $F_{1}, \ldots, F_{2018}$ can be represented as a sum of one or more of the numbers on the blackboard. What is the smallest possible value of $m$ ?
(Here $F_{1}, \ldots, F_{2018}$ are... | Answer: the minimal value for $m$ is 1009 .
Construction: Define $x_{i}=F_{2 i-1}$. This works since $F_{2 k}=F_{1}+F_{3}+\ldots+F_{2 k-1}$, for all $k$, which can easily be proved by induction. Minimality: Again by induction we get that $F_{k+2}=1+F_{1}+F_{2}+\ldots+F_{k}$, for all $k$, which means that
$$
F_{k+2}>F... | 1009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. A chess knight has injured his leg and is limping. He alternates between a normal move and a short move where he moves to any diagonally neighbouring cell.
Normal move
 was given to 4 students. Then each of these 4 students should receive 2 different "supplement... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $f$ be a real-valued function defined on the positive integers satisfying the following condition: For all $n>1$ there exists a prime divisor $p$ of $n$ such that
$$
f(n)=f\left(\frac{n}{p}\right)-f(p)
$$
Given that $f(2001)=1$, what is the value of $f(2002)$ ? | Answer: 2.
For any prime $p$ we have $f(p)=f(1)-f(p)$ and thus $f(p)=\frac{f(1)}{2}$. If $n$ is a product of two primes $p$ and $q$, then $f(n)=f(p)-f(q)$ or $f(n)=f(q)-f(p)$, so $f(n)=0$. By the same reasoning we find that if $n$ is a product of three primes, then there is a prime $p$ such that
$$
f(n)=f\left(\frac{... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the smallest positive odd integer having the same number of positive divisors as 360 ? | Answer: 31185 .
An integer with the prime factorization $p_{1}^{r_{1}} \cdot p_{2}^{r_{2}} \cdot \ldots \cdot p_{k}^{r_{k}}$ (where $p_{1}, p_{2}, \ldots$, $p_{k}$ are distinct primes) has precisely $\left(r_{1}+1\right) \cdot\left(r_{2}+1\right) \cdot \ldots \cdot\left(r_{k}+1\right)$ distinct positive divisors. Sinc... | 31185 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a $40 \times 50$ array of control buttons, each button has two states: $\mathrm{ON}$ and OFF. By touching a button, its state and the states of all buttons in the same row and in the same column are switched. Prove that the array of control buttons may be altered from the all-OFF state to the all-ON state by touchin... | Answer: 2000 .
Altering the state from all-OFF to all-ON requires that the state of each button is changed an odd number of times. This is achieved by touching each button once. We prove that the desired result cannot be achieved if some button is never touched. In order to turn this button ON, the total number of tou... | 2000 | Combinatorics | proof | Yes | Yes | olympiads | false |
Fourteen friends met at a party. One of them, Fredek, wanted to go to bed early. He said goodbye to 10 of his friends, forgot about the remaining 3 , and went to bed. After a while he returned to the party, said goodbye to 10 of his friends (not necessarily the same as before), and went to bed. Later Fredek came back a... | Answer: Fredek returned at least 32 times.
Assume Fredek returned $k$ times, i.e. he was saying good-bye $k+1$ times to his friends. There exists a friend of Fredek, call him $X_{13}$, about whom Fredek forgot $k$ times in a row, starting from the very first time - otherwise Fredek would have come back less than $k$ t... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Two positive integers are written on the blackboard. Initially, one of them is 2000 and the other is smaller than 2000. If the arithmetic mean $m$ of the two numbers on the blackboard is an integer, the following operation is allowed: one of the two numbers is erased and replaced by $m$. Prove that this operation canno... | Each time the operation is performed, the difference between the two numbers on the blackboard will become one half of its previous value (regardless of which number was erased). The mean value of two integers is an integer if and only if their difference is an even number. Suppose the initial numbers were $a=2000$ and... | 10 | Number Theory | proof | Yes | Yes | olympiads | false |
A sequence of positive integers $a_{1}, a_{2}, \ldots$ is such that for each $m$ and $n$ the following holds: if $m$ is a divisor of $n$ and $m<n$, then $a_{m}$ is a divisor of $a_{n}$ and $a_{m}<a_{n}$. Find the least possible value of $a_{2000}$. | Answer: 128.
Let $d$ denote the least possible value of $a_{2000}$. We shall prove that $d=128$.
Clearly the sequence defined by $a_{1}=1$ and $a_{n}=2^{\alpha_{1}+\alpha_{2}+\cdots+\alpha_{k}}$ for $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots \cdots p_{k}^{\alpha_{k}}$ has the required property. Since $2000=2^{4}... | 128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all positive integers $n$ such that $n$ is equal to 100 times the number of positive divisors of $n$. | Answer: 2000 is the only such integer.
Let $d(n)$ denote the number of positive divisors of $n$ and $p \triangleright n$ denote the exponent of the prime $p$ in the canonical representation of $n$. Let $\delta(n)=\frac{n}{d(n)}$. Using this notation, the problem reformulates as follows: Find all positive integers $n$ ... | 2000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let's call a positive integer "interesting" if it is a product of two (distinct or equal) prime numbers. What is the greatest number of consecutive positive integers all of which are "interesting"? | The three consecutive numbers $33=3 \cdot 11,34=2 \cdot 17$ and $35=5 \cdot 7$ are all "interesting". On the other hand, among any four consecutive numbers there is one of the form $4 k$ which is "interesting" only if $k=1$. But then we have either 3 or 5 among the four numbers, neither of which is "interesting". | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Compute the sum of all positive integers whose digits form either a strictly increasing or a strictly decreasing sequence. | Denote by $I$ and $D$ the sets of all positive integers with strictly increasing (respectively, decreasing) sequence of digits. Let $D_{0}, D_{1}, D_{2}$ and $D_{3}$ be the subsets of $D$ consisting of all numbers starting with 9 , not starting with 9 , ending in 0 and not ending in 0 , respectively. Let $S(A)$ denote ... | 25617208995 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Solve the system of equations:
$$
\left\{\begin{array}{l}
x^{5}=y+y^{5} \\
y^{5}=z+z^{5} \\
z^{5}=t+t^{5} \\
t^{5}=x+x^{5} .
\end{array}\right.
$$ | Adding all four equations we get $x+y+z+t=0$. On the other hand, the numbers $x, y, z, t$ are simultaneously positive, negative or equal to zero. Thus, $x=y=z=t=0$ is the only solution. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
There are 13 cities in a certain kingdom. Between some pairs of cities two-way direct bus, train or plane connections are established. What is the least possible number of connections to be established in order that choosing any two means of transportation one can go from any city to any other without using the third k... | An example for 18 connections is shown in Figure 1 (where single, double and dashed lines denote the three different kinds of transportation). On the other hand, a connected graph with 13 vertices has at least 12 edges, so the total number of connections for any two kinds of vehicle is at least 12 . Thus, twice the tot... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
An equilateral triangle $A B C$ is divided into 100 congruent equilateral triangles. What is the greatest number of vertices of small triangles that can be chosen so that no two of them lie on a line that is parallel to any of the sides of the triangle $A B C$ ? | An example for 7 vertices is shown in Figure 2. Now assume we have chosen 8 vertices satisfying the conditions of the problem. Let the height of each small triangle be equal to 1 and denote by $a_{i}, b_{i}, c_{i}$ the distance of the $i$ th point from the three sides of the big triangle. For any $i=1,2, \ldots, 8$ we ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A square is divided into 16 equal squares, obtaining the set of 25 different vertices. What is the least number of vertices one must remove from this set, so that no 4 points of the remaining set are the vertices of any square with sides parallel to the sides of the initial square?
Remark. The proposed solution to thi... | The example in Figure 3a demonstrates that it suffices to remove 8 vertices to "destroy" all squares. Assume now that we have managed to do that by removing only 6 vertices. Denote the horizontal and vertical lines by $A, B, \ldots, E$ and $1,2, \ldots, 5$ respectively. Obviously, one of the removed vertices must be a ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Two circles, both with the same radius $r$, are placed in the plane without intersecting each other. A line in the plane intersects the first circle at the points $A, B$ and the other at the points $C, D$ so that $|A B|=|B C|=|C D|=14 \mathrm{~cm}$. Another line intersects the circles at points $E, F$ and $G, H$ respec... | First, note that the centres $O_{1}$ and $O_{2}$ of the two circles lie on different sides of the line $E H-$ otherwise we have $r<12$ and $A B$ cannot be equal to 14 . Let $P$ be the intersection point of $E H$ and $O_{1} O_{2}$ (see Figure 4). Points $A$ and $D$ lie on the same side of the line $O_{1} O_{2}$ (otherwi... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $\mathbb{N}$ denote the set of positive integers. Let $\varphi: \mathbb{N} \rightarrow \mathbb{N}$ be a bijective function and assume that there exists a finite limit
$$
\lim _{n \rightarrow \infty} \frac{\varphi(n)}{n}=L
$$
What are the possible values of $L$ ? | In this solution we allow $L$ to be $\infty$ as well. We show that $L=1$ is the only possible value. Assume that $L>1$. Then there exists a number $N$ such that for any $n \geq N$ we have $\frac{\varphi(n)}{n}>1$ and thus $\varphi(n) \geq n+1 \geq N+1$. But then $\varphi$ cannot be bijective, since the numbers $1,2, \l... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine all positive integers $n$ with the property that the third root of $n$ is obtained by removing the last three decimal digits of $n$. | Answer: 32768 is the only such integer.
If $n=m^{3}$ is a solution, then $m$ satisfies $1000 m \leqslant m^{3}<1000(m+1)$. From the first inequality, we get $m^{2} \geqslant 1000$, or $m \geqslant 32$. By the second inequality, we then have
$$
m^{2}<1000 \cdot \frac{m+1}{m} \leqslant 1000 \cdot \frac{33}{32}=1000+\fr... | 32768 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A cube with edge length 3 is divided into 27 unit cubes. The numbers $1,2, \ldots, 27$ are distributed arbitrarily over the unit cubes, with one number in each cube. We form the 27 possible row sums (there are nine such sums of three integers for each of the three directions parallel to the edges of the cube). At most ... | Answer: 24.
Since each unit cube contributes to exactly three of the row sums, then the total of all the 27 row sums is $3 \cdot(1+2+\ldots+27)=3 \cdot 14 \cdot 27$, which is even. Hence there must be an even number of odd row sums.
 Let $k(n)$ be the number of positive integers that cover 1998 and have exactly $n$ digits $(n \geqslant 5)$, all different from 0 . What is the remain... | Answer: 1.
Let $1 \leqslant g<h<i<j \leqslant n$ be fixed integers. Consider all $n$-digit numbers $a=\overline{a_{1} a_{2} \ldots a_{n}}$ with all digits non-zero, such that $a_{g}=1, a_{h}=9, a_{i}=9$, $a_{j}=8$ and this quadruple 1998 is the leftmost one in $a$; that is,
$$
\begin{cases}a_{l} \neq 1 & \text { if }... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a certain kingdom, the king has decided to build 25 new towns on 13 uninhabited islands so that on each island there will be at least one town. Direct ferry connections will be established between any pair of new towns which are on different islands. Determine the least possible number of these connections. | Let $a_{1}, \ldots, a_{13}$ be the numbers of towns on each island. Suppose there exist numbers $i$ and $j$ such that $a_{i} \geq a_{j}>1$ and consider an arbitrary town $A$ on the $j$-th island. The number of ferry connections from town $A$ is equal to $25-a_{j}$. On the other hand, if we "move" town $A$ to the $i$-th... | 222 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Prove that there exists a number $\alpha$ such that for any triangle $A B C$ the inequality
$$
\max \left(h_{A}, h_{B}, h_{C}\right) \leq \alpha \cdot \min \left(m_{A}, m_{B}, m_{C}\right)
$$
holds, where $h_{A}, h_{B}, h_{C}$ denote the lengths of the altitudes and $m_{A}, m_{B}, m_{C}$ denote the lengths of the med... | Let $h=\max \left(h_{A}, h_{B}, h_{C}\right)$ and $m=\min \left(m_{A}, m_{B}, m_{C}\right)$. If the longest height and the shortest median are drawn from the same vertex, then obviously $h \leq m$. Now let the longest height and shortest median be $A D$ and $B E$, respectively, with $|A D|=h$ and $|B E|=m$. Let $F$ be ... | 2 | Inequalities | proof | Yes | Yes | olympiads | false |
For a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of real numbers it is known that
$$
a_{n}=a_{n-1}+a_{n+2} \quad \text { for } n=2,3,4, \ldots
$$
What is the largest number of its consecutive elements that can all be positive?
Answer: 5. | The initial segment of the sequence could be $1 ; 2 ; 3 ; 1 ; 1 ;-2 ; 0$. Clearly it is enough to consider only initial segments. For each sequence the first 6 elements are $a_{1} ; a_{2}$; $a_{3} ; a_{2}-a_{1} ; a_{3}-a_{2} ; a_{2}-a_{1}-a_{3}$. As we see, $a_{1}+a_{5}+a_{6}=a_{1}+\left(a_{3}-a_{2}\right)+\left(a_{2}-... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Suppose that the real numbers $a_{i} \in[-2,17], i=1,2, \ldots, 59$, satisfy $a_{1}+a_{2}+\cdots+a_{59}=0$. Prove that
$$
a_{1}^{2}+a_{2}^{2}+\cdots+a_{59}^{2} \leq 2006 .
$$ | For convenience denote $m=-2$ and $M=17$. Then
$$
\left(a_{i}-\frac{m+M}{2}\right)^{2} \leq\left(\frac{M-m}{2}\right)^{2},
$$
because $m \leq a_{i} \leq M$. So we have
$$
\begin{aligned}
\sum_{i=1}^{59}\left(a_{i}-\frac{m+M}{2}\right)^{2} & =\sum_{i} a_{i}^{2}+59 \cdot\left(\frac{m+M}{2}\right)^{2}-(m+M) \sum_{i} a_... | 2006 | Inequalities | proof | Yes | Yes | olympiads | false |
Let $a, b, c, d, e, f$ be non-negative real numbers satisfying $a+b+c+d+e+f=6$. Find the maximal possible value of
$$
a b c+b c d+c d e+d e f+e f a+f a b
$$
and determine all 6-tuples $(a, b, c, d, e, f)$ for which this maximal value is achieved.
Answer: 8 . | If we set $a=b=c=2, d=e=f=0$, then the given expression is equal to 8 . We will show that this is the maximal value. Applying the inequality between arithmetic and geometric mean we obtain
$$
\begin{aligned}
8 & =\left(\frac{(a+d)+(b+e)+(c+f)}{3}\right)^{3} \geq(a+d)(b+e)(c+f) \\
& =(a b c+b c d+c d e+d e f+e f a+f a ... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Determine the maximal size of a set of positive integers with the following properties:
(1) The integers consist of digits from the set $\{1,2,3,4,5,6\}$.
(2) No digit occurs more than once in the same integer.
(3) The digits in each integer are in increasing order.
(4) Any two integers have at least one digit in c... | Associate with any $a_{i}$ the set $M_{i}$ of its digits. By (??), (??) and (??) the numbers are uniquely determined by their associated subsets of $\{1,2, \ldots, 6\}$. By (??) the sets are intersecting. Partition the 64 subsets of $\{1,2, \ldots, 6\}$ into 32 pairs of complementary sets $(X,\{1,2, \ldots, 6\}-X)$. Ob... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The altitudes of a triangle are 12,15 and 20. What is the area of the triangle?
Answer: 150. | Denote the sides of the triangle by $a, b$ and $c$ and its altitudes by $h_{a}, h_{b}$ and $h_{c}$. Then we know that $h_{a}=12, h_{b}=15$ and $h_{c}=20$. By the well known relation $a: b=h_{b}: h_{a}$ it follows $b=\frac{h_{a}}{h_{b}} a=\frac{12}{15} a=\frac{4}{5} a$. Analogously, $c=\frac{h_{a}}{h_{c}} a=\frac{12}{20... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Determine all positive integers $n$ such that $3^{n}+1$ is divisible by $n^{2}$.
Answer: Only $n=1$ satisfies the given condition. | First observe that if $n^{2} \mid 3^{n}+1$, then $n$ must be odd, because if $n$ is even, then $3^{n}$ is a square of an odd integer, hence $3^{n}+1 \equiv 1+1=2(\bmod 4)$, so $3^{n}+1$ cannot be divisible by $n^{2}$ which is a multiple of 4 .
Assume that for some $n>1$ we have $n^{2} \mid 3^{n}+1$. Let $p$ be the sma... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. In a club with 30 members, every member initially had a hat. One day each member sent his hat to a different member (a member could have received more than one hat). Prove that there exists a group of 10 members such that no one in the group has received a hat from another one in the group. | Let $S$ be the given group of 30 people. Consider all subsets $A \subset S$ such that no member of $A$ received a hat from a member of $A$. Among such subsets, let $T$ be a subset of maximal cardinality. The assertion of the problem is that $|T| \geq 10$.
Let $U \subset S$ consist of all people that have received a ha... | 10 | Combinatorics | proof | Yes | Yes | olympiads | false |
. In an acute triangle $A B C$, the segment $C D$ is an altitude and $H$ is the orthocentre. Given that the circumcentre of the triangle lies on the line containing the bisector of the angle $D H B$, determine all possible values of $\angle C A B$. | The value is $\angle C A B=60^{\circ}$.
Denote by $\ell$ the line containing the angle bisector of $D H B$, and let $E$ be the point where the ray $C D \rightarrow$ intersects the circumcircle of the triangle $A B C$ again. The rays $H D \rightarrow$ and $H B \rightarrow$ are symmetric with respect to $\ell$ by the de... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
. The points $M$ and $N$ are chosen on the angle bisector $A L$ of a triangle $A B C$ such that $\angle A B M=\angle A C N=23^{\circ} . X$ is a point inside the triangle such that $B X=C X$ and $\angle B X C=2 \angle B M L$. Find $\angle M X N$. | Answer: $\angle M X N=2 \angle A B M=46^{\circ}$.
Let $\angle B A C=2 \alpha$. The triangles $A B M$ and $A C N$ are similar, therefore $\angle C N L=\angle B M L=$ $\alpha+23^{\circ}$. Let $K$ be the midpoint of the arc $B C$ of the circumcircle of the triangle $A B C$. Then $K$ belongs to the the line $A L$ and $\an... | 46 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
. For a positive integer $k$, let $d(k)$ denote the number of divisors of $k$ (e.g. $d(12)=6$ ) and let $s(k)$ denote the digit sum of $k$ (e.g. $s(12)=3$ ). A positive integer $n$ is said to be amusing if there exists a positive integer $k$ such that $d(k)=s(k)=n$. What is the smallest amusing odd integer greater than... | The answer is 9 . For every $k$ we have $s(k) \equiv k(\bmod 9)$. Calculating remainders modulo 9 we have the following table
| $m$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $m^{2}$ | 0 | 1 | 4 | 0 | 7 | 7 | 0 | 4 | 1 |
| $m^{6}$ | 0 | 1 |... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. For which $k$ do there exist $k$ pairwise distinct primes $p_{1}, p_{2}, \ldots, p_{k}$ such that
$$
p_{1}^{2}+p_{2}^{2}+\cdots+p_{k}^{2}=2010 ?
$$ | We show that it is possible only if $k=7$.
The 15 smallest prime squares are:
$$
4,9,25,49,121,169,289,361,529,841,961,1369,1681,1849,2209
$$
Since $2209>2010$ we see that $k \leq 14$.
Now we note that $p^{2} \equiv 1 \bmod 8$ if $p$ is an odd prime. We also have that $2010 \equiv 2 \bmod 8$. If all the primes are ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Two boxes together contain 65 balls of various sizes. Each ball is white, black, red, or yellow. Each time we take five balls of the same color, at least two are of the same size.
(a) What is the maximum number of types of balls that exist in the boxes? Two balls are considered of different types when they have differ... | (a) There cannot be five balls of the same color and different sizes because every time we pick five balls of the same color, two must be of the same size. Thus, there are at most four sizes for each color. Therefore, there are at most \(4 \times 4=16\) types of balls.
(b) The two boxes together contain 65 balls, and ... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Two irreducible fractions have their denominators equal to 600 and 700. Find the minimum value for the denominator of the sum of the fractions. | Suppose the fractions are a/600 and b/700. Since they are irreducible, a and 600 have no common factor greater than 1, and the same applies to b and 700.
Adding the two fractions, we get
$$
\frac{a}{600}+\frac{b}{700}=\frac{7 a+6 b}{4200}=\frac{7 a+6 b}{2^{3} \times 3 \times 5^{2} \times 7}
$$
Observe that the numer... | 168 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $x_{1}, x_{2}, \ldots, x_{n}$ be a sequence where each term is 0, 1, or -2. If
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=-5 \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=19
\end{array}\right.
$$
determine $x_{1}^{5}+x_{2}^{5}+\cdots+x_{n}^{5}$. | Let $a$ be the number of terms equal to 1 and $b$ be the number of terms equal to -2. We can write:
$$
\left\{\begin{array} { l }
{ a \cdot 1 + b \cdot ( - 2 ) = - 5 } \\
{ a \cdot 1 ^ { 2 } + b \cdot ( - 2 ) ^ { 2 } = 1 9 }
\end{array} \Longleftrightarrow \left\{\begin{array}{l}
a-2 b=-5 \\
a+4 b=19
\end{array}\righ... | -125 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
If $3^{n}=2$ then what is the value of $27^{2 n}$? | We have: $27^{2 n}=\left(3^{3}\right)^{2 n}=3^{6 n}=\left(3^{n}\right)^{6}=2^{6}=64$. | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
If the price of a product increased from $R \$ 5.00$ to $R \$ 5.55$, what was the percentage increase? | The increase in reais was $5.55-5=0.55$; so the percentage increase was
$$
\frac{0.55}{5}=\frac{0.55 \times 20}{5 \times 20}=\frac{11}{100}=11 \%
$$ | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A factory produced an original calculator that performs two operations:
- the usual addition +
- the operation $\circledast$
We know that for any natural number $a$ we have:
$$
\text { (i) } a \circledast a=a \quad \text { and (ii) } a \circledast 0=2 a
$$
and, for any four naturals $a, b, c$ and $d$
$$
\text { (i... | To calculate $(2+3) \circledast(0+3)$, we will use property (iii), and we have:
$$
(2+3) \circledast(0+3)=(2 \circledast 0)+(3 \circledast 3)
$$
Now, by (i) we have $2 \circledast 0=2 \times 2=4$, and by (ii) we have $3 \circledast 3=3$. Therefore,
$$
(2+3) \circledast(0+3)=4+3=7
$$
Now, to calculate $1024 \circled... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The set $\{1,2,3, \ldots, 3000\}$ contains a subset of 2000 elements such that no element is double the other? | Let's construct the requested subset in the following way:
- it contains all odd numbers: $1,3,5, \ldots, 2999$. Here we already have a list with 1500 numbers.
- the set cannot contain numbers of the form $2 \times$ (odd number),
- the set can contain numbers of the form $4 \times$ (odd number), that is,
$$
\underbra... | 2249 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the figure below, $D$ is the midpoint of side $A B, C E: D E=5: 3 \text{ and } B F: E F=1: 3$. If the area of triangle $A B C$ is $192 \text{ cm}^2$, determine the area of triangle $B D F$.
 $x=a+b$ and $y=c+d$ (sums without common addends). Then $a+c, a+d, b+c$ and $b+d$ are, in some order, the numbers 189, 320, 287 and 264. Adding these four sums, we get $a+b+c+d=530$. Thus, $x+y=530$.
II) $x=a+b$ and $y=a+... | 761 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the right isosceles triangle $A O B$, the points $P, Q$, and $S$ are chosen on the sides $O B, O A$, and $A B$, respectively, such that $P Q R S$ is a square. If the lengths of $O P$ and $O Q$ are $a$ and $b$, respectively, and the area of the square $P Q R S$ is $2 / 5$ of the area of the triangle $A O B$, determin... | Solution
Let $C$ be the foot of the perpendicular from point $S$ to segment $O B$. The triangles $S P C$ and $P Q O$ have the same angles, since
$$
\begin{aligned}
\angle C P S & =\angle 180^{\circ}-\angle S P Q-\angle O P Q \\
& =90^{\circ}-\angle O P Q \\
& =\angle P Q O
\end{aligned}
$$
Since $P S=P Q$, these tri... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the following drawing, $\angle C B G=20^{\circ}, \angle G B E=40^{\circ}, \angle E B F=20^{\circ}, \angle B C F=50^{\circ}$ and $\angle F C E=30^{\circ}$.
a) Verify that $B G=B F$.
b) Ve... | Solution
a) We have
$$
\angle C G B=180^{\circ}-\angle B C G-\angle C B G=80^{\circ}=\angle B C G
$$
Therefore, $B C=B G$. On the other hand,
$$
\angle B F C=180^{\circ}-\angle C B F-\ang... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The denominators of two irreducible fractions are 600 and 700. What is the smallest possible value of the denominator of their sum when written as an irreducible fraction?
Note: We say that the fraction $p / q$ is irreducible if the integers $p$ and $q$ do not have any prime factors in common in their factorizations. ... | Solution
Let $a / 600$ and $b / 700$ be the two irreducible fractions. Thus, $m d c(a, 600)=m d c(b, 700)=1$. The sum of the two fractions can be written as
$$
\begin{aligned}
\frac{a}{600}+\frac{b}{700} & =\frac{7 a+6 b}{6 \cdot 7 \cdot 100} \\
& =\frac{7 a+6 b}{3 \cdot 7 \cdot 2^{3} \cdot 5^{2}}
\end{aligned}
$$
S... | 168 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In how many ways can we place 8 digits equal to 1 and 8 digits equal to $0$ on a $4 \times 4$ board so that the sums of the numbers written in each row and column are the same?
| 1 | 0 | 1 | 0 |
| :--- | :--- | :--- | :--- |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | | Solution
Since the sum of the numbers in all the cells of the board is 8, the sum of the numbers in each row and column is $8 / 4=2$. That is, in each row and column, there are exactly two digits equal to 1 and two digits equal to 0. We can choose the position of the first 1 in the first row in 4 ways. Then, we can ch... | 90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
If $n$ is a positive integer, what is the smallest value that the sum of the digits in the decimal representation of $3 n^{2}+n+1$ can take?
# | Solution
If $n=8$, we have that $3 n^{2}+n+1=201$ and the sum of its digits is 3. We will now verify that the sum of the digits of $3 n^{2}+n+1$ cannot be 1 or 2 and conclude that the smallest possible value is 3. Since $n(n+1)$ is the product of two consecutive numbers, it is even, and thus $3 n^{2}+n+1=2 n^{2}+n(n+1... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the following drawing, the chords $D E$ and $B C$ are perpendicular, with $B C$ being a diameter of the circle with center at $A$. Additionally, $\angle C G F=40^{\circ}$ and $G H=2 \text{~cm}$.
a) Determine the value of the angle $\angle C H F$.
b) Find the length of $H J$.
 Since $BC$ is a diameter, it follows that $\angle BFC=90^{\circ}$. Thus, as we also have $\angle CHG=90^{\circ}$, the circle $\Gamma$ with diameter $CG$ passes through $F$ and $H$. In this circle, the angles $\angle CGF$ and $\angle CHF$ are inscribed in the same arc $CF$, so $\angle CHF = \angle CGF = 40^... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $A$ be a subset of $\{1,2,3, \ldots, 2019\}$ having the property that the difference between any two of its elements is not a prime number. What is the largest possible number of elements of $A$?
# | Solution
Suppose $a \in A$. Then, no element of the set $\{a+2, a+3, a+5, a+7\}$ can belong to $A$, and among the elements of $\{a+1, a+4, a+6\}$, at most one of them can belong to $A$. Thus, in every 8 consecutive integers, say the elements of the set $\{a, a+1, a+2, \ldots, a+7\}$, at most two of them can belong to ... | 505 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What is the largest positive integer $n$ for which there is a unique integer $k$ such that
$$
\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13} ?
$$ | Solution
We can write the inequality as
$$
\frac{13}{7}<\frac{n+k}{n}<\frac{15}{8}
$$
Thus, multiplying the members of the inequality by $56n$, we obtain the equivalent inequality
$$
\begin{aligned}
104n & <56n+56k & <105n \\
48n< & 56k & <49n
\end{aligned}
$$
For there to be a unique integer $k$ satisfying the in... | 112 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
a) Given that the decimal representation of $5^{2018}$ has 1411 digits and starts with 3 (the leftmost non-zero digit is 3), for how many integers $1 \leq n \leq 2017$ does the number $5^{n}$ start with 1?
b) The integers $4^{52}$ and $5^{52}$ both start with the digit 2. If the decimal representations of the powers $... | Solution
a) If $5^{k}$ starts with $a$ and has $j$ digits, then
$$
10^{j}5 \cdot 2 \cdot 10^{j}=10^{j+1}
$$
would have at least $j+1$ digits. Therefore, the problem reduces to finding the values of $k \in\{1,2, \ldots, 2017\}$ such that $5^{k}$ and $5^{k+1}$ have the same number of digits. Between two consecutive po... | 607 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Two precocious students from Level 3 participated in a university chess tournament. Each participant plays against all the others exactly once. A win is worth 1 point, a draw is worth 0.5 points, and a loss is worth 0 points. The sum of the scores of the two Level 3 students is 6.5. All university students scored the s... | Solution
Let $x$ be the number of college students and $p$ the common score of all of them. Since exactly 1 point is contested in each game, it follows that the total score of the tournament, which is $6.5 + p x$, coincides with the number of games, which is $\frac{(x+2)(x+1)}{2}$. In addition, the score of each parti... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
a) Verify that for any positive integer $a, \operatorname{with} a>1$, the equation
$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{a}
$$
has at least three solutions of the form $(x, y)$, with $x$ and $y$ positive integers. For example, for $a=3$, the pairs $(6,6),(4,12)$ and $(12,4)$ are solutions.
b) Find the number of pairs ... | Solution
a) We can find an equivalent equation:
$$
\begin{aligned}
\frac{1}{x}+\frac{1}{y} & =\frac{1}{a} \Leftrightarrow \\
(x-a)(y-a) & =a^{2}
\end{aligned}
$$
Since $1 / x$ and $1 / y$ are less than $1 / a$, it follows that $x-a$ and $y-a$ are positive. To find solutions to the last equation, consider the followi... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a certain country, there are exactly 2019 cities and between any two of them, there is exactly one direct flight operated by some airline, that is, given cities $A$ and $B$, there is either a flight from $A$ to $B$ or a flight from $B$ to $A$. Find the smallest number of airlines operating in the country, knowing th... | Solution
The answer is 2019. Since there are 1009 disjoint pairs of cities, each airline can operate on at most 1009 pairs. There are exactly 2019 $\cdot$ 2018/2 direct flights, so the number of airlines is at least $\frac{2019 \cdot 2018}{2 \cdot 1009}=2019$. It remains to exhibit an example to verify that this numbe... | 2019 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A math competition consists of three problems, each of which receives an integer score from 0 to 7. For any two competitors, we know that there is at most one problem on which they obtained the same score. Find the largest possible number of competitors in this competition. | Solution
There are 8 possible scores for each problem, and consequently, $8 \cdot 8=64$ distinct possible scores for the first two problems. Since no two competitors can have exactly the same scores on the first two problems, the total number of competitors cannot be greater than 64. We will now show that this maximum... | 64 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A chicken coop with an area of $240 \mathrm{~m}^{2}$ is to house chickens and chicks, with a desirable free space of $4 \mathrm{~m}^{2}$ for each chicken and $2 \mathrm{~m}^{2}$ for each chick. Additionally, each chick consumes $40 \mathrm{~g}$ of feed per day, and each chicken consumes $160 \mathrm{~g}$ per day, with ... | Let $x$ and $y$ be, respectively, the number of chickens and chicks in the coop.
(a) We have $4 x+2 y=240$, that is, $2 x+y=120$.
Since $8 \text{ kg} = 8000 \text{ g}$, we have: $160 x+40 y \leq 8000$. Thus, $4 x+y \leq 200$. In summary, the number $x$ of chickens and $y$ of chicks satisfy:
$$
(*)\left\{\begin{array... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Starting from her house to get to school, Júlia must walk 8 blocks to the right and 5 blocks up, as indicated in the figure below.
She knows that there are many different ways to make the j... | No matter how Júlia walks from her house to school, she must travel 8 blocks to the right and 5 blocks up. A path connecting her house to the school is then a sequence of "block crossings," with 8 in the horizontal direction (to the right) and 5 in the vertical direction (up). Thus, to define a path, she only needs to ... | 1287 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
If in the fraction $\frac{x}{y}$ we decrease the numerator by $40 \%$ and the denominator $y$ by $60 \%$, then the fraction $\frac{x}{y}$:
(A) decreases by $20 \%$
(B) increases by $20 \%$
(C) decreases by $50 \%$
(D) increases by $50 \%$ | The correct option is (D).
If a number $x$ is decreased by $40\%$, it becomes $60\%$ of $x$, that is: $0.6 x$. Similarly, when a number $y$ is decreased by $60\%$, it becomes $0.4 y$. Therefore, the fraction $\frac{x}{y}$ becomes $\frac{0.6 x}{0.4 y}=\frac{6}{4} \frac{x}{y}=1.5 \frac{x}{y}$. This means that the fracti... | 50 | Algebra | MCQ | Yes | Yes | olympiads | false |
An empty swimming pool was filled with water by two faucets $A$ and $B$, both with constant flow rates. For 4 hours, both faucets were open and filled $50 \%$ of the pool. Then, faucet B was turned off and for 2 hours faucet A filled $15 \%$ of the pool's volume. After this period, faucet A was turned off and faucet $\... | Since taps A and B pour water into the pool at a constant flow rate, the volume of water poured by each tap is proportional to the time it is open. Therefore, if tap A fills $15 \%$ of the pool's volume in 2 hours, then in 4 hours it will fill $30 \%$ of the pool's volume.
However, when taps A and B are both open for ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Davi has a very original calculator; it performs only two operations: the usual addition $(+)$ and another operation, denoted by $*$, which satisfies:
(i) $a * a=a$
(ii) $a * 0=2a$
(iii) $(a * b) + (c * d) = (a * c) + (b * d)$
What are the results of the operations $(2+3) * (0+3)$ and $1024 * 48$? | To calculate $(2 * 3)+(0 * 3)$, we use properties (i), (ii), and (iii). Then
$$
\begin{aligned}
& (2 * 3)+(0 * 3) \quad \stackrel{(\mathrm{iii})}{=} \quad(2 * 0)+(3 * 3) \\
& \text { (i) (ii) } 2 \times 2+3=7 \text {. }
\end{aligned}
$$
To calculate $1024 * 48$, observe that $1024=976+48$. We have:
$$
\begin{aligned... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The rectangular grid in the figure is made of 31 segments of $0.5 \mathrm{~cm}$ and comprises 12 squares. Rosa drew on a rectangular sheet of $21 \mathrm{~cm}$ by $29.7 \mathrm{~cm}$, grid-lined with squares of side $0.5 \mathrm{~cm}$, a large rectangular grid made with 1997 segments. How many squares does this rectang... | Let $m$ and $n$ be, respectively, the number of segments of $0.5 \, \text{cm}$ on two consecutive sides of the rectangle. We know that the total number of segments of $0.5 \, \text{cm}$ in the division of the rectangle into $m \times n$ squares of side $0.5 \, \text{cm}$ is: $m(n+1) + n(m+1)$ (prove this). Thus,
$$
m(... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The radius of the Earth is approximately $6670 \mathrm{~km}$. Suppose a wire is adjusted to fit exactly over the Equator, which is a circle with a radius approximately equal to $6670 \mathrm{~km}$.
 Who will e... | Whoever had $x$ as their monthly grade will receive a discount of $x \%$ on that grade, meaning they will lose
$$
x \% \text { of } x=\frac{x}{100} \times x=\frac{x^{2}}{100}
$$
Thus, after the punishment, the grade becomes $x-\frac{x^{2}}{100}$, where $x$ was the initial grade.
Consider the function "grade after th... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the quadrilateral $A B C D$, we have: $A B=5, B C=17, C D=5, D A=9$, and the measure of the segment $D B$ is an integer. Determine $D B$.
 | Remember that, in a triangle, any side is greater than the difference and less than the sum of the other two. For triangle $A D B$, we have $A D - A B < B D < A D + A B$, and for triangle $C B D$, it follows that $B C - C D < B D < B C + C D$. Substituting the known values
^{2}}{1,001^{k+1}}1000 \Leftrightarrow k>2000
$$
Thus,... | 2001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $p$ and $q$ be positive integers such that $\frac{5}{8}<\frac{p}{q}<\frac{7}{8}$. What is the smallest value of $p$ for which $p+q=2005$? | Given $q=2005-p$, we have
$$
\frac{5}{8}<\frac{p}{2005-p}<\frac{7}{8}
$$
from which it follows that
$$
5(2005-p)<8 p \quad \text { and } 8 p<7(2005-p)
$$
Therefore,
$$
\frac{5 \times 2005}{13}<p<\frac{7 \times 2005}{15} \Rightarrow 771.15<p<935.66
$$
Thus, 772 is the smallest value of $p$ that satisfies the condi... | 772 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A triangle has vertex $A=(3,0), B=(0,3)$ and $C$, where $C$ is on the line $x+y=7$. What is the area of the triangle? | Observe that the height $h$, relative to the side $A B$, of all triangles $A B C$ that have the vertex $C$ on the line $x+y=7$, is the same, since the latter line is parallel to the line passing through $A$ and $B$. Therefore, these triangles all have the same area, namely:
$$
\frac{A B \times h}{2}
$$
We need to det... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Three circles with radii $1 \mathrm{~cm}, 2 \mathrm{~cm}$, and $3 \mathrm{~cm}$ are pairwise externally tangent, as shown in the figure below.
Determine the radius of the circle that is externally tangent to the three circles.
(\sqrt{2}-1)=3-2 \sqrt{2},(\sqrt{2}+1)(\sqrt{2}+1)=3+2 \sqrt{2}$ and $(\sqrt{2}-1)(\sqrt{2}+1)=1$.
Suppose that $a$ products are equal to $3-2 \sqrt{2}$, $b$ products are equal to $3+2 \sqrt{2}$ and $1002-a-b$ pr... | 502 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $A B C D$ be a right trapezoid with bases $A B$ and $C D$, and right angles at $A$ and $D$. Given that the shorter diagonal $B D$ is perpendicular to the side $B C$, determine the smallest possible value for the ratio $\frac{C D}{A D}$. | Let $A \widehat{B} D=B \widehat{D} C=\alpha$. Then we have that $D C=\frac{B D}{\cos \alpha}$ and $A D=B D \sin \alpha$, hence
$\frac{D C}{A D}=\frac{\frac{B D}{\cos \alpha}}{B D \sin \alpha}=\frac{1}{\sin \alpha \cos \alpha}=\frac{2}{\sin 2 \alpha} \geq 2$.
Equality occurs when $\sin 2 \alpha=1$, that is, when $\alp... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The twelve students in an olympiad class went out to play soccer every day after their math class, forming two teams of 6 players each and playing against each other. Each day they formed two different teams from those formed on previous days. By the end of the year, they found that every group of 5 students had played... | For every group of 5 students, there is a unique team formed that contains them. Therefore, we have $C_{12}^{5}=\frac{12.11 .10 .9 .8}{5!}=792$ teams for each 5 students chosen. On the other hand, in each team of 6 players, there are $C_{6}^{5}=6$ ways to choose five players, that is, there are 6 groups of 5 players wh... | 132 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let's denote by $s(n)$ the sum of the digits of the number $n$. For example, $s(2345) = 2 + 3 + 4 + 5 = 14$. Observe that:
$40 - s(40) = 36 = 9 \times 4; 500 - s(500) = 495 = 9 \times 55; 2345 - s(2345) = 2331 = 9 \times 259$.
(a) What can we say about the number $n - s(n)$?
(b) Using the previous item, calculate $s... | (a) Observe these two examples:
$$
\underbrace{2000}_{2 \cdot 10^3}-\underbrace{s(2000)}_{2}=1998, \underbrace{60000}_{6 \cdot 10^4}-\underbrace{s(60000)}_{6}=59994
$$
From these, it is easy to understand that if $a$ is a digit between 1 and 9, then $s\left(a \cdot 10^{k}\right)=a$.
Thus, we have:
$$
a \cdot 10^{k}... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Catarina has 210 numbered cards from 1 to 210.
a) How many of these cards have a number that is a multiple of 3?
b) How many of these cards have an even number that is not a multiple of 3?
c) What is the smallest number of cards Catarina must pick at random to be sure that at least two of them have the number 2 or t... | Solution
a) Since $210 \div 3=70$, there are 70 cards whose numbers are multiples of 3. More precisely, these cards are numbered $3=1 \times 3, 6=2 \times 3, 9=3 \times 3, 12=4 \times 3, \ldots, 204=68 \times 3, 207=69 \times 3$, and $210=70 \times 3$.
b) $1^{st}$ solution: A reasoning identical to that in part a) sh... | 73 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A number is framed when, by adding it to the number obtained by reversing the order of its digits, the result is a perfect square. For example, 164 and 461 are framed, since 164+461 = $625=25^{2}$. How many framed numbers are there between 10 and 100?
A) 5
B) 6
C) 8
D) 9
E) 10 | Solution
## ALTERNATIVE C
Let $n$ be a number between 10 and 100, $a$ its tens digit, and $b$ its units digit; note that $1 \leq a \leq 9$ and $0 \leq b \leq 9$. Then $n=10a+b$ and the number obtained by reversing the digits of $n$ is $10b+a$. Since $n$ is a framed number, we have that $(10a+b) + (10b+a) = 11a + 11b ... | 8 | Number Theory | MCQ | Yes | Yes | olympiads | false |
Patrícia wrote, in ascending order, the positive integers formed only by odd digits: 1, $3,5,7,9,11,13,15,17,19,31,33, \ldots$ What was the $157^{\text{th}}$ number she wrote?
A) 997
B) 999
C) 1111
D) 1113
E) 1115 | Solution
## ALTERNATIVE D
There are five odd digits: 1, 3, 5, 7, and 9. Counting only positive integers, there are then 5 numbers formed by only one odd digit, $5 \times 5=25$ numbers formed by two odd digits, and $5 \times 5 \times 5=125$ numbers formed by three odd digits. Thus, there are $5+25+125=155$ positive in... | 1113 | Number Theory | MCQ | Yes | Yes | olympiads | false |
In the multiplication indicated in the figure on the side, the asterisks represent digits, which may or may not be the same. What is the sum of the numbers that were multiplied?
A) 82
B) 95
C) 110
D) 127
E) 132
 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 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The small squares of the board in the figure must be filled in such a way that:
- in the small squares of each of the regions in the shape of $\hookleftarrow$ appear the numbers 1, 3, 5, and 7 or the numbers 2, 4, 6, and 8;
- in small squares with a common side, consecutive numbers do not appear.
 The sequence is $37 \rightarrow 38 \rightarrow 19 \rightarrow 20 \rightarrow 10 \rightarrow 5 \rightarrow 6 \rightarrow 3 \rightarrow 4 \rightarrow 2 \rightarrow 1$.
b) The only sequence of length 3 is $4 \rightarrow 2 \rightarrow 1$. The sequences of length 4 are $3 \rightarrow 4 \rightarrow 2 \rightarro... | 610 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The square in Figure I is called special because:
- it is divided into 16 equal squares;
- in each row and in each column, the digits 1, 2, 3, and 4 appear;
- in each of the squares \(A, B, C\), and \(D\) (as in Figure II), the digits 1, 2, 3, and 4 appear.
| 4 | 2 | 1 | 3 |
| :--- | :--- | :--- | :--- |
| 1 | 3 | 2 ... | Solution
a) The solution is presented in the figure below:
| $\mathbf{1}$ | 2 | $\mathbf{4}$ | $\mathbf{3}$ |
| :--- | :--- | :--- | :--- |
| 3 | 4 | 2 | 1 |
| $\mathbf{2}$ | $\mathbf{3}$ | 1 | $\mathbf{4}$ |
| $\mathbf{4}$ | $\mathbf{1}$ | $\mathbf{3}$ | 2 |
b) No. Since the small squares in the last column of squa... | 288 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The circles in the figure below have been filled with the numbers 1 to 7, such that all arrows point from a smaller number to a larger one. In this case, we say that the figure is well filled.
 There is only one way to fill in the diagram, as we show below.
- The number 9 cannot be placed below any number, so it must be at the top.
- Above the number 7, we can only place the 9 or the 8. Since the 9 is already at the top, the 8 will be above the 7.
- The number 6 cannot be placed below the 5 or t... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Juliana wants to assign each of the 26 letters $A, B, C, D, \ldots, W, X, Y, Z$ of the alphabet a different non-zero numerical value, such that $A \times C=B, B \times D=C, C \times E=D$, and so on, up to $X \times Z=Y$.
a) If Juliana assigns the values 5 and 7 to $A$ and $B$, respectively, what will be the values of ... | Solution
a) Substituting $A=5$ and $B=7$ into $A \times C=B$, we get $5 \times C=7$ and it follows that $C=\frac{7}{5}$. We can now find $D$ by substituting the values of $B$ and $D$ into $B \times D=C$; we obtain $7 \times D=\frac{7}{5}$ and thus $D=\frac{1}{5}$. Finally, from $C \times E=D$ we have $\frac{7}{5} \tim... | 2010 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The figure is formed by 5 equal isosceles trapezoids. What is the measure of the indicated angle?
A) $72^{\circ}$
B) $74^{\circ}$
C) $76^{\circ}$
D) $78^{\circ}$
E) $80^{\circ}$
# | Solution
## ALTERNATIVE A
We recall that the sum of the interior angles of a polygon with $n$ sides is $(n-2) \times 180^{\circ}$. We can view the figure in the statement as a polygon with 6 sides (in a thicker line in the figure below); the sum of its interior angles is then $(6-2) \times 180^{\circ}=720^{\circ}$. O... | 72 | Geometry | MCQ | Yes | Yes | olympiads | false |
The eight points highlighted in the figure divide the sides of the square into three equal parts. How many right triangles can be drawn with the three vertices at these points?
A) 8
B) 12
C)... | Solution
## ALTERNATIVE D
Let's choose a point among the highlighted points; for example, the first point to the left on the bottom side of the square. Figure 1 shows the three right-angled triangles we can construct with the right angle vertex at this point. Since the same applies to the other highlighted points, we... | 24 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
The figure shows a regular ten-sided polygon with center $O$. What is the measure of angle $a$?
A) $15^{\circ}$
B) $18^{\circ}$
C) $20^{\circ}$
D) $30^{\circ}$
E) $36^{\circ}$
# | Solution
## ALTERNATIVE B
The triangle $A O B$ is isosceles because the sides $O A$ and $O B$ are equal. Therefore, the angles $O A B$ and $O \hat{B} A$ are also equal, that is, both have a measure of $a$. We now notice that the central angle $A \hat{O} B$ measures $\frac{4}{10} \times 360^{\circ}=144^{\circ}$. Since... | 18 | Geometry | MCQ | Yes | Yes | olympiads | false |
In the figure, triangles $A B C$ and $D E F$ are equilateral with sides $14 \mathrm{~cm}$ and $13 \mathrm{~cm}$, respectively, and sides $B C$ and $E F$ are parallel.
a) Calculate the measu... | Solution
a) Since $BC$ and $EF$ are parallel, the angles $EUT$ and $ACB$ are alternate interior angles, hence $E \hat{U} T = A \hat{C} B = 60^{\circ}$. b) From item a) we can conclude that all triangles in the figure are equilateral. Thus, we have $QP = FP$, $UT = UE$, $TS = CS$, and $RQ = RB$. Therefore, the perimete... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A building has three different staircases, all starting at the base of the building and ending at the top. One staircase has 104 steps, another has 117 steps, and the other has 156 steps. Whenever the steps of the three staircases are at the same height, there is a floor. How many floors does the building have? | Solution
Let's call the three staircases $A, B$, and $C$, which have 104, 117, and 156 steps, respectively. Let $a$ be the number of steps in staircase $A$ between each two floors, $b$ be the number of steps in staircase $B$ between each two floors, and $c$ be the number of steps in staircase $C$ between each two floo... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the city of Autolândia, car plate numbers are made up of three digits, ranging from plate 000 to plate 999. To reduce pollution, Mayor Pietro decided to implement a car rotation system, setting the days on which people can use their cars. The rules of the rotation are:
- Monday: only cars with odd-numbered plates;
... | Solution
a) Since the number 729 is odd, the car with the license plate number 729 can circulate on Mondays. Since $7+2+9=18$, the car can also circulate on Tuesdays, but not on Thursdays. Since 729 is a multiple of 3, this car can also circulate on Wednesdays. Since 729 does not have identical digits, is greater than... | 363 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Consider a rectangle $A B C D$ where the lengths of the sides are $\overline{A B}=4$ and $\overline{B C}=8$. On the sides $B C$ and $A D$, points $M$ and $N$ are fixed, respectively, such that the quadrilateral $B M D N$ is a rhombus. Calculate the area of this rhombus. | Solution
Observe the following drawing:
Since $B M D N$ is a rhombus, the lengths of the segments $B M, M D, D N$, and $N B$ are all equal to the same value, say $x$. According to the prob... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A board of size $2013 \times 5$ (i.e., with 2013 rows and 5 columns) must be painted with the colors $A, B, C, D$. Some squares in the first row have already been painted, as shown in the figure below (the squares not represented in the figure have not been painted yet). To continue painting the board, we must follow t... | Solution
a) The house that is missing painting in the first row has neighbors painted with $B$ and $C$. Therefore, it can only be painted with $A$ or $D$. Let's analyze the case where it is painted with $A$, as shown in the figure below.
# | Solution
Let $x$ and $y$ be the measures of the white and gray angles, respectively, as shown in the figure:
Remember that the sum of the interior angles of a triangle is always equal to $1... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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