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2. For the quadratic trinomial $p(x)=(a-1) x^{2}-(a-1) x+2022$, it is known that $-2022 \leq p(x) \leq 2022$ for $x \in[0 ; 1]$. Find the greatest possible value of $a$.
|
Answer: 16177.
Solution. Since $p(0)=p(1)=2022$, the graph of the quadratic trinomial is a parabola symmetric about the line $x=\frac{1}{2}$. From the conditions that $-2022 \leq$ $p(x) \leq 2022$ for $x \in[0 ; 1]$ and $p(0)=p(1)=2022$, it follows that the branches of the parabola are directed upwards. Then the minimum value of $p(x)$ is $p\left(\frac{1}{2}\right)=2022-\frac{(a-1)}{4}$. The maximum possible value of $a$ will be achieved when $p\left(\frac{1}{2}\right)=-2022$. Therefore,
$$
2022-\frac{(a-1)}{4}=-2022 \Rightarrow \frac{(a-1)}{4}=4044 \Rightarrow a-1=16176 \Rightarrow a=16177
$$
|
16177
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the store "Everything for School," chalk is sold in packages of three grades: regular, unusual, and superior. Initially, the ratio of the quantities by grade was 3:4:6. As a result of sales and deliveries from the warehouse, this ratio changed to 2:5:8. It is known that the number of packages of superior chalk increased by $80\%$, and the number of packages of regular chalk decreased by no more than 10 packages. How many packages of chalk were there in the store initially?
(7 points)
|
# Answer: 260 packages.
Solution: Let $x$ be the initial number of packages of regular chalk, then the number of packages of unusual chalk is $\frac{4 x}{3}$. Since the latter number is an integer, then $x=3 n$, where $n \in N$. Therefore, the initial quantities of all three types of packages are $3 n, 4 n, 6 n$ respectively.
After sales and deliveries, the quantity of superior chalk became $(1+0.8) 6 n=\frac{54 n}{5}$, and the quantity of unusual chalk became $\frac{54 n}{5} \cdot \frac{5}{2}=\frac{27 n}{4}$ packages. These numbers are integers, so they are divisible by 4 and 5, that is, $n=20 m$, where $m \in N$. The number of packages of regular chalk will be $\frac{27 n}{20} \cdot 2=\frac{27 n}{10}$. Considering the condition of the problem (the number of packages of regular chalk decreased by no more than 10 packages), we get $0<3 n-\frac{27 n}{10}=(60-54) m=6 m \leq 10$. The only natural number that satisfies this inequality is $m=1$. Therefore, $n=20 m=20$. Thus, the initial number of packages of chalk in the store was
$$
3 n+4 n+6 n=13 n=13 \cdot 20=260
$$
|
260
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the store "Everything for School," chalk is sold in packages of three grades: regular, unusual, and superior. Initially, the ratio of the quantities by grade was 2:3:6. After a certain number of packages of regular and unusual chalk, totaling no more than 100 packages, were delivered to the store, and 40% of the superior chalk packages were sold, the ratio changed to 5:7:4. How many packages of chalk were sold in total in the store?
(7 points)
#
|
# Answer: 24 packs.
Solution: a similar solution to this problem is present in Variant 1 under the same number.
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Determine how many times the number $((2014)^{2^{2014}}-1)$ is larger than the number written in the following form: $\left.\left.\left((2014)^{2^{0}}+1\right) \cdot\left((2014)^{2^{1}}+1\right) \cdot\left((2014)^{2^{2}}+1\right) \cdot \ldots \quad \cdot\right)^{2^{2013}}+1\right)$.
|
Justify the solution.
Answer: 2013.
|
2013
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Find all prime numbers whose decimal representation has the form 101010 ... 101 (ones and zeros alternate).
|
Solution. Let $2n+1$ be the number of digits in the number $A=101010 \ldots 101$.
Let $q=10$ be the base of the numeral system. Then $A=q^{0}+q^{2}+\cdots q^{2n}=\frac{q^{2n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2k \Rightarrow A=\frac{q^{2n+2}-1}{q^{2}-1}=\frac{q^{2k+1}-1}{q-1} \cdot \frac{q^{2k+1}+1}{q+1}$. Thus, the number $A$ is represented as the product of two integer factors (by the theorem of Bezout, the polynomial $q^{2k+1} \pm 1$ is divisible without remainder by the polynomial $q \pm 1$), each of which is different from 1. Therefore, when $n$ is even, the number $A$ is not prime.
- $n=2k-1 \Rightarrow A=\frac{q^{2n+2}-1}{q^{2}-1}=\frac{q^{2k}-1}{q^{2}-1} \cdot (q^{2k}+1)$. For $k>1$, both factors are integers and different from 1; therefore, the number $A$ is composite. It remains to verify that for $k=1$, we get the prime number $A=q^{0}+q^{2}=101$.
Answer: 101.
|
101
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The ordinary fraction $\frac{1}{221}$ is represented as a periodic decimal fraction. Find the length of the period. (For example, the length of the period of the fraction $\frac{25687}{99900}=0.25712712712 \ldots=0.25(712)$ is 3.)
|
Solution. Let's consider an example. We will convert the common fraction $\frac{34}{275}$ to a decimal. For this, we will perform long division (fig.).
As a result, we find $\frac{34}{275}=0.123636363 \ldots=0.123(63)$.
## Interregional Olympiad for Schoolchildren Based on Departmental Educational Organizations in Mathematics
We obtained the non-repeating part 123 and the repeating part 63. Let's discuss why the non-repeating part appears here and show that the fraction $\frac{1}{221}$ does not have it. The matter is that in the decimal representation of the fraction $\frac{34}{275}$, the digit 3 appears every time the remainder from the division by 275 is 100. We see (and this is the key point!), that the same remainder 100 is given by different numbers: 650 and 1750. Where, in turn, did these 650 and 1750 come from? The number 650 was obtained by appending a zero to the number $r_{1}=65$ (the remainder from the division of 340 by 275). That is, $10 r_{1}=650$. Similarly, $10 r_{2}=1750$, where $r_{2}=175$. The numbers 650 and 1750 give the same remainder when divided by 275 because their difference is divisible by 275: $1750-650=10\left(r_{2}-r_{1}\right): 275$. This is possible only because the numbers 10 and 275 are not coprime. Now it is clear why the fraction $\frac{1}{221}$ does not have a non-repeating part: if $r_{1}$ and $r_{2}$ are different remainders from the division by 221, then the product $10\left(r_{2}-r_{1}\right)$ is not divisible by 221 (the number 221, unlike 275, is coprime with 10 - the base of the number system, so there is no non-repeating part).
Thus, the decimal representation of the fraction $\frac{1}{221}$ has the form $\frac{1}{221}=0,\left(a_{1} a_{2} \ldots a_{n}\right)$. Let's find $n$. Denote $A=$ $a_{1} a_{2} \ldots a_{n}$. Then $\frac{1}{221}=10^{-n} \cdot A+10^{-2 n} \cdot A+\cdots$. By the formula for the sum of an infinite geometric series, $\frac{1}{221}=\frac{A}{10^{n}-1}$. From this, $A=\frac{10^{n}-1}{221}$. Since $A$ is a natural number, we need to find (the smallest) natural $n$ for which the number $10^{n}$ gives a remainder of 1 when divided by 221.
Note that $221=13 \cdot 17$. In general, an integer $B$ (in our case $B=10^{n}$) gives a remainder of 1 when divided by 221 if and only if $B$ gives a remainder of 1 when divided by both 13 and 17. The necessity is obvious. Sufficiency: if $B=13 k_{1}+1$ and $B=17 k_{2}+1$, then $13 k_{1}=17 k_{2}$, which means that the number $k_{1}$ is divisible by 17, i.e., $k_{1}=17 m$. Therefore, $B=13 \cdot 17 m+1$, and when divided by 221, the remainder is indeed 1. Now let's find such $n$ that the number $10^{n}$ gives a remainder of 1 when divided by 13. Consider the sequence $b_{n}=10^{n}$. Replace its terms with remainders from division by 13. We get this: $b_{1}=10, b_{2}=9, b_{3}=13, b_{4}=3, b_{5}=4, b_{6}=1, \ldots$ Each subsequent term is uniquely determined by the previous one. Therefore, $\left\{b_{n}\right\}$ is a periodic sequence, in which every sixth term is 1. We will do the same for 17. There, 1 will be equal to every 16th term. Thus, the remainder 1 when divided by both 13 and 17 will be obtained at $n=\text{LCM}(6,16)=48$.
Answer: 48.
|
48
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. It is known that the lengths of the sides of a convex quadrilateral are respectively $a=4, b=5, c=6, d=7$. Find the radius $R$ of the circle circumscribed around this quadrilateral. Provide the integer part of $R^{2}$ as the answer.
|
Solution. By the cosine theorem, we express the length of the diagonal:
$$
l^{2}=a^{2}+b^{2}-2 a b \cos \gamma, l^{2}=c^{2}+d^{2}-2 c d \cos (\pi-\gamma)
$$
From this, we get $\cos \gamma=\frac{a^{2}+b^{2}-c^{2}-d^{2}}{2(a b+c d)}$. Since $R=\frac{l}{2 \sin \gamma^{\prime}}$, we obtain
$$
R^{2}=\frac{l^{2}}{4\left(1-\cos ^{2} \gamma\right)}
$$
For the given side lengths, $R^{2}=\frac{2074799}{131040}$.
Answer: 15.
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. It is known that the equation $x^{4}-8 x^{3}+a x^{2}+b x+16=0$ has (taking into account multiplicity) four positive roots. Find $a-b$.
|
Solution: Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the roots of our equation (some of them may be the same). Therefore, the polynomial on the left side of the equation can be factored as:
$$
x^{4}-8 x^{3}+a x^{2}+b x+16=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)
$$
Expanding the brackets on the right side and equating the coefficients of like powers of $x$, we get:
$$
x_{1}+x_{2}+x_{3}+x_{4}=8, \quad x_{1} x_{2} x_{3} x_{4}=16
$$
It is known that the geometric mean of non-negative numbers does not exceed their arithmetic mean, but in our case, they are equal:
$$
\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}=\sqrt[4]{x_{1} x_{2} x_{3} x_{4}}=2
$$
Therefore, $x_{1}=x_{2}=x_{3}=x_{4}=2$, and
$$
x^{4}-8 x^{3}+a x^{2}+b x+16=(x-2)^{4}
$$
From this, $a=24, b=-32$.
Answer: 56
|
56
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. There is an unlimited number of test tubes of three types - A, B, and C. Each test tube contains one gram of a solution of the same substance. Test tubes of type A contain a $10\%$ solution of this substance, type B $-20\%$ solution, and type C $-90\%$ solution. Sequentially, one after another, the contents of the test tubes are poured into a certain container. In this process, two consecutive pourings cannot use test tubes of the same type. It is known that a $20.17\%$ solution was obtained in the container, performing the minimum number of pourings. What is the maximum number of test tubes of type C that can be used in this process?
|
Solution: Let the number of test tubes of types A, B, and C be \(a\), \(b\), and \(c\) respectively. According to the problem, \(0.1a + 0.2b + 0.9c = 0.2017 \cdot (a + b + c) \Leftrightarrow 1000 \cdot (a + 2b + 9c) = 2017 \cdot (a + b + c)\). The left side of the last equation is divisible by 1000, so the right side must also be divisible by 1000. Therefore, the smallest possible value of the sum \(a + b + c\) is 1000. We will show that this estimate is achievable. That is, we will prove that there exist non-negative integers \(a\), \(b\), and \(c\) such that
\[
\left\{\begin{array}{c}
a + b + c = 1000 \\
a + 2b + 9c = 2017 \\
a \leq 500, b \leq 500, c \leq 500
\end{array}\right.
\]
The last three inequalities are necessary and sufficient conditions to ensure that test tubes of the same type are not used in two consecutive transfers.
From the first two equations of the system (1), we find
\[
a = 7c - 17, \quad b = 1017 - 8c
\]
Substituting these expressions into the last three inequalities of the system (1), we get
\[
7c \leq 517, \quad 8c \geq 518, \quad c \leq 500
\]
From these, the largest value of \(c\) is 73. The corresponding values of \(a\) and \(b\) can be found from (2). They obviously satisfy the inequalities of the system (1). Thus, the solvability of the system (1) in non-negative integers is proven.
Answer: 73
|
73
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Find the sum of the squares of the natural divisors of the number 1800. (For example, the sum of the squares of the natural divisors of the number 4 is $1^{2}+2^{2}+4^{2}=21$ ).
|
Solution: Let $\sigma(N)$ be the sum of the squares of the natural divisors of a natural number $N$. Note that for any two coprime natural numbers $a$ and $b$, the equality $\sigma(a b)=\sigma(a) \cdot \sigma(b)$ holds. Indeed, any divisor of the product $a b$ is the product of a divisor of $a$ and a divisor of $b$. Conversely, multiplying a divisor of $a$ by a divisor of $b$ yields a divisor of the product $a b$. This is clearly true for the squares of divisors as well (the square of a divisor of the product is equal to the product of the squares of the divisors of the factors and vice versa).
Consider the prime factorization of the number $N$: $N=p_{1}^{k_{1}} \cdot \ldots \cdot p_{n}^{k_{n}}$. Here, $p_{i}$ are distinct prime numbers, and all $k_{i} \in N$. Then $\sigma(N)=\sigma\left(p_{1}^{k_{1}}\right) \cdot \ldots \cdot \sigma\left(p_{n}^{k_{n}}\right)$ and $\sigma\left(p^{k}\right)=1+p^{2}+p^{4}+\ldots p^{2 k}$. Since $1800=2^{3} \cdot 3^{2} \cdot 5^{2}$, we have $\sigma(1800)=\left(1+2^{2}+2^{4}+2^{6}\right) \cdot\left(1+3^{2}+3^{4}\right) \cdot\left(1+5^{2}+5^{4}\right)=5035485$.
Answer: 5035485.
|
5035485
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. It is known that the polynomial $f(x)=8+32 x-12 x^{2}-4 x^{3}+x^{4}$ has 4 distinct real roots $\left\{x_{1}, x_{2}, x_{3}, x_{4}\right\}$. The polynomial $\quad$ of the form $g(x)=b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}+x^{4}$ has $\quad$ roots $\left\{x_{1}^{2}, x_{2}^{2}, x_{3}^{2}, x_{4}^{2}\right\}$. Find the coefficient $b_{1}$ of the polynomial $g(x)$.
|
Solution: Let the coefficients of the given polynomial (except the leading one) be denoted by $a_{0}, a_{1}, a_{2}, a_{3}$: $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}$.
Then, according to the problem, we have:
$f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)$.
Together with the polynomial $f(x)$, consider the polynomial $h(x)$, which has roots $\left\{-x_{1},-x_{2},-x_{3},-x_{4}\right\}$:
$$
h(x)=\left(x+x_{1}\right)\left(x+x_{2}\right)\left(x+x_{3}\right)\left(x+x_{4}\right)=a_{0}-a_{1} x+a_{2} x^{2}-a_{3} x^{3}+x^{4}
$$
Consider the polynomial $G(x)=f(x) h(x)$:
$$
\begin{aligned}
& G(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)\left(x+x_{1}\right)\left(x+x_{2}\right)\left(x+x_{3}\right)\left(x+x_{4}\right)= \\
&=\left(x^{2}-x_{1}{ }^{2}\right)\left(x^{2}-x_{2}{ }^{2}\right)\left(x^{2}-x_{3}{ }^{2}\right)\left(x^{2}-x_{4}{ }^{2}\right) .
\end{aligned}
$$
By substituting the variable $y=x^{2}$, we obtain the required polynomial $g(y)$, since
$$
g(y)=\left(y-x_{1}{ }^{2}\right)\left(y-x_{2}{ }^{2}\right)\left(y-x_{3}{ }^{2}\right)\left(y-x_{4}{ }^{2}\right)
$$
In our case:
$$
\begin{gathered}
f(x)=8+32 x-12 x^{2}-4 x^{3}+x^{4} \\
h(x)=8-32 x-12 x^{2}+4 x^{3}+x^{4} \\
g(x)=f(x) h(x)=64-1216 x^{2}+416 x^{4}-40 x^{6}+x^{8} \\
g(y)=64-1216 y+416 y^{2}-40 y^{3}+y^{4}
\end{gathered}
$$
Answer: -1216
|
-1216
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Calculate with an accuracy of one-tenth the value of the expression $\sqrt{86+41 \sqrt{86+41 \sqrt{86+\ldots}}}$
|
Solution: Consider the strictly increasing sequence of values:
$$
\sqrt{86}, \sqrt{86+41 \sqrt{86}}, \sqrt{86+41 \sqrt{86+41 \sqrt{86}}}, \ldots
$$
If this sequence is bounded above, then the value of $F$ is the least upper bound, and thus $F$ is a real number. Therefore, it is sufficient to prove the boundedness of the given sequence. We will prove that this sequence is bounded above by the number 43. Indeed,
$$
\begin{aligned}
\sqrt{86}<43, \Rightarrow 41 \sqrt{86} & <41 \cdot 43 \Rightarrow 86+41 \sqrt{86}<86+41 \cdot 43=43^{2} \Rightarrow \\
& \Rightarrow \sqrt{86+41 \sqrt{86}}<43 \text { and so on. }
\end{aligned}
$$
Interregional Olympiad for Schoolchildren Based on Departmental Educational Organizations in Mathematics
It is clear from the condition that $F$ is the positive root of the equation $F^{2}=86+41 F$. From this, we find $F=43$.
Answer: 43.
|
43
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Calculate with an accuracy of one-tenth the value of the expression $\sqrt{86+41 \sqrt{86+41 \sqrt{86+\ldots}}}$
#
|
# Solution:
Consider the strictly increasing sequence of values:
$$
\sqrt{86}, \sqrt{86+41 \sqrt{86}}, \sqrt{86+41 \sqrt{86+41 \sqrt{86}}}, \ldots
$$
If this sequence is bounded above, then the value of $F$ is the least upper bound, and thus $F$ is a real number. Therefore, it is sufficient to prove the boundedness of the given sequence. We will prove that this sequence is bounded above by the number 43. Indeed,
$$
\sqrt{86}<43, \Rightarrow 41 \sqrt{86}<41 \cdot 43 \Rightarrow 86+41 \sqrt{86}<86+41 \cdot 43=43^{2} \Rightarrow \sqrt{86+41 \sqrt{86}}<43 \text { and so on. }
$$
Obviously, from the condition, $F$ is the positive root of the equation $F^{2}=86+41 F$. From this, we find $F=43$.
Answer: 43.
|
43
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (2 points) A square of 90 by 90 is divided by two horizontal and two vertical lines into 9 rectangles. The sides of the central rectangle are 34 and 42. Find the total area of the four corner rectangles.
|
1. 2688 .
+(2 points) - the solution is correct
-(0 points) - there are errors in the solution, including arithmetic errors
|
2688
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Oleg has 550 rubles, and he wants to give his mother tulips for March 8, and there must be an odd number of them, and no color shade should be repeated. In the store where Oleg came, one tulip costs 49 rubles, and there are eleven shades of flowers available. How many ways are there for Oleg to give his mother flowers? (The answer in the problem should be a compact expression, without summation signs, ellipses, etc.)
|
Solution. From the condition, it is obvious that the maximum number of flowers in a bouquet is 11.
1st method
Using the property of binomial coefficients
$$
\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{3}+\mathrm{C}_{n}^{5}+\cdots=2^{n-1}
$$
and also considering their combinatorial meaning, we get that the number of ways to form a bouquet from an odd number of flowers of no more than 11 shades (with the condition that no shade should be repeated) is:
$$
\mathrm{C}_{11}^{1}+\mathrm{C}_{11}^{3}+\mathrm{C}_{11}^{5}+\cdots+\mathrm{C}_{11}^{11}=2^{10}=1024
$$
2nd method
Consider 10 flowers of 10 different shades. A bouquet can be assembled from these flowers without considering the parity in $2^{10}$ ways. If the bouquet has an odd number of flowers, we keep it; if it has an even number, we add the unused eleventh flower. Thus, the total number of ways to assemble a bouquet is $2^{10}$.
Answer: 1024.
|
1024
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Calculate with an accuracy of one-tenth the value of the expression $\sqrt{86+41 \sqrt{86+41 \sqrt{86+\ldots}}}$
|
Solution. Consider the strictly increasing sequence of values:
$$
\sqrt{86}, \sqrt{86+41 \sqrt{86}}, \sqrt{86+41 \sqrt{86+41 \sqrt{86}}}, \ldots
$$
If this sequence is bounded above, then the value of $F$ is the least upper bound, and thus $F$ is a real number. Therefore, it is sufficient to prove the boundedness of the given sequence. We will prove that this sequence is bounded above by the number 43. Indeed,
$$
\begin{aligned}
\sqrt{86}<43, \Rightarrow & 41 \sqrt{86}<41 \cdot 43 \Rightarrow 86+41 \sqrt{86}<86+41 \cdot 43=43^{2} \Rightarrow \Rightarrow \sqrt{86+41 \sqrt{86}} \\
& <43 \text { and so on. }
\end{aligned}
$$
It is clear from the condition that $F$ is the positive root of the equation $F^{2}=86+41 F$. From this, we find $F=43$.
Answer: 43.
|
43
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A 100-digit number has the form $a=1777 \ldots 76$ (with 98 sevens in the middle). The number $\frac{1}{a}$ is represented as an infinite periodic decimal. Find its period. Justify your answer.
|
# Solution:
Notice that $a=16 \cdot 111 \ldots 11$. The last number $b$ consists of 99 ones. According to the rules for converting a common fraction to a decimal, the number $\frac{1}{b}=0,(00 \ldots 09)$. Its period is 99. Then, when multiplying this fraction by the number $\frac{1}{16}=0.0625$, the period will not change.
Answer: 99.
|
99
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A robot is located in one of the cells of an infinite grid paper, to which the following commands can be given:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent cell to the right).
For example, if the robot executes a sequence of four commands (up, right, down, left), it will obviously return to the initial position, i.e., it will end up in the same cell from which it started. How many different sequences of 4 commands exist that return the robot to its initial position?
|
Solution. For brevity, let's denote the command to move left as L, right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be the number of L commands in the sequence. We will calculate the number $N_{k}$ of the desired sequences for $k$ from 0 to 2.
- $\boldsymbol{k}=\mathbf{0}$. The sequence consists only of U and D commands. Since they are equal in number, U commands must occupy 2 out of 4 positions, and D commands must occupy the remaining 2 positions. The number of ways to choose 2 positions out of 4 is $C_{4}^{2}$. Therefore, $N_{0}=C_{4}^{2}=6$;
- $\boldsymbol{k}=\mathbf{1}$. Each of the commands L, R, U, and D appears in the sequence exactly once. The number of permutations of 4 elements is $4!$. Therefore, $N_{1}=4!=24$;
- $\boldsymbol{k}=\mathbf{2}$. Here, there are two L commands, two R commands, and no U or D commands. The two L commands can be placed in $C_{4}^{2}$ ways. Thus, $N_{2}=C_{4}^{2}=6$.
Therefore, the total number of sequences is $N_{0}+N_{1}+N_{2}=36$.
Answer: 36.
|
36
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Find the value of $f(2019)$, given that $f(x)$ simultaneously satisfies the following three conditions:
1) $f(x)>0$ for any $x>0$
2) $f(1)=1$;
3) $f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}$ for any $a, b \in \mathbb{R}$.
|
Solution. In the identity given in the problem
$$
f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}
$$
let $a=1, b=0$. Then $f(1) \cdot(f(1)+f(0))=2 f(1) \cdot f(0)+1$. Since $f(1)=1$, we find
$$
f(0)=0
$$
Next, by setting $b=-a$ in (1), we get, taking (2) into account, that
$$
f(a) \cdot f(-a)=-a^{2}
$$
Finally, when $b=0$, identity (1) (taking (2) into account) becomes $f(a) \cdot f(a)=a^{2}$.
Thus, it is necessary that $f(a)=a$ for $a>0$, since by condition $f(x)>0$ for $x>0$. Furthermore, according to (3), $f(a)=a$ for $a<0$ as well. Ultimately, $f(x)=x$ for any $x \in \mathbb{R}$. It is easy to verify that such an $f(x)$ indeed satisfies conditions 1), 2), 3) from the problem statement. Therefore, $f(x)=x$.
Answer: 2019.
|
2019
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. In a circle, three chords $A A_{1}, B B_{1}, C C_{1}$ intersect at one point. The angular measures of the arcs $A C_{1}, A B, C A_{1}$, and $A_{1} B_{1}$ are $150^{\circ}, 30^{\circ}, 60^{\circ}$, and $30^{\circ}$, respectively. Find the angular measure of the arc $B_{1} C_{1}$.
|
Solution: Let's formulate several auxiliary statements.
1) Let the angular measure of the arc $AB$ (Fig.1) be $\varphi$. (This means that $\varphi$ is equal to the corresponding central angle $AOB$.) Then the length of the chord
$AB=2R\sin(\varphi/2)$. Here $R$ is the radius of the circle.
2) Let two chords $AA_1$ and $BB_1$ intersect at point $T$ (Fig.2). The angular measures of the arcs $AB$ and $A_1B_1$ are $\varphi$ and $v$. Triangles $ATB$ and $A_1TB_1$ are similar by two angles (equal angles are marked). The similarity ratio
$$
k=AB / A_1B_1=\sin(\varphi/2) / \sin(v/2)
$$
Fig. 1
Fig. 2
Fig. 3
3) Referring to Fig. 3. Three chords intersect at one point, denoted as $T$. The angular measures of the six resulting arcs are marked on the figure. From the similarity of triangles $ATB$ and $A_1TB_1$ it follows (see point 2) the equality $AT / B_1T = \sin(\varphi/2) / \sin(v/2)$. Similarly, $\triangle B T C \sim \Delta B_1 T C_1 \Rightarrow B_1 T / C T = \sin(\psi/2) / \sin(u/2), \Delta C T A_1 \sim \Delta A T C_1 \Rightarrow C T / A T = \sin(\theta/2) / \sin(w/2)$. Multiplying the last three equalities, we get:
$1 = AT / B_1T \cdot B_1T / CT \cdot CT / AT = \sin(\varphi/2) / \sin(v/2) \cdot \sin(\psi/2) / \sin(u/2) \cdot \sin(\theta/2) / \sin(w/2)$.
Thus, the necessary (and in fact sufficient) condition for three chords to intersect at one point is the equality:
$$
\sin(\varphi/2) \sin(\theta/2) \sin(\psi/2) = \sin(u/2) \sin(v/2) \sin(w/2)
$$
Now it is not difficult to obtain the answer to the problem. Substituting the given data into this relation $w=150^\circ, \varphi=30^\circ, \theta=60^\circ, v=30^\circ$, and expressing $u$ from the equality $\varphi + u + \theta + v + \psi + w = 360^\circ$, we get the equation for determining the desired angle $\psi$:
$$
\sin 15^\circ \sin(\psi/2) \sin 30^\circ = \sin 15^\circ \sin \left(\left(90^\circ - \psi\right)/2\right) \sin 75^\circ
$$
From this, it is not difficult to obtain that $\psi = 60^\circ$.
Answer: $60^\circ$.
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Calculate with an accuracy of one-tenth the value of the expression $\sqrt{86+41 \sqrt{86+41 \sqrt{86+\ldots}}}$
#
|
# Solution:
Consider the strictly increasing sequence of values:
$$
\sqrt{86}, \sqrt{86+41 \sqrt{86}}, \sqrt{86+41 \sqrt{86+41 \sqrt{86}}}, \ldots
$$
If this sequence is bounded above, then the value of $F$ is the least upper bound, and thus $F$ is a real number. Therefore, it is sufficient to prove the boundedness of the given sequence. We will prove that this sequence is bounded above by the number 43. Indeed,
$$
\begin{aligned}
\sqrt{86}<43, \Rightarrow & 41 \sqrt{86}<41 \cdot 43 \Rightarrow 86+41 \sqrt{86}<86+41 \cdot 43=43^{2} \Rightarrow \Rightarrow \sqrt{86+41 \sqrt{86}} \\
& <43 \text { and so on. }
\end{aligned}
$$
It is clear from the condition that $F$ is the positive root of the equation $F^{2}=86+41 F$. From this, we find $F=43$.
Answer: 43.
|
43
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find the sum of all even natural numbers $n$ for which the number of divisors (including 1 and $n$ itself) is equal to $\frac{n}{2}$. (For example, the number 12 has 6 divisors: $1,2,3,4,6,12$.)
|
Solution. Let the canonical decomposition of the number $n$ be: $n=2^{t_{1}} \cdot 3^{t_{2}} \cdot 5^{t_{3}} \cdots \cdots \cdot p^{t_{k}}$. Then the number of divisors of the number $n$ is $\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right) \cdots\left(t_{k}+1\right)$. From the condition of the problem, we have the equality
$$
\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right) \cdots\left(t_{k}+1\right)=2^{t_{1}-1} \cdot 3^{t_{2}} \cdot 5^{t_{3}} \cdots \cdots \cdot p^{t_{k}} \cdot\left({ }^{*}\right)
$$
Note that $2^{t_{1}-1}>t_{1}+1$ for $t_{1} \geq 4, 3^{t_{2}}>t_{2}+1$ for $t_{2} \geq 1, \ldots, p^{t_{k}}>t_{k}+1$ for $t_{k} \geq 1$. Therefore, $t_{1}$ can take the values 1, 2, or 3. Substituting the specified values into the equality $(*)$, we find that $n=8$ or $n=12$.
Answer: 20.
|
20
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In how many ways can 4 numbers be chosen from the first 1000 natural numbers $1,2, \ldots, 1000$ to form an increasing arithmetic progression?
|
Solution. Let's find the formula for calculating the number of ways to choose 4 numbers from the first $n$ natural numbers $1,2, \ldots, n$ that form an increasing arithmetic progression. The number of progressions with a difference of 1 is $n-3$ (the first term of the progression can take values from 1 to $n-3$ inclusive), the number of progressions with a difference of 2 is $n-6, \ldots$, the number of progressions with a difference of $d$ is $n-3d$. The difference $d$ satisfies the inequality $1+3d \leq n$ (if the first term of the progression is 1, then its fourth term, $1+3d$, does not exceed $n$). Therefore, the maximum value of the difference is $d_{\max }=\left[\frac{n-1}{3}\right]$ (square brackets denote the integer part of the number). Consequently, the number of progressions that satisfy the condition of the problem is:
$$
(n-3)+(n-6)+\cdots+(n-3k)=\frac{(2n-3k-3)k}{2} \text{, where } k=d_{\max }
$$
For $n=1000$, we have $k=333$ and the number of ways is 166167
Answer: 166167.
|
166167
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. It is known that there exists a natural number $N$ such that $(\sqrt{3}-1)^{N}=4817152-2781184 \cdot \sqrt{3}$. Find $N$.
|
Solution. Suppose that raising the number $a+b \sqrt{3}$ to the power $N$ results in the number $A+B \sqrt{3}$ (where $a, b, A, B$ are integers). Expanding the expression $(a+b \sqrt{3})^{N}$, we get a sum of monomials (with non-essential (for us now) integer coefficients) of the form $a^{N-n}(b \sqrt{3})^{n}$. The terms that contribute to the coefficient $B$ are those with an odd exponent $n$. Therefore, if $(a+b \sqrt{3})^{N}=A+B \sqrt{3}$, then $(a-b \sqrt{3})^{N}=A-B \sqrt{3}$. Multiplying the equations $(\sqrt{3}-1)^{N}=4817152-2781184 \cdot \sqrt{3}$ and $(-\sqrt{3}-1)^{N}=4817152+2781184 \cdot \sqrt{3}$, we get $(-2)^{N}=4817152^{2}-3 \cdot 2781184^{2}$. The exponent $N$ can be found by dividing both sides successively by 2 (for example, we can immediately divide each term on the right by 256).
Answer: $N=16$.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find all prime numbers whose representation in base 14 has the form 101010 ... 101 (ones and zeros alternate).
|
Solution: Let $2n+1$ be the number of digits in the number $A=101010 \ldots 101$. Let $q=$ 14 - the base of the numeral system. Then $A=q^{0}+q^{2}+\cdots q^{2 n}=\frac{q^{2 n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2 k \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k+1}-1}{q-1} \cdot \frac{q^{2 k+1}+1}{q+1}$. Thus, the number $A$ is represented as the product of two integer factors (by the theorem of Bezout, the polynomial $q^{2 k+1} \pm 1$ is divisible without remainder by the polynomial $q \pm 1$), each of which is different from 1. Therefore, when $n$ is even, the number $A$ is not prime.
- $n=2 k-1 \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k}-1}{q^{2}-1} \cdot\left(q^{2 k}+1\right)$. For $k>1$, both factors are integers and different from 1; therefore, the number $A$ is composite. It remains to verify that for $k=1$, the number $A=q^{0}+$ $q^{2}=197$ is prime.
Answer: 197.
|
197
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. It is known that the polynomial $f(x)=8+32 x-12 x^{2}-4 x^{3}+x^{4}$ has 4 distinct real roots $\left\{x_{1}, x_{2}, x_{3}, x_{4}\right\}$. The polynomial of the form $g(x)=b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}+x^{4}$ has roots $\left\{x_{1}^{2}, x_{2}^{2}, x_{3}^{2}, x_{4}^{2}\right\}$. Find the coefficient $b_{1}$ of the polynomial $g(x)$.
|
Solution: Let the coefficients of the given polynomial (except the leading one) be denoted by $a_{0}, a_{1}, a_{2}, a_{3}$
$$
f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}
$$
Then, according to the problem, we have
$$
f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)
$$
Together with the polynomial $f(x)$, consider the polynomial $h(x)$, which has roots $\left\{-x_{1},-x_{2},-x_{3},-x_{4}\right\}$
$$
h(x)=\left(x+x_{1}\right)\left(x+x_{2}\right)\left(x+x_{3}\right)\left(x+x_{4}\right)=a_{0}-a_{1} x+a_{2} x^{2}-a_{3} x^{3}+x^{4}
$$
Consider the polynomial $G(x)=f(x) h(x)$:
$$
\begin{aligned}
& G(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)\left(x+x_{1}\right)\left(x+x_{2}\right)\left(x+x_{3}\right)\left(x+x_{4}\right)= \\
&=\left(x^{2}-x_{1}^{2}\right)\left(x^{2}-x_{2}^{2}\right)\left(x^{2}-x_{3}{ }^{2}\right)\left(x^{2}-x_{4}^{2}\right)
\end{aligned}
$$
By substituting the variable $y=x^{2}$, we obtain the required polynomial $g(y)$, since
$$
g(y)=\left(y-x_{1}{ }^{2}\right)\left(y-x_{2}{ }^{2}\right)\left(y-x_{3}{ }^{2}\right)\left(y-x_{4}{ }^{2}\right)
$$
In our case
$$
\begin{gathered}
f(x)=8+32 x-12 x^{2}-4 x^{3}+x^{4} \\
h(x)=8-32 x-12 x^{2}+4 x^{3}+x^{4} \\
g(x)=f(x) h(x)=64-1216 x^{2}+416 x^{4}-40 x^{6}+x^{8}
\end{gathered}
$$
Interregional Olympiad for Schoolchildren based on Departmental Educational Organizations
$$
g(y)=64-1216 y+416 y^{2}-40 y^{3}+y^{4}
$$
Answer: -1216.
|
-1216
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A robot is located in one of the cells of an infinite grid and can be given the following commands:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent cell to the right).
For example, if the robot executes a sequence of four commands (up, right, down, left), it will obviously return to its initial position, i.e., it will end up in the same cell from which it started. How many different sequences of 4 commands exist that return the robot to its initial position?
|
Solution. For brevity, let's denote the command to the left as L, to the right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be the number of L commands in the sequence. We will calculate the number $N_{k}$ of the desired sequences for $k$ from 0 to 2.
- $\boldsymbol{k}=\mathbf{0}$. The sequence consists only of U and D commands. Since they are equal in number, U commands should be placed on 2 out of 4 positions, and D commands on the remaining 2 positions. The number of ways to choose 2 positions out of 4 is $C_{4}^{2}$. Therefore, $N_{0}=C_{4}^{2}=6$.
- $\boldsymbol{k}=\mathbf{1}$. Each of the commands L, R, U, and D appears in the sequence exactly once. The number of permutations of 4 elements is $4!$. Therefore, $N_{1}=4!=24$;
- $\boldsymbol{k}=\mathbf{2}$. Here, there are two L, two R, and no U or D commands. Two L commands can be placed in $C_{4}^{2}$ ways. Thus, $N_{2}=C_{4}^{2}=6$.
Therefore, the total number of sequences is $N_{0}+N_{1}+N_{2}=36$.
Answer: 36.
|
36
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find all prime numbers whose decimal representation has the form 101010 ... 01.
|
Solution. Let $2 n+1$ be the number of digits in the number $A=101010 \ldots 101$ under investigation. Let $q=$ 10 be the base of the number system. Then $A=q^{0}+q^{2}+\cdots q^{2 n}=\frac{q^{2 n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2 k \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k+1}-1}{q-1} \cdot \frac{q^{2 k+1}+1}{q+1}$. Thus, the number $A$ is represented as the product of two integer factors (by the theorem of Bezout, the polynomial $q^{2 k+1} \pm 1$ is divisible without remainder by the polynomial $q \pm 1$), each of which is different from 1. Therefore, when $n$ is even, the number $A$ is not prime.
- $n=2 k-1 \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k}-1}{q^{2}-1} \cdot\left(q^{2 k}+1\right)$. For $k>1$, both factors are integers and different from 1; therefore, the number $A$ is composite. It remains to verify that for $k=1$, the prime number $A=q^{0}+$ $q^{2}=101$ is obtained.
Answer: 101.
|
101
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. The ordinary fraction $\frac{1}{221}$ is represented as a periodic decimal fraction. Find the length of the period. (For example, the length of the period of the fraction $\frac{25687}{99900}=0.25712712712 \ldots=0.25$ (712) is 3.)
|
Solution. Let's consider an example. We will convert the common fraction $\frac{34}{275}$ to a decimal. For this, we will perform long division (fig.). As a result, we find $\frac{34}{275}=0.123636363 \ldots=0.123(63)$. We obtain the non-repeating part 123 and the repeating part 63. Let's discuss why the non-repeating part appears here, and show that the fraction $\frac{1}{221}$ does not have a non-repeating part. The matter is that in the decimal representation of the fraction $\frac{34}{275}$, the digit 3 appears every time the remainder is 100 when dividing by 275. We see (and this is the key point!), that the same remainder 100 is given by different numbers: 650 and 1750. Where, in turn, did these 650 and 1750 come from? The number 650 was obtained by appending a zero to the number $r_{1}=65$ (the remainder from dividing 340 by 275). That is, $10 r_{1}=650$. Similarly, $10 r_{2}=1750$, where $r_{2}=175$. The numbers 650 and 1750 give the same remainder when divided by 275 because their difference is divisible by 275: $1750-650=10\left(r_{2}-r_{1}\right): 275$. This is possible only because the numbers 10 and 275 are not coprime. Now it is clear why the fraction $\frac{1}{221}$ does not have a non-repeating part: if $r_{1}$ and $r_{2}$ are different remainders from dividing by 221, then the product $10\left(r_{2}-r_{1}\right)$ is not divisible by 221 (the number 221, unlike 275, is coprime with 10 - the base of the number system, so there is no non-repeating part).
Thus, the decimal representation of the fraction $\frac{1}{221}$ has the form $\frac{1}{221}=0,\left(a_{1} a_{2} \ldots a_{n}\right)$. Let's find $n$. Denote $A=a_{1} a_{2} \ldots a_{n}$. Then $\frac{1}{221}=10^{-n} \cdot A+10^{-2 n} \cdot A+\cdots$. By the formula for
825
1750
1650
$-\boxed{1000}$ $\frac{825}{-1750}$ the sum of an infinite geometric series $\frac{1}{221}=\frac{A}{10^{n}-1}$. From this, $A=\frac{10^{n}-1}{221}$. Since $A$ is a natural number, we need to find (the smallest) natural $n$ for which the number $10^{n}$ gives a remainder of 1 when divided by 221.
Note that $221=13 \cdot 17$. In general, an integer $B$ (in our case $B=10^{n}$) gives a remainder of 1 when divided by 221 if and only if $B$ gives a remainder of 1 when divided by both 13 and 17. The necessity is obvious. Sufficiency: if $B=13 k_{1}+1$ and $B=17 k_{2}+1$, then $13 k_{1}=17 k_{2}$, which means that $k_{1}$ is divisible by 17, i.e., $k_{1}=17 m$. Therefore, $B=13 \cdot 17 m+1$, and when divided by 221, the remainder is indeed 1. Now let's find such $n$ that the number $10^{n}$ gives a remainder of 1 when divided by 13. Consider the sequence $b_{n}=10^{n}$. Replace its terms with remainders when divided by 13. We get this: $b_{1}=10, b_{2}=9, b_{3}=13, b_{4}=3, b_{5}=4, b_{6}=1, \ldots$ Each subsequent term is uniquely determined by the previous one. Therefore, $\left\{b_{n}\right\}$ is a periodic sequence, where every sixth term is 1. We will do the same for 17. There, 1 will be equal to every 16th term. Thus, the remainder 1 when divided by both 13 and 17 will be obtained at $n=\operatorname{LCM}(6,16)=48$.
Answer: 48.
|
48
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Anya and Borya are playing "battleship" with the following rules: 29 different points are chosen on a circle, numbered clockwise with natural numbers from 1 to 29. Anya draws a ship - an arbitrary triangle with vertices at these points. We will call a "shot" the selection of two different natural numbers $k$ and $m$ from 1 to 29. If the segment with endpoints at points numbered $k$ and $m$ has at least one common point with Anya's triangle, the ship is considered "wounded." Borya fires a "salvo" - several shots simultaneously. Anya drew a ship and showed it to Borya. And then they noticed that any "salvo" of K different shots will definitely wound Anya's ship. Indicate any position of Anya's ship for which the value of $K$ will be minimal.
|
Solution. The vertices of Anya's triangle divide the circle into three arcs (see figure). Let $x, y$, and $26-x-y$ be the number of points on these arcs (see figure), excluding the vertices of the triangle itself. For a "shot" with endpoints at points $k$ and $m$ to not hit the ship, both these points must lie on one of the arcs. Clearly, two different points can be chosen on the arc containing $x$ points in $C_{x}^{2}$ ways. The same applies to the other arcs. Therefore, the number $N$ of "safe" shots is the sum $N=C_{x}^{2}+C_{y}^{2}+C_{26-x-y}^{2}$. Then the next shot will definitely "hit" the ship, so $K=N+1$.
Thus, we need to find such non-negative integers $x, y$ that satisfy the condition $x+y \leq 26$, for which the value of $N$ is minimal.
Let's write the expression for $N$ in expanded form:
$$
N=\frac{x(x-1)}{2}+\frac{y(y-1)}{2}+\frac{(26-x-y)(25-x-y)}{2}
$$
Expanding the brackets and combining like terms, we get
$$
N=x^{2}-x(26-y)+y^{2}-26 y+325
$$
www.v-olymp.ru
Interregional Olympiad for Schoolchildren Based on Departmental Educational Organizations in Mathematics
For each fixed $y$ from 0 to 26, we will find such a value of $x$ that satisfies the inequality
$$
0 \leq x \leq 26-y
$$
for which the value of $N$ is minimal. If $y$ is fixed, then the right-hand side (1) takes its minimum value at
$$
x=\frac{26-y}{2}
$$
(the vertex of the parabola, belonging to the interval (2)). This minimum value is $13 y+156$. It, in turn, is minimal at $y=\frac{26}{3} \approx 8.6$. From (3), we then find $x=\frac{26}{3}$. Among the points with integer coordinates $(8,8),(8,9),(9,8),(9,9)$ - the nearest integer neighbors of the point of minimum $\left(\frac{26}{3}, \frac{26}{3}\right)$ - we choose the one for which the value of $N$ is the smallest. These are the points $(8,9),(9,8),(9,9)$. For them, $N=100$.
Answer: The ship should be placed such that on the three arcs into which the vertices of the ship divide the circle, there are 8, 9, and 9 points (excluding the vertices of the ship itself).
|
100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (4 points) Numbers are marked on a chessboard (see Fig. 1). How many arrangements of 8 rooks, none of which attack each other, exist such that the numbers on the squares occupied by the rooks include all numbers from 0 to 7? $\left(\begin{array}{llllllll}0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0\end{array}\right)$ Fig. 1
|
# Solution:
Let's number the rows of the board from 1 to 8 from top to bottom. On the first row, the position of the rook can be chosen in 8 ways. For example, we choose the first cell with the number 0. On the second row, there are 7 options left. Six of them (cells from the second to the seventh) have the property that under them in the lower half of the board, there are numbers different from the number occupied by the first rook (in our case - the number 0), and one cell (the last one, with the number 7) is such that under it in the lower half stands the number occupied by the first rook (in our case - 0).
Consider the first group of 6 options and make an arbitrary choice (for example, the second cell with the number 1). After such a choice, it turns out that in the lower half of the board, two different numbers are prohibited for choosing positions (in our case, 7 and 6). Therefore, the positions with these numbers need to be chosen in the upper half of the board, and there are exactly 2 such options (in our case, the 7th and 8th cells in rows 3 and 4). After this choice, in the lower half of the board, there will be 4 columns with different numbers in the columns. That is, there are $4! = 24$ options in the lower half. Thus, we get $8 \times 6 \times 2 \times 24$ options.
Consider the option of choosing a position on the second row when under the chosen cell in the lower half of the board stands the number chosen on the first row (in our case, the last cell with the number 7). After such a choice, in the third row, there are 6 options leading to a single option in the fourth row. After choosing options in the upper half of the board, in the lower half, there are again 4 free columns with four different numbers, i.e., 24 options. Thus, we get $8 \times 1 \times 6 \times 24$ options.
In total, we get:
$8 \times 6 \times 2 \times 24 + 8 \times 1 \times 6 \times 24 = 3456$.
Answer: 3456.
|
3456
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. There is an unlimited number of test tubes of three types - A, B, and C. Each test tube contains one gram of a solution of the same substance. Test tubes of type A contain a $10\%$ solution of this substance, type B $-20\%$ solution, and type C $-90\%$ solution. Sequentially, one after another, the contents of the test tubes are poured into a certain container. In this process, two consecutive pourings cannot use test tubes of the same type. It is known that a $20.17\%$ solution was obtained in the container, performing the minimum number of pourings. What is the maximum number of test tubes of type C that can be used in this process?
|
Solution: Let the number of test tubes of types A, B, and C be $a$, $b$, and $c$ respectively. According to the problem, $0.1a + 0.2b + 0.9c = 0.2017 \cdot (a + b + c) \Leftrightarrow 1000 \cdot (a + 2b + 9c) = 2017 \cdot (a + b + c)$. The left side of the last equation is divisible by 1000, so the right side must also be divisible by 1000. Therefore, the smallest possible value of the sum $a + b + c$ is 1000. We will show that this estimate is achievable. That is, we will prove that there exist non-negative integers $a$, $b$, and $c$ such that
$$
\left\{\begin{array}{c}
a + b + c = 1000 \\
a + 2b + 9c = 2017 \\
a \leq 500, b \leq 500, c \leq 500
\end{array}\right.
$$
The last three inequalities are necessary and sufficient conditions to ensure that test tubes of the same type are not used in two consecutive transfers. From the first two equations of the system (1), we find
$$
a = 7c - 17, \quad b = 1017 - 8c
$$
Substituting these expressions into the last three inequalities of the system (1), we get
$$
7c \leq 517, \quad 8c \geq 518, \quad c \leq 500
$$
From these, the largest value of $c$ is 73. The corresponding values of $a$ and $b$ can be found from (2). They obviously satisfy the inequalities of the system (1). Thus, the solvability of the system (1) in non-negative integers is proven.
## Answer: 73
|
73
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. Find the sum of the squares of the natural divisors of the number 1800. (For example, the sum of the squares of the natural divisors of the number 4 is $1^{2}+2^{2}+4^{2}=21$ ).
|
Solution: Let $\sigma(N)$ be the sum of the squares of the natural divisors of a natural number $N$. Note that for any two coprime natural numbers $a$ and $b$, the equality $\sigma(a b)=\sigma(a) \cdot \sigma(b)$ holds. Indeed, any divisor of the product $a b$ is the product of a divisor of $a$ and a divisor of $b$. Conversely, multiplying a divisor of $a$ by a divisor of $b$ yields a divisor of the product $a b$. This is clearly true for the squares of divisors as well (the square of a divisor of the product is equal to the product of the squares of the divisors of the factors and vice versa).
Consider the prime factorization of the number $N$: $N=p_{1}^{k_{1}} \cdot \ldots \cdot p_{n}^{k_{n}}$. Here, $p_{i}$ are distinct prime numbers, and all $k_{i} \in N$. Then $\sigma(N)=\sigma\left(p_{1}^{k_{1}}\right) \cdot \ldots \cdot \sigma\left(p_{n}^{k_{n}}\right)$ and $\sigma\left(p^{k}\right)=1+p^{2}+p^{4}+\ldots p^{2 k}$. Since $1800=2^{3} \cdot 3^{2} \cdot 5^{2}$, we have $\sigma(1800)=\left(1+2^{2}+2^{4}+2^{6}\right) \cdot\left(1+3^{2}+3^{4}\right) \cdot\left(1+5^{2}+5^{4}\right)=5035485$.
Answer: 5035485.
|
5035485
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. Find the number of matrices that satisfy two conditions:
1) the matrix has the form $\left(\begin{array}{lll}1 & * & * \\ * & 1 & * \\ * & * & 1\end{array}\right)$, where each * can take the value 0 or 1
2) the rows of the matrix do not repeat.
|
Solution: Let $A$ be the set of matrices satisfying condition 1) and $B$ be the subset of $A$ consisting of matrices satisfying condition 2). We need to find the number of elements in the set $B$. Let $A_{ij}, i, j \in\{1,2,3\}, i \neq j$, be the subset of $A$ consisting of matrices in which row $i$ and $j$ are the same. Then $B=A \backslash\left(A_{12} \cup A_{23} \cup A_{13}\right)$ and $|B|=|A|-\left|A_{12} \cup A_{23} \cup A_{13}\right|$. The cardinality $\left|A_{12} \cup A_{23} \cup A_{13}\right|$ is conveniently calculated using the inclusion-exclusion principle:
$\left|A_{12} \cup A_{23} \cup A_{13}\right|=\left|A_{12}\right|+\left|A_{23}\right|+\left|A_{13}\right|-\left|A_{12} \cap A_{23}\right|-\left|A_{13} \cap A_{23}\right|-\left|A_{12} \cap A_{13}\right|+\left|A_{12} \cap A_{23} \cap A_{13}\right|$.
It is easy to calculate the cardinalities of the sets involved in this expression:
$$
\left|A_{12}\right|=\left|A_{23}\right|=\left|A_{13}\right|=2^{3},\left|A_{12} \cap A_{23}\right|=\left|A_{13} \cap A_{23}\right|=\left|A_{12} \cap A_{13}\right|=\left|A_{12} \cap A_{23} \cap A_{13}\right|=1 .
$$
We get $|B|=2^{6}-3 \cdot 2^{3}+3-1=42$.
Answer: 42
|
42
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A motorist set off from point A to point B, the distance between which is 100 km. At the moment when the navigator showed that 30 minutes of travel remained, the motorist first reduced the speed by 10 km/h, and at the moment when the navigator showed that 20 km remained to travel, the motorist reduced the speed by the same 10 km/h again. (The navigator determines the remaining time and distance based on the current speed of movement.) Determine the initial speed of the car, given that the car traveled 5 minutes longer at the reduced speed than at the twice-reduced speed.
NOTE. In the 9th grade, the same problem was given in a slightly different version.
|
Solution. According to the condition, the distance from C to D is 20 km. Let the distance from A to B be denoted as
$x$ (km), then the distance from B to C will be $(80 - x)$ km. Let $v \frac{\text{km}}{4}$ be the initial speed of the car. Then on the segments BC and CD, the speed is $(v-10) \frac{\text{km}}{4}$ and $(v-20) \frac{\text{km}}{4}$, respectively. According to the condition, the journey from B to D would have taken half an hour if the car had continued to move at a speed of $v \frac{\text{km}}{4}$, that is,
$$
100 - x = \frac{v}{2}
$$
Furthermore, the time spent on the journey from B to C is 5 minutes more than the time spent on the journey from C to D:
$$
\frac{80 - x}{v - 10} - \frac{20}{v - 20} = \frac{1}{12}
$$
Expressing $x$ from the first equation and substituting it into the second, we get an equation to determine $v$:
$$
\frac{\frac{v}{2} - 20}{v - 10} - \frac{20}{v - 20} = \frac{1}{12}
$$
which has roots 14 and 100. The root 14 is obviously extraneous, as $v > 20$.
Answer: $100 \frac{\text{km}}{4}$.
|
100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find all roots of the equation $\frac{1}{\cos ^{3} x}-\frac{1}{\sin ^{3} x}=4 \sqrt{2}$, lying in the interval $\left(-\frac{\pi}{2}, 0\right)$. Write the answer in degrees.
|
# Solution:
$\sin ^{3} x-\cos ^{3} x=4 \sqrt{2} \sin ^{3} x \cos ^{3} x \Leftrightarrow(\sin x-\cos x)\left(\sin ^{2} x+\sin x \cos x+\cos ^{2} x\right)=4 \sqrt{2} \sin ^{3} x \cos ^{3} x$. Substitution: $\sin x-\cos x=t, \sin x \cos x=\frac{1-t^{2}}{2}$. Then $t\left(3-t^{2}\right)=\sqrt{2}\left(1-t^{2}\right)^{3}$. Substitution: $t=z \sqrt{2}$. The equation will take the form $z\left(3-2 z^{2}\right)-\left(1-2 z^{2}\right)^{3}=0$. There is a root $z=-1$, and the left side can be factored
$$
(z+1)\left(8 z^{5}-8 z^{4}-4 z^{3}+2 z^{2}+4 z-1\right)=0
$$
Since $x \in\left(-\frac{\pi}{2}, 0\right)$, then $t=\sin x-\cos x=\sqrt{2} \sin \left(x-\frac{\pi}{4}\right)\left|4 z^{3}\right|$ and $\left|8 z^{4}\right|>\left|2 z^{2}\right|$. Therefore, $z=-1$ is the only root of equation (1). It is easy to find that $\sin \left(x-\frac{\pi}{4}\right)=-1$, and $x=-\frac{\pi}{4}$.
Answer: -45
|
-45
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Find the sum of the squares of the natural divisors of the number 1800. (For example, the sum of the squares of the natural divisors of the number 4 is $\left.1^{2}+2^{2}+4^{2}=21\right)$.
|
Solution: Let $\sigma(N)$ be the sum of the squares of the natural divisors of a natural number $N$. Note that for any two coprime natural numbers $a$ and $b$, the equality $\sigma(a b)=\sigma(a) \cdot \sigma(b)$ holds. Indeed, any divisor of the product $a b$ is the product of a divisor of $a$ and a divisor of $b$. Conversely, multiplying a divisor of $a$ by a divisor of $b$ yields a divisor of the product $a b$. This is clearly true for the squares of divisors as well (the square of a divisor of the product is equal to the product of the squares of the divisors of the factors and vice versa).
Consider the prime factorization of the number $N$: $N=p_{1}^{k_{1}} \cdot \ldots \cdot p_{n}^{k_{n}}$. Here, $p_{i}$ are distinct prime numbers, and all $k_{i} \in N$. Then $\sigma(N)=\sigma\left(p_{1}^{k_{1}}\right) \cdot \ldots \cdot \sigma\left(p_{n}^{k_{n}}\right)$ and $\sigma\left(p^{k}\right)=1+p^{2}+p^{4}+\ldots p^{2 k}$. Since $1800=2^{3} \cdot 3^{2} \cdot 5^{2}$, we have $\sigma(1800)=\left(1+2^{2}+2^{4}+2^{6}\right) \cdot\left(1+3^{2}+3^{4}\right) \cdot\left(1+5^{2}+5^{4}\right)=5035485$.
Answer: 5035485.
|
5035485
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. It is known that the polynomial $f(x)=8+32 x-12 x^{2}-4 x^{3}+x^{4}$ has 4 distinct real roots $\left\{x_{1}, x_{2}, x_{3}, x_{4}\right\}$. The polynomial of the form $g(x)=b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}+x^{4}$ has roots $\left\{x_{1}^{2}, x_{2}^{2}, x_{3}^{2}, x_{4}^{2}\right\}$. Find the coefficient $b_{1}$ of the polynomial $g(x)$.
|
Solution: Let the coefficients of the given polynomial (except the leading one) be denoted by $a_{0}, a_{1}, a_{2}, a_{3}$: $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}$.
Then, according to the problem statement, we have:
$f(\quad)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)$.
Together with the polynomial $f(x)$, consider the polynomial $h(x)$, which has roots $\left\{-x_{1},-x_{2},-x_{3},-x_{4}\right\}$:
$$
h(x)=\left(x+x_{1}\right)\left(x+x_{2}\right)\left(x+x_{3}\right)\left(x+x_{4}\right)=a_{0}-a_{1} x+a_{2} x^{2}-a_{3} x^{3}+x^{4}
$$
Consider the polynomial $G($ 可 $)=f(x) h(x)$:
$$
\begin{aligned}
& G(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)\left(x+x_{1}\right)\left(x+x_{2}\right)\left(x+x_{3}\right)\left(x+x_{4}\right)= \\
&=\left(x^{2}-x_{1}{ }^{2}\right)\left(x^{2}-x_{2}{ }^{2}\right)\left(x^{2}-x_{3}{ }^{2}\right)\left(x^{2}-x_{4}{ }^{2}\right) .
\end{aligned}
$$
By substituting the variable $y=x^{2}$, we obtain the required polynomial $g(y)$, since
$$
g(y)=\left(y-x_{1}{ }^{2}\right)\left(y-x_{2}{ }^{2}\right)\left(y-x_{3}{ }^{2}\right)\left(y-x_{4}{ }^{2}\right)
$$
In our case:
$$
\begin{gathered}
f(x)=8+32 x-12 x^{2}-4 x^{3}+x^{4} \\
h(x)=8-32 x-12 x^{2}+4 x^{3}+x^{4} \\
f(x)=f(x) h(x)=64-1216 x^{2}+416 x^{4}-40 x^{6}+x^{8} \\
g(y)=64-1216 y+416 y^{2}-40 y^{3}+y^{4}
\end{gathered}
$$
Answer: -1216
|
-1216
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find the number of matrices that satisfy two conditions:
3) the matrix has the form $\left(\begin{array}{lll}1 & * & * \\ * & 1 & * \\ * & * & 1\end{array}\right)$, where each * can take the value 0 or 1 and the rows of the matrix do not repeat.
|
Solution: Let $A$ be the set of matrices satisfying condition 1) and $B$ be the subset of $A$ consisting of matrices satisfying condition 2). We need to find the number of elements in the set $B$. Let $A_{ij}, i, j \in\{1,2,3\}, i \neq j$, be the subset of $A$ consisting of matrices in which row $i$ and $j$ are the same. Then $B=A \backslash\left(A_{12} \cup A_{23} \cup A_{13}\right)$ and $|B|=|A|-\left|A_{12} \cup A_{23} \cup A_{13}\right|$. The cardinality $\left|A_{12} \cup A_{23} \cup A_{13}\right|$ is conveniently calculated using the inclusion-exclusion principle:
$\left|A_{12} \cup A_{23} \cup A_{13}\right|=\left|A_{12}\right|+\left|A_{23}\right|+\left|A_{13}\right|-\left|A_{12} \cap A_{23}\right|-\left|A_{13} \cap A_{23}\right|-\left|A_{12} \cap A_{13}\right|+\left|A_{12} \cap A_{23} \cap A_{13}\right|$.
It is easy to calculate the cardinalities of the sets involved in this expression:
$$
\left|A_{12}\right|=\left|A_{23}\right|=\left|A_{13}\right|=2^{3},\left|A_{12} \cap A_{23}\right|=\left|A_{13} \cap A_{23}\right|=\left|A_{12} \cap A_{13}\right|=\left|A_{12} \cap A_{23} \cap A_{13}\right|=1
$$
We get $|B|=2^{6}-3 \cdot 2^{3}+3-1=42$.
|
42
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find the value of $f(2019)$, given that $f(x)$ simultaneously satisfies the following three conditions:
1) $f(x)>0$ for any $x>0$
2) $f(1)=1$;
3) $f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}$ for any $a, b \in \mathbb{R}$.
|
Solution. In the identity given in the problem
$$
f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}
$$
let $a=1, b=0$. Then $f(1) \cdot(f(1)+f(0))=2 f(1) \cdot f(0)+1$. Since $f(1)=1$, we find
$$
f(0)=0
$$
Next, by setting $b=-a$ in (1), we get, taking (2) into account, that
$$
f(a) \cdot f(-a)=-a^{2}
$$
Finally, when $b=0$, identity (1) (taking (2) into account) becomes $f(a) \cdot f(a)=a^{2}$. Therefore, it is necessary that $f(a)=a$ for $a>0$, since by condition $f(x)>0$ for $x>0$. Furthermore, according to (3), $f(a)=a$ for $a<0$ as well. Ultimately, $f(x)=x$ for any $x \in \mathbb{R}$. It is easy to verify that such an $f(x)$ indeed satisfies the requirements 1$), 2), 3$) from the problem statement. Thus, $f(x)=x$.
Answer: 2019.
|
2019
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A robot is located in one of the cells of an infinite grid paper, to which the following commands can be given:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent cell to the right).
For example, if the robot executes a sequence of four commands (up, right, down, left), it will obviously return to the initial position, i.e., it will end up in the same cell from which it started. How many different sequences of 8 commands exist that return the robot to the initial position?
|
Solution. For brevity, let's denote the command to move left as L, right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be the number of L commands in the sequence. We will count the number $N_{k}$ of the desired sequences for $k$ from 0 to 4.
- $\boldsymbol{k}=\mathbf{0}$. The sequence consists only of U and D commands. Since their numbers are equal, on 4 out of 8 positions there should be a U, and on the remaining positions, a D. The number of ways to choose 4 positions out of 8 is $C_{8}^{4}$. Therefore, $N_{0}=C_{8}^{4}=70$;
- $\boldsymbol{k}=\mathbf{1}$. The sequence consists of one L command, one R command, and three U and three D commands. The number of ways to place the two commands L and R on 8 positions is $C_{8}^{2} \cdot C_{2}^{1}$: $C_{8}^{2}$ is the number of ways to choose 2 positions out of 8, and $C_{2}^{1}=2$ is the number of ways to place the commands L and R on these two positions. On the remaining 6 positions, 3 U commands can be placed in $C_{6}^{3}$ ways. Therefore, $N_{1}=C_{8}^{2} \cdot C_{2}^{1} \cdot C_{6}^{3}=1120$;
- $\boldsymbol{k}=\mathbf{2}$. Here, there are two L commands, two R commands, and two U and two D commands. For L and R, there are $C_{8}^{4} \cdot C_{4}^{2}$ ways to place them. On the remaining 4 positions, 2 U commands can be placed in $C_{4}^{2}$ ways. Thus, $N_{2}=C_{8}^{4} \cdot C_{4}^{2} \cdot C_{4}^{2}=2520$.
By reasoning similarly, it can be shown that $N_{3}=N_{1}$ and $N_{4}=N_{0}$. Therefore, the desired number of sequences is $2 \cdot\left(N_{0}+N_{1}\right)+N_{2}=4900$.
Answer: 4900.
|
4900
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A robot is located in one of the cells of an infinite grid paper, to which the following commands can be given:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent cell to the right).
For example, if the robot executes a sequence of four commands (up, right, down, left), it will obviously return to the initial position, i.e., it will end up in the same cell from which it started. How many different sequences of 8 commands exist that return the robot to the initial position?
|
Solution. For brevity, let's denote the command to move left as L, right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be the number of L commands in the sequence. We will count the number $N_{k}$ of the desired sequences for $k$ from 0 to 4.
- $\boldsymbol{k}=\mathbf{0}$. The sequence consists only of U and D commands. Since they are equal in number, 4 out of 8 positions should be U, and the rest should be D. The number of ways to choose 4 positions out of 8 is $C_{8}^{4}$. Therefore, $N_{0}=C_{8}^{4}=70$;
## Interregional School Olympiad in Mathematics - $\boldsymbol{k}=\mathbf{1}$. The sequence consists of one L command, one R command, and three U and three D commands. The number of ways to place two L and R commands on 8 positions is $C_{8}^{2} \cdot C_{2}^{1}$: $C_{8}^{2}$ is the number of ways to choose 2 positions out of 8, and $C_{2}^{1}=2$ is the number of ways to place the L and R commands on these two positions. The number of ways to place 3 U commands on the remaining 6 positions is $C_{6}^{3}$. Therefore, $N_{1}=C_{8}^{2} \cdot C_{2}^{1} \cdot C_{6}^{3}=1120$;
- $\boldsymbol{k}=\mathbf{2}$. Here, there are two L commands, two R commands, and two U and two D commands. The number of ways to place L and R commands is $C_{8}^{4} \cdot C_{4}^{2}$. The number of ways to place 2 U commands on the remaining 4 positions is $C_{4}^{2}$. Therefore, $N_{2}=$ $C_{8}^{4} \cdot C_{4}^{2} \cdot C_{4}^{2}=2520$.
By reasoning similarly, it can be shown that $N_{3}=N_{1}$ and $N_{4}=N_{0}$. Thus, the desired number of sequences is $2 \cdot\left(N_{0}+N_{1}\right)+N_{2}=4900$.
Answer: 4900.
|
4900
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Find all prime numbers whose decimal representation has the form 101010 ... 101 (ones and zeros alternate).
|
Solution: Let $2n+1$ be the number of digits in the number $A=101010$...101 under investigation. Let $q=$ 10 be the base of the numeral system. Then $A=q^{0}+q^{2}+\cdots q^{2n}=\frac{q^{2n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2k \Rightarrow A=\frac{q^{2n+2}-1}{q^{2}-1}=\frac{q^{2k+1}-1}{q-1} \cdot \frac{q^{2k+1}+1}{q+1}$. Thus, the number $A$ is represented as the product of two integer factors (by the theorem of Bezout, the polynomial $q^{2k+1} \pm 1$ is divisible without remainder by the polynomial $q \pm 1$), each of which is different from 1. Therefore, when $n$ is even, the number $A$ is not prime.
- $n=2k-1 \Rightarrow A=\frac{q^{2n+2}-1}{q^{2}-1}=\frac{q^{2k}-1}{q^{2}-1} \cdot (q^{2k}+1)$. For $k>1$, both factors are integers and different from 1; thus, the number $A$ is composite. It remains to verify that for $k=1$, the resulting number $A=q^{0}+q^{2}=101$ is prime.
Answer: 101.
|
101
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Anya and Borya are playing "battleship" with the following rules: 29 different points are chosen on a circle, numbered clockwise with natural numbers from 1 to 29. Anya draws a ship - an arbitrary triangle with vertices at these points. We will call a "shot" the selection of two different natural numbers $k$ and $m$ from 1 to 29. If the segment with endpoints at points numbered $k$ and $m$ has at least one common point with Anya's triangle, the ship is considered "wounded." Borya fires a "salvo" - several shots simultaneously. Anya drew a ship and showed it to Borya. And then they noticed that any "salvo" of K different shots will definitely wound Anya's ship. Indicate some placement of Anya's ship for which the value of $K$ will be minimal.
Interregional School Olympiad in Mathematics for Students of Departmental Educational Organizations
|
Solution. The vertices of Anya's triangle divide the circle into three arcs. Let $x, y$ and $26-x-y$ be the number of points on these arcs (see figure), excluding the vertices of the triangle itself. For a "shot" with endpoints at points $k$ and $m$ to not hit the ship, both these points must lie on one of the arcs. Clearly, the number of ways to choose two different points on an arc containing $x$ points is $C_{x}^{2}$. The same applies to the other arcs. Therefore, the number $N$ of "safe" shots is the sum $N=C_{x}^{2}+C_{y}^{2}+C_{26-x-y}^{2}$. Then the next shot will definitely "hit" the ship, so $K=N+1$.
Thus, we need to find such non-negative integers $x, y$ that satisfy the condition $x+y \leq 26$, for which the value of $N$ is minimal. Let's write the expression for $N$ as follows:
$$
N=\frac{x(x-1)}{2}+\frac{y(y-1)}{2}+\frac{(26-x-y)(25-x-y)}{2}
$$
Expanding the brackets and combining like terms, we get
$$
N=x^{2}-x(26-y)+y^{2}-26 y+325
$$
For each fixed $y$ from 0 to 26, we will find such a value of $x$ that satisfies the inequality
$$
0 \leq x \leq 26-y
$$
for which the value of $N$ is minimal. If $y$ is fixed, then the right-hand side (1) takes its minimum value at
$$
x=\frac{26-y}{2}
$$
(the vertex of the parabola, belonging to the interval (2)).
This minimum value is $\frac{3}{4} y^{2}-13 y+156$. It, in turn, is minimal at $y=\frac{26}{3} \approx 8.6$. From (3), we then find $x=\frac{26}{3}$. Among the points with integer coordinates $(8,8),(8,9),(9,8),(9,9)$ - the nearest integer neighbors of the point of minimum $\left(\frac{26}{3}, \frac{26}{3}\right)$ - we choose the one for which the value of $N$ is the smallest. These are the points $(8,9),(9,8),(9,9)$. For them, $N=100$.
Answer: The ship should be placed such that on the three arcs into which the vertices of the ship divide the circle, there are 8, 9, and 9 points (excluding the vertices of the ship itself).
|
100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. It is known that there exists a natural number $N$ such that
$$
(\sqrt{3}-1)^{N}=4817152-2781184 \cdot \sqrt{3}
$$
Find $N$.
|
Solution. Suppose that raising the number $a+b \sqrt{3}$ to the power $N$ results in the number $A+B \sqrt{3}$ (here $a, b, A, B$ are integers). Expanding the expression $(a+b \sqrt{3})^{N}$, we get a sum of monomials (with integer coefficients that are not important to us now) of the form $a^{N-n}(b \sqrt{3})^{n}$. The coefficient $B$ will be contributed by those monomials where the exponent $n$ is odd. Therefore, if $(a+b \sqrt{3})^{N}=A+B \sqrt{3}$, then $(a-b \sqrt{3})^{N}=A-B \sqrt{3}$. Multiplying the equations $(\sqrt{3}-1)^{N}=4817152-2781184 \cdot \sqrt{3}$ and $(-\sqrt{3}-1)^{N}=4817152+2781184 \cdot \sqrt{3}$, we get $(-2)^{N}=4817152^{2}-3 \cdot 2781184^{2}$. The exponent $N$ can be found by successively dividing both sides by 2 (for example, each term on the right can be immediately divided by 256).
Answer: $N=16$.
|
16
|
Number Theory
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math-word-problem
|
Yes
|
Yes
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olympiads
| false
|
2. (2 points) It is known about the numbers $x_{1}$ and $x_{2}$ that $x_{1}+x_{2}=2 \sqrt{1703}$ and $\left|x_{1}-x_{2}\right|=90$. Find $x_{1} \cdot x_{2}$.
|
Solution: Let $A=x_{1}+x_{2}, B=x_{1} \cdot x_{2}, C=\left|x_{1}-x_{2}\right|$. Since $C^{2}=A^{2}-4 \cdot B$, we find $B=-322$.
Answer: -322 .
|
-322
|
Algebra
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math-word-problem
|
Yes
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Yes
|
olympiads
| false
|
3. (3 points) Find the number of integers from 1 to 1000 inclusive that give the same remainder when divided by 11 and by 12.
|
Solution. Let $r_{n}(a)$ be the remainder of the division of the number $a$ by the number $n$. Suppose $a \in [1 ; 1000]$ and $r_{11}(a)=r_{12}(a)=t$. Then $t \in \{0, \ldots, 10\}$ and the following equality holds:
$$
t+11 k=t+12 m=a, \quad k, m \in N_{0}
$$
From the last equality, it follows that $k$ is divisible by 12, and $m$ is divisible by 11. Therefore,
$$
t+11 \cdot 12 \cdot s=t+132 \cdot s=a, \quad s \in N_{0}
$$
It remains to take into account the condition $a=t+132 \cdot s \leq 1000, \quad t \in \{0, \ldots, 10\}$:
$$
\begin{aligned}
& t=0 \Rightarrow 132 s \leq 1000 \Rightarrow s \leq 7.6 \Rightarrow 7 \text{ numbers } \quad(s \neq 0) \\
& t=1 \Rightarrow 1+132 s \leq 1000 \Rightarrow s \leq 7.57 \Rightarrow 8 \text{ numbers } \\
& t=2 \Rightarrow 2+132 s \leq 1000 \Rightarrow s \leq 7.56 \Rightarrow 8 \text{ numbers } \\
& t=3 \Rightarrow 3+132 s \leq 1000 \Rightarrow s \leq 7.55 \Rightarrow 8 \text{ numbers } \\
& t=4 \Rightarrow 4+132 s \leq 1000 \Rightarrow s \leq 7.55 \Rightarrow 8 \text{ numbers } \\
& t=5 \Rightarrow 5+132 s \leq 1000 \Rightarrow s \leq 7.54 \Rightarrow 8 \text{ numbers } \\
& t=6 \Rightarrow 6+132 s \leq 1000 \Rightarrow s \leq 7.54 \Rightarrow 8 \text{ numbers } \\
& t=7 \Rightarrow 7+132 s \leq 1000 \Rightarrow s \leq 7.53 \Rightarrow 8 \text{ numbers } \\
& t=8 \Rightarrow 8+132 s \leq 1000 \Rightarrow s \leq 7.52 \Rightarrow 8 \text{ numbers } \\
& t=9 \Rightarrow 9+132 s \leq 1000 \Rightarrow s \leq 7.51 \Rightarrow 8 \text{ numbers } \\
& t=10 \Rightarrow 10+132 s \leq 1000 \Rightarrow s \leq 7.5 \Rightarrow 8 \text{ numbers }
\end{aligned}
$$
In total, we get 87 numbers. (obviously, the last enumeration could have been shortened)
Answer: 87 numbers.
P.S. An alternative approach consists in enumerating $s$, where $s \leq \frac{1000}{132} \approx 7.57$, and finding those $t \in \{0, \ldots, 10\}$ for which the inequality $a=t+132 \cdot s \leq 1000$ holds.
|
87
|
Number Theory
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math-word-problem
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Yes
|
Yes
|
olympiads
| false
|
1. Find the sum of all natural numbers $n$ that are multiples of three and for which the number of divisors (including 1 and $n$ itself) is equal to $\frac{n}{3}$. (For example, the number 12 has 6 divisors: $1,2,3,4,6,12$.)
|
Solution. Let the canonical decomposition of the number $n$ be: $n=2^{t_{1}} \cdot 3^{t_{2}} \cdot 5^{t_{3}} \cdots \cdot p^{t_{k}}$. Then the number of divisors of the number $n$ is $\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right) \cdots\left(t_{k}+1\right)$. From the condition of the problem, we have the equality:
$\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right) \cdots\left(t_{k}+1\right)=2^{t_{1}} \cdot 3^{t_{2}-1} \cdot 5^{t_{3}} \cdots \cdots \cdot p^{t_{k}} \cdot\left({ }^{*}\right)$
Note that $2^{t_{1}} 3^{t_{2}-1}>\left(t_{1}+1\right)\left(t_{2}+1\right)$ for $t_{1} \geq 4$ and $t_{2} \geq 3, \quad \ldots, \quad p^{t_{k}}>t_{k}+1$ for $t_{k} \geq 1$. Therefore, $t_{1}$ can take the values $0,1,2$ or 3 and $t_{2}$ can take the values 1 or 2. Substituting the specified values into the equality (*), we find that $n=9, n=18$ or $n=24$.
Answer: 51.
|
51
|
Number Theory
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math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (5 points) Numbers are marked on a chessboard (see Fig. 1). How many arrangements of 8 rooks are there such that no rook attacks another and the numbers on the squares occupied by the rooks include all numbers from 0 to 7?
\(\left(\begin{array}{llllllll}0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0\end{array}\right)\)
Fig. 1.
Let's number the rows of the board from 1 to 8 from top to bottom. On the first row, the position of the rook can be chosen in 8 ways. For example, we choose the first cell with the number 0. On the second row, there are 7 options left. Six of them (cells from the second to the seventh) have the property that under them in the lower half of the board are numbers different from the number occupied by the first rook (in our case, 0), and one cell (the last one, with the number 7) is such that under it in the lower half stands the number occupied by the first rook (in our case, 0).
Consider the first group of 6 options and make an arbitrary choice (for example, the second cell with the number 1). After such a choice, it turns out that in the lower half of the board, positions with two different numbers are prohibited (in our case, 7 and 6). Therefore, positions with these numbers need to be chosen in the upper half of the board, and there are exactly 2 such options (in our case, the 7th and 8th cells in rows 3 and 4). After this choice, in the lower half of the board, there will be 4 columns with different numbers in the columns. That is, in the lower half, there are \(4! = 24\) options. Thus, we get \(8 \times 6 \times 2 \times 24\) options.
Consider the option of choosing a position on the second row when under the chosen cell in the lower half of the board stands the number chosen on the first row (in our case, the last cell with the number 7). After such a choice, in the third row, there are 6 options leading to a single option in the fourth row. After choosing options in the upper half of the board, in the lower half, there are again 4 free columns with four different numbers, i.e., 24 options. Thus, we get \(8 \times 1 \times 6 \times 24\) options.
In total, we get:
\(8 \times 6 \times 2 \times 24 + 8 \times 1 \times 6 \times 24 = 3456\).
|
Answer: 3456.
## Variant 2
|
3456
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Combinatorics
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math-word-problem
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Yes
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Yes
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olympiads
| false
|
2. [6] On an island, there live knights, liars, and yes-men; each knows who everyone else is. All 2018 residents were lined up and asked to answer "Yes" or "No" to the question: "Are there more knights than liars on the island?". The residents answered in turn, and everyone could hear their answers. Knights answered truthfully, liars lied. Each yes-man answered the same way as the majority of those who answered before him, and if the number of "Yes" and "No" answers was equal, he gave either of these answers. It turned out that there were exactly 1009 "Yes" answers. What is the maximum number of yes-men that could be among the residents?
|
Answer: 1009 yes-men. Solution: Let's call knights and liars principled people.
Estimate. First method. (Buchaev Abdulqadyr) We will track the balance - the difference between the number of "Yes" and "No" answers. At the beginning and at the end, the balance is zero, and with each answer, it changes by 1. Zero values of the balance divide the row of residents into groups. Within each group, the balance retains its sign. Yes-men always increase the absolute value of the balance. Therefore, to make the balance zero, there must be no fewer principled people than yes-men. This is true for each group, and therefore for all the island's residents.
Second method. A yes-man could not increase the minimum of the current number of "Yes" and "No" answers. Since this minimum increased from 0 to 1009, and with each answer it changed by no more than one unit, there must be at least 1009 principled people.
Example. First, all 1009 yes-men said "No," and then 1009 knights said "Yes."
Note. Any sequence of 1009 principled people can be thinned out by yes-men so that the condition is met.
|
1009
|
Logic and Puzzles
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math-word-problem
|
Yes
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Yes
|
olympiads
| false
|
4. [8] A $7 \times 7$ board is either empty or has an invisible $2 \times 2$ ship placed on it "by cells." It is allowed to place detectors in some cells of the board and then turn them on simultaneously. An activated detector signals if its cell is occupied by the ship. What is the smallest number of detectors needed to guarantee that, based on their signals, it can be determined whether there is a ship on the board, and if so, which cells it occupies?
(R. Zhinodarov)
|
Answer: 16 detectors. Solution: Evaluation. In each $2 \times 3$ rectangle, there should be at least two detectors: the rectangle consists of three $1 \times 2$ dominoes, and if a detector is in the outer domino, we cannot determine whether there is a ship on the other two dominoes, and if a detector is in the middle domino, we cannot determine which of the outer dominoes the ship occupies along with the middle domino. Therefore, there should be no fewer than 16 detectors (see the left image).
Example. On the right image, the black squares indicate the placement of 16 detectors. Any ship intersects with exactly one black $2 \times 2$ square in one cell, two adjacent cells, or all four cells. In any case, the position of the ship or its absence is uniquely determined.
Note. 16 detectors can be placed in only three fundamentally different ways. One is shown in the solution, and the other two are shown in the image below.
|
16
|
Logic and Puzzles
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math-word-problem
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Yes
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Yes
|
olympiads
| false
|
5. [5] In each cell of a strip of length 100, there is a chip. You can swap any two adjacent chips for 1 ruble, and you can also swap any two chips that have exactly 4 chips between them for free. What is the minimum number of rubles needed to rearrange the chips in reverse order?
(Egor Bakaev)
|
Answer: For 61 rubles. Solution: Let's number the chips and cells in order from 0 to 99. The free operation does not change the remainder of the cell number when divided by 5.
Estimate. Let's mentally arrange the piles of chips in a circle. First, the pile of chips with a remainder of 0, then with 1, and so on up to 4. The paid operation swaps a pair of chips from adjacent piles. Chips from the zero pile must participate in at least one such swap to reach the fourth pile. Similarly for chips from the fourth pile. Chips from the first pile must participate in at least two swaps to reach the third pile. Similarly for the third pile. Therefore, at least $(20+20+40+40): 2=60$ rubles are required. But if only 60 rubles are spent, the chips from the first pile will have to go through the second pile, so at least one chip from the second pile will participate in swaps. Therefore, more than 60 rubles are necessary.
Algorithm. It is clear that the free operations can arrange the chips within a pile in any order. Therefore, the correct arrangement of all chips from the zero and fourth piles can be done for 20 rubles. Consider the remaining three piles. Let's mentally leave only one chip $A$ in the second pile. We will swap it with a chip from the first pile. Each time we will move the chip that has come to the second pile further using a new chip. Then for 40 rubles, we will move all the chips from the first pile to the third, and from the third to the first except one: it will remain in the second pile, not reaching the first. We will swap it with $A$, and all the chips will be in the correct piles.
|
61
|
Combinatorics
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math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given a triangular pyramid $S A B C$, the base of which is an equilateral triangle $A B C$, and all plane angles at vertex $S$ are equal to $\alpha$. For what least $\alpha$ can we assert that this pyramid is regular?
M. Malkin
|
Answer: $60^{\circ}$.
We will prove that when $\alpha=60^{\circ}$, the pyramid is regular. Let the sides of the triangle $ABC$ be equal to 1. Note that in any triangle with an angle of $60^{\circ}$, the side opposite this angle is the middle-length side (and if it is strictly less than one of the sides, it is strictly greater than the other). Suppose one of the lateral edges of the pyramid is not equal to 1, for example, $SA>1$. Then in the faces $SAB$ and $SAC$, the edges $SB$ and $SC$ will be less than 1, and thus in the face $SBC$, the edge $BC$ will not be the middle side, which is a contradiction.
Now we will show how to construct an irregular pyramid with dihedral angles $\alpha<60^{\circ}$ at the vertex $S$.
First method. First, we remove the lateral edge $SA$, and rotate the remaining lateral face $SBC$ around its base edge $BC$ until this face lies in the plane of the base and contains the base triangle. During this motion, there will be "current" pyramids, in which two equal dihedral angles are initially equal to $\alpha$, then greater than $\alpha$ (at the moment when the vertex projects onto the vertex of the base - since at this moment the sine of the angle at the vertex $S$ in the lateral faces $SAC$ and $SAB$ is $1: SB$, and in the face $SBC$ it is the ratio of the lateral height of this triangle to $SB$, which is less than 1), and at the end, in the "degenerate" pyramid, they are equal to $\alpha / 2$. Therefore, by continuity, there will be $\alpha$ again.
Second method. Consider the triangle $SAB$ with $SA=SB$ and $\angle S=\alpha$. Since $AB<SB$, there exists a point $C$ on the side $SA$ such that $BC=AB$. Now take a trihedral angle in which all dihedral angles are equal to $\alpha$, and mark segments $SA, SB, SC$ on its edges. The pyramid $SABC$ is the desired one.
Third method. (Sergey Komarov, 11th grade, MSU SUNT) Let $S'ABC$ be a regular pyramid with a dihedral angle $\alpha$ at the vertex $S'$. Let $X, Y$ be points such that $XY=AB$, and construct triangles $XYZ$ and $XYT$ on $XY$ such that $\angle ZXY=\angle ZYX=90^{\circ}-\frac{\alpha}{2}$, $\angle TXY=90^{\circ}+\frac{\alpha}{2}$, $\angle TXY=90^{\circ}-\frac{3\alpha}{2}$, then it is clear that $\angle XTY=\angle XZY=\alpha$. By the Law of Sines for these triangles, we have
$$
\frac{TY}{\sin \left(90^{\circ}+\frac{\alpha}{2}\right)}=\frac{XY}{\sin \alpha}=\frac{ZX}{\sin \left(90^{\circ}-\frac{\alpha}{2}\right)}=\frac{ZY}{\sin \left(90^{\circ}-\frac{\alpha}{2}\right)}
$$
which implies $TY=ZX=ZY$, since the sines of adjacent angles are equal.
Now let $S$ be a point on the extension of the segment $S'B$ beyond point $B$ such that $SB=TX$. Then $\angle SBC=\angle SBA=90^{\circ}+\frac{\alpha}{2}=\angle TXY, BC=BA=XY$, and $SB=SB=TX$, so $\triangle SBC=\triangle SBA=\triangle TXY$, from which $SC=SA=TY=ZX=ZY$, and thus $\triangle SCA=\triangle ZXY$ (by three sides). From all the indicated triangle equalities, it follows that the pyramid $SABC$ has all dihedral angles at the vertex $S$ equal to $\alpha$, so it satisfies the conditions of the problem. But it is not regular, since the line $S'B$ is not perpendicular to $ABC$, and on it only one point projects to the center of the triangle $ABC$, which is the point $S'$, and not the point $S$.
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. [4] In each cell of a strip of length 100, there is a chip. You can swap any two adjacent chips for 1 ruble, and you can also swap any two chips that have exactly three chips between them for free. What is the minimum number of rubles required to rearrange the chips in reverse order?
(Egor Bakaev)
|
Answer. 50 rubles.
Solution. Evaluation. Each chip must change the parity of its number. A free operation does not change the parity, while a paid operation changes the parity of two chips. Therefore, at least 50 rubles will be required.
Example. Let's number the chips in order from 0 to 99. We will color the cells in four colors: $a b c d a b c d . . . d$. A free operation changes the chips in adjacent cells of the same color. Therefore, in cells of the same color, the chips can be freely rearranged in any order. We will swap the chips in all pairs $b c$ and $d a$ - this is 49 paid operations. In cells of color $b$ and $c$, the chips can already be arranged as needed for free. In cells of color $a$ and $d$, we will arrange the chips so that chips 0 and 99 are next to each other. We will swap them with the last paid operation and then arrange all the chips in the required order.
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. [9] Inside an isosceles triangle $A B C$, a point $K$ is marked such that $C K=A B=B C$ and $\angle K A C=30^{\circ}$. Find the angle $A K B$.
E. Bakayev
|
Answer: $150^{\circ}$. Solution 1. Construct an equilateral triangle $B C L$ (see the figure; points $A$ and $L$ are on the same side of the line $B C$). Points $A, C$, and $L$ lie on a circle with radius $B A$ and center at point $B$. Since $K$ lies inside triangle $A B C$, angle $A B C$ is greater than $60^{\circ}$, hence $L$ and $B$ lie on opposite sides of $A C$ and $L$ lies on the smaller arc $A C$. Therefore, the inscribed angle $C A L$ is half of the central angle $C B L$, which is $30^{\circ}$.
Obviously, the point $K$ that satisfies the conditions of the problem is unique, so it coincides with the point symmetric to $L$ with respect to side $A C$. Thus, triangle $A K L$ is equilateral, and point $K$, like point $B$, lies on the perpendicular bisector of segment $A L$, from which $\angle A K B = \angle L K B = 180^{\circ} - 1 / 2 \angle A K L = 150^{\circ}$.
Solution 2. Let the height $B M$ of triangle $A B C$ intersect line $A K$ at point $O$ (see the figure). Then $\angle C O M = \angle A O M = 60^{\circ}$. Therefore, $\angle A O C = 120^{\circ}$ and $\angle C O B = 120^{\circ}$. Consequently, triangles $B O C$ and $K O C$ are congruent by two sides and the angle opposite the larger side (the so-called fourth criterion of triangle congruence). Therefore, $O B = O K$, meaning triangle $B O K$ is isosceles with an angle of $120^{\circ}$ at vertex $O$. Therefore, $\angle O K B = 30^{\circ}$, and $\angle A K B = 150^{\circ}$.
Solution 3. Construct an equilateral triangle $A C L$ on $A C$ such that points $L$ and $B$ lie on the same side of $A C$ (see the figure).
Draw the height $B M$ in triangle $A B C$, which is also the perpendicular bisector of side $A C$. Since $A L C$ is equilateral, point $L$ also lies on line $B M$. Additionally, draw the height $A N$ in triangle $A L C$. Since $A N$ is the angle bisector of $\angle L A C$, point $K$ lies on this line. Note also that $K$ lies on the same side of $B M$ as $A$, because due to $C K = C B$, it cannot lie inside triangle $B M C$; thus, $K$ lies on segment $A N$.
Notice that right triangles $B M C$ and $K N C$ are congruent by a leg and the hypotenuse (since $M C = A C / 2 = L C / 2 = N C$ and $B C = K C$). Therefore, first, $B M = K N$, and second, $B$ lies on segment $L M$ (since $B M = K N 2 \beta > \beta$, then $\angle A K C = 180^{\circ} - \beta$. Thus, $\angle A C K = \beta - 30^{\circ}$, from which $\angle K C B = \angle C - \angle A C K = (90^{\circ} - \beta) - (\beta - 30^{\circ}) = 120^{\circ} - 2 \beta$, $\angle B K C = 1 / 2 (180^{\circ} - \angle K C B) = 30^{\circ} + \beta$, $\angle A K B = 360^{\circ} - \angle B K C - \angle A K C = 360^{\circ} - (30^{\circ} + \beta) - (180^{\circ} - \beta) = 150^{\circ}$.
|
150
|
Geometry
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math-word-problem
|
Yes
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Yes
|
olympiads
| false
|
7. [12] There are 100 piles, each with 400 stones. In one move, Petya selects two piles, removes one stone from each, and earns as many points as the absolute difference in the number of stones in these two piles. Petya must remove all the stones. What is the maximum total number of points he can earn?
M. Diden
|
Answer: 3920000. Solution. Estimation. We will assume that the stones in the piles are stacked on top of each other, and Petya takes the top (at the moment) stones from the selected piles. We will number the stones in each pile from bottom to top with numbers from 1 to 400. Then the number of points Petya gets on each move is equal to the difference of the numbers of the removed stones. As a result, he will get a sum of the form $\left|a_{1}-a_{2}\right|+\left|a_{3}-a_{4}\right|+\ldots+\mid a_{39999}-a_{40000}$, where $a_{i}$ are the numbers of the corresponding stones.
Notice that after expanding the brackets, we get an algebraic sum $S$ of one hundred numbers 400, one hundred numbers 399, ..., one hundred twos, and one hundred ones, with a minus sign in front of exactly half of these numbers.
We will call the numbers from 1 to 200 small, and the rest - large. If it were allowed to take any stones from the piles, then the maximum value of $S$ is obviously achieved when all large numbers enter $S$ with a plus sign, and all small numbers - with a minus sign. Such a sum is equal to $100(400+399+\ldots+201-200-199-\ldots-1)=100((400-200)+(399-199)+\ldots+(201-1))=100 \cdot 200^{2}$.
However, note that each large number will enter the sum with a minus sign at least once: this will happen, for example, the first time Petya removes a stone with this number. Similarly, each of the 200 small numbers will enter the sum with a plus sign at least once (the moment Petya removes the last stone with this number). Therefore, the maximum result Petya can achieve does not exceed $99 \cdot(400+399+\ldots+201)-99 \cdot(200+199+\ldots+1)-(400+399+\ldots+201)+(200+199+\ldots+1)=98 \cdot 200^{2}$.
Example. The specified result can be achieved, for example, as follows. In the first 200 moves, Petya takes 200 stones from the first two piles (at this point, 200 large numbers each get a minus sign once). In the next 200 moves, he removes 200 top stones from the third pile and 200 bottom stones from the first pile, then 200 stones from the second and fourth piles, the third and sixth piles, ..., the 98th and 100th piles (at this point, all numbers enter with the "correct" signs). Finally, there are 200 bottom stones left in the last two piles, which are removed in the last 200 moves (and 200 plus signs appear before the numbers from 200 to 1).
|
3920000
|
Combinatorics
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math-word-problem
|
Yes
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Yes
|
olympiads
| false
|
5. [5] Several natural numbers are written in a row with a sum of 2019. No number and no sum of several consecutive numbers equals 40. What is the maximum number of numbers that could have been written?
A. Shapovalov
|
Answer: 1019 numbers.
Lemma. The sum of any forty consecutive numbers is not less than 80.
Proof. Let the numbers $a_{1}, \ldots, a_{40}$ be written consecutively. Among the numbers $b_{0}=0, b_{1}=a_{1}$, $b_{2}=a_{1}+a_{2}, \ldots, b_{40}=a_{1}+a_{2}+\ldots+a_{40}$, there will be two $b_{i}$ and $b_{j}(i1019$ numbers. By the lemma, the sum of the first $1000=25 \cdot 40$ of them is not less than $25 \cdot 80=2000$. The sum of the remaining numbers (at least 20) is not less than 20. Therefore, the total sum is not less than 2020. Contradiction.
Example. 25 groups 1, $, ., 1,41$ (in each group 39 ones and the number 41) and then 19 ones.
|
1019
|
Number Theory
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math-word-problem
|
Yes
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Yes
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olympiads
| false
|
4. There is a 1001-digit natural number $A$. The 1001-digit number $Z$ is the same number $A$, written from end to beginning (for example, for four-digit numbers, these could be 7432 and 2347). It is known that $A > Z$. For which $A$ will the quotient $A / Z$ be the smallest (but strictly greater than 1)?
|
Answer. For $A$, whose notation (from left to right) is: 501 nines, an eight, and 499 nines.
Solution 1. Let $A=\overline{a_{1000} a_{999 \ldots a_{0}}}$. Since $A>Z$, among the digits $a_{0}, a_{1}, \ldots, a_{499}$ there is at least one non-nine. Therefore, $Z \leqslant Z_{0}=\underbrace{99 \ldots 9}_{499} 8 \underbrace{99 \ldots 9}_{501}$.
We will show that $A-Z \geqslant 10^{501}-10^{499}$. From this it will follow that
$$
\frac{A}{Z}-1 \geqslant \frac{10^{501}-10^{499}}{Z_{0}}
$$
this estimate is achieved when $Z=Z_{0}$, which gives the answer.
We have
$$
\begin{aligned}
A-Z & =\left(a_{1000}-a_{0}\right)\left(10^{1000}-1\right)+\left(a_{999}-a_{1}\right)\left(10^{999}-10\right)+\cdots+\left(a_{501}-a_{499}\right)\left(10^{501}-10^{499}\right)= \\
& =\varphi_{499} \Delta_{499}+\varphi_{498} \Delta_{498}+\cdots+\varphi_{0} \Delta_{0}
\end{aligned}
$$
where $\varphi_{i}=a_{501+i}-a_{499-i}$ and $\Delta_{i}=10^{501+i}-10^{499-i}$ for $i=0,1, \ldots, 499$. Note that $\Delta_{i+1}>10 \Delta_{i}$.
Let $j$ be the largest index for which $\varphi_{j} \neq 0$. Then
$$
\begin{aligned}
\left|\varphi_{j} \Delta_{j}+\varphi_{j-1} \Delta_{j-1}+\cdots+\varphi_{0} \Delta_{0}\right| & \geqslant\left|\varphi_{j} \Delta_{j}\right|-\left|\varphi_{j-1} \Delta_{j-1}\right|-\cdots-\left|\varphi_{0} \Delta_{0}\right| \geqslant \\
& \geqslant \Delta_{j}\left(1-\frac{9}{10}-\frac{9}{100}-\cdots-\frac{9}{10^{j}}\right)=\frac{\Delta_{j}}{10^{j}} \geqslant \Delta_{0}
\end{aligned}
$$
which is what we needed.
Solution 2. It is clear that we can minimize (the positive) number $\frac{A}{Z}-1=\frac{A-Z}{Z}$.
Number the digits in $A$ from left to right as $a_{1}, a_{2}, \ldots, a_{1001}$. Let $k$ be the smallest index for which $a_{k} \neq a_{1002-k}$ (then $k \leqslant 500$ and $a_{k}>a_{1002-k}$, since $A>Z$).
Consider an arbitrary optimal example. Replace the first and last $k-1$ digits with nines. $A-Z$ will not change, and $Z$ will not decrease, so our fraction will not increase. For the same reason, $a_{501}$ can be replaced with 9.
Replace $a_{k}$ with 9 and $a_{1002-k}$ with 8. In this case, $A-Z$ will not increase, and $Z$ will not decrease.
Replace all digits $a_{k+1}, \ldots, a_{500}$ with zeros, and $a_{502}, \ldots, a_{1001-k}$ with nines. Then $A-Z$ will not increase, and $Z$ will not decrease (this will only happen if the second half was already nines!). Since in the optimal example $A-Z<Z$ (simply fewer digits), it is clear that $\frac{A-Z}{Z}$ will not increase.
Thus, we can assume that $A$ has the form
$$
\underbrace{99 \ldots 9}_{k} \underbrace{00 \ldots 0}_{500-k} 9 \underbrace{99 \ldots 9}_{500-k} 8 \underbrace{99 \ldots 9}_{k-1} .
$$
In this case,
$$
A-Z=10^{501}+10^{500}-10^{k}-10^{k-1}
$$
This expression reaches its minimum when $k=500$, and for this same $k$, the maximum value of the considered $Z$ is achieved. Therefore, this is the answer.
|
501
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. [4] In each cell of a strip of length 100, there is a chip. You can swap any two adjacent chips for 1 ruble, and you can also swap any two chips that have exactly three chips between them for free. What is the minimum number of rubles needed to rearrange the chips in reverse order?
(Egor Bakaev)
|
Answer. 50 rubles.
Solution. Evaluation. Each chip must change the parity of its number. A free operation does not change the parity, while a paid operation changes the parity of two chips. Therefore, at least 50 rubles will be required.
Example. Let's number the chips in order from 0 to 99. We will color the cells in four colors: $a b c d a b c d . . . d$. A free operation changes the chips in adjacent cells of the same color. Therefore, in cells of the same color, the chips can be freely rearranged in any order. We will swap the chips in all pairs $b c$ and $d a$ - this is 49 paid operations. In cells of color $b$ and $c$, the chips can already be arranged as needed for free. In cells of color $a$ and $d$, we will arrange the chips so that chips 0 and 99 are next to each other. We will swap them with the last paid operation and then arrange all the chips in the required order.
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. [5] On an island, there live knights, liars, and yes-men; each knows who everyone else is. All 2018 inhabitants were lined up and asked to answer "Yes" or "No" to the question: "Are there more knights than liars on the island?". The inhabitants answered in turn, and everyone could hear their answers. Knights answered truthfully, liars lied. Each yes-man answered the same way as the majority of those who answered before him, and if the number of "Yes" and "No" answers was equal, he gave either of these answers. It turned out that there were exactly 1009 "Yes" answers. What is the maximum number of yes-men that could be among the inhabitants of the island?
(M. Kuznev)
|
Answer: 1009 yes-men. Solution: Let's call knights and liars principled people.
Estimate. First method. (Buchaev Abdulqadir) We will track the balance - the difference between the number of "Yes" and "No" answers. At the beginning and at the end, the balance is zero, and with each answer, it changes by 1. Zero values of the balance divide the row of residents into groups. Within each group, the balance retains its sign. Yes-men always increase the absolute value of the balance. Therefore, to make the balance zero, there must be no fewer principled people than yes-men. This is true for each group, and therefore for all the island's residents.
Second method. A yes-man could not increase the minimum of the current number of "Yes" and "No" answers. Since this minimum increased from 0 to 1009, and with each answer, it changed by no more than one, there must be at least 1009 principled people.
Example. First, all 1009 yes-men said "No," and then 1009 knights said "Yes."
Note. Any sequence of 1009 principled people can be thinned out by yes-men so that the condition is met.
|
1009
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. [5] Pentagon $A B C D E$ is circumscribed around a circle. The angles at its vertices $A, C$ and $E$ are $100^{\circ}$. Find the angle $A C E$.
|
Answer: $40^{\circ}$. It is not hard to understand that such a pentagon exists.
Solution 1. The lines connecting the vertices with the center $O$ of the inscribed circle $\omega$ are the angle bisectors of the pentagon. Therefore, $\angle O A E=\angle O E A=50^{\circ}, \angle A O E=80^{\circ}$. Let $\omega$ touch the sides $B C$ and $A E$ at points $K$ and $M$ respectively. Then $\angle O C K=50^{\circ}$, and the right triangles $O M A, O K C$ and $O M E$ are equal by the leg and the opposite angle. Thus, $O A=O C=O E$, i.e., points $A, C$ and $E$ lie on a circle with center $O$. Therefore, $\angle A C E=\frac{1}{2} \angle A O E=40^{\circ}$. Solution 2. Since the angles $A, C, E$ are equal, all tangents to the circle from these vertices are equal. Since the tangents from vertex $B$ are also equal, triangle $A B C$ is isosceles, and triangle $C D E$ is also (analogously). The sum of angles $B$ and $D$ is $540^{\circ}-3 \cdot 100^{\circ}=240^{\circ}$, so $\angle A C B+\angle E C D=\left(2 \cdot 180^{\circ}-240^{\circ}\right): 2=$ $=60^{\circ}, \mathrm{a} \angle A C E=100^{\circ}-60^{\circ}=40^{\circ}$.
|
40
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.
It is known that three numbers $a_{1}, a_{2}, a_{3}$ were obtained as follows. First, a natural number $A$ was chosen and the numbers $A_{1}=[A]_{16}, A_{2}=$ $[A / 2]_{16}, A_{3}=[A / 4]_{16}$ were found, where $[X]_{16}$ is the remainder of the integer part of the number $X$ divided by 16 (for example, $[53 / 2]_{16}=10$). Then an integer $B$ was chosen such that 0 $\leq \mathrm{B} \leq 15$. The numbers $A_{1}, A_{2}, A_{3}$ and $B$ are written in binary, i.e., each is represented as a string of 0s and 1s of length 4, adding the necessary number of zeros on the left. We agree to add such strings symbol by symbol "in a column" without carrying to the next digit according to the rule: $1+1=0+0=0$ and $0+1=1+0=0$, and denote the operation of symbol-by-symbol addition by the symbol $\diamond$. For example, $3 \diamond 14=(0011) \diamond$ $(1110)=(1101)=13$. Let $a_{1}=A_{1} \diamond B, a_{2}=A_{2} \diamond B, a_{3}=A_{3} \diamond B$. Find the sum of all possible values of the number $a_{3}$, given that $a_{1}=2, a_{2}=$ 9. Choose the correct answer:
#
|
# Answer: 16.
## Conditions and answers to the problems of the final stage of the 2012-13 academic year
#
|
16
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
5. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The arrangement of digits on the keyboard changes after entering the code depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $ik$. The user enters digits from the left column with the left hand, and the rest with the right hand. Restore the lock code if it is known that when entering the code, the user entered the digits as follows:
when $a_{3}=3$ - right, right, left, right;
when $a_{3}=9$ - left, left, left, left;
when $a_{3}=1$ - right, right, right, right;
when $a_{3}=7$ - left, right, right, right.
In your answer, specify the obtained code.
| $\mathrm{a}_{1}$ | $\mathrm{a}_{2}$ | $\mathrm{a}_{3}$ | | |
| :--- | :--- | :--- | :---: | :---: |
| $\mathrm{a}_{4}$ | $\mathrm{a}_{5}$ | $\mathrm{a}_{6}$ | | |
| $\mathrm{a}_{7}$ | $\mathrm{a}_{8}$ | $\mathrm{a}_{9}$ | | |
| | | | | $\mathrm{a}_{0}$ |
| | | | | |
| | | | | |
|
5. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The layout of the digits on the keyboard changes after entering the code based on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $i k$. The user enters digits from the left column with the left hand, and the rest with the right hand. Restore the lock code if it is known that when entering the code, the user entered the digits as follows:
when $a_{3}=3$ - right, right, left, right;
when $a_{3}=9$ - left, left, left, left;
when $a_{3}=1$ - right, right, right, right;
when $a_{3}=7$ - left, right, right, right.
In the answer, specify the obtained code.
| 1 | 2 | 3 |
| :--- | :--- | :--- |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
| 0 | | |
| | | |
| $a_{1}$ | $a_{2}$ | $\mathrm{a}_{3}$ |
| :---: | :---: | :---: |
| $\mathrm{a}_{4}$ | $a_{5}$ | $\mathrm{a}_{6}$ |
| $a_{7}$ | $a_{8}$ | $\mathrm{a}_{9}$ |
| | $\mathrm{a}_{0}$ | |
Answer: 3212
|
3212
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. At the robot running competition, a certain number of mechanisms were presented. The robots were released on the same distance in pairs. The protocol recorded the differences in the finish times of the winner and the loser in each of the races. All of them turned out to be different: 1 sec., 2 sec., 3 sec., 4 sec., 5 sec., 7 sec. It is known that during the races, each robot competed with each other exactly once, and that each robot always ran at the same speed. In your answer, indicate the time in seconds of the slowest mechanism, if the best time to complete the distance was 30 seconds.
|
3. At the robot running competition, a certain number of mechanisms were presented. The robots were released on the same distance in pairs. The protocol recorded the differences in the finish times of the winner and the loser in each of the races. All of them turned out to be different: 1 sec., 2 sec., 3 sec., 4 sec., 5 sec., 7 sec. It is known that during the races, each robot competed with each other exactly once, and that each robot always ran at the same speed. In the answer, indicate the time in seconds of the slowest mechanism, if the best time to complete the distance was 30 seconds.
Answer: 37.
|
37
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The arrangement of digits on the keyboard changes after entering the code depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $i k$. The user enters digits from the left column with the left hand, and the rest with the right hand. Restore the lock code if it is known that when entering the code, the user entered the digits as follows:
when $a_{3}=3$ - right, right, left, right;
when $a_{3}=9$ - left, left, left, left;
when $a_{3}=1$ - right, right, right, right;
when $a_{3}=7$ - left, right, right, right.
In your answer, specify the obtained code.
| 1 | 2 | 3 | |
| :--- | :--- | :--- | :---: |
| 4 | 5 | 6 | |
| 7 | 8 | 9 | |
| 0 | | | |
| | | | |
| $\mathrm{a}_{1}$ | $\mathrm{a}_{2}$ | $\mathrm{a}_{3}$ |
| :---: | :---: | :---: |
| $\mathrm{a}_{4}$ | $\mathrm{a}_{5}$ | $\mathrm{a}_{6}$ |
| $\mathrm{a}_{7}$ | $\mathrm{a}_{8}$ | $\mathrm{a}_{9}$ |
| | | $\mathrm{a}_{0}$ |
| | | |
| | | |
|
6. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The arrangement of digits on the keyboard changes after entering the code depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $i k$. The user enters digits from the left column with the left hand, and the rest with the right hand. Restore the lock code if it is known that when entering the code, the user entered the digits as follows:
when $a_{3}=3$ - right, right, left, right;
when $a_{3}=9$ - left, left, left, left;
when $a_{3}=1$ - right, right, right, right;
when $a_{3}=7$ - left, right, right, right.
In the answer, specify the obtained code.
| 1 | 2 | 3 | |
| :--- | :--- | :--- | :---: |
| 4 | 5 | 6 | |
| 7 | 8 | 9 | |
| 0 | | | |
| | | | |
| $\mathrm{a}_{1}$ | $\mathrm{a}_{2}$ | $\mathrm{a}_{3}$ | | |
| :--- | :--- | :--- | :---: | :---: |
| $\mathrm{a}_{4}$ | $\mathrm{a}_{5}$ | $\mathrm{a}_{6}$ | | |
| $\mathrm{a}_{7}$ | $\mathrm{a}_{8}$ | $\mathrm{a}_{9}$ | | |
| | | | | $\mathrm{a}_{0}$ |
| | | | | |
Answer: 3212.
|
3212
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Workers at an aluminum plant played the following game during their break. They drew cards from a deck (36 cards) until they had cards of all 4 suits among the drawn cards. What is the minimum number of cards that must be drawn to ensure that all 4 suits are among them?
a) 4
b) 9
c) 18
d) 28
e) 36
|
8. answer: g. There are 9 cards of each suit. If you draw 27 cards, they could all be from only three suits. Suppose you draw 28 cards and among the first 27 cards, they are only from three suits, then in the deck, only cards of the fourth suit will remain. And the 28th card will already be of the 4th suit.
|
28
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
8. During the lunch break on the day of the "13th Element. ALchemy of the Future" competition, a cafeteria worker mixed 2 liters of one juice with a $10 \%$ sugar content and 3 liters of another juice with a $15 \%$ sugar content. What is the sugar content of the resulting mixture?
a) $5 \%$
б) $12.5 \%$
в) $12.75 \%$
г) $13 \%$
д) $25 \%$
|
8. Answer g. In the first juice, there is $0.1 \cdot 2=0.2$ liters of sugar, and in the second juice, there is $0.15 \cdot 3=0.45$ liters of sugar. Then, in the resulting mixture, there will be $0.2+0.45=0.65$ liters of sugar. The total volume is 5 liters. Then the percentage of sugar in the mixture is $\frac{0.65}{5} \cdot 100 \% = 13 \%$.
|
13
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
15. Hydrogen was passed over a heated powder (X1). The resulting red substance (X2) was dissolved in concentrated sulfuric acid. The resulting solution of the substance blue (X3) was neutralized with potassium hydroxide - a blue precipitate (X4) formed, which upon heating turned into a black powder (X1). What substances are involved in the described process? Indicate the molar mass of the initial and final substance (X1).
|
15. 80 g/mol; $\mathrm{m}(\mathrm{CuO})=80$ g/mol; $\mathrm{X}_{1}-\mathrm{CuO} ; \mathrm{X}_{2}-\mathrm{Cu} ; \mathrm{X}_{3}-\mathrm{CuSO}_{4 ;} \mathrm{X}_{4}-\mathrm{Cu}(\mathrm{OH})_{2}$
|
80
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. What is the mass fraction (%) of oxygen in aluminum oxide $\mathrm{Al}_{2} \mathrm{O}_{3}$?
a) $53 \%$
b) $27 \%$
c) $16 \%$
d) $102 \%$
|
7. $53 \% . \operatorname{Mr}(\mathrm{Al} 2 \mathrm{O} 3)=27 * 2+16 * 3=102 \quad \mathrm{~W}(\mathrm{Al})=27 * 2 / 102 * 100 \%=53 \%$
7. $53 \% . \operatorname{Mr}(\mathrm{Al} 2 \mathrm{O} 3)=27 * 2+16 * 3=102 \quad \mathrm{~W}(\mathrm{Al})=27 * 2 / 102 * 100 \%=53 \%$
The text is already in a format that is a mix of mathematical notation and chemical formula, which is the same in English. Therefore, the translation is identical to the original text.
|
53
|
Other
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
16. Every sixth bus in the bus park of the aluminum plant is equipped with an air conditioner. After the plant director ordered to install it on 5 more buses, a quarter of the buses had an air conditioner. How many buses are there in the park of the plant, if each bus is equipped with only one air conditioner?
|
16. 60. Let the number of buses in the park be denoted by $x$. Then, $\frac{1}{6} x + 5$ buses are equipped with air conditioning. This constitutes a quarter of all the buses in the park. We have the equation: $\left(\frac{1}{6} x + 5\right) \cdot 4 = x$ or $\frac{1}{3} x = 20$. From this, $x = 60$.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three is to the sum of all terms except the last three as $4: 3$. Find the number of terms in this progression.
|
Answer: 20.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all terms except the first three is $-\frac{(n-3) \cdot(a+3 d+a+(n-1) d)}{2}$, and the sum of all terms except the last three is $-\frac{(n-3) \cdot(a+a+(n-4) d)}{2}$. From the condition, we have the system:
$$
\left\{\begin{aligned}
2 \cdot 13(a+6 d) & =13(a+(n-7) d) \\
3 \cdot(n-3) \cdot(2 a+(n+2) d) & =4 \cdot(n-3) \cdot(2 a+(n-4) d)
\end{aligned}\right.
$$
or, after transformations,
$$
\left\{\begin{aligned}
a & =(n-19) d \\
-2 a & =(n-22) d
\end{aligned}\right.
$$
(since otherwise the sum of all terms except the first three would equal the sum of all terms except the last three), we get $n=20$.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number written with the same digits in reverse order is equal to $N$. It turns out that the number $N$ is divisible by 100. Find $N$.
|
Answer: 11000.
Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d, Y=\overline{d c b a}=1000 d+100 c+10 b+a$, where $a$, $b, c, d$ are digits and $a \neq 0$.
According to the condition, $X+Y$ is divisible by 100, i.e., $1001(a+d)+110(b+c) \vdots 100$.
We have $1001(a+d) \vdots 10$, i.e., $a+d \vdots 10$, from which, since $a$ and $d$ are digits and $a \neq 0, 1 \leq a+d \leq 18$, thus $a+d=10$. Further, $1001 \cdot 10+110(b+c) \vdots 100$, i.e., $b+c+1 \vdots 10$, from which, since $b$ and $c$ are digits, $1 \leq b+c+1 \leq 19$, thus, $b+c=9$.
Therefore, $N=X+Y=1001 \cdot 10+110 \cdot 9=11000$.
|
11000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 65 and 31, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$.
|
Answer: 2015.
Solution. Let $O$ be the intersection point of diagonals $AC$ and $BD$. From the similarity of triangles $AOB$ and $COD$, it follows that $\overrightarrow{OC} = \frac{31}{65} \overrightarrow{AO}$, and $\overrightarrow{OD} = \frac{31}{65} \overrightarrow{BO}$.
Let the vector $\overrightarrow{AO}$ be denoted by $\vec{a}$, and the vector $\overrightarrow{BO}$ by $\vec{b}$. Then, from the condition, it follows that $(\vec{a}, \vec{b}) = 0$ and
$$
\overrightarrow{AD} = \overrightarrow{AO} + \overrightarrow{OD} = \vec{a} + \frac{31}{65} \vec{b}, \quad \overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = \vec{b} + \frac{31}{65} \vec{a}
$$
From which
$$
(\overrightarrow{AD}, \overrightarrow{BC}) = \left(\vec{a} + \frac{31}{65} \vec{b}, \vec{b} + \frac{31}{65} \vec{a}\right) = \frac{31}{65} \left(|\vec{a}|^2 + |\vec{b}|^2\right) + (\ldots) \cdot (\vec{a}, \vec{b}) = \frac{31}{65} |AB|^2 = 2015
$$
where the penultimate equality follows from the fact that triangle $AOB$ is a right triangle.
|
2015
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10. A cylinder of volume 9 is inscribed in a cone. The plane of the upper base of this cylinder cuts off a frustum of volume 63 from the original cone. Find the volume of the original cone.
|
Answer: 64.
Solution. Let the height and radius of the original cone be $H$ and $R$, and the height and radius of the cylinder be $h$ and $r$. We use the formula for the volume of a frustum of a cone: $\frac{1}{3} \pi\left(R^{2}+R r+r^{2}\right) h=63$. We also know that $\pi r^{2} h=9$. Dividing the corresponding parts of the equations, we get $\left(\frac{R}{r}\right)^{2}+\left(\frac{R}{r}\right)+1=$ $\frac{63 \cdot 3}{9}=21$. Solving the quadratic equation, we get the roots 4 and -5, only the positive root has geometric meaning. $R / r=4, \frac{H-h}{H}=4, \frac{h}{H}=\frac{3}{4}$, from which we obtain for the original cone:
$$
V=\frac{1}{3} \pi R^{2} H=\frac{1}{3}\left(\pi r^{2} h\right)\left(\frac{R}{r}\right)^{2} \frac{H}{h}=\frac{1}{3} \cdot 9 \cdot 4^{2} \cdot \frac{4}{3}=64
$$
## Variant II
|
64
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three terms is to the sum of all terms except the last three as $5: 4$. Find the number of terms in this progression.
|
Answer: 22.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all terms except the first three is $-\frac{(n-3) \cdot(a+3 d+a+(n-1) d)}{2}$, and the sum of all terms except the last three is $-\frac{(n-3) \cdot(a+a+(n-4) d)}{2}$. From the condition, we have the system:
$$
\left\{\begin{aligned}
2 \cdot 13(a+6 d) & =13(a+(n-7) d) \\
4 \cdot(n-3) \cdot(2 a+(n+2) d) & =5 \cdot(n-3) \cdot(2 a+(n-4) d)
\end{aligned}\right.
$$
or, after transformations,
$$
\left\{\begin{aligned}
a & =(n-19) d \\
-2 a & =(n-28) d
\end{aligned}\right.
$$
Multiplying the first equation by 2 and adding it to the second, we get $(3 n-66) d=0$. Since $d \neq 0$ (otherwise, the sum of all terms except the first three would equal the sum of all terms except the last three), we obtain $n=22$.
|
22
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number obtained from $X$ by swapping its second and third digits is divisible by 900. Find the remainder when the number $X$ is divided by 90.
|
Answer: 45.
Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d, Y=\overline{a c b d}=1000 a+100 c+10 b+d$, where $a$, $b, c, d$ are digits and $a \neq 0, d \neq 0$.
According to the condition, $X+Y$ is divisible by 900, i.e., $2000 a+110(b+c)+2 d \vdots 900$.
We have, $2 d \vdots 10$, i.e., $d \vdots 5$, so since $d \neq 0$ and $d$ is a digit, $d=5$. Next, $110(b+c)+10 \vdots 100$, i.e., $b+c+1 \vdots 10$, from which, since $b$ and $c$ are digits, $1 \leq b+c+1 \leq 19, b+c=9$. Finally, $2000 a+110 \cdot 9+10 \vdots 9$, i.e., $2 a+1 \vdots 9$, from which, since $a$ is a digit, $a=4$.
Thus, $X=4000+90 b+90+5=90 q+45$.
|
45
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 155 and 13, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$.
|
Answer: 2015.
Solution. Let $O$ be the intersection point of diagonals $AC$ and $BD$. From the similarity of triangles $AOB$ and $COD$, it follows that $\overrightarrow{OC} = \frac{13}{155} \overrightarrow{AO}$, and $\overrightarrow{OD} = \frac{13}{155} \overrightarrow{BO}$.
Let the vector $\overrightarrow{AO}$ be denoted by $\vec{a}$, and the vector $\overrightarrow{BO}$ by $\vec{b}$. Then, from the condition, it follows that $(\vec{a}, \vec{b}) = 0$ and
$$
\overrightarrow{AD} = \overrightarrow{AO} + \overrightarrow{OD} = \vec{a} + \frac{13}{155} \vec{b}, \quad \overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = \vec{b} + \frac{13}{155} \vec{a}
$$
From which
$$
(\overrightarrow{AD}, \overrightarrow{BC}) = \left(\vec{a} + \frac{13}{155} \vec{b}, \vec{b} + \frac{13}{155} \vec{a}\right) = \frac{13}{155} \left(|\vec{a}|^2 + |\vec{b}|^2\right) + (\ldots) \cdot (\vec{a}, \vec{b}) = \frac{13}{155} |AB|^2 = 2015
$$
where the penultimate equality follows from the fact that triangle $AOB$ is a right triangle.
|
2015
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three terms is to the sum of all terms except the last three as $3: 2$. Find the number of terms in this progression.
|
Answer: 18.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all terms except the first three is $-\frac{(n-3) \cdot(a+3 d+a+(n-1) d)}{2}$, and the sum of all terms except the last three is $-\frac{(n-3) \cdot(a+a+(n-4) d)}{2}$. From the condition, we have the system:
$$
\left\{\begin{aligned}
2 \cdot 13(a+6 d) & =13(a+(n-7) d) \\
2 \cdot(n-3) \cdot(2 a+(n+2) d) & =3 \cdot(n-3) \cdot(2 a+(n-4) d)
\end{aligned}\right.
$$
or, after transformations,
$$
\left\{\begin{aligned}
a & =(n-19) d \\
-2 a & =(n-16) d
\end{aligned}\right.
$$
(since otherwise the sum of all terms except the first three would equal the sum of all terms except the last three), we get $n=18$.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 65 and 31, respectively, and its lateral sides are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$.
|
Answer: -2015.
Solution. Let $O$ be the point of intersection of the lines containing the lateral sides $AD$ and $BC$. From the similarity of triangles $AOB$ and $DOC$, it follows that $\overrightarrow{OC} = -\frac{31}{65} \overrightarrow{BO}$, and $\overrightarrow{OD} = -\frac{31}{65} \overrightarrow{AO}$.
Let the vector $\overrightarrow{AO}$ be denoted by $\vec{a}$, and the vector $\overrightarrow{BO}$ by $\vec{b}$. Then, from the condition, it follows that $(\vec{a}, \vec{b}) = 0$ and
$$
\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = \vec{a} - \frac{31}{65} \vec{b}, \quad \overrightarrow{BD} = \overrightarrow{BO} + \overrightarrow{OD} = \vec{b} - \frac{31}{65} \vec{a}
$$
From which
$$
(\overrightarrow{AC}, \overrightarrow{BD}) = \left(\vec{a} - \frac{31}{65} \vec{b}, \vec{b} - \frac{31}{65} \vec{a}\right) = -\frac{31}{65}\left(|\vec{a}|^2 + |\vec{b}|^2\right) + (\ldots) \cdot (\vec{a}, \vec{b}) = -\frac{31}{65}|AB|^2 = -2015
$$
where the penultimate equality follows from the fact that triangle $AOB$ is a right triangle.
|
-2015
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three terms is in the ratio $6: 5$ to the sum of all terms except the last three. Find the number of terms in this progression.
|
Answer: 24.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all terms except the first three is $-\frac{(n-3) \cdot(a+3 d+a+(n-1) d)}{2}$, and the sum of all terms except the last three is $-\frac{(n-3) \cdot(a+a+(n-4) d)}{2}$. From the condition, we have the system:
$$
\left\{\begin{aligned}
2 \cdot 13(a+6 d) & =13(a+(n-7) d) \\
5 \cdot(n-3) \cdot(2 a+(n+2) d) & =6 \cdot(n-3) \cdot(2 a+(n-4) d)
\end{aligned}\right.
$$
or, after transformations,
$$
\left\{\begin{aligned}
a & =(n-19) d \\
-2 a & =(n-34) d
\end{aligned}\right.
$$
(since otherwise the sum of all terms except the first three would equal the sum of all terms except the last three), we get $n=24$.
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. If the sum of the digits of a four-digit number $X$ is subtracted from $X$, the result is a natural number $N=K^{2}$, where $K$ is a natural number that gives a remainder of 5 when divided by 20 and a remainder of 3 when divided by 21. Find the number $N$.
|
Answer: 2025.
Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d$, where $a, b, c, d$ are digits and $a \neq 0$.
According to the condition, $X-a-b-c-d=999 a+99 b+9 c=K^{2}$, where $K=20 u+5, K=21 v+3$.
Notice that $999 a+99 b+9 c \vdots 9$, i.e., $K^{2} \vdots 9$, which means $K=3 M$, and $M^{2}=111 a+11 b+c \leq$ $111 \cdot 9+11 \cdot 9+9=1107, M \leq 33, K=3 M \leq 99$.
Thus, we need to find among the numbers from 1 to 99 all those numbers that are divisible by 3, give a remainder of 5 when divided by 20, and a remainder of 3 when divided by 21. This can be done in various ways, for example, by simple enumeration: list all numbers from 1 to 99 that give a remainder of 3 when divided by 21:
$$
3,24,45,66,87
$$
Among them, only one number - 45 gives a remainder of 5 when divided by 20. That is, $K=45$, and $N=2025$.
|
2025
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 155 and 13, respectively, and its lateral sides are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$.
|
Answer: -2015.
Solution. Let $O$ be the point of intersection of the lines containing the lateral sides $AD$ and $BC$. From the similarity of triangles $AOB$ and $DOC$, it follows that $\overrightarrow{OC} = -\frac{13}{155} \overrightarrow{BO}$, and $\overrightarrow{OD} = -\frac{13}{155} \overrightarrow{AO}$.
Let the vector $\overrightarrow{AO}$ be denoted by $\vec{a}$, and the vector $\overrightarrow{BO}$ by $\vec{b}$. Then, from the condition, it follows that $(\vec{a}, \vec{b}) = 0$ and
$$
\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = \vec{a} - \frac{13}{155} \vec{b}, \quad \overrightarrow{BD} = \overrightarrow{BO} + \overrightarrow{OD} = \vec{b} - \frac{13}{155} \vec{a}
$$
From which
$$
(\overrightarrow{AC}, \overrightarrow{BD}) = \left(\vec{a} - \frac{13}{155} \vec{b}, \vec{b} - \frac{13}{155} \vec{a}\right) = -\frac{13}{155}\left(|\vec{a}|^{2} + |\vec{b}|^{2}\right) + (\ldots) \cdot (\vec{a}, \vec{b}) = -\frac{13}{155}|AB|^{2} = -2015
$$
where the penultimate equality follows from the fact that triangle $AOB$ is a right triangle.
|
-2015
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Dasha wrote the numbers $9,10,11, \ldots, 22$ on the board, and then erased one or several of them. It turned out that the remaining numbers on the board cannot be divided into several groups such that the sums of the numbers in the groups are equal. What is the greatest possible value of the sum of the remaining numbers on the board?
|
Answer: 203.
Solution. The sum of numbers from 9 to 22 is 217. If at least one number is erased, the sum of the remaining numbers does not exceed 208. Let's sequentially consider the options:
- if the sum is 208, then Dasha could have erased only the number 9; then the remaining numbers can be divided into two groups with a sum of 104:
$$
22+21+20+19+12+10=18+17+16+15+14+13+11
$$
- if the sum is 207, then Dasha could have erased only the number 10; then the remaining numbers can be divided into three groups with a sum of 69:
$$
22+21+17+9=20+19+18+12=16+15+14+13+11
$$
- if the sum is 206, then Dasha could have erased only the number 11; then the remaining numbers can be divided into two groups with a sum of 103:
$$
22+21+20+19+12+9=18+17+16+15+14+13+10
$$
- if the sum is 205, then Dasha could have erased only the number 12; then the remaining numbers can be divided into five groups with a sum of 41:
$$
22+19=21+20=18+13+10=17+15+9=16+14+11
$$
- if the sum is 204, then Dasha could have erased only the number 13; then the remaining numbers can be divided into two groups with a sum of 102:
$$
22+21+20+19+11+9=18+17+16+15+14+12+10
$$
- if Dasha erased the number 14, then the numbers left on the board have a sum of 203: they could be divided into 7 groups with a sum of 29, or 29 groups with a sum of 7, or 203 groups with a sum of 1; one of these groups will contain the number 22; since we have no options with a sum of 22, at least one more number will be added to this group; therefore, the sum in this group will be at least 31; thus, in this case, it is impossible to divide the numbers into groups with the same sum.
|
203
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. Calculate
$$
\operatorname{tg} \frac{\pi}{43} \cdot \operatorname{tg} \frac{2 \pi}{43}+\operatorname{tg} \frac{2 \pi}{43} \cdot \operatorname{tg} \frac{3 \pi}{43}+\ldots+\operatorname{tg} \frac{k \pi}{43} \cdot \operatorname{tg} \frac{(k+1) \pi}{43}+\ldots+\operatorname{tg} \frac{2019 \pi}{43} \cdot \operatorname{tg} \frac{2020 \pi}{43}
$$
|
Answer: -2021.
Solution. From the formula
$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$
we express the product of tangents:
$$
\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{\operatorname{tg}(\alpha-\beta)}-1
$$
Then
$$
\operatorname{tg} \frac{k \pi}{43} \cdot \operatorname{tg} \frac{(k+1) \pi}{43}=\frac{\operatorname{tg} \frac{(k+1) \pi}{43}-\operatorname{tg} \frac{k \pi}{43}}{\operatorname{tg}\left(\frac{(k+1) \pi}{43}-\frac{k \pi}{43}\right)}-1=\left(\operatorname{tg} \frac{(k+1) \pi}{43}-\operatorname{tg} \frac{k \pi}{43}\right) \cdot\left(\operatorname{tg} \frac{\pi}{43}\right)^{-1}-1
$$
Adding these equalities for all $k$ from 1 to 2019, we get that the expression in the condition equals
$$
\left(\operatorname{tg} \frac{2020 \pi}{43}-\operatorname{tg} \frac{\pi}{43}\right) \cdot\left(\operatorname{tg} \frac{\pi}{43}\right)^{-1}-2019
$$
Notice that
$$
\operatorname{tg} \frac{2020 \pi}{43}=\operatorname{tg}\left(47 \pi-\frac{\pi}{43}\right)=-\operatorname{tg} \frac{\pi}{43}
$$
which means $(*)$ equals -2021.
|
-2021
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. Given an isosceles triangle $K L M(K L=L M)$ with the angle at the vertex equal to $114^{\circ}$. Point $O$ is located inside triangle $K L M$ such that $\angle O M K=30^{\circ}$, and $\angle O K M=27^{\circ}$. Find the measure of angle $\angle L O M$.
|
Answer: $150^{\circ}$.
Solution. Let $L H$ be the height/median/bisector of the triangle. Let $S$ be the intersection of ray $M O$ and segment $L H$. Note that $K S=S M$. For example, since in triangle $K S M$ the median SH coincides with the height.
Let's calculate the angles:
|
150
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Sasha wrote the numbers $7, 8, 9, \ldots, 17$ on the board, and then erased one or several of them. It turned out that the remaining numbers on the board cannot be divided into several groups such that the sums of the numbers in the groups are equal. What is the greatest possible value of the sum of the remaining numbers on the board?
|
Answer: 121.
Solution. The sum of numbers from 7 to 17 is 132. If at least one number is erased, the sum of the remaining numbers does not exceed 125. Let's sequentially consider the options:
- if the sum is 125, then Sasha could have erased only the number 7; then the remaining numbers can be divided into five groups with a sum of 25:
$$
8+17=9+16=10+15=11+14=12+13
$$
- if the sum is 124, then Sasha could have erased only the number 8; then the remaining numbers can be divided into two groups with a sum of 62:
$$
17+16+15+14=13+12+11+10+9+7
$$
- if the sum is 123, then Sasha could have erased only the number 9; then the remaining numbers can be divided into three groups with a sum of 41:
$$
17+16+8=15+14+12=13+11+10+7
$$
- if the sum is 122, then Sasha could have erased only the number 10; then the remaining numbers can be divided into two groups with a sum of 61:
$$
17+16+15+13=14+12+11+9+8+7
$$
- if Sasha erased the number 11, then the numbers left on the board have a sum of 121: they could be divided either into 11 groups with a sum of 11, or into 121 groups with a sum of 1; but some group will include the number 17, so the sum in this group will be at least 17; therefore, in this case, it is impossible to divide the numbers into groups with the same sum.
|
121
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. In the surgical department, there are 4 operating rooms: 1, 2, 3, and 4. In the morning, they were all empty. At some point, a surgery began in operating room 1, after some time - in operating room 2, then after some more time - in operating room 3, and finally in operating room 4.
All four surgeries ended simultaneously, and the total duration of all surgeries was 2 hours and 7 minutes. 18 minutes before the completion of all surgeries, the total duration of the ongoing surgeries was 60 minutes, and 15 minutes before that - 25 minutes. Which operating rooms' surgery durations can be determined from these data, and which cannot?
|
Answer: Only the duration of the operation in operating room 4 can be determined.
Solution. First, let's prove that the durations of operations in operating rooms 1, 2, and 3 cannot be determined uniquely. Indeed, it is not difficult to verify that if the durations of the operations are 58, 29, 27, 13 minutes or 46, 45, 23, 13 minutes, then all conditions of the problem are satisfied. However, in these two cases, the durations of the operations in operating rooms 1, 2, and 3 are different.
Now let's prove that the duration of the operation in operating room 4 can be uniquely restored. For this, let's note that the total duration of the operations 33 and 18 minutes before the end of the operations increased by 35 minutes. This means that 18 minutes before the end of the operations, the operations in operating rooms 1, 2, and 3 were already in progress, otherwise, the total duration would have increased by no more than 30 minutes. Then, by the end of all operations, their total duration is \(60 + 18 \cdot 3 = 114\) minutes. Therefore, the operation in operating room 4 lasted \(127 - 114 = 13\) minutes.
Comment. The durations of the operations can be \(58; 28+s; 28-s; 13\) minutes, where \(s \in [0; 5]\) or \((45.5+t; 45.5-t; 23; 13)\), where \(t \in [0; 12.5]\).
|
13
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. Calculate
$$
\operatorname{tg} \frac{\pi}{47} \cdot \operatorname{tg} \frac{2 \pi}{47}+\operatorname{tg} \frac{2 \pi}{47} \cdot \operatorname{tg} \frac{3 \pi}{47}+\ldots+\operatorname{tg} \frac{k \pi}{47} \cdot \operatorname{tg} \frac{(k+1) \pi}{47}+\ldots+\operatorname{tg} \frac{2021 \pi}{47} \cdot \operatorname{tg} \frac{2022 \pi}{47}
$$
|
Answer: -2021.
Solution. From the formula
$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$
we express the product of tangents:
$$
\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{\operatorname{tg}(\alpha-\beta)}-1
$$
Then
$$
\operatorname{tg} \frac{k \pi}{47} \cdot \operatorname{tg} \frac{(k+1) \pi}{47}=\frac{\operatorname{tg} \frac{(k+1) \pi}{47}-\operatorname{tg} \frac{k \pi}{47}}{\operatorname{tg}\left(\frac{(k+1) \pi}{47}-\frac{k \pi}{47}\right)}-1=\left(\operatorname{tg} \frac{(k+1) \pi}{47}-\operatorname{tg} \frac{k \pi}{47}\right) \cdot\left(\operatorname{tg} \frac{\pi}{47}\right)^{-1}-1
$$
Adding these equalities for all $k$ from 1 to 2021, we get that the expression in the condition equals
$$
\left(\operatorname{tg} \frac{2022 \pi}{47}-\operatorname{tg} \frac{\pi}{47}\right) \cdot\left(\operatorname{tg} \frac{\pi}{47}\right)^{-1}-2021
$$
Notice that
$$
\operatorname{tg} \frac{2022 \pi}{47}=\operatorname{tg}\left(43 \pi+\frac{\pi}{47}\right)=\operatorname{tg} \frac{\pi}{47}
$$
which means ( ${ }^{*}$ ) equals -2021.
|
-2021
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Masha wrote the numbers $4,5,6, \ldots, 16$ on the board, and then erased one or several of them. It turned out that the remaining numbers on the board cannot be divided into several groups such that the sums of the numbers in the groups are equal. What is the greatest possible value of the sum of the remaining numbers on the board?
|
Answer: 121.
Solution. The sum of numbers from 4 to 16 is 130. If at least one number is erased, the sum of the remaining numbers does not exceed 126. Let's sequentially consider the options:
- if the sum is 126, then Masha could have erased only the number 4; then the remaining numbers can be divided into two groups with a sum of 63:
$$
16+15+14+13+5=12+11+10+9+8+7+6
$$
- if the sum is 125, then Masha could have erased only the number 5; then the remaining numbers can be divided into five groups with a sum of 25:
$$
16+9=15+10=14+11=13+12=8+7+6+4
$$
- if the sum is 124, then Masha could have erased only the number 6; then the remaining numbers can be divided into two groups with a sum of 62:
$$
16+15+14+13+4=12+11+10+9+8+7+5
$$
- if the sum is 123, then Masha could have erased only the number 7; then the remaining numbers can be divided into three groups with a sum of 41:
$$
16+15+10=14+13+9+5=12+11+8+6+4
$$
- if the sum is 122, then Masha could have erased only the number 8; then the remaining numbers can be divided into two groups with a sum of 61:
$$
16+15+14+12+4=13+11+10+9+7+6+5
$$
- if Masha erased the number 9, then the numbers left on the board have a sum of 121: they could be divided either into 11 groups with a sum of 11, or into 121 groups with a sum of 1; but some group will include the number 16, so the sum in this group will be at least 16; therefore, in this case, it is impossible to divide the numbers into groups with the same sum.
|
121
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. In a surgical department, there are 4 operating rooms: A, B, V, and G. In the morning, they were all empty. At some point, an operation began in operating room A, after some time - in operating room B, then after some more time - in V, and then in $\Gamma$.
All four operations ended simultaneously, and the total duration of all operations was 3 hours and 5 minutes. 36 minutes before the completion of all operations, the total duration of the ongoing operations was 46 minutes, and 10 minutes before that - 19 minutes. Which operating rooms' operation durations can be determined from these data, and which cannot?
|
Answer: Only the duration of the operation in operating room $\Gamma$ can be determined.
Solution. First, let's prove that the durations of operations in operating rooms A, B, and C cannot be determined uniquely. Indeed, it is easy to verify that if the durations of the operations are $65, 45, 44, 31$ or $56, 55, 43, 31$ minutes, then all conditions of the problem are satisfied. However, in these two cases, the durations of the operations in operating rooms A, B, and C are different.
Now let's prove that the duration of the operation in operating room Г can be uniquely determined. For this, let's note that the total duration of the operations 46 and 36 minutes before the end of the operations increased by 27 minutes. This means that 36 minutes before the end of the operations, the operations in operating rooms A, B, and C were already in progress, otherwise, the total duration would have increased by no more than 20 minutes. Then, by the end of all operations, their total duration is $46 + 36 \cdot 3 = 154$ minutes. Therefore, the operation in operating room Г lasted $185 - 154 = 31$ minutes.
Comment. The durations of the operations can be ( $65 ; 44.5+s ; 44.5-s ; 31$ ), where $s \in [0 ; 1.5]$ or ( $55.5+t ; 55.5-t ; 43 ; 31$ ), where $t \in [0 ; 9.5]$.
|
31
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. Calculate
$$
\operatorname{tg} \frac{\pi}{43} \cdot \operatorname{tg} \frac{2 \pi}{43}+\operatorname{tg} \frac{2 \pi}{43} \cdot \operatorname{tg} \frac{3 \pi}{43}+\ldots+\operatorname{tg} \frac{k \pi}{43} \cdot \operatorname{tg} \frac{(k+1) \pi}{43}+\ldots+\operatorname{tg} \frac{2021 \pi}{43} \cdot \operatorname{tg} \frac{2022 \pi}{43}
$$
|
Answer: -2021.
Solution. From the formula
$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$
we express the product of tangents:
$$
\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{\operatorname{tg}(\alpha-\beta)}-1
$$
Then
$$
\operatorname{tg} \frac{k \pi}{43} \cdot \operatorname{tg} \frac{(k+1) \pi}{43}=\frac{\operatorname{tg} \frac{(k+1) \pi}{43}-\operatorname{tg} \frac{k \pi}{43}}{\operatorname{tg}\left(\frac{(k+1) \pi}{43}-\frac{k \pi}{43}\right)}-1=\left(\operatorname{tg} \frac{(k+1) \pi}{43}-\operatorname{tg} \frac{k \pi}{43}\right) \cdot\left(\operatorname{tg} \frac{\pi}{43}\right)^{-1}-1
$$
Adding these equalities for all $k$ from 1 to 2021, we get that the expression in the condition equals
$$
\left(\operatorname{tg} \frac{2022 \pi}{43}-\operatorname{tg} \frac{\pi}{43}\right) \cdot\left(\operatorname{tg} \frac{\pi}{43}\right)^{-1}-2021
$$
Notice that
$$
\operatorname{tg} \frac{2022 \pi}{43}=\operatorname{tg}\left(47 \pi+\frac{\pi}{43}\right)=\operatorname{tg} \frac{\pi}{43}
$$
which means $(*)$ equals -2021.
|
-2021
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. For what least value of the parameter $a$ will the coefficient of $x^{4}$ in the expansion of the polynomial $\left(1-2 x+a x^{2}\right)^{8}$ be equal to $-1540?$
|
Answer: -19.
Solution. Applying the polynomial formula, we get
$$
\left(1-2 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-2 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-2)^{n_{2}} \cdot a^{n_{3}} \cdot x^{n_{2}+2 n_{3}}
$$
To determine which terms in the sum contain $x^{4}$, we need to solve the system of equations in non-negative integers:
$$
\left\{\begin{array}{l}
n_{1}+n_{2}+n_{3}=8 \\
n_{2}+2 n_{3}=4
\end{array}\right.
$$
From the second equation, it follows that $n_{2}$ is even. Due to the non-negativity of the variables, $n_{2}$ can take the values 0, 2, and 4. Solving the system for each of these $n_{2}$, we will have three cases:
|
-19
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. Given an isosceles triangle $A B C(A B=B C)$ with the angle at the vertex equal to $102^{\circ}$. Point $O$ is located inside triangle $A B C$ such that $\angle O C A=30^{\circ}$, and $\angle O A C=21^{\circ}$. Find the measure of angle $\angle B O A$.
|
Answer: $81^{\circ}$.
Solution. Let $B H$ be the height/median/bisector of the triangle. Let $S$ be the intersection of ray $C O$ and segment $B H$. Note that $A S=S C$. For example, in triangle $A S C$, the median $S H$ coincides with the height.
Let's calculate the angles:
|
81
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Pasha wrote the numbers $4,5,6, \ldots, 14$ on the board, and then erased one or several of them. It turned out that the remaining numbers on the board cannot be divided into several groups such that the sums of the numbers in the groups are equal. What is the greatest possible value of the sum of the remaining numbers on the board?
|
# Answer: 91.
Solution. The sum of the numbers from 4 to 14 is 99. If at least one number is erased, the sum of the remaining numbers does not exceed 95. Let's sequentially consider the options:
- if the sum is 95, then Pasha could have erased only the number 4; then the remaining numbers can be divided into five groups with a sum of 19:
$$
14+5=13+6=12+7=11+8=10+9
$$
- if the sum is 94, then Pasha could have erased only the number 5; then the remaining numbers can be divided into two groups with a sum of 47:
$$
14+13+12+8=11+10+9+7+6+4.
$$
- if the sum is 93, then Pasha could have erased only the number 6; then the remaining numbers can be divided into three groups with a sum of 31:
$$
14+13+4=12+11+8=10+9+7+5
$$
- if the sum is 92, then Pasha could have erased only the number 7; then the remaining numbers can be divided into two groups with a sum of 46:
$$
14+13+11+8=12+10+9+6+5+4.
$$
- if Pasha erased the number 8, then the numbers left on the board have a sum of 91: they could be divided either into 7 groups with a sum of 13, or into 13 groups with a sum of 7, or into 91 groups with a sum of 1; but some group will include the number 14, so the sum in this group will be at least 14; therefore, in this case, it is impossible to divide the numbers into groups with the same sum.
|
91
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. In a surgical department, there are 4 operating rooms: A, B, C, and D. In the morning, they were all empty. At some point, a surgery began in operating room A, after some time - in operating room B, then after some more time - in C, and finally in D.
All four surgeries ended simultaneously, and the total duration of all surgeries was 2 hours and 38 minutes. Twenty-four minutes before the completion of all surgeries, the total duration of the ongoing surgeries was 1 hour and 9 minutes, and fifteen minutes before that, it was 33 minutes. Which operating rooms' surgery durations can be determined from these data, and which cannot?
|
Answer: Only the duration of the operation in operating room D can be determined.
Solution. First, let's prove that the durations of the operations in operating rooms A, B, and C cannot be determined uniquely. Indeed, it is not difficult to verify that if the durations of the operations are $72, 35, 34, 17$ or $56, 55, 30, 17$ minutes, then all conditions of the problem are satisfied. However, in these two cases, the durations of the operations in operating rooms A, B, and C are different.
Now let's prove that the duration of the operation in operating room D can be uniquely restored. For this, let's note that the total duration of the operations 39 and 24 minutes before the end of the operations increased by 36 minutes. This means that 24 minutes before the end of the operations, the operations in operating rooms A, B, and C were already in progress, otherwise, the total duration would have increased by no more than 30 minutes. Then, by the end of all operations, their total duration is $69 + 24 \cdot 3 = 141$ minutes. Therefore, the operation in operating room D lasted $158 - 141 = 17$ minutes.
Comment. The durations of the operations can be $(72; 35.5 + s; 35.5 - s; 17)$, where $s \in [0; 3.5]$ or $(55.5 + t; 55.5 - t; 30; 17)$, where $t \in [0; 16.5]$.
|
17
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. Calculate
$$
\operatorname{tg} \frac{\pi}{47} \cdot \operatorname{tg} \frac{2 \pi}{47}+\operatorname{tg} \frac{2 \pi}{47} \cdot \operatorname{tg} \frac{3 \pi}{47}+\ldots+\operatorname{tg} \frac{k \pi}{47} \cdot \operatorname{tg} \frac{(k+1) \pi}{47}+\ldots+\operatorname{tg} \frac{2019 \pi}{47} \cdot \operatorname{tg} \frac{2020 \pi}{47}
$$
|
Answer: -2021.
Solution. From the formula
$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$
we express the product of tangents:
$$
\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{\operatorname{tg}(\alpha-\beta)}-1
$$
Then
$$
\operatorname{tg} \frac{k \pi}{47} \cdot \operatorname{tg} \frac{(k+1) \pi}{47}=\frac{\operatorname{tg} \frac{(k+1) \pi}{47}-\operatorname{tg} \frac{k \pi}{47}}{\operatorname{tg}\left(\frac{(k+1) \pi}{47}-\frac{k \pi}{47}\right)}-1=\left(\operatorname{tg} \frac{(k+1) \pi}{47}-\operatorname{tg} \frac{k \pi}{47}\right) \cdot\left(\operatorname{tg} \frac{\pi}{47}\right)^{-1}-1
$$
Adding these equalities for all $k$ from 1 to 2019, we get that the expression in the condition equals
$$
\left(\operatorname{tg} \frac{2020 \pi}{47}-\operatorname{tg} \frac{\pi}{47}\right) \cdot\left(\operatorname{tg} \frac{\pi}{47}\right)^{-1}-2019
$$
Notice that
$$
\operatorname{tg} \frac{2020 \pi}{47}=\operatorname{tg}\left(43 \pi-\frac{\pi}{47}\right)=-\operatorname{tg} \frac{\pi}{47}
$$
which means $(*)$ equals -2021.
|
-2021
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. For what least value of the parameter $a$ will the coefficient of $x^{4}$ in the expansion of the polynomial $\left(1-3 x+a x^{2}\right)^{8}$ be equal to $70 ?$
|
Answer: -50.
Solution. Applying the polynomial formula, we get
$$
\left(1-3 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-3 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-3)^{n_{2}} \cdot a^{n_{3}} \cdot x^{n_{2}+2 n_{3}}
$$
To determine which terms in the sum contain $x^{4}$, we need to solve the system of equations in non-negative integers:
$$
\left\{\begin{array}{l}
n_{1}+n_{2}+n_{3}=8 \\
n_{2}+2 n_{3}=4
\end{array}\right.
$$
From the second equation, it follows that $n_{2}$ is even. Due to the non-negativity of the variables, $n_{2}$ can take the values 0, 2, and 4. Solving the system for each of these $n_{2}$, we will have three cases:
|
-50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. Given an isosceles triangle $X Y Z(X Y=Y Z)$ with the angle at the vertex equal to $96^{\circ}$. Point $O$ is located inside triangle $X Y Z$ such that $\angle O Z X=30^{\circ}$, and $\angle O X Z=18^{\circ}$. Find the measure of the angle $\angle Y O X$.
|
Answer: $78^{\circ}$.
Solution. Let $Y H$ be the height/median/bisector of the triangle. Let $S$ be the intersection of ray $Z O$ and segment $Y H$. Note that $X S=S Z$. For example, since in triangle $X S Z$ the median $S H$ coincides with the height.
Let's calculate the angles:
|
78
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. The number 27 is written on the board. Every minute, the number is erased from the board and replaced with the product of its digits, increased by 12. For example, after one minute, the number on the board will be $2 \cdot 7+12=26$. What will be on the board after an hour?
|
# Answer: 14
Solution. Let's find the next few numbers that will appear on the board. After 26, it will be 24, then $20, 12, 14, 16, 18$ and again 20. Notice that the sequence has looped and each subsequent number will coincide with the one that is 5 positions earlier. Therefore, after an hour, that is, 60 minutes, the number written will be the same as the one written 55 minutes earlier, which in turn coincides with the one written 50 minutes earlier, and so on, down to the one written 5 minutes earlier, which is the number 14.
|
14
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 2. Angelica wants to choose a three-digit code for her suitcase. To make it easier to remember, Angelica wants all the digits in her code to be in non-decreasing order. How many different options does Angelica have for choosing the code?
|
Problem 2. Angelica wants to choose a three-digit code for her suitcase. To make it easier to remember, Angelica wants all the digits in her code to be in non-decreasing order. How many different ways can Angelica choose her code? Answer. 220.
|
220
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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