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Problem 2. Let $n$ three-digit numbers satisfy the following properties:
(1) No number contains the digit 0 .
(2) The sum of the digits of each number is 9 .
(3) The units digits of any two numbers are different.
(4) The tens digits of any two numbers are different.
(5) The hundreds digits of any two numbers are ... |
Solution. Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0 . We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of 6 A's (which means that we add 1 to the current digit) and $2 G$ 's (w... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. For any set $A=\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}$ of five distinct positive integers denote by $S_{A}$ the sum of its elements, and denote by $T_{A}$ the number of triples $(i, j, k)$ with $1 \leqslant i<j<k \leqslant 5$ for which $x_{i}+x_{j}+x_{k}$ divides $S_{A}$.
Find the largest possibl... |
Solution. We will prove that the maximum value that $T_{A}$ can attain is 4 . Let $A=$ $\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}$ be a set of five positive integers such that $x_{1}x_{4}$ and $x_{3}>x_{2}$. Analogously we can show that any triple of form $(x, y, 5)$ where $y>2$ isn't good.
By above, the numbe... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression
$$
A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}
$$
$19^{\text {th ... | ## Solution:
We can rewrite $A$ as follows:
$$
\begin{aligned}
& A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^{2}-b^{2}-c^{2}= \\
& 2\left(\frac{a b+b c+c a}{a b c}\right)-\left(a^{2}+b^{2}+c^{2}\right)=2\left(\frac{a b+b c+c a}{a b c}\right)-\left((a+b+c... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
C1. Consider a regular $2 n+1$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in... |
Solution. Answer: $n=1$ is clearly a solution, we can just color each side of the equilateral triangle in a different color, and the conditions are satisfied. We prove there is no larger $n$ that fulfills the requirements.
Lemma 1. Given a regular $2 n+1$-gon in the plane, and a sequence of $n+1$ consecutive sides $s... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
NT5. Find all positive integers $n$ such that there exists a prime number $p$, such that
$$
p^{n}-(p-1)^{n}
$$
is a power of 3 .
Note. A power of 3 is a number of the form $3^{a}$ where $a$ is a positive integer.
|
Solution. Suppose that the positive integer $n$ is such that
$$
p^{n}-(p-1)^{n}=3^{a}
$$
for some prime $p$ and positive integer $a$.
If $p=2$, then $2^{n}-1=3^{a}$ by $(1)$, whence $(-1)^{n}-1 \equiv 0(\bmod 3)$, so $n$ should be even. Setting $n=2 s$ we obtain $\left(2^{s}-1\right)\left(2^{s}+1\right)=3^{a}$. It ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A5. Find the largest positive integer $n$ for which the inequality
$$
\frac{a+b+c}{a b c+1}+\sqrt[n]{a b c} \leq \frac{5}{2}
$$
holds for all $a, b, c \in[0,1]$. Here $\sqrt[1]{a b c}=a b c$.
|
Solution. Let $n_{\max }$ be the sought largest value of $n$, and let $E_{a, b, c}(n)=\frac{a+b+c}{a b c+1}+\sqrt[n]{a b c}$. Then $E_{a, b, c}(m)-E_{a, b, c}(n)=\sqrt[m]{a b c}-\sqrt[n]{a b c}$ and since $a . b c \leq 1$ we clearly have $E_{a, b, c}(m) \geq$ $E_{a, b, c}(n)$ for $m \geq n$. So if $E_{a, b, c}(n) \geq... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
## A3 MNE
Let $a, b, c$ be positive real numbers. Prove that
$$
\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2
$$
| ## Solution:
Starting from the double expression on the left-hand side of given inequality, and applying twice the Arithmetic-Geometric mean inequality, we find that
$$
\begin{aligned}
2 \frac{a}{b}+2 \sqrt{\frac{b}{c}}+2 \sqrt[3]{\frac{c}{a}} & =\frac{a}{b}+\left(\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt{\frac{b}{c}}\rig... | 4 | Inequalities | proof | Yes | Yes | olympiads | false |
## C4
Let $A=1 \cdot 4 \cdot 7 \cdot \ldots \cdot 2014$ be the product of the numbers less or equal to 2014 that give remainder 1 when divided by 3 . Find the last non-zero digit of $A$.
|
Solution. Grouping the elements of the product by ten we get:
$$
\begin{aligned}
& (30 k+1)(30 k+4)(30 k+7)(30 k+10)(30 k+13)(30 k+16) \\
& (30 k+19)(30 k+22)(30 k+25)(30 k+28)= \\
& =(30 k+1)(15 k+2)(30 k+7)(120 k+40)(30 k+13)(15 k+8) \\
& (30 k+19)(15 k+11)(120 k+100)(15 k+14)
\end{aligned}
$$
(We divide all even ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem N2. Find all positive integers $n$ such that $36^{n}-6$ is a product of two or more consecutive positive integers.
|
Solution. Answer: $n=1$.
Among each four consecutive integers there is a multiple of 4 . As $36^{n}-6$ is not a multiple of 4 , it must be the product of two or three consecutive positive integers.
Case I. If $36^{n}-6=x(x+1)$ (all letters here and below denote positive integers), then $4 \cdot 36^{n}-23=(2 x+1)^{2}... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A1. Let $a, b, c$ be positive real numbers such that $a b c=8$. Prove that
$$
\frac{a b+4}{a+2}+\frac{b c+4}{b+2}+\frac{c a+4}{c+2} \geq 6
$$
|
Solution. We have $a b+4=\frac{8}{c}+4=\frac{4(c+2)}{c}$ and similarly $b c+4=\frac{4(a+2)}{a}$ and $c a+4=\frac{4(b+2)}{b}$. It follows that
$$
(a b+4)(b c+4)(c a+4)=\frac{64}{a b c}(a+2)(b+2)(c+2)=8(a+2)(b+2)(c+2)
$$
so that
$$
\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}=8
$$
Applying AM-GM, we conclude:
$$
\... | 6 | Inequalities | proof | Yes | Yes | olympiads | false |
A3. Determine the number of pairs of integers $(m, n)$ such that
$$
\sqrt{n+\sqrt{2016}}+\sqrt{m-\sqrt{2016}} \in \mathbb{Q}
$$
|
Solution. Let $r=\sqrt{n+\sqrt{2016}}+\sqrt{m-\sqrt{2016}}$. Then
$$
n+m+2 \sqrt{n+\sqrt{2016}} \cdot \sqrt{m-\sqrt{2016}}=r^{2}
$$
and
$$
(m-n) \sqrt{2106}=\frac{1}{4}\left(r^{2}-m-n\right)^{2}-m n+2016 \in \mathbb{Q}
$$
Since $\sqrt{2016} \notin \mathbb{Q}$, it follows that $m=n$. Then
$$
\sqrt{n^{2}-2016}=\fra... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
C1. Let $S_{n}$ be the sum of reciprocal values of non-zero digits of all positive integers up to (and including) $n$. For instance, $S_{13}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{1}{1... | ## Solution.
We will first calculate $S_{999}$, then $S_{1999}-S_{999}$, and then $S_{2016}-S_{1999}$.
Writing the integers from 1 to 999 as 001 to 999, adding eventually also 000 (since 0 digits actually do not matter), each digit appears exactly 100 times in each position(as unit, ten, or hundred). Hence
$$
S_{999... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
N2. Find the maximum number of natural numbers $x_{1}, x_{2}, \ldots, x_{m}$ satisfying the conditions:
a) No $x_{i}-x_{j}, 1 \leq i<j \leq m$ is divisible by 11 ; and
b) The sum $x_{2} x_{3} \ldots x_{m}+x_{1} x_{3} \ldots x_{m}+\cdots+x_{1} x_{2} \ldots x_{m-1}$ is divisible by 11 .
|
Solution. The required maximum is 10 .
According to a), the numbers $x_{i}, 1 \leq i \leq m$, are all different $(\bmod 11)$ (1)
Hence, the number of natural numbers satisfying the conditions is at most 11.
If $x_{j} \equiv 0(\bmod 11)$ for some $j$, then
$$
x_{2} x_{3} \ldots x_{m}+x_{1} x_{3} \ldots x_{m}+\cdots... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A 2. Find the maximum positive integer $k$ such that for any positive integers $m, n$ such that $m^{3}+n^{3}>$ $(m+n)^{2}$, we have
$$
m^{3}+n^{3} \geq(m+n)^{2}+k
$$
|
Solution. We see that for $m=3$ and $n=2$ we have $m^{3}+n^{3}>(m+n)^{2}$, thus
$$
3^{3}+2^{3} \geq(3+2)^{2}+k \Rightarrow k \leq 10
$$
We will show that $k=10$ is the desired maximum. In other words, we have to prove that
$$
m^{3}+n^{3} \geq(m+n)^{2}+10
$$
The last inequality is equivalent to
$$
(m+n)\left(m^{2}... | 10 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
A 4. Let $k>1, n>2018$ be positive integers, and let $n$ be odd. The nonzero rational numbers $x_{1}$, $x_{2}, \ldots, x_{n}$ are not all equal and satisfy
$$
x_{1}+\frac{k}{x_{2}}=x_{2}+\frac{k}{x_{3}}=x_{3}+\frac{k}{x_{4}}=\cdots=x_{n-1}+\frac{k}{x_{n}}=x_{n}+\frac{k}{x_{1}}
$$
Find:
a) the product $x_{1} x_{2} \... |
Solution. a) If $x_{i}=x_{i+1}$ for some $i$ (assuming $x_{n+1}=x_{1}$ ), then by the given identity all $x_{i}$ will be equal, a contradiction. Thus $x_{1} \neq x_{2}$ and
$$
x_{1}-x_{2}=k \frac{x_{2}-x_{3}}{x_{2} x_{3}}
$$
Analogously
$$
x_{1}-x_{2}=k \frac{x_{2}-x_{3}}{x_{2} x_{3}}=k^{2} \frac{x_{3}-x_{4}}{\left... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
C 2. A set $T$ of $n$ three-digit numbers has the following five properties:
(1) No number contains the digit 0 .
(2) The sum of the digits of each number is 9 .
(3) The units digits of any two numbers are different.
(4) The tens digits of any two numbers are different.
(5) The hundreds digits of any two numbers ... |
Solution. Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0 . We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of $6 A$ 's (which means that we add 1 to the current digit) and 2 G's (w... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
C2 Five players $(A, B, C, D, E)$ take part in a bridge tournament. Every two players must play (as partners) against every other two players. Any two given players can be partners not more than once per day. What is the least number of days needed for this tournament?
|
Solution: A given pair must play with three other pairs and these plays must be in different days, so at three days are needed. Suppose that three days suffice. Let the pair $A B$ play against $C D$ on day $x$. Then $A B-D E$ and $C D-B E$ cannot play on day $x$. Then one of the other two plays of $D E$ (with $A C$ an... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
G2 In a right trapezoid $A B C D(A B \| C D)$ the angle at vertex $B$ measures $75^{\circ}$. Point $H$ is the foot of the perpendicular from point $A$ to the line $B C$. If $B H=D C$ and $A D+A H=8$, find the area of $A B C D$.
|
Solution: Produce the legs of the trapezoid until they intersect at point $E$. The triangles $A B H$ and $E C D$ are congruent (ASA). The area of $A B C D$ is equal to area of triangle $E A H$ of hypotenuse
$$
A E=A D+D E=A D+A H=8
$$
Let $M$ be the midpoint of $A E$. Then
$$
M E=M A=M H=4
$$
and $\angle A M H=30^... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
G6. A point $O$ and the circles $k_{1}$ with center $O$ and radius $3, k_{2}$ with center $O$ and radius 5, are given. Let $A$ be a point on $k_{1}$ and $B$ be a point on $k_{2}$. If $A B C$ is equilateral triangle, find the maximum value of the distance $O C$.
| ## Solution
It is easy to see that the points $O$ and $C$ must be in different semi-planes with respect to the line $A B$.
Let $O P B$ be an equilateral triangle ( $P$ and $C$ on the same side of $O B$ ). Since $\angle P B C$ $=60^{\circ}-\angle A B P$ and $\angle O B A=60^{\circ}-\angle A B P$, then $\angle P B C=\a... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
NT2. Find all natural numbers $n$ such that $5^{n}+12^{n}$ is perfect square.
| ## Solution
By checking the cases $n=1,2,3$ we get the solution $n=2$ and $13^{2}=5^{2}+12^{2}$.
If $n=2 k+1$ is odd, we consider the equation modulo 5 and we obtain
$$
\begin{aligned}
x^{2} & \equiv 5^{2 k+1}+12^{2 k+1}(\bmod 5) \equiv 2^{2 k} \cdot 2(\bmod 5) \\
& \equiv(-1)^{k} \cdot 2(\bmod 5) \equiv \pm 2(\bmod... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A4 Let $x, y$ be positive real numbers such that $x^{3}+y^{3} \leq x^{2}+y^{2}$. Find the greatest possible value of the product $x y$.
| ## Solution 1
We have $(x+y)\left(x^{2}+y^{2}\right) \geq(x+y)\left(x^{3}+y^{3}\right) \geq\left(x^{2}+y^{2}\right)^{2}$, hence $x+y \geq x^{2}+y^{2}$. Now $2(x+y) \geq(1+1)\left(x^{2}+y^{2}\right) \geq(x+y)^{2}$, thus $2 \geq x+y$. Because $x+y \geq 2 \sqrt{x y}$, we will obtain $1 \geq x y$. Equality holds when $x=y... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
A7 Let $a, b, c$ be positive real numbers with $a b c=1$. Prove the inequality:
$$
\frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1}+\frac{2 b^{2}+\frac{1}{b}}{c+\frac{1}{b}+1}+\frac{2 c^{2}+\frac{1}{c}}{a+\frac{1}{c}+1} \geq 3
$$
| ## Solution 1
By $A M-G M$ we have $2 x^{2}+\frac{1}{x}=x^{2}+x^{2}+\frac{1}{x} \geq 3 \sqrt[3]{\frac{x^{4}}{x}}=3 x$ for all $x>0$, so we have:
$\sum_{\text {cyc }} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \sum_{c y c} \frac{3 a}{1+b+b c}=3\left(\sum_{c y c} \frac{a^{2}}{1+a+a b}\right) \geq \frac{3(a+b+c)^{... | 3 | Inequalities | proof | Yes | Yes | olympiads | false |
G3 The vertices $A$ and $B$ of an equilateral $\triangle A B C$ lie on a circle $k$ of radius 1 , and the vertex $C$ is inside $k$. The point $D \neq B$ lies on $k, A D=A B$ and the line $D C$ intersects $k$ for the second time in point $E$. Find the length of the segment $C E$.
| ## Solution
As $A D=A C, \triangle C D A$ is isosceles. If $\varangle A D C=\varangle A C D=\alpha$ and $\varangle B C E=\beta$, then $\beta=120^{\circ}-\alpha$. The quadrilateral $A B E D$ is cyclic, so $\varangle A B E=180^{\circ}-\alpha$. Then $\varangle C B E=$ $120^{\circ}-\alpha$ so $\varangle C B E=\beta$. Thus... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
NT3 Let $s(a)$ denote the sum of digits of a given positive integer $a$. The sequence $a_{1}, a_{2}, \ldots a_{n}, \ldots$ of positive integers is such that $a_{n+1}=a_{n}+s\left(a_{n}\right)$ for each positive integer $n$. Find the greatest possible $n$ for which it is possible to have $a_{n}=2008$.
| ## Solution
Since $a_{n-1} \equiv s\left(a_{n-1}\right)$ (all congruences are modulo 9 ), we have $2 a_{n-1} \equiv a_{n} \equiv 2008 \equiv 10$, so $a_{n-1} \equiv 5$. But $a_{n-1}<2008$, so $s\left(a_{n-1}\right) \leq 28$ and thus $s\left(a_{n-1}\right)$ can equal 5,14 or 23 . We check $s(2008-5)=s(2003)=5, s(2008-1... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT4 Find all integers $n$ such that $n^{4}+8 n+11$ is a product of two or more consecutive integers.
|
Solution
We will prove that $n^{4}+8 n+11$ is never a multiple of 3 . This is clear if $n$ is a multiple of 3 . If
$n$ is not a multiple of 3 , then $n^{4}+8 n+11=\left(n^{4}-1\right)+12+8 n=(n-1)(n+1)\left(n^{2}+1\right)+12+8 n$, where $8 n$ is the only term not divisible by 3 . Thus $n^{4}+8 n+11$ is never the prod... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT6 Let $f: \mathbb{N} \rightarrow \mathbb{R}$ be a function, satisfying the following condition:
for every integer $n>1$, there exists a prime divisor $p$ of $n$ such that $f(n)=f\left(\frac{n}{p}\right)-f(p)$. If
$$
f\left(2^{2007}\right)+f\left(3^{2008}\right)+f\left(5^{2009}\right)=2006
$$
determine the value o... | ## Solution
If $n=p$ is prime number, we have
$$
f(p)=f\left(\frac{p}{p}\right)-f(p)=f(1)-f(p)
$$
i.e.
$$
f(p)=\frac{f(1)}{2}
$$
If $n=p q$, where $p$ and $q$ are prime numbers, then
$$
f(n)=f\left(\frac{n}{p}\right)-f(p)=f(q)-f(p)=\frac{f(1)}{2}-\frac{f(1)}{2}=0
$$
If $n$ is a product of three prime numbers, we... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
ALG 5. Let $A B C$ be a scalene triangle with $B C=a, A C=b$ and $A B=c$, where $a_{r} b, c$ are positive integers. Prove that
$$
\left|a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a\right| \geq 2
$$
|
Solution. Denote $E=a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a$. We have
$$
\begin{aligned}
E= & \left(a b c-c^{2} a\right)+\left(c a^{2}-a^{2} b\right)+\left(b c^{2}-b^{2} c\right)+\left(a b^{2}-a b c\right)= \\
& (b-c)\left(a c-a^{2}-b c+a b\right)=(b-c)\left(a a^{2}-b\right)(c-a)
\end{aligned}
$$
So, $|E|=|a... | 2 | Algebra | proof | Yes | Yes | olympiads | false |
87.3. Let $f$ be a strictly increasing function defined in the set of natural numbers satisfying the conditions $f(2)=a>2$ and $f(m n)=f(m) f(n)$ for all natural numbers $m$ and $n$. Determine the smallest possible value of $a$.
|
Solution. Since $f(n)=n^{2}$ is a function satisfying the conditions of the problem, the smallest posiible $a$ is at most 4. Assume $a=3$. It is easy to prove by induction that $f\left(n^{k}\right)=f(n)^{k}$ for all $k \geq 1$. So, taking into account that $f$ is strictly increasing, we get
$$
\begin{gathered}
f(3)^{... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
91.1. Determine the last two digits of the number
$$
2^{5}+2^{5^{2}}+2^{5^{3}}+\cdots+2^{5^{1991}}
$$
written in decimal notation.
|
Solution. We first show that all numbers $2^{5^{k}}$ are of the form $100 p+32$. This can be shown by induction. The case $k=1$ is clear $\left(2^{5}=32\right)$. Assume $2^{5^{k}}=100 p+32$. Then, by the binomial formula,
$$
2^{5^{k+1}}=\left(2^{5^{k}}\right)^{5}=(100 p+32)^{5}=100 q+32^{5}
$$
and
$$
\begin{gathere... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
92.4. Peter has many squares of equal side. Some of the squares are black, some are white. Peter wants to assemble a big square, with side equal to $n$ sides of the small squares, so that the big square has no rectangle formed by the small squares such that all the squares in the vertices of the rectangle are of equal... |
Solution. We show that Peter only can make a $4 \times 4$ square. The construction is possible, if $n=4$ :
Now consider the case $n=5$. We may assume that at least 13 of the 25 squares are ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
96.4. The real-valued function $f$ is defined for positive integers, and the positive integer a satisfies
$$
\begin{gathered}
f(a)=f(1995), \quad f(a+1)=f(1996), \quad f(a+2)=f(1997) \\
f(n+a)=\frac{f(n)-1}{f(n)+1} \quad \text { for all positive integers } n
\end{gathered}
$$
(i) Show that $f(n+4 a)=f(n)$ for all po... |
Solution. To prove (i), we the formula $f(n+a)=\frac{f(n)-1}{f(n)+1}$ repeatedly:
$$
\begin{gathered}
f(n+2 a)=f((n+a)+a)=\frac{\frac{f(n)-1}{f(n)+1}-1}{\frac{f(n)-1}{f(n)+1}+1}=-\frac{1}{f(n)} \\
f(n+4 a)=f((n+2 a)+2 a)=-\frac{1}{-\frac{1}{f(n)}}=f(n)
\end{gathered}
$$
(ii) If $a=1$, then $f(1)=f(a)=f(1995)=f(3+498... | 3 | Algebra | proof | Yes | Yes | olympiads | false |
97.1. Let A be a set of seven positive numbers. Determine the maximal number of triples $(x, y, z)$ of elements of A satisfying $x<y$ and $x+y=z$.
|
Solution. Let $0<a_{1}<a_{2}<\ldots<a_{7}$ be the elements of the set $A$. If $\left(a_{i}, a_{j}, a_{k}\right)$ is a triple of the kind required in the problem, then $a_{i}<a_{j}<a_{i}+a_{j}=a_{k}$. There are at most $k-1$ pairs $\left(a_{i}, a_{j}\right)$ such that $a_{i}+a_{j}=a_{k}$. The number of pairs satisfying... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
99.2. Consider 7-gons inscribed in a circle such that all sides of the 7-gon are of different length. Determine the maximal number of $120^{\circ}$ angles in this kind of a 7-gon.
|
Solution. It is easy to give examples of heptagons $A B C D E F G$ inscribed in a circle with all sides unequal and two angles equal to $120^{\circ}$. These angles cannot lie on adjacent vertices of the heptagon. In fact, if $\angle A B C=\angle B C D=120^{\circ}$, and arc $B C$ equals $b^{\circ}$, then arcs $A B$ and... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
01.3. Determine the number of real roots of the equation
$$
x^{8}-x^{7}+2 x^{6}-2 x^{5}+3 x^{4}-3 x^{3}+4 x^{2}-4 x+\frac{5}{2}=0
$$
|
Solution. Write
$$
\begin{gathered}
x^{8}-x^{7}+2 x^{6}-2 x^{5}+3 x^{4}-3 x^{3}+4 x^{2}-4 x+\frac{5}{2} \\
=x(x-1)\left(x^{6}+2 x^{4}+3 x^{2}+4\right)+\frac{5}{2}
\end{gathered}
$$
If $x(x-1) \geq 0$, i.e. $x \leq 0$ or $x \geq 1$, the equation has no roots. If $0x(x-1)=\left(x-\frac{1}{2}\right)^{2}-\frac{1}{4} \ge... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
06.3. A sequence of positive integers $\left\{a_{n}\right\}$ is given by
$$
a_{0}=m \quad \text { and } \quad a_{n+1}=a_{n}^{5}+487
$$
for all $n \geq 0$. Determine all values of $m$ for which the sequence contains as many square numbers as possible.
|
Solution. Consider the expression $x^{5}+487$ modulo 4. Clearly $x \equiv 0 \Rightarrow x^{5}+487 \equiv 3$, $x \equiv 1 \Rightarrow x^{5}+487 \equiv 0 ; x \equiv 2 \Rightarrow x^{5}+487 \equiv 3$, and $x \equiv 3 \Rightarrow x^{5}+487 \equiv 2$. Square numbers are always $\equiv 0$ or $\equiv 1 \bmod 4$. If there is ... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
09.2. On a faded piece of paper it is possible, with some effort, to discern the following:
$$
\left(x^{2}+x+a\right)\left(x^{15}-\ldots\right)=x^{17}+x^{13}+x^{5}-90 x^{4}+x-90
$$
Some parts have got lost, partly the constant term of the first factor of the left side, partly the main part of the other factor. It wo... |
Solution. We denote the polynomial $x^{2}+x+a$ by $P_{a}(x)$, the polynomial forming the other factor of the left side by $Q(x)$ and the polynomial on the right side by $R(x)$. The polynomials are integer valued for every integer $x$. For $x=0$ we get $P_{a}(0)=a$ and $R(0)=-90$, so $a$ is a divisor of $90=2 \cdot 3 \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. Laura has 2010 lamps connected with 2010 buttons in front of her. For each button, she wants to know the corresponding lamp. In order to do this, she observes which lamps are lit when Richard presses a selection of buttons. (Not pressing anything is also a possible selection.) Richard always presses the buttons ... |
Solution. a) Let us say that two lamps are separated, if one of the lamps is turned on while the other lamp remains off. Laura can find out which lamps belong to the buttons if every two lamps are separated. Let Richard choose two arbitrary lamps. To begin with, he turns both lamps on and then varies all the other lam... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.4. A positive integer is called simple if its ordinary decimal representation consists entirely of zeroes and ones. Find the least positive integer $k$ such that each positive integer $n$ can be written as $n=a_{1} \pm a_{2} \pm a_{3} \pm \cdots \pm a_{k}$, where $a_{1}, \ldots, a_{k}$ are simple.
|
Solution. We can always write $n=a_{l}+a_{2}+\cdots+a_{9}$ where $a_{j}$ has 1 's in the places where $n$ has digits greater or equal to $j$ and 0 's in the other places. So $k \leq 9$. To show that $k \geq 9$, consider $n=10203040506070809$. Suppose $n=a_{l}+a_{2}+\cdots+a_{j}-a_{j+l}-a_{j+2}-\cdots-a_{k}$, where $a_... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
PROBLEM 1. The real numbers $a, b, c$ are such that $a^{2}+b^{2}=2 c^{2}$, and also such that $a \neq b, c \neq-a, c \neq-b$. Show that
$$
\frac{(a+b+2 c)\left(2 a^{2}-b^{2}-c^{2}\right)}{(a-b)(a+c)(b+c)}
$$
is an integer.
|
SolUTiON. Let us first note that
$$
\frac{a+b+2 c}{(a+c)(b+c)}=\frac{(a+c)+(b+c)}{(a+c)(b+c)}=\frac{1}{a+c}+\frac{1}{b+c}
$$
Further we have
$$
2 a^{2}-b^{2}-c^{2}=2 a^{2}-\left(2 c^{2}-a^{2}\right)-c^{2}=3 a^{2}-3 c^{2}=3(a+c)(a-c)
$$
and
$$
2 a^{2}-b^{2}-c^{2}=2\left(2 c^{2}-b^{2}\right)-b^{2}-c^{2}=3 c^{2}-3 b... | 3 | Algebra | proof | Yes | Yes | olympiads | false |
Problem 3. Find the smallest positive integer $n$, such that there exist $n$ integers $x_{1}, x_{2}, \ldots, x_{n}$ (not necessarily different), with $1 \leq x_{k} \leq n, 1 \leq k \leq n$, and such that
$$
x_{1}+x_{2}+\cdots+x_{n}=\frac{n(n+1)}{2}, \quad \text { and } \quad x_{1} x_{2} \cdots x_{n}=n!
$$
but $\left... |
Solution. If it is possible to find a set of numbers as required for some $n=k$, then it will also be possible for $n=k+1$ (choose $x_{1}, \ldots, x_{k}$ as for $n=k$, and
let $x_{k+1}=k+1$ ). Thus we have to find a positive integer $n$ such that a set as required exists, and prove that such a set does not exist for $... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXIV OM - I - Problem 10
Find the smallest natural number $ n > 1 $ with the following property: there exists a set $ Z $ consisting of $ n $ points in the plane such that every line $ AB $ ($ A, B \in Z $) is parallel to some other line $ CD $ ($ C, D \in Z $). | We will first prove that the set $ Z $ of vertices of a regular pentagon has the property given in the problem, that is, $ n \leq 5 $. We will show that each side of the regular pentagon is parallel to a certain diagonal and vice versa, each diagonal is parallel to a corresponding side.
It suffices to prove that $ AB \... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LVII OM - III - Problem 2
Determine all positive integers $ k $ for which the number $ 3^k+5^k $ is a power of an integer with an exponent greater than 1. | If $ k $ is an even number, then the numbers $ 3^k $ and $ 5^k $ are squares of odd numbers, giving a remainder of 1 when divided by 4. Hence, the number $ 3^k + 5^k $ gives a remainder of 2 when divided by 4, and thus is divisible by 2 but not by $ 2^2 $. Such a number cannot be a power of an integer with an exponent ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXXIV OM - II - Problem 6
For a given number $ n $, let $ p_n $ denote the probability that when a pair of integers $ k, m $ satisfying the conditions $ 0 \leq k \leq m \leq 2^n $ is chosen at random (each pair is equally likely), the number $ \binom{m}{k} $ is even. Calculate $ \lim_{n\to \infty} p_n $. | om34_2r_img_10.jpg
The diagram in Figure 10 shows Pascal's triangle written modulo $2$, meaning it has zeros and ones in the places where the usual Pascal's triangle has even and odd numbers, respectively. Just like in the usual Pascal's triangle, each element here is the sum of the elements directly above it, accordin... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
IX OM - II - Task 2
Six equal disks are placed on a plane in such a way that their centers lie at the vertices of a regular hexagon with a side equal to the diameter of the disks. How many rotations will a seventh disk of the same size make while rolling externally on the same plane along the disks until it returns to... | Let circle $K$ with center $O$ and radius $r$ roll without slipping on a circle with center $S$ and radius $R$ (Fig. 16). The rolling without slipping means that different points of one circle successively coincide with different points of the other circle, and in this correspondence, the length of the arc between two ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXXVI OM - III - Problem 1
Determine the largest number $ k $ such that for every natural number $ n $ there are at least $ k $ natural numbers greater than $ n $, less than $ n+17 $, and coprime with the product $ n(n+17) $. | We will first prove that for every natural number $n$, there exists at least one natural number between $n$ and $n+17$ that is coprime with $n(n+17)$.
In the case where $n$ is an even number, the required property is satisfied by the number $n+1$. Of course, the numbers $n$ and $n+1$ are coprime. If a number $d > 1$ we... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
III OM - I - Task 4
a) Given points $ A $, $ B $, $ C $ not lying on a straight line. Determine three mutually parallel lines passing through points $ A $, $ B $, $ C $, respectively, so that the distances between adjacent parallel lines are equal.
b) Given points $ A $, $ B $, $ C $, $ D $ not lying on a plane. Deter... | a) Suppose that the lines $a$, $b$, $c$ passing through points $A$, $B$, $C$ respectively and being mutually parallel satisfy the condition of the problem, that is, the distances between adjacent parallel lines are equal. Then the line among $a$, $b$, $c$ that lies between the other two is equidistant from them. Let th... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LV OM - III - Task 5
Determine the maximum number of lines in space passing through a fixed point and such that any two intersect at the same angle. | Let $ \ell_1,\ldots,\ell_n $ be lines passing through a common point $ O $. A pair of intersecting lines determines four angles on the plane containing them: two vertical angles with measure $ \alpha \leq 90^\circ $ and the other two angles with measure $ 180^\circ - \alpha $. According to the assumption, the value of ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XVI OM - II - Task 4
Find all prime numbers $ p $ such that $ 4p^2 +1 $ and $ 6p^2 + 1 $ are also prime numbers. | To solve the problem, we will investigate the divisibility of the numbers \( u = 4p^2 + 1 \) and \( v = 6p^2 + 1 \) by \( 5 \). It is known that the remainder of the division of the product of two integers by a natural number is equal to the remainder of the division of the product of their remainders by that number. B... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
L OM - I - Task 5
Find all pairs of positive integers $ x $, $ y $ satisfying the equation $ y^x = x^{50} $. | We write the given equation in the form $ y = x^{50/x} $. Since for every $ x $ being a divisor of $ 50 $, the number on the right side is an integer, we obtain solutions of the equation for $ x \in \{1,2,5,10,25,50\} $. Other solutions of this equation will only be obtained when $ x \geq 2 $ and for some $ k \geq 2 $,... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXXIX OM - I - Problem 1
For each positive number $ a $, determine the number of roots of the polynomial $ x^3+(a+2)x^2-x-3a $. | Let's denote the considered polynomial by $ F(x) $. A polynomial of the third degree has at most three real roots. We will show that the polynomial $ F $ has at least three real roots - and thus has exactly three real roots (for any value of the parameter $ a > 0 $).
It is enough to notice that
If a continuous func... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
V OM - I - Task 2
Investigate when the sum of the cubes of three consecutive natural numbers is divisible by $18$. | Let $ a - 1 $, $ a $, $ a + 1 $, be three consecutive natural numbers; the sum of their cubes
can be transformed in the following way:
Since one of the numbers $ a - 1 $, $ a $, $ a + 1 $ is divisible by $ 3 $, then one of the numbers $ a $ and $ (a + 1) (a - 1) + 3 $ is also divisible by $ 3 $. Therefore, the sum $ ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
LIX OM - II - Task 1
Determine the maximum possible length of a sequence of consecutive integers, each of which can be expressed in the form $ x^3 + 2y^2 $ for some integers $ x, y $. | A sequence of five consecutive integers -1, 0, 1, 2, 3 satisfies the conditions of the problem: indeed, we have
On the other hand, among any six consecutive integers, there exists a number, say
$ m $, which gives a remainder of 4 or 6 when divided by 8. The number $ m $ is even; if there were a representation
in... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXIV OM - III - Task 2
Let $ p_n $ be the probability that a series of 100 consecutive heads will appear in $ n $ coin tosses. Prove that the sequence of numbers $ p_n $ is convergent and calculate its limit. | The number of elementary events is equal to the number of $n$-element sequences with two values: heads and tails, i.e., the number $2^n$. A favorable event is a sequence containing 100 consecutive heads. We estimate the number of unfavorable events from above, i.e., the number of sequences not containing 100 consecutiv... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false |
XV OM - II - Task 3
Prove that if three prime numbers form an arithmetic progression with a difference not divisible by 6, then the smallest of these numbers is $3$. | Suppose that the prime numbers $ p_1 $, $ p_2 $, $ p_3 $ form an arithmetic progression with a difference $ r > 0 $ not divisible by $ 6 $, and the smallest of them is $ p_1 $. Then
Therefore, $ p_1 \geq 3 $, for if $ p_1 = 2 $, the number $ p_3 $ would be an even number greater than $ 2 $, and thus would not be a pri... | 3 | Number Theory | proof | Yes | Yes | olympiads | false |
XLIII OM - I - Problem 2
In square $ABCD$ with side length $1$, point $E$ lies on side $BC$, point $F$ lies on side $CD$, the measures of angles $EAB$ and $EAF$ are $20^{\circ}$ and $45^{\circ}$, respectively. Calculate the height of triangle $AEF$ drawn from vertex $A$. | The measure of angle $ FAD $ is $ 90^\circ - (20^\circ + 45^\circ) = 25^\circ $. From point $ A $, we draw a ray forming angles of $ 20^\circ $ and $ 25^\circ $ with rays $ AE $ and $ AF $, respectively, and we place a segment $ AG $ of length $ 1 $ on it (figure 2).
From the equality $ |AG| =|AB| = 1 $, $ | \measureda... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XLVI OM - III - Problem 2
The diagonals of a convex pentagon divide this pentagon into a pentagon and ten triangles. What is the maximum possible number of triangles with equal areas? | om46_3r_img_12.jpg
Let's denote the considered pentagon by $ABCDE$, and the pentagon formed by the intersection points of the diagonals by $KLMNP$ so that the following triangles are those mentioned in the problem:
$\Delta_0$: triangle $LEM$; $\quad \Delta_1$: triangle $EMA$;
$\Delta_2$: triangle $MAN$; $\quad \Delta... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LIX OM - I - Task 9
Determine the smallest real number a with the following property:
For any real numbers $ x, y, z \geqslant a $ satisfying the condition $ x + y + z = 3 $
the inequality holds | Answer: $ a = -5 $.
In the solution, we will use the following identity:
Suppose that the number $ a \leqslant 1 $ has the property given in the problem statement. The numbers
$ x = a, y = z = 2\cdot (\frac{3-a}{2}) $ satisfy the conditions $ x, y, z \geqslant a $ and $ x+y+z = 3 $, thus
by virtue of (1),... | -5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
XV OM - I - Problem 7
Given a circle and points $ A $ and $ B $ inside it. Find a point $ P $ on this circle such that the angle $ APB $ is subtended by a chord $ MN $ equal to $ AB $. Does the problem have a solution if the given points, or only one of them, lie outside the circle? | Suppose that point $ P $ of a given circle $ C $ with radius $ r $ is a solution to the problem (Fig. 7).
Since points $ A $ and $ B $ lie inside the circle $ C $, points $ M $ and $ N $ lie on the rays $ PA $ and $ PB $ respectively, and angle $ APB $ coincides with angle $ MPN $. Triangles $ APB $ and $ MPN $ have eq... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. By what natural number can the numerator and denominator of the ordinary fraction of the form $\frac{4 n+3}{5 n+2}$ be reduced? For which integers $n$ can this occur? | Answer: can be reduced by 7 when $n=7 k+1, k \in Z$. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Let $A=\overline{a b c b a}$ be a five-digit symmetric number, $a \neq 0$. If $1 \leq a \leq 8$, then the last digit of the number $A+11$ will be $a+1$, and therefore the first digit in the representation of $A+11$ should also be $a+1$. This is possible only with a carry-over from the digit, i.e., when $b=c=9$. Then... | Answer: eight numbers of the form $\overline{a 999 a}$, where $a=1,2, \ldots, 8$. | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. A plot of land in the form of a right-angled triangle with legs of 4 and 3 needs to be divided by a line $L$ into two plots such that 1) the plots have equal area; 2) the length of the common boundary (fence) of the plots is minimized. Indicate the points on the sides of the triangle through which the desired line $... | Problem 6. Answer: 1) The line intersects the larger leg $B C$ (angle $\measuredangle C=90^{\circ}$) at point $M: B M=\sqrt{10}$ and the hypotenuse $B A$ at point $N: B N=\sqrt{10}$.
2) $L_{\text {min }}=2$ | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. According to the properties of logarithms, after transformations we get
$$
\begin{aligned}
& \log _{2}\left(b_{2} b_{3} \ldots b_{n}\right)=\log _{2} b_{1}^{2} \\
& b_{2} b_{3} \ldots b_{n}=b_{1}^{2}
\end{aligned}
$$
Using the formula for the general term of a geometric progression, we get
$$
b_{2}=b_{1} q, b_{3}... | Answer: -12 when $n=4$ and $n=9$. | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The border of a square with a side of 9, cut out of white cardboard, is painted red. It is necessary to cut the square into 6 equal-area parts, the boundaries of which contain segments painted red with the same total length.
## Solutions
Option 1
Problem 1 | Answer: 2 km.
Solution
$S-$ the length of the path, $S_{1}$ - the length of the path on descents, $S_{2}$ - the length of the path on ascents, $S / 2=S_{1}+S_{2}$
$t=15-8-6=1-$ the time of the journey there and back. Then
$1=\frac{S_{1}}{6}+\frac{S_{2}}{3}+\frac{S}{2 \cdot 4}+\frac{S_{1}}{3}+\frac{S_{2}}{6}+\frac{S... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. The real numbers $x_{1}, x_{2}$, and $x_{3}$ are the three roots of the equation $x^{3}-3 x^{2}+2(1-p) x+4=0$, considering their possible multiplicities. Find the smallest value of the expression $\left(x_{1}-1\right)^{2}+\left(x_{2}-1\right)^{2}+\left(x_{3}-1\right)^{2}$ under these conditions. For which $p$ is it ... | The smallest value 6 is achieved when $p=1$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The quadratic trinomial $p(x)=a x^{2}+b x+c, a>0$ when divided by ( $x-1$ ) gives a remainder of 4, and when divided by ( $x-2$ ) - a remainder of 15. Find the maximum possible value of the ordinate of the vertex of the parabola $y=p(x)$ under these conditions. For what value of $x$ is it achieved? | 4. Solution. By the Remainder Theorem, the remainder of the division of a polynomial $p(x)$ by $x-a$ is $p(a)$.
Therefore, the conditions of the problem are equivalent to the system:
$\left\{\begin{array}{l}4 a+2 b+c=15 \\ a+b+c=4\end{array} \Rightarrow\left\{\begin{array}{l}b=11-3 a \\ c=2 a-7\end{array}\right.\righ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Pasha, Masha, Tolya, and Olya ate 88 candies, and each of them ate at least one candy. Masha and Tolya ate 57 candies, but Pasha ate the most candies. How many candies did Olya eat? | Solution. Either Masha or Tolya ate no less than 29 candies, then Pasha ate no less than 30 candies. Then the number of candies eaten by Pasha, Masha, and Tolya is no less than $57+30=87$. Since there are a total of 88 candies, and Olya did not refuse any candies, she gets one candy.
Answer: 1 candy. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Let point $M$ divide edge $A B$ in the ratio $A M: M B=\lambda$, point $N$ divide edge $D C$ in the ratio $D N: N C=\mu$, and point $P$ divide edge $D B$ in the ratio $D P: P B=\theta$. We need to find the ratio $A Q: Q C$.
 x^{2}+\left(6 n^{2}-31 n-106\right) x-6(n-8)(n+2)=0 \text { greater than }-1 \text { ? }
$$ | Solution. Rewrite the equation as
$$
x^{3}-(5 n-9) x^{2}+\left(6 n^{2}-31 n-106\right) x-6 n^{2}+36 n+96=0 .
$$
Notice that $x_{1}=1>-1$ is a solution for all $n$ (the sum of the coefficients of the polynomial on the left side of the equation is zero for all $n$). Divide the equation by $x-1$:
$$
x^{2}-5(n-2) x+6\le... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The midline of a trapezoid is 4. A line parallel to the bases of the trapezoid and dividing its area in half intersects the lateral sides at points $M$ and $N$. Find the smallest possible length of the segment $M N$. | # Solution.
Notations: $A D=a, B C=b, C K=H, N E=h$ - heights of triangles $N C P$ and $D N Q$, $M N=x$
$$
S_{\text {AMND }}=S_{\text {MBCN }} \rightarrow \frac{x+b}{2} H=\frac{a+x}{2} h \r... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find all $x$ that satisfy the inequality $n^{2} x^{2}-\left(2 n^{2}+n\right) x+n^{2}+n-6 \leq 0$ for any natural $n$. | Solution. The roots of the quadratic trinomial $n^{2} x^{2}-\left(2 n^{2}+n\right) x+n^{2}+n-6$ are
$$
x_{1}=1-\frac{2}{n}, x_{2}=1+\frac{3}{n}
$$
Factorize the left side of the inequality
$$
n^{2}\left(x-\left(1-\frac{2}{n}\right)\right)\left(x-\left(1+\frac{3}{n}\right)\right) \leq 0
$$
Solving the inequality usi... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. On the edges of a trihedral angle with vertex at point $S$, there are points $M, N$, and $K$ such that $S M^{2}+S N^{2}+S K^{2} \leq 12$. Find the area of triangle $S M N$, given that the angle $M S N$ is $30^{\circ}$, and the volume of the pyramid $S M N K$ is maximally possible. | Solution. Let's introduce the notations: $S M=m, S N=n, S K=k$.
The volume of the pyramid $S M N K$ is
$$
V=\frac{1}{3} S_{S M N} \cdot h=\frac{1}{6} m n \sin \alpha \cdot h=\frac{1}{6} m n... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Quadrilateral $ABCD$ is inscribed in a circle, and its diagonals intersect at point $P$. Points $K, L$, and $M$ are the midpoints of sides $AB, BC$, and $CD$ respectively. The radius of the circle circumscribed around triangle $KLP$ is 1. Find the radius of the circle circumscribed around triangle $LMP$.
Problem 1 ... | Solution. Let $y_{k}$ be the number of passengers in the car with number $k, k=1,2,3, \ldots, 10$. According to the problem, $\sum_{k=1}^{10} y_{k}=270$. Additionally, it is stated that $y_{2} \geq y_{1}+2, y_{3} \geq y_{1}+4, \ldots, y_{9} \geq y_{1}+16, \quad y_{10} \geq y_{1}+18$. Adding these inequalities,
$$
y_{2... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.
when transferring money abroad
Correct answers: when using the Fast Payment System for amounts up to 100 thousand rubles per month, when transferring funds between one's own accounts in the same bank
Question 17
Score: 7.00
Last year, a beauty salon offered a $20 \%$ discount on facial massage when purchasing ... | Answer:
The correct answer: 1
Question 18
Score: 3.00
Select all true statements regarding digital financial assets $(DFA)$.
Select one or more answers: | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (mathematics) There are scales with two pans, 4 weights of 2 kg each, 3 weights of 3 kg each, and two weights of 5 kg each. In how many different ways can a 12 kg load be balanced on the scales, if the weights are allowed to be placed on both pans? | Answer: 7 ways
Solution: Let $x$ be the number of 2 kg weights used in weighing, $y$ be the number of 3 kg weights, and $z$ be the number of 5 kg weights. Then the equilibrium condition is g... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Express z from the first equation and substitute into the second:
$x^{2}-2 x+y^{2}-2 \sqrt{3} y=-4 \rightarrow(x-1)^{2}+(y-\sqrt{3})^{2}=0 \rightarrow\left\{\begin{array}{c}x=1 \\ y=\sqrt{3}\end{array} \rightarrow z=x^{2}+y^{2}+2 x=6\right.$ | Answer: $x=1, y=\sqrt{3}, z=6$ | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. The administration divided the region into several districts based on the principle: the population of a large district exceeds $8 \%$ of the region's population and for any large district, there are two non-large districts with a combined population that is larger. Into what minimum number of districts was the... | Answer: 8 districts.
Solution. The number of "small" districts is no less than 2 according to the condition, and their population does not exceed $8 \%$ of the total population of the region. We will show that the number of districts in the region is no less than 8. If the number of districts in the region is no more ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Point $M$ is located on side $CD$ of a square such that $CM: MD=1: 3$. Line $AM$ intersects the circumcircle of the square at point $E$. The area of triangle $ACE$ is 14. Find the side length of the square. | Answer: 10.
Solution.
Triangles $A M D$ and $C M E$ are similar with a similarity coefficient $k=5$. Then
$$
C E=\frac{4 x}{5}, M E=\frac{3 x}{5} \rightarrow A E=5 x+\frac{3 x}{5}=\frac{28... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. Seven students in the class receive one two every two days of study, and nine other students receive one two every three days each. The rest of the students in the class never receive twos. From Monday to Friday, 30 new twos appeared in the journal. How many new twos will appear in the class journal on Satur... | Answer: 9
Solution. Over the period from Monday to Saturday (six days), in the journal, there will be 3 new twos from each student of the first group (seven people) and 2 new twos from each of the 9 students of the second group. The total number of new twos for the school week is $7 \cdot 3 + 9 \cdot 2 = 39$. Then, on... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Through the vertex $B$ of an equilateral triangle $ABC$, a line $L$ is drawn, intersecting the extension of side $AC$ beyond point $C$. On line $L$, segments $BM$ and $BN$ are laid out, each equal in length to the side of triangle $ABC$. The lines $MC$ and $NA$ intersect at a common point $D$ and intersect t... | # Answer: 1.
Solution. Triangles $M B C$ and $A B N$ are isosceles, therefore
$$
\begin{gathered}
\angle B M C = \angle B C M = \alpha \rightarrow \angle N B C = 2 \alpha, \angle B A N = \angle B N A = \beta \rightarrow \angle A B M = 2 \beta \\
2 \alpha + 2 \beta + 60^{\circ} = 180^{\circ} \rightarrow \alpha + \beta... | 1 | Geometry | proof | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane that have natural coordinates $(x, y)$ and lie on the parabola $y=-\frac{x^{2}}{4}+3 x+\frac{253}{4}$. | Answer: 11.
Solution. Let's find those values of $x$ for which $y$ is positive: $-\frac{x^{2}}{4}+3 x+\frac{253}{4}>0 \Leftrightarrow-\frac{1}{4}(x+11)(x-23)>0$, from which $-11<x<23$. On this interval, there are 22 natural values of $x: x=1, x=2, \ldots, x=22$. During this interval, $y$ takes integer values only for ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{4}+5 x+39$. | Answer: 12.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{4}+5 x+39>0 \Leftrightarrow-\frac{1}{4}(x+6)(x-26)>0$, from which $-6<x<26$. On this interval, there are 25 natural values of $x: x=1, x=2, \ldots, x=25$. In this interval, $y$ takes integer values only for even $x$ - a total ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{3}+7 x+54$. | Answer: 8.
Solution. Let's find those values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+7 x+54>0 \Leftrightarrow-\frac{1}{3}(x+6)(x-27)>0$, from which $-6<x<27$. On this interval, there are 26 natural values of $x: x=1, x=2, \ldots, x=26$. In this interval, $y$ takes integer values only when $x$ is divisible ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{3}+5 x+72$. | Answer: 7.
Solution. Let's find those values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+5 x+72>0 \Leftrightarrow-\frac{1}{3}(x+9)(x-24)>0$, from which $-9<x<24$. On this interval, there are 23 natural values of $x: x=1, x=2, \ldots, x=23$. During this time, $y$ takes integer values only when $x$ is divisible ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{9}+50$. | Answer: 7.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{9}+50>0 \Leftrightarrow x^{2}<450$, from which $-\sqrt{450}<x<\sqrt{450}$. On this interval, there are 21 natural values of $x: x=1, x=2, \ldots, x=21$. During this time, $y$ takes integer values only when $x$ is divisible by 3... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane that have natural coordinates $(x, y)$ and lie on the parabola $y=-\frac{x^{2}}{3}+70$. | Answer: 4.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+70>0 \Leftrightarrow x^{2}<210$, from which $-\sqrt{210}<x<\sqrt{210}$. On this interval, there are 14 natural values of $x: x=1, x=2, \ldots, x=14$. During this time, $y$ takes integer values only when $x$ is divisible by 3... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{9}+33$. | Answer: 5.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{9}+33>0 \Leftrightarrow x^{2}<297$, from which $-\sqrt{297}<x<\sqrt{297}$. On this interval, there are 17 natural values of $x: x=1, x=2, \ldots, x=17$. At the same time, $y$ takes integer values only when $x$ is divisible by 3... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the plane $x O y$ that have natural coordinates $(x, y)$ and lie on the parabola $y=-\frac{x^{2}}{3}+98$ | Answer: 5.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+98>0 \Leftrightarrow x^{2}<294$, from which $-\sqrt{294}<x<\sqrt{294}$. On this interval, there are 17 natural values of $x: x=1, x=2, \ldots, x=17$. At the same time, $y$ takes integer values only when $x$ is divisible by 3... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
17. It is known that the number $a$ satisfies the equation param1, and the number $b$ satisfies the equation param2. Find the greatest possible value of the sum $a+b$.
| param1 | param2 | |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-3 x^{2}+5 x+11=0$ | |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}+3 x^{2}+... | 17. It is known that the number $a$ satisfies the equation param 1, and the number $b$ satisfies the equation param2. Find the greatest possible value of the sum $a+b$.
| param 1 | param2 | Answer |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-3 x^{2}+5 x+11=0$ | 2 |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
20. Find param 1 given param 2.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| maximum $2 x+y$ | $\|4 x-3 y\|+5 \sqrt{x^{2}+y^{2}-20 y+100}=30$ | |
| maximum $x+2 y$ | $\|4 y-3 x\|+5 \sqrt{x^{2}+y^{2}+20 y+100}=40$ | |
| maximum $2 y-x$ | $\|4 y+3 x\|+5 \sqrt{x^{2}+y^{2}+10 x+25}=15$ | |
| maximum $x-5 y$... | 20. Find param 1 given param2.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| maximum $2 x+y$ | $\|4 x-3 y\|+5 \sqrt{x^{2}+y^{2}-20 y+100}=30$ | 16 |
| maximum $x+2 y$ | $\|4 y-3 x\|+5 \sqrt{x^{2}+y^{2}+20 y+100}=40$ | -12 |
| maximum $2 y-x$ | $\|4 y+3 x\|+5 \sqrt{x^{2}+y^{2}+10 x+25}=15$ | 8 |
| maximum $x... | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
|x-1|+|5-x| \leqslant 4 \\
\frac{x^{2}-6 x+2 y+7}{y+x-4} \leqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 4.
Consider the first inequality. To open the absolute values, we consider three possible cases.
1) $x<1$. Then $1-x+5-x \leqslant 4 \Leftrightarrow x \geqslant 1$, i.e., there are no solutions.
2) $1 \leqslant x \leqslant 5$. Then $x-1+5-x \leqslant 4 \Leftrightarrow 4 \leqslant 4$, which is always true, so ... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
|y|+|4-y| \leqslant 4 \\
\frac{y^{2}+x-4 y+1}{2 y+x-7} \leqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 8.
Solution. Consider the first inequality. To open the absolute values, we consider three possible cases.
1) $y < 0$. Then $-y-4+y \leqslant 4 \Leftrightarrow -4 \leqslant 4$, which is always true, so $y \in (-\infty, 0)$.
2) $0 \leqslant y \leqslant 4$. Then $y-4+y \leqslant 4 \Leftrightarrow 2y \leqslant 8... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
y-x \geqslant|x+y| \\
\frac{x^{2}+8 x+y^{2}+6 y}{2 y-x-8} \leqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 8.
Solution. The first inequality is equivalent to the system ${ }^{1}$ $\left\{\begin{array}{l}x+y \leqslant y-x, \\ x+y \geqslant x-y\end{array} \Leftrightarrow\left\{\begin{array}{l}x \leqslant 0, \\ y \geqslant 0 .\end{array}\right.\right.$
Consider the second inequality. It can be written as $\frac{(x+4)... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
x+y+|x-y| \leqslant 0 \\
\frac{x^{2}+6 x+y^{2}-8 y}{x+3 y+6} \geqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 3.
Solution. The first inequality is equivalent to the system $\left\{\begin{array}{l}x-y \leqslant-x-, \\ x-y \geqslant x+y\end{array} \Leftrightarrow\left\{\begin{array}{l}x \leqslant 0, \\ y \leqslant 0 .\end{array}\right.\right.$
Consider the second inequality. It can be written as $\frac{(x+3)^{2}+(y-4)^... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. Two parabolas param 1 and param 2 touch at a point lying on the $O x$ axis. Through point $D$, the second intersection point of the first parabola with the $O x$ axis, a vertical line is drawn, intersecting the second parabola at point $A$ and the common tangent to the parabolas at point $B$. Find the ratio $D A: D ... | 1. Two parabolas param 1 and param 2 touch at a point lying on the $O x$ axis. Through point $D$, the second intersection point of the first parabola with the $O x$ axis, a vertical line is drawn, intersecting the second parabola at point $A$ and the common tangent to the parabolas at point $B$. Find the ratio $D A: D ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. It is known that the number $a$ satisfies the equation param1, and the number $b$ satisfies the equation param2. Find the smallest possible value of the sum $a+b$.
| param1 | param2 | |
| :---: | :---: | :--- |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-6 x^{2}+14 x+2=0$ | |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}+6 x^{2}+15... | 7. It is known that the number $a$ satisfies the equation param1, and the number $b$ satisfies the equation param2. Find the smallest possible value of the sum $a+b$.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-6 x^{2}+14 x+2=0$ | 3 |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}+6 ... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
17. It is known that the number $a$ satisfies the equation param1, and the number $b$ satisfies the equation param2. Find the maximum possible value of the sum $a+b$.
| param1 | param2 | |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-3 x^{2}+5 x+11=0$ | |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}+3 x^{2}+6... | 17. It is known that the number $a$ satisfies the equation param 1, and the number $b$ satisfies the equation param2. Find the greatest possible value of the sum $a+b$.
| param 1 | param2 | Answer |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-3 x^{2}+5 x+11=0$ | 2 |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. Find the minimum value of the function $(g(x))^{2}+$ $2 f(x)$, if the minimum value of the function $(f(x))^{2}+2 g(x)$ is 5. | Answer: -7.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. Consider $h(x)=(f(x))^{2}+2 g(x)$. Expanding the brackets, we get $h(x)=(a x+b)^{2}+2(a x+c)=a^{2} x^{2}+2 a(b+1) x+b^{2}+2 c$. The graph of $y=$ $h(x)$ is a parabola opening upwards, and the minimum value is attained at the vertex. The x-coordinate... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The equation $x^{2}+a x+5=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,
$$
x_{1}^{2}+\frac{250}{19 x_{2}^{3}}=x_{2}^{2}+\frac{250}{19 x_{1}^{3}}
$$
Find all possible values of $a$. | Answer: $a=10$.
Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-20>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=5$. Then $x_{1}^{2}+$ $x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-5$.
Transform the given equality:
$$
x_{1... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. Find the minimum value of the function $(g(x))^{2}+$ $8 f(x)$, if the minimum value of the function $(f(x))^{2}+8 g(x)$ is -29. | Answer: -3.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. Consider $h(x)=(f(x))^{2}+8 g(x)$. Expanding the brackets, we get $h(x)=(a x+b)^{2}+8(a x+c)=a^{2} x^{2}+2 a(b+4) x+b^{2}+8 c$. The graph of $y=$ $h(x)$ is a parabola opening upwards, and the minimum value is attained at the vertex. The x-coordinate... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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