problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
Ten men sit side by side at a long table, all facing the same direction. Each of them is either a knight (and always tells the truth) or a knave (and always lies). Each of the people announces: "There are more knaves on my left than knights on my right." How many knaves are in the line? | Number the people from 1 to 10, from left to right according to their own perspective.
Person 1 is certainly lying, since he has no one to his left, and hence is a knave.
Person 10 is therefore a knight, since he has no one to his right and at least one knave (person 1) to his left.
Person 2 has exactly one knave to... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
At a party with 100 people, everyone is either a knight, who always tells the truth, or a knave, who always lies. Each person says they shook hands with a different number of knights at the party, from 0 to 99 . Each pair of people shook hands at most once, and everyone knows whether each other person is a knight or kn... | Call the person who said they shook hands with $i$ people person $i$ for each $i$ from 0 to 99 . Now, if person 99 is telling the truth, person 99 must have shaken hands with everyone else, and all the other people must be knights. But then person 0 would be lying, since they said they shook hands with no knights but m... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A castle has infinitely many rooms labeled $1,2,3, \ldots$, which are divided into several halls. Suppose room $n$ is on the same hall as rooms $3 n+1$ and $n+10$ for every $n$. Determine the maximum possible number of different halls in the castle. | There are at most three different halls in the castle. Because rooms $n$ and $n+10$ are on the same hall, any two rooms with the same units digit must be on the same hall.
Now, repeatedly using the rule that rooms $n$ and $3 n+1$ are on the same hall, we find that room 1 is on the same hall as rooms 4, 13, and 40, and... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
You are out walking and see a group of rhinoceroses (which each have two horns and four legs) and triceratopses (which each have three horns and four legs). If you count 31 horns and 48 legs, how many triceratopses are there? | Since each animal has 4 legs, there must be $48 / 4=12$ animals. Each triceratops has 3 horns and each rhinoceros has 2, so if there are $t$ triceratopses and $r$ rhinoceroses we get $t+r=12$ and $3 t+2 r=31$. Subtracting twice the first equation from the second gives $t=7$. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Prove that every positive real number $y$ satisfies
$$
2 y \geq 3-\frac{1}{y^{2}}
$$
When does equality occur? | (Problem source: Aops) Since $y$ is positive, we can multiply both sides by $y$ and rearrange to get the equivalent inequality
$$
2 y^{3}-3 y^{2}+1 \geq 0
$$
This factors as
$$
(2 y+1)(y-1)^{2} \geq 0 .
$$
The first factor is positive since $y>0$, and the second is nonnegative since it is a real number squared. Thu... | 1 | Inequalities | proof | Yes | Yes | olympiads | false |
It's a week before Thanksgiving, and a family is trying to find their turkey. There are 5 boxes in a row, and the turkey is hiding in one of the 5 boxes.
Every day, the family is allowed to check one box to try to find the turkey, and every night, the turkey moves to a box right next to the box it was in. For example,... | One strategy is as follows:
First assume the turkey starts in an even numbered box (box 2 or 4). On day 1, check box 2 . If the turkey is not there, he must have been in box 4 to begin with, so tomorrow he will be in box 3 or box 5 . On day 2 , check box 3 . If the turkey is not there, then he must currently be in box... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Prove that
$$
A=\sqrt{4-2 \sqrt{3}}-\frac{\sqrt{3}+1}{\sqrt{3}-1}
$$
is an integer. | $\sqrt{4-2 \sqrt{3}}=\sqrt{3^{2}-2 \sqrt{3}+\sqrt{1}^{2}}=\sqrt{(\sqrt{3}-1)^{2}}=|\sqrt{3}-1|=\sqrt{3}-1$. If multiply both numerator and the denominator of $\frac{\sqrt{3}+1}{\sqrt{3}-1}$ by $\sqrt{3}+1$ we get:
$$
\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{(\sqrt{3}+1)^{2}}{\sqrt{3}^{2}-1^{2}}=\frac{4+2 \sqrt{3}}{2}=2+\sq... | -3 | Algebra | proof | Yes | Yes | olympiads | false |
A $2012 \times 2012$ table is to be filled with integers in such a way that each of the 4026 rows, columns, and main diagonals has a different sum. What is the smallest number of distinct values that must be used in the table? | Answer: 3.
If at most two numbers are used, say $x$ and $y$, the sum of every row and column is completely determined by the number of $y$ 's it has, which ranges from 0 to 2012 . Thus there are only 2013 possible sums, not enough for the 4026 rows, columns, and diagonals.
On the other hand, if $n$ is a large integer... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5 married couples gather at a party. As they come in and greet each other, various people exchange handshakes - but, of course, people never shake hands with themselves or with their own respective spouses. At the end of the party, one woman goes around asking people how many hands they shook, and she gets nine differe... | Suppose that there were $n$ couples, and the woman asked all $2 n-1$ other attendees how many hands they shook and received $2 n-1$ different answers. We will show that she herself shook $n-1$ hands; hence, in our particular case, the answer is 4 .
We work by induction. When $n=1$, there is one couple, and no handshak... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A circle is inscribed in a sector that is one sixth of a circle of radius 6 . (That is, the circle is tangent to both segments and the arc forming the sector.) Find, with proof, the radius of the small circle.
. Therefore, since $\angle B A C$... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the hold of a pirate ship are ten treasure chests lying on pedestals in a circular arrangement. The captain would like to move each chest clockwise by one pedestal. However, the chests are so heavy that the captain and his assistant can only switch two chests at a time. What is the minimum number of switches needed ... | The answer is 9. It is easy to see that nine moves are sufficient; they can move a single chest counterclockwise one pedestal at a time, until after nine moves all of the other chests have been moved clockwise one pedestal.
Suppose that, at each stage of the game, we draw an arrow from each pedestal to the pedestal on... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Aerith repeatedly flips a fair coin.
a) Find the expected number of flips to get two heads in a row.
b) Find the expected number of flips to get heads followed by tails.
# | a) Let $x$ be the expected number of flips. There is a $\frac{1}{4}$ chance that she gets two heads right away. There is a $\frac{1}{2}$ chance the first flip is tails, in which case she is basically starting over after the first flip, so it will take an expected $x+1$ flips total. Finally, there is a $\frac{1}{4}$ cha... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find all composite positive integers $n$ such that all the divisors of $n$ can be written in the form $a^{r}+1$, where $a$ and $r$ are integers with $a \geq 0$ and $r \geq 2$. | The only such number is $n=10$. It is easy to see that $n=10$ indeed satisfies the conditions. Call $n$ "good" if every divisor of $n$ has the form $a^{r}+1, a \geq 0, r \geq 2$ (a good $n$ may be prime or composite).
First, it is easy to check that 4 is not good, and so 4 does not divide any good number.
Second, we ... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
With two properly chosen weights and a balance scale, it is possible to determine the weight of an unknown object known to weigh an integer number of pounds from 1 to $n$. Find the largest possible value of $n$.
Remark. The balance scale tells whether the weights placed on each side are equal and, if not, which side i... | Let $a$ and $b$ be the known weights. The balance scale allows one to compare the unknown weight with four known weights: $a, b, a+b$, and $a-b$ (the last of these is gotten by balancing $x+b$ on one side with $a$ on the other). After the comparisons are done, there are at most 4 values that $x$ can be known to equal a... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Evaluate the sum
$$
\sum_{k=1}^{\infty}\left(\prod_{i=1}^{k} \frac{P_{i}-1}{P_{i+1}}\right)=\frac{1}{3}+\frac{1}{3} \cdot \frac{2}{5}+\frac{1}{3} \cdot \frac{2}{5} \cdot \frac{4}{7}+\frac{1}{3} \cdot \frac{2}{5} \cdot \frac{4}{7} \cdot \frac{6}{11}+\ldots
$$
where $P_{n}$ denotes the $n^{\text {th }}$ prime number. | Rewrite the given sum as
$$
\begin{gathered}
\frac{1}{3}+\frac{2}{3} \cdot \frac{1}{5}+\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{1}{7}+\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7}+\cdot \frac{1}{11}+\ldots \\
=\frac{1}{3}+\left(1-\frac{1}{3}\right) \frac{1}{5}+\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right) \f... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Given a fixed triangle $\triangle A B C$ and a point $P$, find the maximum value of
$$
\frac{A B^{2}+B C^{2}+C A^{2}}{P A^{2}+P B^{2}+P C^{2}}
$$ | We use the following lemma.
Lemma. Given $a, b, c, p \in \mathbb{R}$ such that not all of $a, b, c$ are equal,
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \leq 3\left((p-a)^{2}+(p-b)^{2}+(p-c)^{2}\right)
$$
with equality if and only if $p=(a+b+c) / 3$.
Proof. For fixed $a, b, c$, the right hand side is a quadratic in $p$, nam... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Find all positive prime numbers $p$ such that $p+2$ and $p+4$ are prime as well.
Hint. Show that for most prime numbers $p$, either $p+2$ or $p+4$ is divisible by 3 . | For $p=3, p+2=5, p+4=7$ and these are obviously prime. For $p>3$, we know that $p$ is not divisible by 3 . The remainder of $p$ when divided by 3 can be either 1 or 2 . If it is one, then $p+2$ is divisible by 3 , if it is 2 , then $p+4$ is divisible by 3 . Hence $p=3$ is the only solution. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the units digit of $17^{2021}$. | The units digits of powers of 17 cycle: $7,9,3,1,7,9,3,1, \ldots$, so the units digit of $17^{n}$ is 1 whenever $n$ is a multiple of 4 . Since 2020 is a multiple of 4 , $17^{2020}$ has units digit 1 , so $17^{2021}$ has units digit 7 . | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Carl computes the number
$$
N=5^{555}+6^{666}+7^{777}
$$
and writes it in decimal notation. What is the last digit of $N$ that Carl writes? | We look at the last digit of each term.
- The last digit of $5^{\bullet}$ is always 5 .
- The last digit of $6^{\bullet}$ is always 6 .
- The last digit of $7^{\bullet}$ cycles $7,9,3,1,7,9,3, \ldots$.
So the last digits are $5,6,7$ in that order. Since $5+6+7=18$, the answer is 8 . | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
If $a, b, c, d$ are positive real numbers such that $\frac{5 a+b}{5 c+d}=\frac{6 a+b}{6 c+d}$ and $\frac{7 a+b}{7 c+d}=9$, calculate $\frac{9 a+b}{9 c+d}$. | Let $\frac{5 a+b}{5 c+d}=\frac{6 a+b}{6 c+d}=k$. Then $5 a+b=k(5 c+d)$ and $6 a+b=k(6 c+d)$. Subtracting these two equations gives $a=k c$. Now we can easily get that $b=k d$. From $\frac{7 a+b}{7 c+d}=\frac{7 k c+k d}{7 c+d}=k=9$ we get $\frac{9 a+b}{9 c+d}=9$ as well. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A triangle, two of whose sides are 3 and 4 , is inscribed in a circle. Find the minimal possible radius of the circle. | Since the circle has a chord of length 4 , its diameter is at least 4 and so its radius is at least 2 . To achieve equality, choose a right triangle with hypotenuse 4 and one leg 3 (the other leg will, by the Pythagorean theorem, have length $\sqrt{7}$ ). Then the midpoint of the hypotenuse is the center of a circle of... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
There are three boxes of stones. Each hour, Sisyphus moves a stones from one box to another. For each transfer of a stone, he receives from Zeus a number of coins equal to the number of stones in the box from which the stone is drawn minus the number of stones in the recipient box, with the stone Sisyphus just carried ... | The myth of Sisyphus suggests the answer: 0 .
Let $x$ be the amount of money Sisyphus has and $a, b, c$ the sizes of the boxes. The key observation is that the quantity
$$
N=2 x+a^{2}+b^{2}+c^{2}
$$
does not change; for example, after one operation from the first box to the second the quantity becomes
$N^{\prime}=2... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A $23 \times 23$ square is divided into smaller squares of dimensions $1 \times 1,2 \times 2$, and $3 \times 3$. What is the minimum possible number of $1 \times 1$ squares? | Color the rows of the square black and white alternately, so the top and bottom rows are black. Then each $2 \times 2$ tile covers two cells of each color, and each $3 \times 3$ tile covers six of one color and three of the other. In particular, if only $2 \times 2$ and $3 \times 3$ tiles are used, the difference betwe... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $a_{1}=5$ and $a_{n+1}=a_{n}^{3}-2 a_{n}^{2}+2$ for all $n \geq 1$. Prove that if $p$ is a prime divisor of $a_{2014}+1$ and $p \equiv 3 \bmod 4$, then $p=3$. | Observe that $a_{n+1}-2=a_{n}^{2}\left(a_{n}-2\right)$ for all $n \geq 1$. By induction on $n$ we obtain
$$
a_{n+1}-2=3 a_{n}^{2} a_{n-1}^{2} \cdots a_{1}^{2}
$$
for all $n \geq 1$. Therefore
$$
a_{2014}+1=3\left(a_{2013}^{2} a_{2012}^{2} \cdots a_{1}^{2}+1\right)=3\left[\left(a_{2013} a_{2012} \cdots a_{1}\right)^{... | 3 | Number Theory | proof | Yes | Yes | olympiads | false |
a) One Sunday, Zvezda wrote 14 numbers in a circle, so that each number is equal to the sum of its two neighbors. Prove that the sum of all 14 numbers is 0 .
b) On the next Sunday, Zvezda wrote 21 numbers in a circle, and this time each number was equal to half the sum of its two neighbors. What is the sum of all 21 n... | a) Denoting the numbers $a_{1}, a_{2}, \ldots, a_{14}$, and their sum as $S$ we have $a_{i}=$ $a_{i-1}+a_{i+1}$ for $i=1, \ldots, 14$ (we take $a_{15}=a_{1}, a_{0}=a_{14}$ ). Summing all these equalities we get $S=2 S$ (since each $a_{i}$ appears exactly once on the left and exactly twice on the right). Therefor $S=0$.... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For which positive integers $n$ is $n^{4}+4$ equal to a prime number? | For $n=1$ we get $1^{4}+4=5$, which works.
For all other values of $n$, the key idea is that
$$
n^{4}+4=n^{4}+4 n^{2}+4-4 n^{2}=\left(n^{2}+2\right)^{2}-(2 n)^{2}=\left(n^{2}+2 n+2\right)\left(n^{2}-2 n+2\right)
$$
which is the product of two integers greater than 1, and hence cannot be prime. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A set $S$ of irrational real numbers has the property that among any subset of five numbers in $S$, one can find two with irrational sum. How large can $|S|$ be? | The answer is $|S| \leq 8$. An example is $S=\{n \pm \sqrt{2} \mid n=1,2,3,4\}$ (and any of its subsets).
In general, construct a graph with vertex set $S$ in which we join two numbers with rational sum. We claim this graph is bipartite; indeed if $a_{1}+a_{2}, a_{2}+a_{3}, \ldots, a_{n}+a_{1}$ are all rational for so... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Given three squares of dimensions $2 \times 2,3 \times 3$, and $6 \times 6$, choose two of them and cut each into 2 figures, such that it is possible to make another square from the obtained 5 figures. | The solution is shown in the picture below
 | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Find all positive integers $p$ such that $p, p+4$, and $p+8$ are all prime. | If $p=3$, then $p+4=7$ and $p+8=11$, both prime. If $p \neq 3$, then $p$ is not a multiple of 3 and is therefore of one of the forms $3 k+1,3 k+2(k \geq 0)$ If $p=3 k+1$, then $p+8=3 k+9=3(k+3)$, which is not prime since $k+3>1$. If $p=3 k+2$, then $p+4=3 k+6=3(k+2)$, which is not prime since $k+2>1$. Thus $p=3$ is the... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The figure below consists of five congruent squares. If the area of the shaded triangle equals the area outside the shaded triangle, calculate the ratio $A B / B C$.
 | Let $s$ be the side length of the square. As the total area of the figure is $5 s^{2}$, the triangle should have area $5 s^{2} / 2$. Here $A B$ is a base of the triangle and the height is $3 s$, hence $A B=5 s / 3$. Since $A C=2 s$, we get $B C=s / 3$ and we obtain the final answer $A B: B C=5$. | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $a, b, c$ be positive real numbers such that $a b c=1$. Simplify
$$
\frac{1}{1+a+a b}+\frac{1}{1+b+b c}+\frac{1}{1+c+c a} .
$$ | We may let $a=y / x, b=z / y, c=x / z$ for some real numbers $x, y, z$. Then
$$
\begin{aligned}
\frac{1}{1+a+a b}+\frac{1}{1+b+b c}+\frac{1}{1+c+c a} & =\frac{1}{1+y / x+z / x}+\frac{1}{1+z / y+x / y}+\frac{1}{1+x / z+y / z} \\
& =\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z} \\
& =1 .
\end{aligned}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Lisa considers the number
$$
x=\frac{1}{1^{1}}+\frac{1}{2^{2}}+\cdots+\frac{1}{100^{100}} .
$$
Lisa wants to know what $x$ is when rounded to the nearest integer. Help her determine its value. | The answer is 1 . Indeed, note that
$$
x \leq 1+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{100}} .
$$
By the formula for the sum of a geometric series, we see that
$$
x \leq 1+\frac{1}{2}-\frac{1}{2^{101}} \text {. }
$$
Thus $x<3 / 2$, and the closest integer to $x$ is 1 . | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For which prime numbers $p$ is $p^{2}+2$ also prime? Prove your answer. | The answer is $p=3$. This indeed works, since $3^{2}+2=11$.
Consider any other prime number $p \neq 3$. Then it follows that $p^{2} \equiv 1(\bmod 3)$; i.e. that $p$ leaves remainder 1 when divided by 3 . Consequently, $p^{2}+2$ is divisible by 3 . Since $p \geq 2$, we have $p^{2}+2 \geq 7$ as well, thus $p^{2}+2$ can... | 3 | Number Theory | proof | Yes | Yes | olympiads | false |
Find the minimal natural number $n$ with the following property: It is possible to tile the plane with squares whose side lengths belong to the set $\{1,2, \ldots, n\}$ so that no two squares with the same side length touch along a segment of an edge.
Remark. Squares with the same side length can touch at a vertex, ho... | The answer is $n=5$. The desired tiling is shown in Figure 1. It is formed by translation from the L-shaped region in bold borders. Since none of the five squares in this region border on squares of like side length, neither does any square in the infinite tiling.
To show that $n \leq 4$ does not work, it is necessary... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A point $P$ lies inside a regular hexagon $A B C D E F$. The distances from $P$ to the sides $A B, B C, C D, D E, E F$, and $F A$ are respectively $1,2,5,7,6$, and $x$. Find $x$. | Opposite sides of a regular hexagon are parallel. The sum of the distances from $P$ to $A B$ and to $D E$ is simply the distance between $A B$ and $D E$, which must therefore equal $1+7=8$. Because of the symmetry of the hexagon, the distance between any pair of opposite sides must be 8. Applying this to sides $C D$ an... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Two villages, $A$ and $B$, lie on opposite sides of a straight river in the positions shown. There will be a market in village $B$, and residents of village $A$ wish to attend. The people of village $A$ would like to build a bridge across and perpendicular to the river so that the total route walked by residents of $A$... | For convenience, we label the point $X$, the foot of the perpendicular from $A$ to the river. We also label $Y$, the point where the perpendicular $B Y$ to the river through $B$ meets the parallel $A Y$ to the river through $A$. The key construction is to translate $B$ southward $1 \mathrm{~km}$ to point $E$. Let $A C ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Find all prime numbers $p$ such that $p^{2}+8$ is prime number, as well.
Remark. A number $p$ is prime if it has exactly 2 divisors: 1 and $p$. Numbers $2,3,5,7,11,13, \ldots$ are prime, while 4 and 2007 are not.
Hint. Write down first several prime numbers (hint - you can copy them from the paragraph above), calcula... | For $p=3$ we have $p^{2}+8=17$, which is prime. If $p \neq 3$ then $p$ is not divisible by 3 . The remainder of $p$ when divided by 3 is either 1 or 2 . This means that $p=3 k+1$ for some integer $k$, or $p=3 l+1$ for some integer $l$. In the first case we get $p^{2}=9 k^{2}+6 k+1$ and in the second, $p^{2}=9 l^{2}-6 l... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$p$ is a prime number such that the period of its decimal reciprocal is 200 . That is,
$$
\frac{1}{p}=0 . X X X X \ldots
$$
for some block of 200 digits $X$, but
$$
\frac{1}{p} \neq 0 . Y Y Y Y \ldots
$$
for all blocks $Y$ with less than 200 digits. Find the 101st digit, counting from the left, of $X$. | Let $X$ be a block of $n$ digits and let $a=0 . X \ldots$ Then $10^{n} a=X . X \ldots$. Subtracting the previous two equalities gives us $\left(10^{n}-1\right) a=X$, i.e. $a=\frac{X}{10^{n}-1}$.
Then the condition that $a=\frac{1}{p}$ reduces to $\frac{1}{p}=\frac{X}{10^{n}-1}$ or $p X=10^{n}-1$. For a given $p$ and $... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The country of Squareland is shaped like a square and is divided into 64 congruent square cities. We want to divide Squareland into states and assign to each state a capital city so that the following rules are satisfied:
(a) Every city lies entirely within one state.
(b) Given any two states, the numbers of cities i... | In the diagram below, no city shares a corner with any two of the cities marked X. Therefore the nine X's are
in nine different states. The diagram at right shows that nine states are also su... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A palindrome is a positive integer that reads the same forward and backward, like 2552 or 1991. Find a positive integer greater than 1 that divides all four-digit palindromes. | We claim that 11 divides all four digit palindromes. Note that any four digit palindrome $a b b a$ is the sum
$$
a b b a=a 00 a+b b 0=a \times 1001+b \times 110 .
$$
Now, 110 is a multiple of 11 , since it is $11 \times 10$, and 1001 is a multiple of 11 , since it is $11 \times 91$. Thus, $a 00 a$ and $b b 0$ are mul... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the following figure-not drawn to scale!- $E$ is the midpoint of $B C$, triangle $F E C$ has area 7, and quadrilateral $D B E G$ has area 27. Triangles $A D G$ and $G E F$ have the same area, $x$. Find $x$.
$ choices of the sequences $a_{1}, a_{2}, \ldots, a_{7}$ and $b_{1}, b_{2}, \ldots, b_{7}$ for which $2 k-1$ is an $a_{j}$ with $a_{j}<b_{j}$. Indeed, the $a$ 's can be any of the 7 ! permutations of the 7 odd integers; then $j$ is the subscript such that $a_{j}=2 k-1$, and ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
. Prove that $1993^{1993}+1994^{1994}+1995^{1995}+1996^{1996}$ is divisible by 10 . | Since $5^{2}=25$ ends in 5 again, and $6^{2}=36$ ends in 6 again, no matter to what power we raise 5 or 6 , the resulting numbers will end in 5 or 6 , respectively. Thus, $1995^{1995}$ ends in 5 and $1996^{1996}$ ends in 6 . Further, $4^{1}=1,4^{2}=16,4^{3}=\ldots 4,4^{4}=\ldots 6$, etc, i.e. an odd power of 4 ends in ... | 0 | Number Theory | proof | Yes | Yes | olympiads | false |
Determine all positive integers $n$ for which the equation
$$
x^{n}+(2+x)^{n}+(2-x)^{n}=0
$$
has an integer as a solution.
Answer: $n=1$.
# | If $n$ is even, $x^{n}+(2+x)^{n}+(2-x)^{n}>0$, so $n$ is odd.
For $n=1$, the equation reduces to $x+(2+x)+(2-x)=0$, which has the unique solution $x=-4$.
For $n>1$, notice that $x$ is even, because $x, 2-x$, and $2+x$ have all the same parity. Let $x=2 y$, so the equation reduces to
$$
y^{n}+(1+y)^{n}+(1-y)^{n}=0 \t... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Given a permutation $\left(a_{0}, a_{1}, \ldots, a_{n}\right)$ of the sequence $0,1, \ldots, n$. A transposition of $a_{i}$ with $a_{j}$ is called legal if $a_{i}=0$ for $i>0$, and $a_{i-1}+1=a_{j}$. The permutation $\left(a_{0}, a_{1}, \ldots, a_{n}\right)$ is called regular if after a number of legal transpositions i... | A legal transposition consists of looking at the number immediately before 0 and exchanging 0 and its successor; therefore, we can perform at most one legal transposition to any permutation, and a legal transposition is not possible only and if only 0 is preceded by $n$.
If $n=1$ or $n=2$ there is nothing to do, so $n... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. let $A, B, C$ and $D$ be four points lying in this order on a circle. Assume that there is a point $K$ on the line $A B$ such that $B D$ bisects the line $K C$ and $A C$ bisects the line $K D$. Determine the smallest possible value that $\left|\frac{A B}{C D}\right|$ can take. | Response: The minimum is 2.
Solution: We only use non-oriented (i.e., positive) distances.
We obtain a construction if $A B C D$ is an isosceles trapezoid with base $A B=2 C D$ and $K$ the midpoint of $A B$.
The condition on the midpoints implies that $[A D C]=[A K C]$ where the bracket notation denotes the area of ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Determine the smallest possible value of the expression
$$
\frac{a b+1}{a+b}+\frac{b c+1}{b+c}+\frac{c a+1}{c+a}
$$
where $a, b, c \in \mathbb{R}$ satisfy $a+b+c=-1$ and $a b c \leq-3$. | Solution: The minimum is 3 , which is obtained for $(a, b, c)=(1,1,-3)$ and permutations of this triple.
As $a b c$ is negative, the triple $(a, b, c)$ has either exactly one negative number or three negative numbers. Also, since $|a b c| \geq 3$, at least one of the three numbers has absolute value greater than 1 .
... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Determine all integers $n \geq 3$ such that
$$
n!\mid \prod_{\substack{p<q \leq n \\ p, q \text { prime }}}(p+q)
$$
Remark: The expression on the right-hand side denotes the product over all sums of two distinct primes less than or equal to $n$. For $n=6$, this is equal to $(2+3)(2+5)(3+5)$. | Solution:For a fixed $n \geq 3$, let us denote the product on the right by $P(n)$ and let $r$ be the largest prime less than or equal to $n$. We now observe that
$$
r|n!| P(n)=P(r)
$$
and therefore there are primes $p<q \leq r$ with $r \mid p+q$. But $0<p+q<2 r$ implies $p+q=r$. Since $r \geq 3$ is an odd prime, we m... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Z1) Find all integer values that the expression
$$
\frac{p q+p^{p}+q^{q}}{p+q}
$$
where $p$ and $q$ are prime numbers.
Answer: The only integer value is 3 . | Solution: If $p$ and $q$ are both odd, then the numerator is odd and the denominator is even. Because an even number never divides an odd number, this does not yield an integer value, so we can assume that at least one of our primes is even, and therefore equal to 2. Because the expression is symmetric in $p$ and $q$, ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
N1) Find all integer values that the expression
$$
\frac{p q+p^{p}+q^{q}}{p+q}
$$
can take, where $p$ and $q$ are prime numbers. | Answer: The only attainable relative integer is 3. Solution: If $p$ and $q$ are odd, the numerator is odd but the denominator is even. Since an even number never divides an odd number, the expression is not a natural integer. Therefore, we can assume that at least one of the two primes is even. Since the expression is ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
N1) Determine all integer values that the expression
$$
\frac{p q+p^{p}+q^{q}}{p+q}
$$
can take, where $p$ and $q$ are both prime numbers.
Answer: The only possible integer value is 3 . | Solution: If both $p$ and $q$ are odd, then the numerator is odd while the denominator is even. Since an even number never divides an odd number, this does not lead to an integer value. Hence we can assume that one of our primes is even and therefore equal to 2 . Since the expression is symmetric in $p$ and $q$, we can... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. the plane is divided into unit squares. Each square is to be colored with one of $n$ colors, so that the following applies: If four squares can be covered with an L-tetromino, then these squares have four different colors (the L-tetromino may be rotated and mirrored). Determine the smallest value of $n$ for which th... | ## Solution
Two of the seven fields within the figure on the left in Figure 1 can be covered simultaneously with an L-tetromino. These fields therefore have different colors, in particular $n \geq 7$. Assume that $n=7$ is possible and color the fields in the area as in Figure 1. The two fields to the right and left be... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. let $a, b, c$ be positive real numbers with $a b c=1$. Determine all possible values that the expression
$$
\frac{1+a}{1+a+a b}+\frac{1+b}{1+b+b c}+\frac{1+c}{1+c+c a}
$$
can accept.
## Solution | Let $A$ be the given expression. The following applies because $a b c=1$
$$
\begin{aligned}
A & =\frac{1+a}{1+a+a b}+\frac{a(1+b)}{a(1+b+b c)}+\frac{a b(1+c)}{a b(1+c+c a)} \\
& =\frac{1+a}{1+a+a b}+\frac{a+a b}{1+a+a b}+\frac{a b+1}{1+a+a b} \\
& =\frac{2(1+a+a b)}{1+a+a b}=2
\end{aligned}
$$
The expression is there... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. The real numbers $a, b, c, d$ are positive and satisfy $(a+c)(b+d)=a c+b d$. Find the minimum of
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}
$$
Answer: 8. | Solution(Raphael):
After trying out some things one might suspect that the minimum is reached when $a=c$ and $b=d$. So we'll try to use an inequality with equality case $a=c$ and $b=d$. Going for the easiest such inequality one comes across:
$$
a c+b d=(a+c)(b+d) \geq 2 \sqrt{a c} \cdot 2 \sqrt{b d} \Leftrightarrow \... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. For which integers $n \geq 2$ can we arrange the numbers $1,2, \ldots, n$ in a row, such that for all integers $1 \leq k \leq n$ the sum of the first $k$ numbers in the row is divisible by $k$ ?
Answer: This is only possible for $n=3$. | Solution (Valentin): If we let $k=n$ we find that
$$
n \left\lvert\, \frac{n(n+1)}{2}\right.
$$
which implies that $n$ has to be odd. For $n=3$ the arrangement 1,3,2 satisfies the condition. From now on assume that $a_{1}, a_{2}, \ldots, a_{n}$ is a valid arrangement for an odd $n \geq 5$. If we now let $k=n-1$ we fi... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. consider a $7 \times 7$ field that is divided into 49 unit squares. Into this field we want to place tiles in the shape of a Swiss cross, consisting of 5 unit squares. The edges of the crosses should lie on the lines of the field. Determine the smallest possible number of squares that must be marked on the field so ... | Insert coordinates. It is easy to see that each cross represents one of the 7 squares
$$
(2,5),(3,2),(3,3),(4,6),(5,4),(6,2),(6,5)
$$
covered. This shows that 7 markings are sufficient. We now show that 7 marks are necessary. Assume not. The crosses with centers
$$
(2,2),(2,6),(3,4),(5,2),(5,6),(6,4)
$$
are disjoin... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. A natural number $n \geq 2$ is called resistant if it is coprime with the sum of all its divisors (1 and $n$ included). What is the maximum length of a sequence of consecutive resistant numbers? | Response: The maximum length of such a sequence is 5.
Solution: (Louis) In this type of problem, it is important to try small values of $n$ in the hope that it will give us ideas to solve the problem. We see that among the numbers between 2 and 30, the most resistant numbers are $2,3,4,5,7,8,9,11,13$, $16,17,19,21,23,... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. let $q(n)$ be the sum of the digits of the natural number $n$. Determine the value of
$$
q\left(q\left(q\left(2000^{2000}\right)\right)\right)
$$
## Solution | It applies $q\left(2000^{2000}\right)=q\left(2^{2000}\right)$, because the two numbers only differ by appended zeros. We now estimate. The following applies $2^{2000}=4 \cdot 8^{666}<10^{667}$ and therefore
$$
q\left(2^{2000}\right) \leq 9 \cdot 667=6003
$$
Furthermore
$$
q\left(q\left(2^{2000}\right)\right) \leq 5+... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. let $m$ be a natural number greater than 1 . the sequence $x_{0}, x_{1}, x_{2}, \ldots$ is defined by
$$
x_{i}= \begin{cases}2^{i}, & \text { for } \quad 0 \leq i \leq m-1 \\ \sum_{j=1}^{m} x_{i-j} & \text { for } \quad i \geq m\end{cases}
$$
Find the largest $k$ such that there are $k$ consecutive sequential elem... | ## Solution
Let $r_{i}$ be the remainder of $x_{i}$ when divided by $m$. Instead of the sequence $\left(x_{i}\right)$, we consider the sequence $\left(r_{i}\right)$. The recursion formula can be used to calculate the next, but also the previous one, from $m$ consecutive sequence members. Such $m$ consecutive links the... | -1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. let $M$ be a finite set of real numbers with the following property: From every three different elements of $M$, two can always be selected whose sum lies in $M$. What is the maximum number of elements $M$ can have?
## Solution | Answer: 7 .
Assume that $M$ contains more than three positive elements and denote the four largest with $a>b>c>d>0$. According to the prerequisite, you can choose two of the three elements $a, b, c$ whose sum lies in $M$. Because of $a+b>a$ and $a+c>a$ and the maximality of $a$, these two sums are not in $M$. Conseque... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. the polynomial $P(x)=x^{3}-2 x^{2}-x+1$ has the three real zeros $a>b>c$. Find the value of the expression
$$
a^{2} b+b^{2} c+c^{2} a
$$
## Solution | We set $A=a^{2} b+b^{2} c+c^{2} a$ and $B=a^{2} c+c^{2} b+b^{2} a$. Because $a>b>c$ applies
$$
A-B=(a-c)(c-b)(b-a)>0
$$
so $A$ is greater than $B$. In the following, we use the terms
$$
\begin{aligned}
u & =a+b+c \\
v & =a b+b c+c a \\
w & =a b c
\end{aligned}
$$
according to Vieta's theorem, $u=2, v=-1, w=-1$. Now... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. find all natural numbers $k$ such that $3^{k}+5^{k}$ is a power of a natural number with exponent $\geq 2$.
## 1st solution | Assume that $3^{k}+5^{k}=n^{t}$ with $t \geq 2$. If $k$ is even, then $3^{k}+5^{k} \equiv 1+1=2$ $(\bmod 4)$, so the left side is even but not divisible by 4, a contradiction to $t \geq 2$. If $k$ is odd, then the following applies
$$
3^{k}+5^{k}=(3+5)\left(3^{k-1}-3^{k-2} \cdot 5+\ldots-3 \cdot 5^{k-2}+5^{k-1}\right)... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. a regular 2008-vertical is somehow divided into many triangles with 2005 non-intersecting diagonals. Determine the smallest possible number of non-isosceles triangles that can occur in such a decomposition.
## 1st solution | We call an isosceles triangle good and a non-isosceles triangle bad. For a natural number $n$, let $n^{(2)}$ denote the number of ones in the binary representation of $n$. We will show more generally that in every triangulation of a regular $n$ vertex there are at least $n^{(2)}-2$ bad triangles. To do this, first cons... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. determine all odd natural numbers of the form
$$
\frac{p+q}{p-q}
$$
where $p>q$ are prime numbers.
## 1st solution | Let $n$ be this number. According to the prerequisite, $n(p-q)=p+q=(p-q)+2 q$, therefore $p-q$ is a divisor of $2 q$. Because $\operator name{ggT}(p-q, q)=1$, $p-q=1$ or $p-q=2$ therefore applies. In the first case, $p$ or $q$ is even, i.e. $p=3, q=2$ and $n=5$. In the second case, $p$ and $q$ are odd and therefore $n=... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Sei $A B C$ ein spitzwinkliges Dreieck mit $A B \neq B C$ und Umkreis $k$. Seien $P$ und $Q$ die Schnittpunkte von $k$ mit der Winkelhalbierenden beziehungsweise der Aussenwinkelhalbierenden von $\angle C B A$. Sei $D$ der Schnittpunkt von $A C$ und $P Q$. Bestimme das Verhältnis $A D: D C$.
## 1st solution: | By the inscribed angle theorem and the fact that $B P$ is the bisector of $\angle A B C, $\angle P C A=$ $\angle P B A=\angle P B C=\angle P A C$, so $\triangle A P C$ is isosceles. Similarly, let $R$ be a point on the line $A B$ such that $B$ lies between $A$ and $R$. Then $\angle Q A C=\angle Q B C=\angle Q B R=180^{... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. We have an $8 \times 8$ board. An inner edge is an edge between two $1 \times 1$ fields. We cut the board into $1 \times 2$ dominoes. For an inner edge $k$, $N(k)$ denotes the number of ways to cut the board such that the cut goes along the edge $k$. Calculate the last digit of the sum we get when we add all $N(k)$,... | Solution: First, we calculate along how many inner edges a cut is made during a decomposition. In total, there are $7 \cdot 8$ horizontal and vertical inner edges each. During a decomposition, the board is cut along all inner edges except the 32 within a domino tile. Therefore, in each decomposition, cuts are made alon... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Let $a, b, c$ be real numbers such that:
$$
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1
$$
Determine all values that the following expression can take:
$$
\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}
$$ | ## Solution:
$$
\begin{aligned}
& a+b+c=a+b+c \\\
& \overbrace{\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)}^{1} \cdot(a+b+c)=a+b+c \\
& \frac{a^{2}}{b+c}+\frac{a(b+c)}{b+c}+\frac{b^{2}}{a+c}+\frac{b(a+c)}{a+c}+\frac{c^{2}}{a+b}+\frac{c(a+b)}{a+b}=a+b+c \\
& \frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. The Brazilian IMO leader chooses two natural numbers $n$ and $k$ with $n>k$, and then tells these to his deputy and a participant. The leader then whispers a binary sequence of length $n$ into the deputy's ear. The deputy writes down all binary sequences of length $n$ that differ from the leader's sequence at exactl... | Remark: A binary sequence of length $n$ is a sequence of length $n$ that consists only of 0 and 1.
Solution: If we consider a fixed position in the solution sequence $L$, then $\binom{n-1}{k}$ of the listed sequences match $L$ at this position, and $\binom{n-1}{k-1}$ of the listed sequences differ from $L$ at this pos... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. determine the largest natural number $k$ with the following property: The set of natural numbers can be divided into $k$ disjoint subsets $A_{1}, \ldots, A_{k}$ in such a way that every natural number $n \geq 15$ can be written as the sum of two different elements from $A_{i}$ for every $i \in\{1, \ldots, k\}$. | Solution For $k=3$ you can divide the natural numbers as follows:
$$
\begin{aligned}
& A_{1}=\{1,2,3\} \cup\{3 m \mid m \geq 4\} \\
& A_{2}=\{4,5,6\} \cup\{3 m-1 \mid m \geq 4\} \\
& A_{3}=\{7,8,9\} \cup\{3 m-2 \mid m \geq 4\}
\end{aligned}
$$
In $A_{1}$ all numbers $n \geq 12+1=13$ can be represented as a sum, in $A... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Z1) If $p \geq 5$ is a prime number, let $q$ be the smallest prime number such that $q>p$ and let $n$ be the number of positive divisors of $p+q$ ( 1 and $p+q$ included).
a) Show that no matter which prime number $p$ is chosen, the number $n$ is greater than or equal to 4.
b) Find the smallest possible value $m$ that... | Solution: Since $p \geq 5$, the two prime numbers $p, q$ are both odd and therefore $p+q$ even, so $p+q$ is divisible by $1,2, \frac{p+q}{2}$ and $p+q$. In addition, $p+q>4$ and therefore the divisors mentioned are all different. This proves part a).
for part $\mathrm{b}_{1}$ ) we can test small cases of $p$ and see t... | 6 | Number Theory | proof | Yes | Yes | olympiads | false |
6 Given a real number $a$ such that there is only one real number $x$ satisfying the inequality $\left|x^{2}+2 a x+3 a\right| \leqslant 2$, then the number of all real numbers $a$ that satisfy the condition is ().
(A) 1
(B) 2
(C) 3
(D) infinitely many | 6 Let $f(x)=x^{2}+2 a x+3 a$, from $f\left(-\frac{3}{2}\right)=\frac{9}{4}$ we know, the graph of function $f(x)$ passes through the point $\left(-\frac{3}{2}, \frac{9}{4}\right)$. To make the inequality $\left|x^{2}+2 a x+3 a\right| \leqslant 2$ have only one solution, the graph of the parabola $f(x)=x^{2}+2 a x+3 a$ ... | 2 | Inequalities | MCQ | Yes | Yes | olympiads | false |
8. (5 points) Person A and Person B stand facing each other 30 meters apart. They play "Rock-Paper-Scissors," with the winner moving forward 8 meters, the loser moving back 5 meters, and in the case of a tie, both moving forward 1 meter. After 10 rounds, they are 7 meters apart. How many rounds did they tie? | 【Answer】Solution: Assume all games end in a draw, then the distance between the two people should be: $30-10 \times 2=10$ (meters), $(10-7) \div(8-5-2)=3$ times (win)
Draws are: $10-3=7$ (times),
Answer: The number of draws is 7.
Therefore, the answer is: 7. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) On the clock tower of a railway station, there is an electronic clock. On the boundary of the circular clock face, there is a small colored light at each minute mark. At 9:35:20 PM, there are $\qquad$ small colored lights within the acute angle formed by the minute and hour hands. | 【Analysis】First, find the angle between the hour hand and the minute hand at 9:35:20 PM; then, according to the clock face being divided into 60 small segments, with each large segment corresponding to an angle of $30^{\circ}$ and each small segment corresponding to an angle of $6^{\circ}$, we can find the minute marks... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Given $\mathbf{Z}$ is the set of integers, the set $A=\{x|| x-3 |<\pi, x \in \mathbf{Z}\}, B=\left\{x \mid x^{2}-11 x+5<0\right.$, $x \in \mathbf{Z}\}, C=\left\{x|| 2 x^{2}-11 x+10|\geqslant| 3 x-2 \mid, x \in \mathbf{Z}\right\}, \bar{C}$ is the complement of $C$ in $\mathbf{Z}$, then the number of proper subsets of... | 1. A Obviously, the sets are $A=\{0,1,2,3,4,5,6\} ; B=\{0,1,2,3,4,5,6,7,8,9,10\}$; for set $C$, $\left|2 x^{2}-11 x+10\right| \geqslant|3 x-2| \Leftrightarrow\left(2 x^{2}-11 x+10\right)^{2}-(3 x-2)^{2} \geqslant 0 \Leftrightarrow\left(2 x^{2}-8 x+8\right)\left(2 x^{2}-14 x+12\right)$ $\geqslant 0 \Leftrightarrow(x-2)^... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
3.70 Can integers be written in each cell of an infinitely large grid paper so that in every rectangle consisting of $4 \times 6$ cells and bounded by grid lines, the sum of all numbers is equal to (1) $10, (2) 1$? | ［Solution］The filling principle shown in the right figure is: select two sets of diagonal lines in different directions. The first set is to select a line every 4 cells moved, and fill in 1 and 0 alternately on the line; the second set is to select a line every 6 cells moved, and fill in -1 and 0 alternately on the lin... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$33 \cdot 1$. The units digit of the number $2137^{753}$ is
(A) 1.
(B) 3.
(C) 5.
(D) 7.
(E) 9.
(12th American High School Mathematics Examination, 1961) | [Solution] Let $\langle a\rangle$ denote the unit digit of the integer $a$. Clearly, if the unit digit of $a$ is $b$, then $\langle a\rangle=\langle b\rangle$, and $\left\langle a^{n}\right\rangle=\left\langle b^{n}\right\rangle$ (where $n$ is a natural number).
Also, $\left\langle 7^{0}\right\rangle=1,\left\langle 7^{... | 7 | Number Theory | MCQ | Yes | Yes | olympiads | false |
5. Select some numbers without repetition from $1,2, \ldots 15$, such that the sum of any two numbers is not a perfect square of a natural number, then the maximum number of numbers that can be selected is ( ). | Answer: 8
Explanation: First, we argue that the first digit can just form the square number 16, with the groups being $(1,15) \quad(2,14) \ldots(7,9)$, (8)
There are 8 groups in total, and from each group, at most 1 number can be selected, so the maximum number of selections is 8.
Second, construction: if the maximum ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. (15 points) Insert 2 " $\div$ " and 2 "+" between the 9 "1"s below to make the calculation result an integer. The smallest integer is $\qquad$
\begin{tabular}{|lllllllllll|}
\hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & $=$ & $?$ \\
\hline
\end{tabular} | 【Answer】Solution: According to the analysis, we have
$$
\begin{array}{l}
111 \div 111+1 \div 1+1 \\
=1+1+1 \\
=3
\end{array}
$$
Answer: The smallest integer is 3.
Therefore, the answer is: 1. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$. The sequence $\left\{a_{n}\right\}$ satisfies:
$$
x_{1}=1, x_{n+1}=4 x_{n}+\left[\sqrt{11} x_{n}\right] \text {. }
$$
Then the units digit of $x_{2021}$ is | 8.9.
Given $x_{2}=7$.
Notice, for any $x_{n} \in \mathbf{Z}_{+}$,
$$
\begin{array}{l}
4 x_{n}+\sqrt{11} x_{n}-1<x_{n+1}<4 x_{n}+\sqrt{11} x_{n} \\
\Rightarrow 5 x_{n}-(4-\sqrt{11})<(4-\sqrt{11}) x_{n+1}<5 x_{n} \\
\Rightarrow\left\{\begin{array}{l}
4 x_{n+1}-5 x_{n}<\sqrt{11} x_{n+1}, \\
\sqrt{11} x_{n+1}+(4-\sqrt{11}... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. As shown in the figure, $\angle A C B=90^{\circ}, A C=B C, A D \perp C E$, $B E \perp C E$, with the feet of the perpendiculars being $D, E$ respectively. Given $A D=8$, $B E=3$, then $D E=$ $\qquad$ | In Rt $\triangle A C D$ and Rt $\triangle C B E$, $A C=B C, \angle C A D=\angle B C E$, then $\triangle A C D \cong \triangle C B E \Rightarrow C E=A D=8, C D=B E=3 \Rightarrow D E=5$. | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
15. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{n+1}=\frac{1}{8} a_{n}^{2}+m\left(n \in \mathbf{N}^{*}\right)$, if for any positive integer $n$, we have $a_{n}<4$, find the maximum value of the real number $m$. | 15. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{n+1}=\frac{1}{8} a_{n}^{2}+m\left(n \in \mathbf{N}^{*}\right)$, if for any positive integer $n$, we have $a_{n}<4$, note that as $n \rightarrow+\infty$, $(m-2)(n-1) \rightarrow+\infty$,
Therefore, there exists a sufficiently large $n$ such that $1+(m... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given the complex numbers $z_{1}=\sin \alpha+2 \mathrm{i}, z_{2}=1+\mathrm{i} \cos \alpha$ (where $\alpha$ is a real number, $\mathrm{i}$ is the imaginary unit). Then the minimum value of $\frac{13-\left|z_{1}+\mathrm{i} z_{2}\right|^{2}}{\left|z_{1}-\mathrm{i} z_{2}\right|}$ is $\qquad$ | 4. 2 .
Let
$$
\left|z_{1}-\mathrm{i} z_{2}\right|=\sqrt{(\sin \alpha+\cos \alpha)^{2}+1^{2}}=\sqrt{2+\sin 2 \alpha}=t,
$$
then $t \in[1, \sqrt{3}]$, and at this time we have
$$
\left|z_{1}+\mathrm{i} z_{2}\right|=(\sin \alpha-\cos \alpha)^{2}+3^{2}=10-\sin 2 \alpha=12-t^{2},
$$
so
$$
\frac{13-\left|z_{1}+\mathrm{i} ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
25-13 In the same Cartesian coordinate system, there are infinitely many lines with the expression $y=k x+b$ (where $k, b$ are real numbers, and $k \neq 0$). Among these lines, no matter how they are selected, it is necessary to ensure that there are two lines passing through exactly the same quadrants. What is the min... | [Solution] According to the quadrants in which the lines are located, the lines on the plane can be divided into the following six categories:
$$
k>0,\left\{\begin{array}{l}
b>0, \\
b=0, \\
b<0 ;
\end{array}\right.\right.
$$
$$
k<0,\left\{\begin{array}{l}
b>0, \\
b=0, \\
b<0 .
\end{array}\right.\right.
$$
To ensure th... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 3 Given that $\boldsymbol{a}$ and $\boldsymbol{b}$ are two mutually perpendicular unit vectors, and $|\boldsymbol{c}|=13, \boldsymbol{c} \cdot \boldsymbol{a}=3, \boldsymbol{c} \cdot \boldsymbol{b}=4$, then for any real numbers $t_{1}, t_{2},\left|\boldsymbol{c}-t_{1} \boldsymbol{a}-t_{2} \boldsymbol{b}\right|$ ... | Analyze According to the problem, we get $|\boldsymbol{a}|=|\boldsymbol{b}|=1$, and $\boldsymbol{a} \cdot \boldsymbol{b}=0$, thus $\left|\boldsymbol{c}-t_{1} \boldsymbol{a}-t_{2} \boldsymbol{b}\right|^{2}=c^{2}+t_{1}^{2} a^{2}+$ $t_{2}^{2} \boldsymbol{b}^{2}-2 t_{1} \boldsymbol{c} \cdot \boldsymbol{a}-2 t_{2} \boldsymb... | 12 | Algebra | MCQ | Yes | Yes | olympiads | false |
9. (10 points) Four black $1 \times 1 \times 1$ cubes and four white $1 \times 1 \times 1$ cubes can form $\qquad$ different $2 \times 2 \times 2$ cubes (considering the same cube after rotation as one situation). | 【Analysis】First, analyze the case where a color is on the same face. Then, enumerate the cases where the white on the same face becomes 3 and then 2.
【Solution】Solution: According to the problem:
(1) White at the bottom positions 5, 6, 7, 8 is 1 case (same face).
(2) White at the bottom positions 5, 6, 7, the fourth b... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$12 \cdot 35$ If two sets are given as
$$
\begin{array}{l}
M=\left\{z \left\lvert\, z=\frac{t}{1+t}+i \frac{1+t}{t}\right., \quad t \in R, \quad t \neq-1, \quad t \neq 0\right\}, \\
N=\{z|z=\sqrt{2}[\cos (\arcsin t)+i \cos (\arccos t)], t \in R,| t \mid \leqslant 1\},
\end{array}
$$
then the number of elements in $M \... | [Solution] In the Cartesian coordinate system, the elements of sets $M$ and $N$ are points on the curves
$M:\left\{\begin{array}{l}x=\frac{t}{1+t}, \\ y=\frac{1+t}{t} ;\end{array}\right.$
$(t \in R, \quad t \neq 0, \quad t \neq-1)$,
$N:\left\{\begin{array}{l}x=\sqrt{2\left(1-t^{2}\right)}, \\ y=\sqrt{2} t .\end{array}\... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 6 If there exists a permutation $a_{1}, a_{2}, \cdots, a_{n}$ of $1,2, \cdots, n$, such that $k+a_{k} (k=1,2, \cdots, n)$ are all perfect squares, then $n$ is called a "good number". Among the set $\{11,13,15,17,19\}$, which are "good numbers" and which are not "good numbers"? Explain your reasoning!
(2004 Chin... | Analysis: Since $\{11,13,15,17,19\}$ is a finite set and contains only 5 elements, we can discuss these elements one by one.
Solution (1) 11 is not a "good number," because 4 can only be added to 5 to get $3^{2}$, and 11 can only be added to 5 to get $4^{2}$, thus there does not exist a permutation that meets the requ... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Let a set of three real numbers be represented both as $\left\{a, \frac{b}{a}, 1\right\}$ and as $\left\{a^{2}, a+b, 0\right\}$, then the value of $a^{2002}+b^{2003}$ is $\qquad$. | 7. 1 .
$$
\left\{\begin{array} { l }
{ a + \frac { b } { a } + 1 = a ^ { 2 } + ( a + b ) + 0 } \\
{ a \cdot \frac { b } { a } \cdot 1 = a ^ { 2 } \cdot ( a + b ) \cdot 0 , }
\end{array} \text { thus } \left\{\begin{array}{l}
a=-1 \text { (discard } 1) \\
b=0,
\end{array} \text { hence } a^{2002}+b^{2003}=1\right.\righ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (5 points) A set of Go costs 24 yuan, and a set of Chinese chess costs 18 yuan. With 300 yuan, you can exactly buy a total of 14 sets of the two types of chess, among which there are $\qquad$ sets of Chinese chess. | 【Analysis】Assuming all are Go boards, then there would be $24 \times 14=336$ yuan, which is $336-300$ $=36$ yuan more than the known 300 yuan. Since a Go board is $24-18=6$ yuan more expensive than a Chinese chess set, we can thus find the number of Chinese chess sets.
【Solution】Solution: Assuming all are Go boards, t... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 4 Let $x, y \in \mathbf{R}$, find the minimum value of the function $f(x, y)=x^{2}+$ $6 y^{2}-2 x y-14 x-6 y+72$.
| Solution: $f(x, y)=(x-y-7)^{2}+5(y-2)^{2}+3$, so when $x=9, y=2$, $f_{\text {min }}=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4 Given $f(x)=a \cos ^{2} x-b \sin x \cos x-\frac{a}{2}$ has a maximum value of $\frac{1}{2}$, and $f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{4}$, then the value of $f\left(-\frac{\pi}{3}\right)$ is
A. $\frac{1}{2}$
B. $-\frac{\sqrt{3}}{4}$
C. $-\frac{1}{2}$ or $\frac{\sqrt{3}}{4}$
D. 0 or $-\frac{\sqrt{3}}{4}$ | 4 D.
$$
\begin{aligned}
f(x) & =\frac{a}{2}(1+\cos 2 x)-\frac{b}{2} \sin 2 x-\frac{a}{2} \\
& =\frac{1}{2}(a \cos 2 x-b \sin 2 x)
\end{aligned}
$$
The maximum value is $\frac{1}{2} \sqrt{a^{2}+b^{2}}$, so
$$
a^{2}+b^{2}=1
$$
From $f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{4}$ we get
$$
a+\sqrt{3} b=-\sqrt{3}
$$
So... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
$7 n$ teams are to hold a double round-robin match (each pair of teams plays two matches, one home and one away match). Each team can play multiple matches within a week (from Sunday to Saturday). If a team has $1:1$ matches in a week, no away matches can be scheduled for this team in the same week. If all matches can ... | (1) As shown in the table: The table has “*”, indicating that the team has a match in that week. It is easy to verify that, according to the schedule in the table, 6 teams can complete the matches in that week.
(2) The following proves that 7 teams cannot complete the matches in that week. Let $S_{i}(i=1, 2, 3, 4, 5, 6... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Given a constant $a \in(0,1)$, $|x|+|y| \leqslant 1$, the maximum value of the function $f(x, y)=a x+y$ is $\qquad$ . | 6. 1 .
Solution: $f(x, y)=a x+y \leqslant|a x|+|y| \leqslant a|x|+(1$ $-|x|)=(a-1)|x|+1$ The first equality holds if $x \geqslant$ $0, y \geqslant 0$. When $0 \leqslant x \leqslant 1, 0<a<1$, $g(x)=(a-1) x+$ 1 is a decreasing function. When $x=0$, $g(x)_{\max }=1$, so $\left\{\begin{array}{l}x=0 \\ y=1\end{array}\righ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
23. There are 2000 identical circular paper pieces on the table, some of which are externally tangent to each other. How many different colors are needed at minimum to color these paper pieces so that no two tangent pieces have the same color?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 23. As shown in the figure: If only 3 colors are used, assume there are 11 circular paper pieces, and let's say 6 of them are colored as indicated by the numbers in the figure. Thus, the three circular pieces $A$, $B$, and $C$ can only be colored with 2 or 3, and $A$ and $C$ must be the same color, but $A$ and $C$ must... | 4 | Number Theory | proof | Yes | Yes | olympiads | false |
7. Five contestants $A, B, C, D, E$ participate in the "The Voice" competition, and the five of them stand in a row for a group appearance. They each have a contestant number on their chest, and the sum of the five numbers is 35. It is known that the sum of the numbers of the contestants standing to the right of $\math... | 【Answer】 11
【Analysis】According to the problem, $D$ is on the far left, so $D=35-31=4, B$ is on the far right, so $B=7,7+4=11$. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Points $A, B, C, D$ lie on the circumference of a circle, and $B C=C D=4, E$ is the intersection of $A C$ and $B D$, and $A E=6$. The lengths of segments $B E$ and $D E$ are both integers. What is the length of $B D$?
(1988 National Junior High School League Question) | 1. From $\triangle C D E \sim \triangle B A E$ and $\triangle C B E \sim \triangle D A E$, we have $A B=\frac{4 B E}{C E}, A D=\frac{4 D E}{C E}$. Applying Ptolemy's theorem to quadrilateral $A B C D$, we get $B D \cdot(A E+C E)=4(A B+A D)=16 \cdot \frac{B E+D E}{C E}$. Let $C E=x$, we obtain the equation $x^{2}+6 x-16... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In the Cartesian coordinate system, $A(1,2)$, $B(3,0)$, and $P$ is any point on the circle $(x-3)^{2}+(y-2)^{2}=1$. Let
$$
\overrightarrow{O P}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}(\lambda, \mu \in \mathbf{R}) \text {. }
$$
Then the minimum value of $11 \lambda+9 \mu$ is | 4. 12 .
Let $P(3+\cos \theta, 2+\sin \theta)(\theta \in[0,2 \pi))$.
Then by the given conditions,
$$
\lambda+3 \mu=3+\cos \theta, 2 \lambda=2+\sin \theta \text {. }
$$
Thus, $11 \lambda+9 \mu=3(\lambda+3 \mu)+4(2 \lambda)$
$$
\begin{array}{l}
=3(3+\cos \theta)+4(2+\sin \theta) \\
=17+3 \cos \theta+4 \sin \theta \\
=1... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. For the cubic function $y=x^{3}+a x^{2}+b x+c$, its graph intersects the $x$-axis at points $A, T$, and $B$ in sequence. The lines $A P$ and $B Q$ are tangent to the cubic curve at points $P$ and $Q$ (where $P$ does not coincide with $A$, and $Q$ does not coincide with $B$). Then the ratio of the projections of $\ov... | 8. -2
Let the coordinates of $A, B, P$ be $(\alpha, 0), (\beta, 0), \left(x_{1}, y_{1}\right)$, and let $T(\gamma, 0)$, then
$$
y=(x-a)(x-\beta)(x-\gamma)
$$
For the line $y=m(x-\beta)$ passing through $A$ and intersecting (1) at two points, the x-coordinates of these two points are given by $(x-\beta)(x-\gamma)-m$. ... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(1) If the circumcenter of $\triangle A B O$ lies on the ellipse, find the value of the real number $p$;
(2) If the circumcircle of $\triangle A B O$ passes through the point $N\left(0, \frac{13}{2}\right)$, find the value of the real number $p$.
11. (20 points) As shown in Figure 4, the ellipse $C_{1}: \frac{x^{2}}{4}... | (1) From the symmetry of the parabola, ellipse, and circle, we know that the circumcenter of $\triangle A B O$ is the upper vertex $M(0,1)$ of the ellipse. Therefore,
$$
M A=M B=M O=1 \text {. }
$$
Let $B\left(x_{0}, y_{0}\right)\left(x_{0}>0\right)$. Hence,
$$
\left\{\begin{array} { l }
{ x _ { 0 } ^ { 2 } = 2 p y _... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.