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Problem 4. A $5 \times 100$ table is divided into 500 unit square cells, where $n$ of them are coloured black and the rest are coloured white. Two unit square cells are called adjacent if they share a common side. Each of the unit square cells has at most two adjacent black unit square cells. Find the largest possible value of $n$.
|
Solution. If we colour all the cells along all edges of the board together with the entire middle row except the second and the last-but-one cell, the condition is satisfied and there are 302 black cells. The figure below exhibits this colouring for the $5 \times 8$ case.
We can cover the table by one fragment like the first one on the figure below, 24 fragments like the middle one, and one fragment like the third one.
In each fragment, among the cells with the same letter, there are at most two coloured black, so the total number of coloured cells is at most $(5+24 \cdot 6+1) \cdot 2+2=302$.
| 302 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$.
Thus we have
$$
180^{\circ}-\angle N B M=\angle B N M+\angle B M N=\angle B A N+\angle B A P=\angle N A P=\angle N P A
$$
so the quadrangle $M B N P$ is cyclic (since the points $B$ and $P$ lie on different sides of $M N$ ). Hence $\angle A P B=\angle M P B=\angle M N B$ and the triangles $A P B$ and $M N B$ are congruent ( $M N=2 A M=A M+M P=A P$ ). From that we get $A B=M B$, i.e. the triangle $A M B$ is isosceles, and since $t$ is tangent to $k_{1}$ and perpendicular to $A M$, the centre of $k_{1}$ is on $A M$, hence $A M B$ is a right-angled triangle. From the last two statements we infer $\angle A M B=45^{\circ}$, and so $\angle N M B=90^{\circ}-\angle A M B=45^{\circ}$.
Figure 1
| 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Let $M$ be a subset of the set of 2021 integers $\{1,2,3, \ldots, 2021\}$ such that for any three elements (not necessarily distinct) $a, b, c$ of $M$ we have $|a+b-c|>10$. Determine the largest possible number of elements of $M$.
|
Solution. The set $M=\{1016,1017, \ldots, 2021\}$ has 1006 elements and satisfies the required property, since $a, b, c \in M$ implies that $a+b-c \geqslant 1016+1016-2021=11$. We will show that this is optimal.
Suppose $M$ satisfies the condition in the problem. Let $k$ be the minimal element of $M$. Then $k=|k+k-k|>10 \Rightarrow k \geqslant 11$. Note also that for every $m$, the integers $m, m+k-10$ cannot both belong to $M$, since $k+m-(m+k-10)=10$.
Claim 1: $M$ contains at most $k-10$ out of any $2 k-20$ consecutive integers.
Proof: We can partition the set $\{m, m+1, \ldots, m+2 k-21\}$ into $k-10$ pairs as follows:
$$
\{m, m+k-10\},\{m+1, m+k-9\}, \ldots,\{m+k-11, m+2 k-21\}
$$
It remains to note that $M$ can contain at most one element of each pair.
Claim 2: $M$ contains at most $[(t+k-10) / 2]$ out of any $t$ consecutive integers.
Proof: Write $t=q(2 k-20)+r$ with $r \in\{0,1,2 \ldots, 2 k-21\}$. From the set of the first $q(2 k-20)$ integers, by Claim 1 at most $q(k-10)$ can belong to $M$. Also by claim 1, it follows that from the last $r$ integers, at $\operatorname{most} \min \{r, k-10\}$ can belong to $M$.
Thus,
- If $r \leqslant k-10$, then at most
$$
q(k-10)+r=\frac{t+r}{2} \leqslant \frac{t+k-10}{2} \text { integers belong to } M
$$
- If $r>k-10$, then at most
$$
q(k-10)+k-10=\frac{t-r+2(k-10)}{2} \leqslant \frac{t+k-10}{2} \text { integers belong to } M
$$
By Claim 2, the number of elements of $M$ amongst $k+1, k+2, \ldots, 2021$ is at most
$$
\left[\frac{(2021-k)+(k-10)}{2}\right]=1005
$$
Since amongst $\{1,2, \ldots, k\}$ only $k$ belongs to $M$, we conclude that $M$ has at most 1006 elements as claimed.
| 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
G4. Let $A B C$ be an acute-angled triangle with circumcircle $\Gamma$, and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A^{\prime}$ be the point where the angle bisector of angle $B A C$ meets $\Gamma$. If $A^{\prime} H=A H$, find the measure of angle $B A C$.
Figure 4: Exercise G4.
|
Solution. The segment $A A^{\prime}$ bisects $\angle O A H$ : if $\angle B C A=y$ (Figure 4), then $\angle B O A=$ $2 y$, and since $O A=O B$, it is $\angle O A B=\angle O B A=90^{\circ}-y$. Also since $A H \perp B C$, it is
$\angle H A C=90^{\circ}-y=\angle O A B$ and the claim follows.
Since $A . A^{\prime}$ bisects $\angle O A H$ and $A^{\prime} H=A H . O A^{\prime}=O A$, we have that the isosceles triangles $O A A^{\prime}, H A A^{\prime}$ are equal. Thus
$$
A H=O A=R
$$
where $R$ is the circumradius of triangle $A B C$.
Call $\angle A C H=a$ and recall by the law of sines that $A H=2 R^{\prime} \sin a$, where $R^{\prime}$ is the circumradius of triangle $A H C$. Then (4) implies
$$
R=2 R^{\prime} \sin a
$$
But notice that $R=R^{\prime}$ because $\frac{A C}{\sin (A H C)}=2 R^{\prime}, \frac{A C}{\sin (A B C)}=2 R$ and $\sin (A H C)=$ $\sin \left(180^{\circ}-A B C\right)=\sin (A B C)$. So (5) gives $1=2 \sin a$, and $a$ as an acuite angle can only be $30^{\circ}$. Finally, $\angle B A C=90^{\circ}-a=60^{\circ}$.
Remark. The steps in the above proof can be traced backwards making the converse also true, that is: If $\angle B A C=60^{\circ}$ then $A^{\prime} H=A H$.
- G5. Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$ that touches them at $M$ and, $N$ respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
Figure 5: Exercise G5.
| 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## A1 MLD
Let $x, y, z$ be real numbers, satisfying the relations
$$
\left\{\begin{array}{l}
x \geq 20 \\
y \geq 40 \\
z \geq 1675 \\
x+y+z=2015
\end{array}\right.
$$
Find the greatest value of the product $P=x \cdot y \cdot z$.
| ## Solution 1:
By virtue of $z \geq 1675$ we have
$$
y+z<2015 \Leftrightarrow y<2015-z \leq 2015-1675<1675
$$
It follows that $(1675-y) \cdot(1675-z) \leq 0 \Leftrightarrow y \cdot z \leq 1675 \cdot(y+z-1675)$.
By using the inequality $u \cdot v \leq\left(\frac{u+v}{2}\right)^{2}$ for all real numbers $u, v$ we obtain
$$
\begin{gathered}
P=x \cdot y \cdot z \leq 1675 \cdot x \cdot(y+z-1675) \leq 1675 \cdot\left(\frac{x+y+z-1675}{2}\right)^{2}= \\
1675 \cdot\left(\frac{2015-1675}{2}\right)^{2}=1675 \cdot 170^{2}=48407500
\end{gathered}
$$
$$
\text { We have } P=x \cdot y \cdot z=48407500 \Leftrightarrow\left\{\begin{array} { l }
{ x + y + z = 2 0 1 5 , } \\
{ z = 1 6 7 5 , } \\
{ x = y + z - 1 6 7 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=170 \\
y=170 \\
z=1675
\end{array}\right.\right.
$$
So, the greatest value of the product is $P=x \cdot y \cdot z=48407500$.
| 48407500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## A2 ALB
3) If $x^{3}-3 \sqrt{3} x^{2}+9 x-3 \sqrt{3}-64=0$, find the value of $x^{6}-8 x^{5}+13 x^{4}-5 x^{3}+49 x^{2}-137 x+2015$.
|
Solution
$x^{3}-3 \sqrt{3} x^{2}+9 x-3 \sqrt{3}-64=0 \Leftrightarrow(x-\sqrt{3})^{3}=64 \Leftrightarrow(x-\sqrt{3})=4 \Leftrightarrow x-4=\sqrt{3} \Leftrightarrow x^{2}-8 x+16=3 \Leftrightarrow$ $x^{2}-8 x+13=0$
$x^{6}-8 x^{5}+13 x^{4}-5 x^{3}+49 x^{2}-137 x+2015=\left(x^{2}-8 x+13\right)\left(x^{4}-5 x+9\right)+1898=0+1898=1898$
| 1898 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
NT1 SAU
What is the greatest number of integers that can be selected from a set of 2015 consecutive numbers so that no sum of any two selected numbers is divisible by their difference?
| ## Solution:
We take any two chosen numbers. If their difference is 1 , it is clear that their sum is divisible by their difference. If their difference is 2 , they will be of the same parity, and their sum is divisible by their difference. Therefore, the difference between any chosen numbers will be at least 3 . In other words, we can choose at most one number of any three consecutive numbers. This implies that we can choose at most 672 numbers.
Now, we will show that we can choose 672 such numbers from any 2015 consecutive numbers. Suppose that these numbers are $a, a+1, \ldots, a+2014$. If $a$ is divisible by 3 , we can choose $a+1, a+4, \ldots, a+2014$. If $a$ is not divisible by 3 , we can choose $a, a+3, \ldots, a+$ 2013.
| 672 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## C3 ALB
Positive integers are put into the following table
| 1 | 3 | 6 | 10 | 15 | 21 | 28 | 36 | | |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 | | |
| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 | | |
| 7 | 12 | 18 | 25 | 33 | 42 | | | | |
| 11 | 17 | 24 | 32 | 41 | | | | | |
| 16 | 23 | | | | | | | | |
| $\ldots$ | | | | | | | | | |
| $\ldots$ | | | | | | | | | |
Find the number of the line and column where the number 2015 stays.
| ## Solution 1:
We shall observe straights lines as on the next picture. We can call these lines diagonals.
| 1 | $\sqrt{3}$ | 6 | 10 | 15 | 21 | 28 | 36 | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 | |
| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 | |
| | 12 | 18 | 25 | 33 | 42 | | | |
| 11 | 17 | 24 | 32 | 41 | | | | |
On the first diagonal is number 1 .
On the second diagonal are two numbers: 2 and 3 .
On the 3rd diagonal are three numbers: 4,5 and 6
.
On the $n$-th diagonal are $n$ numbers. These numbers are greater then $\frac{(n-1) n}{2}$ and not greater than $\frac{n(n+1)}{2}$ (see the next sentence!).
On the first $n$ diagonals are $1+2+3+\ldots+n=\frac{n(n+1)}{2}$ numbers.
If $m$ is in the $k$-th row $l$-th column and on the $n$-th diagonal, then it is $m=\frac{(n-1) n}{2}+l$ and $n+1=k+l$. So, $m=\frac{(k+l-2)(k+l-1)}{2}+l$.
We have to find such numbers $n, k$ and $l$ for which:
$$
\begin{gathered}
\frac{(n-1) n}{2}<2015 \leq \frac{n(n+1)}{2} \\
n+1=k+l \\
2015=\frac{(k+l-2)(k+l-1)}{2}+l
\end{gathered}
$$
(1), (2), (3) $\Rightarrow n^{2}-n<4030 \leq n^{2}+n \Rightarrow n=63, k+l=64,2015=\frac{(64-2)(64-1)}{2}+l \Rightarrow$ $t=2015-31 \cdot 63=62, k=64-62=2$
Therefore 2015 is located in the second row and 62 -th column.
| 2015 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
C2 Consider 50 points in the plane, no three of them belonging to the same line. The points have been colored into four colors. Prove that there are at least 130 scalene triangles whose vertices are colored in the same color.
| ## Solution
Since $50=4 \cdot 12+2$, according to the pigeonhole principle we will have at least 13 points colored in the same color. We start with the:
Lemma. Given $n>8$ points in the plane, no three of them collinear, then there are at least $\frac{n(n-1)(n-8)}{6}$ scalene triangles with vertices among the given points.
Proof. There are $\frac{n(n-1)}{2}$ segments and $\frac{n(n-1)(n-2)}{6}$ triangles with vertices among the given points. We shall prove that there are at most $n(n-1)$ isosceles triangles. Indeed, for every segment $A B$ we can construct at most two isosceles triangles (if we have three $A B C, A B D$ and $A B E$, than $C, D, E$ will be collinear). Hence we have at least
$$
\frac{n(n-1)(n-2)}{6}-n(n-1)=\frac{n(n-1)(n-8)}{6} \text { scalene triangles. }
$$
For $n=13$ we have $\frac{13 \cdot 12 \cdot 5}{6}=130$, QED.
| 130 | Combinatorics | proof | Yes | Yes | olympiads | false |
G2 Let $A B C D$ be a convex quadrilateral with $\varangle D A C=\varangle B D C=36^{\circ}, \varangle C B D=18^{\circ}$ and $\varangle B A C=72^{\circ}$. If $P$ is the point of intersection of the diagonals $A C$ and $B D$, find the measure of $\varangle A P D$.
| ## Solution
On the rays ( $D A$ and ( $B A$ we take points $E$ and $Z$, respectively, such that $A C=A E=$ $A Z$. Since $\varangle D E C=\frac{\varangle D A C}{2}=18^{\circ}=\varangle C B D$, the quadrilateral $D E B C$ is cyclic.
Similarly, the quadrilateral $C B Z D$ is cyclic, because $\varangle A Z C=\frac{\varangle B A C}{2}=36^{\circ}=\varangle B D C$. Therefore the pentagon $B C D Z E$ is inscribed in the circle $k(A, A C)$. It gives $A C=A D$ and $\varangle A C D=\varangle A D C=\frac{180^{\circ}-36^{\circ}}{2}=72^{\circ}$, which gives $\varangle A D P=36^{\circ}$ and $\varangle A P D=108^{\circ}$.
| 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## A1
For any real number a, let $\lfloor a\rfloor$ denote the greatest integer not exceeding a. In positive real numbers solve the following equation
$$
n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor=2014
$$
|
Solution1. Obviously $n$ must be positive integer. Now note that $44^{2}=19362000$ than $2014=n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor>2000+44+12=2056$, a contradiction!
So $1950 \leq n \leq 2000$, therefore $\lfloor\sqrt{n}\rfloor=44$ and $\lfloor\sqrt[3]{n}\rfloor=12$. Plugging that into the original equation we get:
$$
n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor=n+44+12=2014
$$
From which we get $n=1956$, which is the only solution.
| 1956 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## C2
In a country with $n$ cities, all direct airlines are two-way. There are $r>2014$ routes between pairs of different cities that include no more than one intermediate stop (the direction of each route matters). Find the least possible $n$ and the least possible $r$ for that value of $n$.
|
Solution. Denote by $X_{1}, X_{2}, \ldots X_{n}$ the cities in the country and let $X_{i}$ be connected to exactly $m_{i}$ other cities by direct two-way airline. Then $X_{i}$ is a final destination of $m_{i}$ direct routes and an intermediate stop of $m_{i}\left(m_{i}-1\right)$ non-direct routes. Thus $r=m_{1}^{2}+\ldots+m_{n}^{2}$. As each $m_{i}$ is at most $n-1$ and $13 \cdot 12^{2}<2014$, we deduce $n \geq 14$.
Consider $n=14$. As each route appears in two opposite directions, $r$ is even, so $r \geq 2016$. We can achieve $r=2016$ by arranging the 14 cities uniformly on a circle connect (by direct two-way airlines) all of them, except the diametrically opposite pairs. This way, there are exactly $14 \cdot 12^{2}=2016$ routes.
| 2016 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A3. Let $A$ and $B$ be two non-empty subsets of $X=\{1,2, \ldots, 11\}$ with $A \cup B=X$. Let $P_{A}$ be the product of all elements of $A$ and let $P_{B}$ be the product of all elements of $B$. Find the minimum and maximum possible value of $P_{A}+P_{B}$ and find all possible equality cases.
|
Solution. For the maximum, we use the fact that $\left(P_{A}-1\right)\left(P_{B}-1\right) \geqslant 0$, to get that $P_{A}+P_{B} \leqslant P_{A} P_{B}+1=11!+1$. Equality holds if and only if $A=\{1\}$ or $B=\{1\}$.
For the minimum observe, first that $P_{A} \cdot P_{B}=11!=c$. Without loss of generality let $P_{A} \leqslant P_{B}$. In this case $P_{A} \leqslant \sqrt{c}$. We write $P_{A}+P_{B}=P_{A}+\frac{c}{P_{A}}$ and consider the function $f(x)=x+\frac{c}{x}$ for $x \leqslant \sqrt{c}$. Since
$$
f(x)-f(y)=x-y+\frac{c(y-x)}{y x}=\frac{(x-y)(x y-c)}{x y}
$$
then $f$ is decreasing for $x \in(0, c]$.
Since $x$ is an integer and cannot be equal with $\sqrt{c}$, the minimum is attained to the closest integer to $\sqrt{c}$. We have $\lfloor\sqrt{11!}\rfloor=\left\lfloor\sqrt{2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7 \cdot 11}\right\rfloor=\lfloor 720 \sqrt{77}\rfloor=6317$ and the closest integer which can be a product of elements of $X$ is $6300=2 \cdot 5 \cdot 7 \cdot 9 \cdot 10$.
Therefore the minimum is $f(6300)=6300+6336=12636$ and it is achieved for example for $A=\{2,5,7,9,10\}, B=\{1,3,4,6,8,11\}$.
Suppose now that there are different sets $A$ and $B$ such that $P_{A}+P_{B}=402$. Then the pairs of numbers $(6300,6336)$ and $\left(P_{A}, P_{B}\right)$ have the same sum and the same product, thus the equality case is unique for the numbers 6300 and 6336. It remains to find all possible subsets $A$ with product $6300=2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 7$. It is immediate that $5,7,10 \in A$ and from here it is easy to see that all posibilities are $A=\{2,5,7,9,10\},\{1,2,5,7,9,10\},\{3,5,6,7,10\}$ and $\{1,3,5,6,7,10\}$.
| 12636 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
C3. In a $5 \times 100$ table we have coloured black $n$ of its cells. Each of the 500 cells has at most two adjacent (by side) cells coloured black. Find the largest possible value of $n$.
|
Solution. If we colour all the cells along all edges of the board together with the entire middle row except the second and the last-but-one cell, the condition is satisfied and there are 302 black cells. The figure below exhibits this colouring for the $5 \times 8$ case.
We can cover the table by one fragment like the first one on the figure below, 24 fragments like the middle one, and one fragment like the third one.
In each fragment, among the cells with the same letter, there are at most two coloured black, so the total number of coloured cells is at most $(5+24 \cdot 6+1) \cdot 2+2=302$.
| 302 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
C4. We have a group of $n$ kids. For each pair of kids, at least one has sent a message to the other one. For each kid $A$, among the kids to whom $A$ has sent a message, exactly $25 \%$ have sent a message to $A$. How many possible two-digit values of $n$ are there?
|
Solution. If the number of pairs of kids with two-way communication is $k$, then by the given condition the total number of messages is $4 k+4 k=8 k$. Thus the number of pairs of kids is $\frac{n(n-1)}{2}=7 k$. This is possible only if $n \equiv 0,1 \bmod 7$.
- In order to obtain $n=7 m+1$, arrange the kids in a circle and let each kid send a message to the first $4 m$ kids to its right and hence receive a message from the first $4 m$ kids to its left. Thus there are exactly $m$ kids to which it has both sent and received messages.
- In order to obtain $n=7 m$, let kid $X$ send no messages (and receive from every other kid). Arrange the remaining $7 m-1$ kids in a circle and let each kid on the circle send a message to the first $4 m-1$ kids to its right and hence receive a message from the first $4 m-1$ kids to its left. Thus there are exactly $m$ kids to which it has both sent and received messages.
There are 26 two-digit numbers with remainder 0 or 1 modulo 7 . (All numbers of the form $7 m$ and $7 m+1$ with $2 \leqslant m \leqslant 14$.)
| 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
G2 Let $A D, B F$ and $C E$ be the altitudes of $\triangle A B C$. A line passing through $D$ and parallel to $A B$ intersects the line $E F$ at the point $G$. If $H$ is the orthocenter of $\triangle A B C$, find the angle $\widehat{C G H}$.
| ## Solution 1
We can see easily that points $C, D, H, F$ lies on a circle of diameter $[C H]$.
Take $\left\{F, G^{\prime}\right\}=\odot(C H F) \cap E F$. We have $\widehat{E F H}=\widehat{B A D}=\widehat{B C E}=\widehat{D F H}$ since the quadrilaterals $A E D C, A E H F, C D H F$ are cyclic. Hence $[F B$ is the bisector of $\widehat{E F D}$, so $H$ is the midpoint of the arc $D G^{\prime}$. It follows that $D G^{\prime} \perp C H$ since $[C H]$ is a diameter. Therefore $D G^{\prime} \| A B$ and $G \equiv G^{\prime}$. Finally $G$ lies on the circle $\odot(C F H)$, so $\widehat{H G C}=90^{\circ}$.
| 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
G3 Let $A B C$ be a triangle in which ( $B L$ is the angle bisector of $\widehat{A B C}(L \in A C), A H$ is an altitude of $\triangle A B C(H \in B C)$ and $M$ is the midpoint of the side $[A B]$. It is known that the midpoints of the segments $[B L]$ and $[M H]$ coincides. Determine the internal angles of triangle $\triangle A B C$.
| ## Solution
Let $N$ be the intersection of the segments $[B L]$ and $[M H]$. Because $N$ is the midpoint of both segments $[B L]$ and $[M H]$, it follows that $B M L H$ is a parallelogram. This implies that $M L \| B C$ and $L H \| A B$ and hence, since $M$ is the midpoint of $[A B]$, the angle bisector [ $B L$ and the altitude $A H$ are also medians of $\triangle A B C$. This shows that $\triangle A B C$ is an equilateral one with all internal angles measuring $60^{\circ}$.
| 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
C2. Given $m \times n$ table, each cell signed with "-". The following operations are
(i) to change all the signs in entire row to the opposite, i. e. every "-" to "+", and every "+" to "-";
(ii) to change all the signs in entire column to the opposite, i. e. every "-" to "+" and every "+" to " -".
(a) Prove that if $m=n=100$, using the above operations one can not obtain 2004 signs "t".
(b) If $m=1004$, find the least $n>100$ for which 2004 signs " + " can be obtained.
|
Solution. If we apply (i) to $l$ rows and (ii) to $k$ columns we obtain $(m-k) l+(n-l) k$
(a) We have equation $(100-k) l+(100-l) k=2004$, or $100 l+100 k-2 l k=2004$, le
$$
50 l+50 k-1 k=1002
$$
Rewrite the lasc equation as
$$
(50-l)(50-h)=2.500-100.2=1498
$$
Since $1498=2 \cdot 7 \cdot 107$, this equation has no solitions in natural numbers.
(b) Let $n=101$. Then we have
$$
(100-k) l+(101-l) k=2004
$$
OI
$$
100 l+101 k-2 l k=2004
$$
l.e.
$$
101 k=2004-100 l+2 l k \div 2(1002-50 l+l k)
$$
Hence $s=2 t$ and we have $101 t=501-25 l+2 l t$. From here we have
$$
t=\frac{501-25 l}{101-2 l}=4+\frac{97-17 l}{101-2 l}
$$
Since $t$ is natural number and $97-17 l<101-2 l$, this is a contradiction, Hence $n \neq 101$. Let $n=1.02$. Then we have
$$
(100-k) l+(102-l) k=2004
$$
or
$$
100 l+102 k-2 l k=2004
$$
$$
50 l+51 k-l k=1002
$$
Rewrite the last equation as
$$
(51-l)(50-k)=25.50-1002=1.548
$$
Since $145 S=2 \cdot 2 \cdot 3 \cdot 3 \cdot 43$ we have $51-l=36$ and $50-k=43$. From here obtain $l=15$ and $k=7$. Indeed,
$$
(100-\bar{\imath}) \cdot 15+(102-1.5) \cdot \overline{7}=93 \cdot 15+87 \cdot 7=1395+609=2004
$$
Hence, the least $n$ is 102 .
| 102 | Combinatorics | proof | Yes | Yes | olympiads | false |
Problem A2. Determine all four digit numbers $\overline{a b c d}$ such that
$$
a(a+b+c+d)\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\left(a^{6}+2 b^{6}+3 c^{6}+4 d^{6}\right)=\overline{a b c d}
$$
|
Solution. From $\overline{a b c d}\overline{1 b c d}=(1+b+c+d)\left(1+b^{2}+c^{2}+d^{2}\right)\left(1+2 b^{6}+3 c^{6}+4 d^{6}\right) \geq$ $(b+1)\left(b^{2}+1\right)\left(2 b^{6}+1\right)$, so $b \leq 2$. Similarly one gets $c\overline{2 b c d}=2(2+b+c+d)\left(4+b^{2}+c^{2}+d^{2}\right)\left(64+2 b^{6}+3 c^{6}+4 d^{6}\right) \geq$ $2(b+2)\left(b^{2}+4\right)\left(2 b^{6}+64\right)$, imposing $b \leq 1$. In the same way one proves $c<2$ and $d<2$. By direct check, we find out that 2010 is the only solution.
| 2010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem G2. Consider a triangle $A B C$ and let $M$ be the midpoint of the side $B C$. Suppose $\angle M A C=\angle A B C$ and $\angle B A M=105^{\circ}$. Find the measure of $\angle A B C$.
|
Solution. The angle measure is $30^{\circ}$.
Let $O$ be the circumcenter of the triangle $A B M$. From $\angle B A M=105^{\circ}$ follows $\angle M B O=15^{\circ}$. Let $M^{\prime}, C^{\prime}$ be the projections of points $M, C$ onto the line $B O$. Since $\angle M B O=15^{\circ}$, then $\angle M O M^{\prime}=30^{\circ}$ and consequently $M M^{\prime}=\frac{M O}{2}$. On the other hand, $M M^{\prime}$ joins the midpoints of two sides of the triangle $B C C^{\prime}$, which implies $C C^{\prime}=M O=A O$.
The relation $\angle M A C=\angle A B C$ implies $C A$ tangent to $\omega$, hence $A O \perp A C$. It follows that $\triangle A C O \equiv \triangle O C C^{\prime}$, and furthermore $O B \| A C$.
Therefore $\angle A O M=\angle A O M^{\prime}-\angle M O M^{\prime}=90^{\circ}-30^{\circ}=60^{\circ}$ and $\angle A B M=$ $\frac{\angle A O M}{2}=30^{\circ}$.
| 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
C2. The natural numbers from 1 to 50 are written down on the blackboard. At least how many of them should be deleted, in order that the sum of any two of the remaining numbers is not a prime?
|
Solution. Notice that if the odd, respectively even, numbers are all deleted, then the sum of any two remaining numbers is even and exceeds 2 , so it is certainly not a prime. We prove that 25 is the minimal number of deleted numbers. To this end, we group the positive integers from 1 to 50 in 25 pairs, such that the sum of the numbers within each pair is a prime:
$$
\begin{aligned}
& (1,2),(3,4),(5,6),(7,10),(8,9),(11,12),(13,16),(14,15),(17,20) \\
& (18,19),(21,22),(23,24),(25,28),(26,27),(29,30),(31,36),(32,35) \\
& (33,34),(37,42),(38,41),(39,40),(43,46),(44,45),(47,50),(48,49)
\end{aligned}
$$
Since at least one number from each pair has to be deleted, the minimal number is 25 .
| 25 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
C3. Consider any four pairwise distinct real numbers and write one of these numbers in each cell of a $5 \times 5$ array so that each number occurs exactly once in every $2 \times 2$ subarray. The sum over all entries of the array is called the total sum of that array. Determine the maximum number of distinct total sums that may be obtained in this way.
|
Solution. We will prove that the maximum number of total sums is 60 .
The proof is based on the following claim.
Claim. Either each row contains exactly two of the numbers, or each column contains exactly two of the numbers.
Proof of the Claim. Indeed, let $R$ be a row containing at least three of the numbers. Then, in row $R$ we can find three of the numbers in consecutive positions, let $x, y, z$ be the numbers in consecutive positions(where $\{x, y, s, z\}=\{a, b, c, d\}$ ). Due to our hypothesis that in every $2 \times 2$ subarray each number is used exactly once, in the row above $\mathrm{R}$ (if there is such a row), precisely above the numbers $x, y, z$ will be the numbers $z, t, x$ in this order. And above them will be the numbers $x, y, z$ in this order. The same happens in the rows below $R$ (see at the following figure).
$$
\left(\begin{array}{lllll}
\bullet & x & y & z & \bullet \\
\bullet & z & t & x & \bullet \\
\bullet & x & y & z & \bullet \\
\bullet & z & t & x & \bullet \\
\bullet & x & y & z & \bullet
\end{array}\right)
$$
Completing all the array, it easily follows that each column contains exactly two of the numbers and our claim has been proven.
Rotating the matrix (if it is necessary), we may assume that each row contains exactly two of the numbers. If we forget the first row and column from the array, we obtain a $4 \times 4$ array, that can be divided into four $2 \times 2$ subarrays, containing thus each number exactly four times, with a total sum of $4(a+b+c+d)$. It suffices to find how many different ways are there to put the numbers in the first row $R_{1}$ and the first column $C_{1}$.
Denoting by $a_{1}, b_{1}, c_{1}, d_{1}$ the number of appearances of $a, b, c$, and respectively $d$ in $R_{1}$ and $C_{1}$, the total sum of the numbers in the entire $5 \times 5$ array will be
$$
S=4(a+b+c+d)+a_{1} \cdot a+b_{1} \cdot b+c_{1} \cdot c+d_{1} \cdot d
$$
If the first, the third and the fifth row contain the numbers $x, y$, with $x$ denoting the number at the entry $(1,1)$, then the second and the fourth row will contain only the numbers $z, t$, with $z$ denoting the number at the entry $(2,1)$. Then $x_{1}+y_{1}=7$ and $x_{1} \geqslant 3$, $y_{1} \geqslant 2, z_{1}+t_{1}=2$, and $z_{1} \geqslant t_{1}$. Then $\left\{x_{1}, y_{1}\right\}=\{5,2\}$ or $\left\{x_{1}, y_{1}\right\}=\{4,3\}$, respectively $\left\{z_{1}, t_{1}\right\}=\{2,0\}$ or $\left\{z_{1}, t_{1}\right\}=\{1,1\}$. Then $\left(a_{1}, b_{1}, c_{1}, d_{1}\right)$ is obtained by permuting one of the following quadruples:
$$
(5,2,2,0),(5,2,1,1),(4,3,2,0),(4,3,1,1)
$$
There are a total of $\frac{4!}{2!}=12$ permutations of $(5,2,2,0)$, also 12 permutations of $(5,2,1,1)$, 24 permutations of $(4,3,2,0)$ and finally, there are 12 permutations of $(4,3,1,1)$. Hence, there are at most 60 different possible total sums.
We can obtain indeed each of these 60 combinations: take three rows ababa alternating
with two rows $c d c d c$ to get $(5,2,2,0)$; take three rows ababa alternating with one row $c d c d c$ and a row $(d c d c d)$ to get $(5,2,1,1)$; take three rows $a b a b c$ alternating with two rows $c d c d a$ to get $(4,3,2,0)$; take three rows abcda alternating with two rows $c d a b c$ to get $(4,3,1,1)$. By choosing for example $a=10^{3}, b=10^{2}, c=10, d=1$, we can make all these sums different. Hence, 60 is indeed the maximum possible number of different sums.
| 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
N1. Determine the largest positive integer $n$ that divides $p^{6}-1$ for all primes $p>7$.
|
Solution. Note that
$$
p^{6}-1=(p-1)(p+1)\left(p^{2}-p+1\right)\left(p^{2}+p+1\right)
$$
For $p=11$ we have
$$
p^{6}-1=1771560=2^{3} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 19 \cdot 37
$$
For $p=13$ we have
$$
p^{6}-1=2^{3} \cdot 3^{2} \cdot 7 \cdot 61 \cdot 157
$$
From the last two calculations we find evidence to try showing that $p^{6}-1$ is divisible by $2^{3} \cdot 3^{2} \cdot 7=504$ and this would be the largest positive integer that divides $p^{6}-1$ for all primes greater than 7 .
By Fermat's theorem, $7 \mid p^{6}-1$.
Next, since $p$ is odd, $8 \mid p^{2}-1=(p-1)(p+1)$, hence $8 \mid p^{6}-1$.
It remains to show that $9 \mid p^{6}-1$.
Any prime number $p, p>3$ is 1 or -1 modulo 3 .
In the first case both $p-1$ and $p^{2}+p+1$ are divisible by 3 , and in the second case, both $p+1$ and $p^{2}-p+1$ are divisible by 3 .
Consequently, the required number is indeed 504
| 504 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
N5. Determine all four-digit numbers $\overline{a b c d}$ such that
$$
(a+b)(a+c)(a+d)(b+c)(b+d)(c+d)=\overline{a b c d}
$$
|
Solution. Depending on the parity of $a, b, c, d$, at least two of the factors $(a+b),(a+c)$, $(a+d),(b+c),(b+d),(c+d)$ are even, so that $4 \mid \overline{a b c d}$.
We claim that $3 \mid \overline{a b c d}$.
Assume $a+b+c+d \equiv 2(\bmod 3)$. Then $x+y \equiv 1(\bmod 3)$, for all distinct $x, y \in\{a, b, c, d\}$. But then the left hand side in the above equality is congruent to $1(\bmod 3)$ and the right hand side congruent to $2(\bmod 3)$, contradiction.
Assume $a+b+c+d \equiv 1(\bmod 3)$. Then $x+y \equiv 2(\bmod 3)$, for all distinct $x, y \in\{a, b, c, d\}$, and $x \equiv 1(\bmod 3)$, for all $x, y \in\{a, b, c, d\}$. Hence, $a, b, c, d \in\{1,4,7\}$, and since $4 \mid \overline{a b c d}$, we have $c=d=4$. Therefore, $8 \mid \overline{a b 44}$, and since at least one more factor is even, it follows that $16 \overline{a b 44}$. Then $b \neq 4$, and the only possibilities are $b=1$, implying $a=4$, which is impossible because 4144 is not divisible by $5=1+4$, or $b=7$, implying $11 \mid \overline{a 744}$, hence $a=7$, which is also impossible because 7744 is not divisible by $14=7+7$.
We conclude that $3 \mid \overline{a b c d}$, hence also $3 \mid a+b+c+d$. Then at least one factor $x+y$ of $(a+b),(a+c),(a+d),(b+c),(b+d),(c+d)$ is a multiple of 3 , implying that also $3 \mid a+b+c+d-x-y$, so $9 \mid \overline{a b c d}$. Then $9 \mid a+b+c+d$, and $a+b+c+d \in\{9,18,27,36\}$. Using the inequality $x y \geq x+y-1$, valid for all $x, y \in \mathbb{N}^{*}$, if $a+b+c+d \in\{27,36\}$, then
$$
\overline{a b c d}=(a+b)(a+c)(a+d)(b+c)(b+d)(c+d) \geq 26^{3}>10^{4}
$$
which is impossible.
Using the inequality $x y \geq 2(x+y)-4$ for all $x, y \geq 2$, if $a+b+c+d=18$ and all two-digit sums are greater than 1 , then $\overline{a b c d} \geq 32^{3}>10^{4}$. Hence, if $a+b+c+d=18$, some two-digit sum must be 1 , hence the complementary sum will be 17 , and the digits are $\{a, b, c, d\}=\{0,1,8,9\}$. But then $\overline{a b c d}=1 \cdot 17 \cdot 8 \cdot 9^{2} \cdot 10>10^{4}$.
We conclude that $a+b+c+d=9$. Then among $a, b, c, d$ there are either three odd or three even numbers, and $8 \mid \overline{a b c d}$.
If three of the digits are odd, then $d$ is even and since $c$ is odd, divisibility by 8 implies that $d \in\{2,6\}$. If $d=6$, then $a=b=c=1$. But 1116 is not divisible by 7 , so this is not a solution. If $d=2$, then $a, b, c$ are either $1,1,5$ or $1,3,3$ in some order. In the first case $2 \cdot 6^{2} \cdot 3^{2} \cdot 7=4536 \neq \overline{a b c d}$. The second case cannot hold because the resulting number is not a multiple of 5 .
Hence, there has to be one odd and three even digits. At least one of the two-digits sums of even digits is a multiple of 4 , and since there cannot be two zero digits, we have either $x+y=4$ and $z+t=5$, or $x+y=8$ and $z+t=1$ for some ordering $x, y, z, t$ of $a, b, c, d$. In the first case we have $d=0$ and the digits are $0,1,4,4$, or $0,2,3,4$, or $0,2,2,5$. None of these is a solution because $1 \cdot 4^{2} \cdot 5^{2} \cdot 8=3200,2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7=5040$ and $2^{2} \cdot 5 \cdot 4 \cdot 7^{2}=3920$. In the second case two of the digits are 0 and 1 , and the other two have to be either 4 and 4 , or 2 and 6 . We already know that the first possibility fails. For the second, we get
$$
(0+1) \cdot(0+2) \cdot(0+6) \cdot(1+2) \cdot(1+6) \cdot(2+6)=2016
$$
and $\overline{a b c d}=2016$ is the only solution.
| 2016 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT 3. Find the largest integer $k(k \geq 2)$, for which there exists an integer $n(n \geq k)$ such that from any collection of $n$ consecutive positive integers one can always choose $k$ numbers, which verify the following conditions:
1. each chosen number is not divisible by 6 , by 7 and by 8 ;
2. the positive difference of any two different chosen numbers is not divisible by at least one of the numbers 6,7 or 8 .
|
Solution. An integer is divisible by 6,7 and 8 if and only if it is divisible by their Least Common Multiple, which equals $6 \times 7 \times 4=168$.
Let $n$ be a positive integer and let $A$ be an arbitrary set of $n$ consecutive positive integers. Replace each number $a_{i}$ from $A$ with its remainder $r_{i}$ ( mod 168). The number $a_{i}$ is divisible by 6 ( 7 or 8 ) if and only if its remainder $r_{i}$ is divisible by 6 (respectively 7 or 8 ). The difference $\left|a_{i}-a_{j}\right|$ is divisible by 168 if and only if their remainders $r_{i}=r_{j}$.
Choosing $k$ numbers from the initial set $A$, which verify the required conditions, is the same as choosing $k$ their remainders ( mod 168) such that:
1. each chosen remainder is not divisible by 6,7 and 8 ;
2. all chosen remainders are different.
Suppose we have chosen $k$ numbers from $A$, which verify the conditions. Therefore, all remainders are different and $k \leq 168$ (otherwise, there would be two equal remainders).
Denote by $B=\{0,1,2,3, \ldots, 167\}$ the set of all possible remainders ( $\bmod 168)$ and by $B_{m}$ the subset of all elements of $B$, which are divisible by $m$. Compute the number of elements of the following subsets:
$$
\begin{gathered}
\left|B_{6}\right|=168: 6=28, \quad\left|B_{7}\right|=168: 7=24, \quad\left|B_{8}\right|=168: 8=21 \\
\left|B_{6} \cap B_{7}\right|=\left|B_{42}\right|=168: 42=4, \quad\left|B_{6} \cap B_{8}\right|=\left|B_{24}\right|=168: 24=7 \\
\left|B_{7} \cap B_{8}\right|=\left|B_{56}\right|=168: 56=3, \quad\left|B_{6} \cap B_{7} \cap B_{8}\right|=\left|B_{168}\right|=1
\end{gathered}
$$
Denote by $D=B_{6} \cup B_{7} \cup B_{8}$, the subset of all elements of $B$, which are divisible by at least one of the numbers 6,7 or 8 . By the Inclusion-Exclusion principle we got
$$
\begin{gathered}
|D|=\left|B_{6}\right|+\left|B_{7}\right|+\left|B_{8}\right|-\left(\left|B_{6} \cap B_{7}\right|+\left|B_{6} \cap B_{8}\right|+\left|B_{7} \cap B_{8}\right|\right)+\left|B_{6} \cap B_{7} \cap B_{8}\right|= \\
28+24+21-(4+7+3)+1=60 .
\end{gathered}
$$
Each chosen remainder belongs to the subset $B \backslash D$, since it is not divisible by 6,7 and 8 . Hence, $k \leq|B \backslash D|=168-60=108$.
Let us show that the greatest possible value is $k=108$. Consider $n=168$. Given any collection $A$ of 168 consecutive positive integers, replace each number with its remainder ( $\bmod 168$ ). Choose from these remainders 108 numbers, which constitute the set $B \backslash D$. Finally, take 108 numbers from the initial set $A$, having exactly these remainders. These $k=108$ numbers verify the required conditions.
| 108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A 5. Let $a, b, c, d$ and $x, y, z, t$ be real numbers such that
$$
0 \leq a, b, c, d \leq 1, \quad x, y, z, t \geq 1 \text { and } a+b+c+d+x+y+z+t=8
$$
Prove that
$$
a^{2}+b^{2}+c^{2}+d^{2}+x^{2}+y^{2}+z^{2}+t^{2} \leq 28
$$
When does the equality hold?
|
Solution. We observe that if $u \leq v$ then by replacing $(u, v)$ with $(u-\varepsilon, v+\varepsilon)$, where $\varepsilon>0$, the sum of squares increases. Indeed,
$$
(u-\varepsilon)^{2}+(v+\varepsilon)^{2}-u^{2}-v^{2}=2 \varepsilon(v-u)+2 \varepsilon^{2}>0
$$
Then, denoting
$$
E(a, b, c, d, x, y, z, t)=a^{2}+b^{2}+c^{2}+d^{2}+x^{2}+y^{2}+z^{2}+t^{2}
$$
and assuming without loss of generality that $a \leq b \leq c \leq d$ and $x \leq y \leq z \leq t$, we have
$$
\begin{aligned}
E(a, b, c, d, x, y, z, t) & \leq E(0,0,0,0, a+x, b+y, c+z, d+t) \\
& \leq E(0,0,0,0,1, b+y, c+z, a+d+x+t-1) \\
& \leq E(0,0,0,0,1,1, c+z, a+b+d+x+y+t-2) \\
& \leq E(0,0,0,0,1,1,1,5)=28
\end{aligned}
$$
Note that if $(a, b, c, d, x, y, z, t) \neq(0,0,0,0,1,1,1,5)$, at least one of the above inequalities, obtained by the $\epsilon$ replacement mentioned above, should be a strict inequality. Thus, the maximum value of $E$ is 28 , and it is obtained only for $(a, b, c, d, x, y, z, t)=(0,0,0,0,1,1,1,5)$ and permutations of $a, b, c, d$ and of $x, y, z, t$.
| 28 | Inequalities | proof | Yes | Yes | olympiads | false |
A 7. Let $A$ be a set of positive integers with the following properties:
(a) If $n$ is an element of $A$ then $n \leqslant 2018$.
(b) If $S$ is a subset of $A$ with $|S|=3$ then there are two elements $n, m$ of $S$ with $|n-m| \geqslant \sqrt{n}+\sqrt{m}$.
What is the maximum number of elements that $A$ can have?
|
Solution. Assuming $n>m$ we have
$$
\begin{aligned}
|n-m| \geqslant \sqrt{n}+\sqrt{m} & \Leftrightarrow(\sqrt{n}-\sqrt{m})(\sqrt{n}+\sqrt{m}) \geqslant \sqrt{n}+\sqrt{m} \\
& \Leftrightarrow \sqrt{n} \geqslant \sqrt{m}+1 .
\end{aligned}
$$
Let $A_{k}=\left\{k^{2}, k^{2}+1, \ldots,(k+1)^{2}-1\right\}$. Note that each $A_{k}$ can contain at most two elements of since if $n, m \in$ with $n>m$ then
$$
\sqrt{n}-\sqrt{m} \leqslant \sqrt{(k+1)^{2}-1}-\sqrt{k^{2}}<(k+1)-k=1
$$
In particular, since $\subseteq A_{1} \cup \cdots \cup A_{44}$, we have $|S| \leqslant 2 \cdot 44=88$.
On the other hand we claim that $A=\left\{m^{2}: 1 \leqslant m \leqslant 44\right\} \cup\left\{m^{2}+m: 1 \leqslant m \leqslant 44\right\}$ satisfies the properties and has $|A|=88$. We check property (b) as everything else is trivial.
So let $r, s, t$ be three elements of $A$ and assume $r<s<t$. There are two cases for $r$.
(i) If we have that $r=m^{2}$, then $t \geqslant(m+1)^{2}$ and so $\sqrt{t}-\sqrt{r} \geq 1$ verifying (b).
(ii) If we have that $r=m^{2}+m$, then $t \geqslant(m+1)^{2}+(m+1)$ and
$$
\begin{aligned}
\sqrt{t} \geqslant \sqrt{r}+1 & \Leftrightarrow \sqrt{(m+1)^{2}+(m+1)} \geqslant \sqrt{m^{2}+m}+1 \\
& \Leftrightarrow m^{2}+3 m+2 \geqslant m^{2}+m+1+2 \sqrt{m^{2}+m} \\
& \Leftrightarrow 2 m+1 \geqslant 2 \sqrt{m^{2}+m} \\
& \Leftrightarrow 4 m^{2}+4 m+1 \geqslant 4 m^{2}+4 m .
\end{aligned}
$$
So property (b) holds in this case as well.
## COMBINATORICS
| 88 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
C3 a) In how many ways can we read the word SARAJEVO from the table below, if it is allowed to jump from cell to an adjacent cell (by vertex or a side) cell?
b) After the letter in one cell was deleted, only 525 ways to read the word SARAJEVO remained. Find all possible positions of that cell.
|
Solution: In the first of the tables below the number in each cell shows the number of ways to reach that cell from the start (which is the sum of the quantities in the cells, from which we can come), and in the second one are the number of ways to arrive from that cell to the end (which is the sum of the quantities in the cells, to which we can go).
a) The answer is 750 , as seen from the second table.
b) If we delete the letter in a cell, the number of ways to read SARAJEVO will decrease by the product of the numbers in the corresponding cell in the two tables. As $750-525=225$, this product has to be 225. This happens only for two cells on the third row. Here is the table with the products:
| 750 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
NT2 A group of $n>1$ pirates of different age owned total of 2009 coins. Initially each pirate (except for the youngest one) had one coin more than the next younger.
a) Find all possible values of $n$.
b) Every day a pirate was chosen. The chosen pirate gave a coin to each of the other pirates. If $n=7$, find the largest possible number of coins a pirate can have after several days.
| ## Solution:
a) If $n$ is odd, then it is a divisor of $2009=7 \times 7 \times 41$. If $n>49$, then $n$ is at least $7 \times 41$, while the average pirate has 7 coins, so the initial division is impossible. So, we can have $n=7, n=41$ or $n=49$. Each of these cases is possible (e.g. if $n=49$, the average pirate has 41 coins, so the initial amounts are from $41-24=17$ to $41+24=65$ ).
If $n$ is even, then 2009 is multiple of the sum $S$ of the oldest and the youngest pirate. If $S<7 \times 41$, then $S$ is at most 39 and the pairs of pirates of sum $S$ is at least 41 , so we must have at least 82 pirates, a contradiction. So we can have just $S=7 \times 41=287$ and $S=49 \times 41=2009$; respectively, $n=2 \times 7=14$ or $n=2 \times 1=2$. Each of these cases is possible (e.g. if $n=14$, the initial amounts are from $144-7=137$ to $143+7=150$ ). In total, $n$ is one of the numbers $2,7,13,41$ and 49 .
b) If $n=7$, the average pirate has $7 \times 41=287$ coins, so the initial amounts are from 284 to 290; they have different residues modulo 7. The operation decreases one of the amounts by 6 and increases the other ones by 1 , so the residues will be different at all times. The largest possible amount in one pirate's possession will be achieved if all the others have as little as possible, namely $0,1,2,3,4$ and 5 coins (the residues modulo 7 have to be different). If this happens, the wealthiest pirate will have $2009-14=1994$ coins. Indeed, this can be achieved e.g. if every day (until that moment) the coins are given by the second wealthiest: while he has more than 5 coins, he can provide the 6 coins needed, and when he has no more than five, the coins at the poorest six pirates have to be $0,1,2,3,4,5$. Thus, $n=1994$ can be achieved.
| 1994 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
C4 In a group of $n$ people, each one had a different ball. They performed a sequence of swaps; in each swap, two people swapped the ball they had at that moment. Each pair of people performed at least one swap. In the end each person had the ball he/she had at the start. Find the least possible number of swaps, if: $a$ ) $n=5$; b) $n=6$.
| ## Solution
We will denote the people by $A, B, C, \ldots$ and their initial balls by the corresponding small letters. Thus the initial state is $A a, B b, C c, D d, E e(, F f)$. A swap is denoted by the (capital) letters of the people involved.
a) Five people form 10 pairs, so at least 10 swaps are necessary.
In fact, 10 swaps are sufficient:
Swap $A B$, then $B C$, then $C A$; the state is now $A a, B c, C b, D d, E e$.
Swap $A D$, then $D E$, then $E A$; the state is now $A a, B c, C b, D e, E d$.
Swap $B E$, then $C D$; the state is now $A a, B d, C e, D b, E c$.
Swap $B D$, then $C E$; the state is now $A a, B b, C c, D d, E e$.
All requirements are fulfilled now, so the answer is 10 .
b) Six people form 15 pairs, so at least 15 swaps are necessary. We will prove that the final number of swaps must be even. Call a pair formed by a ball and a person inverted if letter of the ball lies after letter of the person in the alphabet. Let $T$ be the number of inverted pairs; at the start we have $T=0$. Each swap changes $T$ by 1 , so it changes the parity of $T$. Since in the end $T=0$, the total number of swaps must be even. Hence, at least 16 swaps are necessary. In fact 16 swaps are sufficient:
Swap $A B$, then $B C$, then $C A$; the state is now $A a, B c, C b, D d, E e, F f$. Swap $A D$, then $D E$, then $E A$; the state is now $A a, B c, C b, D e, E d, F f$. Swap $F B$, then $B E$, then $E F$; the state is now $A a, B d, C b, D e, E c, F f$. Swap $F C$, then $C D$, then $D F$; the state is now $A a, B d, C e, D b, E c, F f$. Swap $B D$, then $C E$, then twice $A F$, the state is now $A a, B b, C c, D d, E e, F f$. All requirements are fulfilled now, so the answer is 16 .
| 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
C5 A set $S$ of natural numbers is called good, if for each element $x \in S, x$ does not divide the sum of the remaining numbers in $S$. Find the maximal possible number of elements of a good set which is a subset of the set $A=\{1,2,3, \ldots, 63\}$.
|
Solution
Let set $B$ be the good subset of $A$ which have the maximum number of elements. We can easily see that the number 1 does not belong to $B$ since 1 divides all natural numbers. Based on the property of divisibility, we know that $x$ divides the sum of the remaining numbers if and only if $x$ divides the sum of all numbers in the set $B$. If $B$ has exactly 62 elements, than $B=\{2,3,4, \ldots, 62\}$, but this set can't be good since the sum of its elements is 2015 which is divisible by 5 . Therefore $B$ has at most 61 elements. Now we are looking for the set, whose elements does not divide their sum, so the best way to do that is making a sum of elements be a prime number. $2+3+4+\ldots+63=2015$ and if we remove the number 4, we will obtain the prime number 2011. Hence the set $B=\{2,3,5,6,7, \ldots, 63\}$ is a good one. We conclude that our number is 61 .
| 61 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
G3. Let $A B C D E F$ be a regular hexagon. The points $\mathrm{M}$ and $\mathrm{N}$ are internal points of the sides $\mathrm{DE}$ and $\mathrm{DC}$ respectively, such that $\angle A M N=90^{\circ}$ and $A N=\sqrt{2} \cdot C M$. Find the measure of the angle $\angle B A M$.
| ## Solution
Since $A C \perp C D$ and $A M \perp M N$ the quadrilateral $A M N C$ is inscribed. So, we have
$$
\angle M A N=\angle M C N
$$
Let $P$ be the projection of the point $M$ on the line $C D$. The triangles $A M N$ and $C P M$ are similar implying
$$
\frac{A M}{C P}=\frac{M N}{P M}=\frac{A N}{C M}=\sqrt{2}
$$
So, we have
$$
\frac{M P}{M N}=\frac{1}{\sqrt{2}} \Rightarrow \angle M N P=45^{\circ}
$$
Figure 4
Hence we have
$$
\angle C A M=\angle M N P=45^{\circ}
$$
and finally, we obtain
$$
\angle B A M=\angle B A C+\angle C A M=75^{\circ}
$$
| 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A6 Let $x_{i}>1$, for all $i \in\{1,2,3, \ldots, 2011\}$. Prove the inequality $\sum_{i=1}^{2011} \frac{x_{i}^{2}}{x_{i+1}-1} \geq 8044$ where $x_{2012}=x_{1}$. When does equality hold?
| ## Solution 1
Realize that $\left(x_{i}-2\right)^{2} \geq 0 \Leftrightarrow x_{i}^{2} \geq 4\left(x_{i}-1\right)$. So we get:
$\frac{x_{1}^{2}}{x_{2}-1}+\frac{x_{2}^{2}}{x_{3}-1}+\ldots+\frac{x_{2011}^{2}}{x_{1}-1} \geq 4\left(\frac{x_{1}-1}{x_{2}-1}+\frac{x_{2}-1}{x_{3}-1}+\ldots+\frac{x_{2011}-1}{x_{1}-1}\right)$. By $A M-G M$ :
$\frac{x_{1}-1}{x_{2}-1}+\frac{x_{2}-1}{x_{3}-1}+\ldots+\frac{x_{2011}-1}{x_{1}-1} \geq 2011 \cdot \sqrt[2011]{\frac{x_{1}-1}{x_{2}-1} \cdot \frac{x_{2}-1}{x_{3}-1} \cdot \ldots \cdot \frac{x_{2011}-1}{x_{1}-1}}=2011$
Finally, we obtain that $\frac{x_{1}^{2}}{x_{2}-1}+\frac{x_{2}^{2}}{x_{3}-1}+\ldots+\frac{x_{2011}^{2}}{x_{1}-1} \geq 8044$.
Equality holds when $\left(x_{i}-2\right)^{2}=0,(\forall) i=\overline{1,2011}$, or $x_{1}=x_{2}=\ldots=x_{2011}=2$.
| 8044 | Inequalities | proof | Yes | Yes | olympiads | false |
A9 Consider an integer $n \geq 4$ and a sequence of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$. An operation consists in eliminating all numbers not having the rank of the form $4 k+3$, thus leaving only the numbers $x_{3}, x_{7}, x_{11}, \ldots$ (for example, the sequence $4,5,9,3,6,6,1,8$ produces the sequence 9,1 . Upon the sequence $1,2,3, \ldots, 1024$ the operation is performed successively for 5 times. Show that at the end only 1 number remains and find this number.
| ## Solution
After the first operation 256 number remain; after the second one, 64 are left, then 16, next 4 and ultimately only one number.
Notice that the 256 numbers left after the first operation are $3,7, \ldots, 1023$, hence they are in arithmetical progression of common difference 4. Successively, the 64 numbers left after the second operation are in arithmetical progression of ratio 16 and so on.
Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be the first term in the 5 sequences obtained after each of the 5 operations. Thus $a_{1}=3$ and $a_{5}$ is the requested number. The sequence before the fifth operation has 4 numbers, namely
$$
a_{4}, a_{4}+256, a_{4}+512, a_{4}+768
$$
and $a_{5}=a_{4}+512$. Similarly, $a_{4}=a_{3}+128, a_{3}=a_{2}+32, a_{2}=a_{1}+8$.
Summing up yields $a_{5}=a_{1}+8+32+128+512=3+680=683$.
### 2.2 Combinatorics
| 683 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
NT7 Determine the minimal prime number $p>3$ for which no natural number $n$ satisfies
$$
2^{n}+3^{n} \equiv 0(\bmod p)
$$
| ## Solution
We put $A(n)=2^{n}+3^{n}$. From Fermat's little theorem, we have $2^{p-1} \equiv 1(\bmod p)$ and $3^{p-1} \equiv 1(\bmod p)$ from which we conclude $A(n) \equiv 2(\bmod p)$. Therefore, after $p-1$ steps
at most, we will have repetition of the power. It means that in order to determine the minimal prime number $p$ we seek, it is enough to determine a complete set of remainders $S(p)=\{0,1, \ldots, p-1\}$ such that $2^{n}+3^{n} \not \equiv 0(\bmod p)$, for every $n \in S(p)$.
For $p=5$ and $n=1$ we have $A(1) \equiv 0(\bmod 5)$.
For $p=7$ and $n=3$ we have $A(3) \equiv 0(\bmod 7)$.
For $p=11$ and $n=5$ we have $A(5) \equiv 0(\bmod 11)$.
For $p=13$ and $n=2$ we have $A(2) \equiv 0(\bmod 13)$.
For $p=17$ and $n=8$ we have $A(8) \equiv 0(\bmod 17)$.
For $p=19$ we have $A(n) \not \equiv 0(\bmod 19)$, for all $n \in S(19)$.
Hence the minimal value of $p$ is 19 .
| 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
88.1. The positive integer $n$ has the following property: if the three last digits of $n$ are removed, the number $\sqrt[3]{n}$ remains. Find $n$.
|
Solution. If $x=\sqrt[3]{n}$, and $y, 0 \leq y1000$, and $x>31$. On the other hand, $x^{3}<1000 x+1000$, or $x\left(x^{2}-1000\right)<1000$. The left hand side of this inequality is an increasing function of $x$, and $x=33$ does not satisfy the inequality. So $x<33$. Since $x$ is an integer, $x=32$ and $n=32^{3}=32768$.
| 32768 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
95.2. Messages are coded using sequences consisting of zeroes and ones only. Only sequences with at most two consecutive ones or zeroes are allowed. (For instance the sequence 011001 is allowed, but 011101 is not.) Determine the number of sequences consisting of exactly 12 numbers.
|
Solution 1. Let $S_{n}$ be the set of acceptable sequences consisting of $2 n$ digits. We partition $S_{n}$ in subsets $A_{n}, B_{n}, C_{n}$, and $D_{n}$, on the basis of the two last digits of the sequence. Sequences ending in 00 are in $A_{n}$, those ending in 01 are in $B_{n}$, those ending in 10 are in $C_{n}$, and those ending in 11 are in $D_{n}$. Denote by $x_{n}, a_{n}, b_{n}, c_{n}$, and $d_{n}$ the number of elements in $S_{n}, A_{n}, B_{n}, C_{n}$, and $D_{n}$. We compute $x_{6}$. Because $S_{1}=\{00,01,10,11\}$, $x_{1}=4$ and $a_{1}=b_{1}=c_{1}=d_{1}=1$. Every element of $A_{n+1}$ can be obtained in a unique manner from an element of $B_{n}$ or $D_{n}$ by adjoining 00 to the end. So $a_{n+1}=b_{n}+d_{n}$. The elements of $B_{n+1}$ are similarly obtained from elements of $B_{n}, C_{n}$, and $D_{n}$ by adjoining 01 to the end. So $b_{n+1}=b_{n}+c_{n}+d_{n}$. In a similar manner we obtain the recursion formulas $c_{n+1}=a_{n}+b_{n}+c_{n}$ and $d_{n+1}=a_{n}+c_{n}$. So $a_{n+1}+d_{n+1}=\left(b_{n}+d_{n}\right)+\left(a_{n}+c_{n}\right)=x_{n}$ and $x_{n+1}=2 a_{n}+3 b_{n}+3 c_{n}+2 d_{n}=3 x_{n}-\left(a_{n}+b_{n}\right)=3 x_{n}-x_{n-1}$. Starting from the initial values $a_{1}=b_{1}=c_{1}=d_{1}=1$, we obtain $a_{2}=d_{2}=2, b_{2}=c_{2}=3$, and $x_{2}=10$. So $x_{3}=3 x_{2}-x_{1}=3 \cdot 10-4=26, x_{4}=3 \cdot 26-10=68, x_{5}=3 \cdot 68-26=178$, and $x_{6}=3 \cdot 178-68=466$.
| 466 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
00.1. In how many ways can the number 2000 be written as a sum of three positive, not necessarily different integers? (Sums like $1+2+3$ and $3+1+2$ etc. are the same.)
|
Solution. Since 3 is not a factor of 2000 , there has to be at least two different numbers among any three summing up to 2000 . Denote by $x$ the number of such sums with three different summands and by $y$ the number of sums with two different summands. Consider 3999 boxes consequtively numbered fron 1 to 3999 such that all boxes labelled by an odd number contain a red ball. Every way to put two blue balls in the even-numbered boxes produces a partition of 2000 in three summands. There are $\binom{1999}{2}=999 \cdot 1999$ ways to place the blue balls. But htere are $3!=6$ different placements, which produce the same partition of 2000 into three different summands, and $\frac{3!}{2}=3$ different placements, which produce the same partition of 2000 into summands two which are equal. Thus $6 x+3 y=$ 1999.999. But $y=999$, because the number appering twice in the partition can be any of the numbers $1,2, \ldots, 999$. This leads to $x=998 \cdot 333$, so $x+y=1001 \cdot 333=333333$.
| 333333 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Problem 2
Let $A B C D$ be a cyclic quadrilateral satisfying $A B=A D$ and $A B+B C=C D$.
Determine $\angle C D A$.
|
Solution 2 Answer: $\angle C D A=60^{\circ}$.
Choose the point $E$ on the segment $C D$ such that $D E=A D$. Then $C E=C D-A D=$ $C D-A B=B C$, and hence the triangle $C E B$ is isosceles.
Now, since $A B=A D$ then $\angle B C A=\angle A C D$. This shows that $C A$ is the bisector of $\angle B C D=\angle B C E$. In an isosceles triangle, the bisector of the apex angle is also the perpendicular bisector of the base. Hence $A$ is on the perpendicular bisector of $B E$, and $A E=A B=A D=D E$. This shows that triangle $A E D$ is equilateral, and thus $\angle C D A=60^{\circ}$.
| 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem 4
King George has decided to connect the 1680 islands in his kingdom by bridges. Unfortunately the rebel movement will destroy two bridges after all the bridges have been built, but not two bridges from the same island.
What is the minimal number of bridges the King has to build in order to make sure that it is still possible to travel by bridges between any two of the 1680 islands after the rebel movement has destroyed two bridges?
|
Solution 4 Answer: 2016
An island cannot be connected with just one bridge, since this bridge could be destroyed. Consider the case of two islands, each with only two bridges, connected by a bridge. (It is not possible that they are connected with two bridges, since then they would be isolated from the other islands no matter what.) If they are also connected to two separate islands, then they would be isolated if the rebel movement destroys the two bridges from these islands not connecting the two. So the two bridges not connecting them must go to the same island. That third island must have at least two other bridges, otherwise the rebel movement could cut off these three islands.
Suppose there is a pair of islands with exactly two bridges that are connected to each other. From the above it is easy to see that removing the pair (and the three bridges connected to them) must leave a set of islands with the same properties. Continue removing such pairs, until there are none left. (Note that the reduced set of islands could have a new such pair and that also needs to be removed.) Suppose we are left with $n$ islands and since two islands are removed at a time, $n$ must be an even number. And from the argument above it is clear that $n \geq 4$.
Consider the remaining set of islands and let $x$ be the number of islands with exactly two bridges (which now are not connected to each other). Then $n-x$ islands have at least three bridges each. Let $B^{\prime}$ be the number of bridges in the reduced set. Now $B^{\prime} \geq 2 x$ and $2 B^{\prime} \geq 2 x+3(n-x)=3 n-x$. Hence $2 B^{\prime} \geq \max (4 x, 3 n-x) \geq 4 \cdot \frac{3 n}{5}$, and thus $B^{\prime} \geq \frac{6 n}{5}$. Now let $B$ be the number of bridges in the original set. Then
$$
B=B^{\prime}+3 \cdot \frac{1680-n}{2} \geq \frac{6 n}{5}+\frac{6(1680-n)}{4} \geq \frac{6 \cdot 1680}{5}=2016
$$
It is possible to construct an example with exactly 2016 bridges: Take 672 of the islands and number them $0,1,2, \ldots 671$. Connect island number $i$ with the islands numbered $i-1$, $i+1$ and $i+336$ (modulo 672). This gives 1008 bridges. We now have a circular path of 672 bridges: $0-1-2-\cdots-671-0$. If one of these 672 bridges are destroyed, the 672 islands are still connected. If two of these bridges are destroyed, the path is broken into two parts. Let $i$ be an island on the shortest path (if they have the same length, just pick a random one). Then island $i+336$ (modulo 672) must be on the other part of the path, and the bridge connecting these two islands will connect the two paths. Hence no matter which two bridges the rebel movement destroys, it is possible to travel between any of the 672 islands.
Now for every of the 1008 bridges above, replace it with two bridges with a new island between the two. This increases the number of bridges to 2016 and the number of islands to $672+1008=1680$ completing the construction. Since the rebel movement does not destroy two bridges from the same island, the same argument as above shows that with this construction it is possible to travel between any of the 1680 islands after the destruction of the two bridges.
| 2016 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXV - I - Task 1
During World War I, a battle took place near a certain castle. One of the shells destroyed a statue of a knight with a spear standing at the entrance to the castle. This happened on the last day of the month. The product of the day of the month, the month number, the length of the spear expressed in feet, half the age of the battery commander firing at the castle expressed in years, and half the time the statue stood expressed in years equals 451,066. In which year was the statue erected? | The last day of the month can only be $28$, $29$, $30$, or $31$. Of these numbers, only $29$ is a divisor of the number $451,066 = 2 \cdot 7 \cdot 11 \cdot 29 \cdot 101$. Therefore, the battle took place on February $29$ in a leap year. During World War I, only the year $1916$ was a leap year. From the problem statement, it follows that a divisor of the number $7 \cdot 11 \cdot 101$ is half the age of the battery commander. Only the number $11$ satisfies this condition, so the battery commander was $22$ years old. The length of the pike is a divisor of the number $7 \cdot 101$. Therefore, the pike was $7$ feet long. Thus, half the time the statue stood is $101$ years. It was erected $202$ years before the year $1916$, i.e., in the year $1714$. | 1714 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
LVII OM - I - Problem 4
Participants in a mathematics competition solved six problems, each graded with one of the scores 6, 5, 2, 0. It turned out that
for every pair of participants $ A, B $, there are two problems such that in each of them $ A $ received a different score than $ B $.
Determine the maximum number of participants for which such a situation is possible. | We will show that the largest number of participants for which such a situation is possible is 1024. We will continue to assume that the permissible ratings are the numbers 0, 1, 2, 3 (instead of 5 points, we give 4, and then divide each rating by 2).
Let $ P = \{0,1,2,3\} $ and consider the set
Set $ X $ obviously has 4096 elements. We will consider subsets $ A $ of set $ X $ with the following property (*):
(*) If $ (a_1,a_2,\dots,a_6) $, $ (b_1,b_2,\dots,b_6) \in A $, then there exist $i, j$ such that
It suffices to show that the largest number of elements in set $ A $ with property (*) is 1024.
First, we show that if set $ A $ has property (*), then it has at most 1024 elements. Assume, therefore, that we have a subset $ A $ of set $ X $ with property (*) and suppose that it has at least 1025 elements. Since there are exactly 1024 sequences of length 5 with terms from the four-element set $ P $, it follows from the pigeonhole principle that in set $ A $ there are at least two sequences that have the same terms from the first to the fifth. These sequences differ, therefore, only in one term—the sixth, which contradicts property (*). Therefore, set $ A $ has at most 1024 elements.
Now we show that there exists a set $ A $ with at least 1024 elements and having property (*). It suffices to take the following set:
First, we show that set $ A $ has at least 1024 elements. Take any numbers $ a_1,a_2,\dots,a_5 \in P $. We can make such a choice in 1024 ways. Let $ r $ be the remainder of the division of the sum $ a_1+a_2+\dots+a_5 $ by 4, and let $ a_6 = 4 - r $. Then, of course, $ (a_1,a_2,\dots,a_6) \in A $, so we have indicated at least 1024 different sequences in set $ A $.
Finally, we show that set $ A $ has property (*). Suppose that
and sequences $ (a_1,a_2,\dots,a_6) $ and $ (b_1,b_2,\dots,b_6) $ differ in only one term, say the term with index $ k: \; a_k \neq b_k $, where $ 1 \leq k \leq 6 $ and $ a_i = b_i $ for $ i \neq k $. Since the numbers $ a_1 +a_2 +\dots +a_6 $ and $ b_1 +b_2 +\dots+b_6 $ are divisible by 4, their difference is also divisible by 4. But
Thus, the number $ a_k - b_k $ is divisible by 4. Since $ a_k, b_k \in P $, then
In the set $ \{-3,-2, -1,0,1,2,3\} $, there is only one number divisible by 4, namely 0. Therefore, $ a_k = b_k $, contrary to the assumption that sequences $ (a_1,a_2,\dots,a_6) $ and $ (b_1,b_2,\dots,b_6) $ differ in the term with index $ k $. This contradiction proves that set $ A $ has property (*), which completes the proof. | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXIII OM - I - Problem 7
A broken line contained in a square with a side length of 50 has the property that the distance from any point of this square to it is less than 1. Prove that the length of this broken line is greater than 1248. | Let the broken line $ A_1A_2 \ldots A_n $ have the property given in the problem. Denote by $ K_i $ ($ i= 1, 2, \ldots, n $) the circle with center at point $ A_i $ and radius of length $ 1 $, and by $ F_i $ ($ i= 1, 2, \ldots, n-1 $) the figure bounded by segments parallel to segment $ \overline{A_iA_{i+1}} $ and at a distance of $ 1 $ from it, as well as by arcs of circles $ K_i $ and $ K_{i+1} $ (in Fig. 5, the figure $ F_i $ is shaded).
The set of points in the plane at a distance of less than $ 1 $ from some point of segment $ \overline{A_i A_{i+1}} $ is contained in the union of circles $ K_i $ and $ K_{i+1} $ and figure $ F_i $. From the conditions of the problem, it follows that the given square is contained in the set
The area of the figure $ K_i \cup F_i $ is not less than $ 2A_iA_{i+1} $ (Fig. 5), and the area of the circle $ K_n $ is equal to $ \pi $. Therefore, the area of the given square does not exceed the sum of the areas of these figures, i.e.,
Hence the length of the broken line $ \displaystyle = \sum_{i=1}^{n-1} A_iA_{i+1} \geq 1250 - \frac{\pi}{2} > 1248 $. | 1248 | Geometry | proof | Yes | Yes | olympiads | false |
VIII OM - I - Task 6
Find a four-digit number, whose first two digits are the same, the last two digits are the same, and which is a square of an integer. | If $ x $ is the number sought, then
where $ a $ and $ b $ are integers satisfying the inequalities $ 0 < a \leq 9 $, $ 0 \leq b \leq 9 $. The number $ x $ is divisible by $ 11 $, since
Since $ x $ is a perfect square, being divisible by $ 11 $ it must be divisible by $ 11^2 $, so the number
is divisible by $ 11 $. It follows that $ a + b $ is divisible by $ 11 $, and since $ 0 < a + b \leq 18 $, then $ a + b = 11 $. Therefore,
from which we infer that $ 9a + 1 $ is the square of some natural number $ m $:
Since $ 9a + 1 \leq 82 $, then $ m \leq 9 $.
From the above,
It follows from this equality that the product $ (m + 1) (m - 1) $ is divisible by $ 9 $, and since at most one of the numbers $ m + 1 $ and $ m - 1 $ is divisible by $ 3 $, then one of them is divisible by $ 9 $. Considering that the natural number $ m $ is less than $ 10 $, we conclude from this that $ m + 1 = 9 $, so $ m = 8 $. In this case, $ a = 7 $, $ b = 4 $, and the sought number is $ 7744 = (88)^2 $. | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
LI OM - II - Problem 4
Point $ I $ is the center of the circle inscribed in triangle $ ABC $, where $ AB \neq AC $. Lines $ BI $ and $ CI $ intersect sides $ AC $ and $ AB $ at points $ D $ and $ E $, respectively. Determine all possible measures of angle $ BAC $ for which the equality $ DI = EI $ can hold. | We will show that the only value taken by angle $ BAC $ is $ 60^\circ $.
By the Law of Sines applied to triangles $ ADI $ and $ AEI $, we obtain $ \sin \measuredangle AEI = \sin \measuredangle ADI $. Hence,
om51_2r_img_6.jpg
First, suppose that the equality $ \measuredangle AEI = \measuredangle ADI $ holds (Fig. 1). Then also $ \measuredangle AIE = \measuredangle AID $, which means that triangles $ AEI $ and $ ADI $ are congruent (angle-side-angle criterion). Therefore, $ AD = AE $. This proves that triangles $ ADB $ and $ AEC $ are also congruent (angle-side-angle criterion). Hence, we obtain $ AB = AC $, which contradicts the assumptions made in the problem statement.
om51_2r_img_7.jpg
The remaining case to consider is when $ \measuredangle AEI + \measuredangle ADI = 180^\circ $ (Fig. 2). Then points $ A $, $ E $, $ I $, $ D $ lie on a single circle. Therefore,
From this, we obtain
which means $ \measuredangle BAC = 60^\circ $.
To complete the solution, it remains to show that there exists a triangle $ ABC $ in which $ AB \neq AC $, $ \measuredangle BAC = 60^\circ $, and $ DI = EI $. We will show more: in any triangle $ ABC $ with $ \measuredangle BAC = 60^\circ $, the equality $ DI = EI $ holds.
If $ \measuredangle BAC = 60^\circ $, then
Therefore, a circle can be circumscribed around quadrilateral $ AEID $. Since $ AI $ is the angle bisector of $ \angle EAD $, then $ DI = EI $. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
VI OM - II - Task 3
What should be the angle at the vertex of an isosceles triangle so that a triangle can be constructed with sides equal to the height, base, and one of the remaining sides of this isosceles triangle? | We will adopt the notations indicated in Fig. 9. A triangle with sides equal to $a$, $c$, $h$ can be constructed if and only if the following inequalities are satisfied:
Since in triangle $ADC$ we have $a > h$, $\frac{c}{2} + h > a$, the first two of the above inequalities always hold, so the necessary and sufficient condition for the existence of a triangle with sides $a$, $c$, $h$ is the inequality
From triangle $ADC$ we have $h = a \cos \frac{x}{2}$, $\frac{c}{2} = a \sin \frac{x}{2}$; substituting into inequality (1) gives
or
and since $\frac{x}{4} < 90^\circ$, the required condition takes the form
or
Approximately, $4 \arctan -\frac{1}{2} \approx 106^\circ$ (with a slight deficit). | 106 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXXVIII OM - III - Zadanie 5
Wyznaczyć najmniejszą liczbę naturalną $ n $, dla której liczba $ n^2-n+11 $ jest iloczynem czterech liczb pierwszych (niekoniecznie różnych).
|
Niech $ f(x) = x^2-x+11 $. Wartości przyjmowane przez funkcję $ f $ dla argumentów całkowitych są liczbami całkowitymi niepodzielnymi przez $ 2 $, $ 3 $, $ 5 $, $ 7 $. Przekonujemy się o tym badając reszty z dzielenia $ n $ i $ f(n) $ przez te cztery początkowe liczby pierwsze:
\begin{tabular}{lllll}
&\multicolumn{4}{l}{Reszty z dzielenia:}\\
&przez 2&przez 3&przez 5&przez 7\\
$ n $&0 1&0 1 2 &0 1 2 3 4&0 1 2 3 4 5 6\\
$ f(n) $&1 1&2 2 1&1 1 3 2 3& 4 4 6 3 2 3 6
\end{tabular}
Zatem dowolna liczba $ N $ będąca wartością $ f $ dla argumentu naturalnego i spełniająca podany w zadaniu warunek musi mieć postać $ N = p_1p_2p_3p_4 $, gdzie czynniki $ p_i $ są liczbami pierwszymi $ \geq 11 $.
Najmniejsza z takich liczb $ N= 11^4 $ prowadzi do równania kwadratowego $ x^2-x+11=11^4 $ o pierwiastkach niewymiernych. Ale już druga z kolei $ N = 11^3 \cdot 13 $ jest równa wartości $ f(132) $. Funkcja $ f $ jest ściśle rosnąca w przedziale $ \langle 1/2; \infty) $ wobec czego znaleziona minimalna możliwa wartość $ N $ wyznacza minimalną możliwą wartość $ n $. Stąd odpowiedź: $ n = 132 $.
| 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XII OM - II - Task 4
Find the last four digits of the number $ 5^{5555} $. | \spos{1} We will calculate a few consecutive powers of the number $ 5 $ starting from $ 5^4 $:
It turned out that $ 5^8 $ has the same last four digits as $ 5^4 $, and therefore the same applies to the numbers $ 5^9 $ and $ 5^5 $, etc., i.e., starting from $ 5^4 $, two powers of the number $ 5 $, whose exponents differ by a multiple of $ 4 $, have the same last four digits. The number $ 5^{5555} = 5^{4\cdot 1388+3} $ therefore has the same last $ 4 $ digits as the number $ 5^7 $, i.e., the digits $ 8125 $. | 8125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
L OM - I - Task 3
In an isosceles triangle $ ABC $, angle $ BAC $ is a right angle. Point $ D $ lies on side $ BC $, such that $ BD = 2 \cdot CD $. Point $ E $ is the orthogonal projection of point $ B $ onto line $ AD $. Determine the measure of angle $ CED $. | Let's complete the triangle $ABC$ to a square $ABFC$. Assume that line $AD$ intersects side $CF$ at point $P$, and line $BE$ intersects side $AC$ at point $Q$. Since
$ CP= \frac{1}{2} CF $. Using the perpendicularity of lines $AP$ and $BQ$ and the above equality, we get $ CQ= \frac{1}{2} AC $, and consequently $ CP=CQ $. Points $C$, $Q$, $E$, $P$ lie on the same circle, from which $ \measuredangle CED =\measuredangle CEP =\measuredangle CQP = 45^\circ $. | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LX OM - III - Zadanie 2
Let $ S $ be the set of all points in the plane with both coordinates being integers. Find
the smallest positive integer $ k $ for which there exists a 60-element subset of the set $ S $
with the following property: For any two distinct elements $ A $ and $ B $ of this subset, there exists a point
$ C \in S $ such that the area of triangle $ ABC $ is equal to $ k $. | Let $ K $ be a subset of the set $ S $ having for a given number $ k $ the property given in the problem statement.
Let us fix any two different points $ (a, b), (c, d) \in K $. Then for some integers
$ x, y $ the area of the triangle with vertices $ (a, b) $, $ (c, d) $, $ (x, y) $ is $ k $, i.e., the equality
$ \frac{1}{2}|(a - c)(y - d) - (b - d)(x - c)| = k $ holds. From this, we obtain the condition that the equation
has for any fixed and different $ (a, b), (c, d) \in K $ a solution in integers $ x, y $.
We will prove that if the number $ m $ does not divide $ 2k $, then the set $ K $ has no more than $ m^2 $ elements.
To this end, consider pairs $ (a \mod m, b \mod m) $ of residues of the coordinates of the points of the set $ K $ modulo $ m $.
There are $ m^2 $ of them, so if $ |K| > m^2 $, then by the pigeonhole principle, we will find two different points
$ (a, b) \in K $ and $ (c, d) \in K $ such that $ a \equiv c \mod m $ and $ b \equiv d \mod m $. For such points,
the equation (1) has no solution, since the left side of the equation for any $ x $ and $ y $ is divisible by $ m $,
while the right side is not. Therefore, $ |K| \leqslant m^2 $ for any $ m $ that is not a divisor of $ 2k $.
From the above considerations, it follows that if $ |K| = 60 $, then $ 2k $ must be divisible by all numbers
$ m \leqslant 7 $, since $ 60 > 7^2 $. It is easy to check that the smallest natural number divisible by
2, 3, 4, 5, 6, 7 is $ 2^2 \cdot 3 \cdot 5 \cdot 7 = 420 $, so $ 210|k $.
We will show that for every $ k $ such that $ 210|k $, a 60-element set $ K $ having the property given in the problem statement can be constructed. Let $ K $ be the set of all elements of the set $ S $,
whose both coordinates are in the set $ \{0, 1,..., 7\} $. Fix any two different points
$ A =(a, b) $ and $ B=(c, d) $ from the set $ K $. Then $ a - c, b - d \in \{-7, -6,..., 6, 7\} $, so if
$ a \neq c $, then $ a-c|420 $ and if $ b = d $, then $ b-d|420 $. Without loss of generality, we can assume that $ b = d $.
Then $ b - d|2k $, since $ 420|2k $. The point $ C =(c + \frac{2k}{d-b} ,d) $ is therefore an element of the set $ S $,
and the area of the triangle $ ABC $ is
Moreover, $ |K| = 60 $. As the set $ K $, we can take any 60-element subset of the set $ K $.
Thus, we have shown that a 60-element set having the desired property exists only for positive integers $ k $
that are multiples of 210. The smallest such number is 210. | 210 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LII OM - I - Task 4
Determine whether 65 balls with a diameter of 1 can fit into a cubic box with an edge of 4. | Answer: It is possible.
The way to place the balls is as follows.
At the bottom of the box, we place a layer consisting of 16 balls. Then we place a layer consisting of 9 balls, each of which is tangent to four balls of the first layer (Fig. 1 and 2). The third layer consists of 16 balls that are tangent to the balls of the second layer (Fig. 4 and 5). Similarly, we place two more layers (Fig. 6).
om52_1r_img_2.jpg
om52_1r_img_3.jpg
om52_1r_img_4.jpg
In total, we have placed $ 16 + 9 + 16 + 9 + 16 = 66 $ balls. It remains to calculate how high the fifth layer reaches.
om52_1r_img_5.jpg
om52_1r_img_6.jpg
om52_1r_img_7.jpg
Let's choose any ball from the second layer; this ball is tangent to four balls of the first layer. The centers of these five balls are the vertices of a regular square pyramid, each edge of which has a length of 1 (Fig. 3). By the Pythagorean theorem, the height of this pyramid is $ \frac{\sqrt{2}}{2} $. Therefore, the highest point that the fifth layer reaches is at a distance of $ \frac{1}{2} + 4 \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} = 1 + 2\sqrt{2} < 4 $ from the base plane. The 66 balls placed in this way fit into a cubic box with an edge of 4. | 66 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XX OM - II - Task 2
Find all four-digit numbers in which the thousands digit is equal to the hundreds digit, and the tens digit is equal to the units digit, and which are squares of integers. | Suppose the number $ x $ satisfies the conditions of the problem and denote its consecutive digits by the letters $ a, a, b, b $. Then
The number $ x $ is divisible by $ 11 $, so as a square of an integer, it is divisible by $ 11^2 $, i.e., $ x = 11^2 \cdot k^2 $ ($ k $ - an integer), hence
Therefore,
The number $ a+b $ is thus divisible by $ 11 $. Since $ 0 < a \leq 9 $, $ 0 \leq b \leq 9 $, then $ 0 < a+b \leq 18 $, hence
Therefore, we conclude that $ b \ne 0 $, $ b \ne 1 $; since $ b $ is the last digit of the square of an integer, it cannot be any of the digits $ 2, 3, 7, 8 $. Thus, $ b $ is one of the digits $ 4, 5, 6, 9 $. The corresponding values of $ a $ are $ 7, 6, 5, 2 $, so the possible values of $ x $ are only the numbers $ 7744 $, $ 6655 $, $ 5566 $, $ 2299 $. Only the first one is a square of an integer.
The problem has one solution, which is the number $ 7744 $. | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XV OM - I - Problem 11
In triangle $ ABC $, angle $ A $ is $ 20^\circ $, $ AB = AC $. On sides $ AB $ and $ AC $, points $ D $ and $ E $ are chosen such that $ \measuredangle DCB = 60^\circ $ and $ \measuredangle EBC = 50^\circ $. Calculate the angle $ EDC $. | Let $ \measuredangle EDC = x $ (Fig. 9). Notice that $ \measuredangle ACB = \measuredangle $ABC$ = 80^\circ $, $ \measuredangle CDB = 180^\circ-80^\circ-60^\circ = 40^\circ $, $ \measuredangle CEB = 180^\circ - 80^\circ-50^\circ = \measuredangle EBC $, hence $ EC = CB $. The ratio $ \frac{DC}{CE} $ of the sides of triangle $ EDC $ equals the ratio of the sides $ \frac{DC}{CB} $ of triangle $ BDC $, so the ratios of the sines of the angles opposite the corresponding sides in these triangles are equal:
The right side of the obtained equation can be transformed:
We need to find the convex angle $ x $ that satisfies the equation
or the equation
By transforming the products of sines into differences of cosines, we obtain an equivalent equation
Considering the condition $ 0 < x < 180^\circ $, we get $ x = 30^\circ $.
Note. The last part of the solution can be slightly shortened. Specifically, from the form of equation (1), it is immediately clear that it has a root $ x = 30^\circ $. No other convex angle satisfies this equation; if
then
so
thus, if $ 0 < x < 180^\circ $ and $ 0 < y < 180^\circ $, then $ x = y $. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXVIII - II - Task 3
In a hat, there are 7 slips of paper. On the $ n $-th slip, the number $ 2^n-1 $ is written ($ n = 1, 2, \ldots, 7 $). We draw slips randomly until the sum exceeds 124. What is the most likely value of this sum? | The sum of the numbers $2^0, 2^1, \ldots, 2^6$ is $127$. The sum of any five of these numbers does not exceed $2^2 + 2^3 + 2^4 + 2^5 + 2^6 = 124$. Therefore, we must draw at least six slips from the hat.
Each of the events where we draw six slips from the hat, and the seventh slip with the number $2^{n-1}$ ($n = 1, 2, \ldots, 7$) remains in the hat, is equally likely. The probability of such an event is thus $\displaystyle \frac{1}{7}$.
The sum of the numbers on the drawn slips is equal to $127 - 2^{n-1}$. If $n = 1$, this sum is $126$; if $n = 2$, it is $125$; if $n = 3, 4, 5, 6$ or $7$, the sum is less than $124$ and we must draw a seventh slip. In this last case, the sum of the numbers on all the drawn slips will be $127$. Therefore, the probability that the sum of the numbers on all the slips drawn according to the conditions of the problem is $125$, $126$, or $127$, is $\displaystyle \frac{1}{7}$, $\displaystyle \frac{1}{7}$, $\displaystyle \frac{5}{7}$, respectively.
Thus, the most probable value of the sum is $127$. | 127 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XLII OM - I - Problem 8
Determine the largest natural number $ n $ for which there exist in space $ n+1 $ polyhedra $ W_0, W_1, \ldots, W_n $ with the following properties:
(1) $ W_0 $ is a convex polyhedron with a center of symmetry,
(2) each of the polyhedra $ W_i $ ($ i = 1,\ldots, n $) is obtained from $ W_0 $ by a translation,
(3) each of the polyhedra $ W_i $ ($ i = 1,\ldots, n $) has a point in common with $ W_0 $,
(4) the polyhedra $ W_0, W_1, \ldots, W_n $ have pairwise disjoint interiors. | Suppose that polyhedra $W_0, W_1, \ldots, W_n$ satisfy the given conditions. Polyhedron $W_1$ is the image of $W_0$ under a translation by a certain vector $\overrightarrow{\mathbf{v}}$ (condition (2)). Let $O_0$ be the center of symmetry of polyhedron $W_0$ (condition (1)); the point $O_1$, which is the image of $O_0$ under the translation by $\overrightarrow{\mathbf{v}}$, is the center of symmetry of $W_1$. Figure 3 illustrates a planar variant of the considered problem (a representation of the spatial configuration would obscure this illustration); polyhedra $W_0$ and $W_1$ are depicted as centrally symmetric polygons.
om42_1r_img_3.jpg
By condition (3), polyhedra $W_0$ and $W_1$ have common points (possibly many). Let $K$ be a common point of $W_0$ and $W_1$ (arbitrarily chosen). Denote by $L$ the image of point $K$ under the central symmetry with respect to $O_1$; thus, $L \in W_1$. Let $N$ be a point such that $\overrightarrow{NL} = \overrightarrow{\mathbf{v}}$ and let $M$ be the midpoint of segment $NK$ (Figure 4). Therefore, $N \in W_0$. According to condition (1), the set $W_0$ is convex; this means that with any two points belonging to $W_0$, the entire segment connecting these points is contained in $W_0$. Since $K \in W_0$ and $N \in W_0$, it follows that $M \in W_0$. Segment $MO_1$ connects the midpoints of segments $KN$ and $KL$, and thus $\overrightarrow{MO_1} = \frac{1}{2} \overrightarrow{NL} = \frac{1}{2} \overrightarrow{\mathbf{v}}$, which means $M$ is the midpoint of segment $O_0O_1$.
Let $U$ be the image of polyhedron $W_0$ under a homothety with center $O_0$ and scale factor 3. We will show that $W_1 \subset U$. Take any point $P \in W_1$: let $Q \in W_0$ be a point such that $\overrightarrow{QP} = \overrightarrow{\mathbf{v}}$ and let $S$ be the center of symmetry of parallelogram $O_0O_1PQ$. The medians $O_0S$ and $QM$ of triangle $O_0O_1Q$ intersect at a point $G$ such that $\overrightarrow{O_0G} = \frac{2}{3}\overrightarrow{O_0S} = \frac{1}{3}\overrightarrow{O_0P}$ (Figure 5). This means that $P$ is the image of point $G$ under the considered homothety. Since $G$ is a point on segment $QM$ with endpoints in the set (convex) $W_0$, it follows that $G \in W_0$. Therefore, $P \in U$ and from the arbitrariness of the choice of point $P \in W_1$ we conclude that $W_1 \subset U$.
om42_1r_img_4.jpg
om42_1r_img_5.jpg
In the same way, we prove that each of the sets $W_i (i=1, \ldots, n)$ is contained in $U$. Of course, also $W_0 \subset U$. Thus, the set $W_0 \cup W_1 \cup \ldots \cup W_n$ is a polyhedron contained in $U$. By conditions (2) and (4), its volume equals the volume of $W_0$ multiplied by $n+1$. On the other hand, the volume of $U$ equals the volume of $W_0$ multiplied by 27. Therefore, $n \leq 26$.
It remains to note that the value $n = 26$ can be achieved (example realization: 27 cubes $W_0, \ldots, W_{26}$ arranged like a Rubik's cube). Thus, the sought number is $26$.
Note. We obtain the same result assuming that $W_0, \ldots, W_n$ are any bounded, closed convex bodies (not necessarily polyhedra), with non-empty interiors, satisfying conditions (1)-(4); the reasoning carries over without any changes. Moreover, condition (4) turns out to be unnecessary. This was proven by Marcin Kasperski in the work 27 convex sets without a center of symmetry, awarded a gold medal at the Student Mathematical Paper Competition in 1991; a summary of the work is presented in Delta, issue 3 (1992). | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXII OM - III - Problem 5
Find the largest integer $ A $ such that for every permutation of the set of natural numbers not greater than 100, the sum of some 10 consecutive terms is at least $ A $. | The sum of all natural numbers not greater than $100$ is equal to $1 + 2 + \ldots + 100 = \frac{1 + 100}{2} \cdot 100 = 5050$. If $a_1, a_2, \ldots, a_{100}$ is some permutation of the set of natural numbers not greater than $100$ and the sum of any $10$ terms of this permutation is less than some number $B$, then in particular
By adding these inequalities side by side, we get that $a_1 + a_2 + \ldots + a_{100} = 1 + 2 + \ldots + 100 < 10B$, which means $505 < B$.
Thus, the number $A$ defined in the problem satisfies the inequality
On the other hand, consider the following permutation $a_1, a_2, \ldots, a_{100}$ of the set of natural numbers not greater than $100$
This permutation can be defined by the formulas:
We will prove that the sum of any $10$ consecutive terms of this permutation is not greater than $505$.
Indeed, if the first of the considered $10$ terms has an even number $2k$, then
If, however, the first of the considered terms has an odd number $2k + 1$, then
Thus, the sum of any $10$ consecutive terms of this permutation is not greater than $505$. Therefore, the number $A$ defined in the problem satisfies the inequality
From (1) and (2), it follows that $A = 505$.
Note 1. The problem can be generalized as follows: Find the largest integer $A$ such that for any permutation of the set of natural numbers not greater than an even number $n = 2t$, the sum of some $m = 2r$ (where $r$ is a divisor of $t$) consecutive terms is at least $A$.
By making minor changes in the solution provided above, consisting in replacing the number $100$ with $2t$ and the number $10$ with $2r$, it can be proved that $A = \frac{1}{2} m(n + 1)$.
Note 2. In the case where $m \leq n$ are any natural numbers, it is generally not true that every permutation of the set of natural numbers not greater than $n$ contains $m$ consecutive terms with a sum not less than $\frac{1}{2} m(n + 1)$. For example, for $n = 6$, $m = 4$, the permutation $6, 4, 1, 2, 3, 5$ does not contain four consecutive terms with a sum not less than $14$. | 505 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
L OM - I - Problem 11
In an urn, there are two balls: a white one and a black one. Additionally, we have 50 white balls and 50 black balls at our disposal. We perform the following action 50 times: we draw a ball from the urn, and then return it to the urn along with one more ball of the same color as the drawn ball. After completing these actions, we have 52 balls in the urn. What is the most probable number of white balls in the urn? | Let $ P(k,n) $, where $ 1 \leq k\leq n-1 $, denote the probability of the event that when there are $ n $ balls in the urn, exactly $ k $ of them are white. Then
Using the above relationships, we prove by induction (with respect to $ n $) that $ P(k,n) = 1/(n-1) $ for $ k = 1,2,\ldots,n-1 $. In particular
Therefore, each possible number of white balls after $ 50 $ draws (from $ 1 $ to $ 51 $) is equally likely. | 51 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XLIV OM - I - Problem 11
In six different cells of an $ n \times n $ table, we place a cross; all arrangements of crosses are equally probable. Let $ p_n $ be the probability that in some row or column there will be at least two crosses. Calculate the limit of the sequence $ (np_n) $ as $ n \to \infty $. | Elementary events are determined by six-element subsets of the set of $n^2$ cells of the table; there are $\binom{n^2}{6}$ of them. Let $\mathcal{Z}$ be the complementary event to the event considered in the problem. The configurations favorable to event $\mathcal{Z}$ are obtained as follows: we place the first cross in any arbitrarily chosen cell: here we have $n^2$ possibilities. We then "cross out" the entire horizontal row and the entire vertical row intersecting at the cell where we placed the first cross, and place the second cross in any of the remaining cells: thus, we now have $(n-1)^2$ possibilities. We repeat this scheme four more times and obtain the number of possibilities equal to $n^2 (n -1 )^2 (n - 2)^2 (n - 3)^2 (n - 4 )^2 (n - 5)^2$. This number must still be divided by $6!$ (the number of permutations of a six-element set) to make the reasoning independent of the order of placing the crosses.
Thus, the probability of event $\mathcal{Z}$ (equal to $1 - p_n$) is
One should not multiply all these factors! Let us denote the numerator of the obtained fraction by $L_n$, and the denominator by $M_n$. It is enough to notice that
where $\phi(n)$ and $f(n)$ are polynomials (in the variable $n$), of degree (at most) 4 and 10, respectively; similarly,
where $\psi(n)$ and $g(n)$ are polynomials (in the variable $n$), of degree (at most) 8 and 10, respectively. Therefore,
Since the polynomials $f(n)$ and $g(n)$ are of degree (at most) tenth, then
and consequently, $\displaystyle \lim_{n\to \infty} np_n = 30$. | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXVI - I - Task 1
At the ball, there were 42 people. Lady $ A_1 $ danced with 7 gentlemen, Lady $ A_2 $ danced with 8 gentlemen, ..., Lady $ A_n $ danced with all the gentlemen. How many gentlemen were at the ball? | The number of ladies at the ball is $ n $, so the number of gentlemen is $ 42-n $. The lady with number $ k $, where $ 1 \leq k \leq n $, danced with $ k+6 $ gentlemen. Therefore, the lady with number $ n $ danced with $ n+ 6 $ gentlemen. These were all the gentlemen present at the ball. Thus, $ 42-n = n + 6 $. Solving this equation, we get $ n = 18 $. The number of gentlemen at the ball is therefore $ 42-18 = 24 $. | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
L OM - II - Task 5
Let $ S = \{1, 2,3,4, 5\} $. Determine the number of functions $ f: S \to S $ satisfying the equation $ f^{50} (x) = x $ for all $ x \in S $.
Note: $ f^{50}(x) = \underbrace{f \circ f \circ \ldots \circ f}_{50} (x) $. | Let $ f $ be a function satisfying the conditions of the problem. For numbers $ x \neq y $, we get $ f^{49}(f(x)) = x \neq y = f^{49}(f(y)) $, hence $ f(x) \neq f(y) $. Therefore, $ f $ is a permutation of the set $ S $. Denote by $ r(x) $ ($ x \in S $) the smallest positive integer such that $ f^{r(x)}(x) = x $. Then $ r(x) \leq 5 $ and $ r(x) \mid 50 $, so $ r(x) \in \{1, 2, 5\} $.
If there exists a number $ a \in S $ such that $ r(a) = 5 $, then the numbers $ a $, $ f(a) $, $ f^2(a) $, $ f^3(a) $, $ f^4(a) $ are distinct; they thus exhaust the set $ S $. Then for any number $ x \in S $, $ r(x) = 5 $. The function $ f $ is thus uniquely determined by the permutation $ (f(1), f^2(1), f^3(1), f^4(1)) $ of the set $ \{2, 3, 4, 5\} $; hence it can be defined in $ 4! = 24 $ ways.
If for all $ x \in S $, $ r(x) = 1 $, then $ f $ is the identity function. Such a function is unique.
The remaining case to consider is when the maximum value attained by the function $ r $ is $ 2 $. Let $ a $ be an element of the set $ S $ such that $ r(a) = 2 $. Then also $ r(b) = 2 $, where $ b = f(a) $.
If $ r(x) = 1 $ for all $ x \in S \setminus \{a, b\} $, then $ f $ is determined by the choice of a two-element subset $ \{a, b\} $ of the set $ S $, which can be done in $ {5 \choose 2} = 10 $ ways.
If, however, there exists a number $ c \in S \setminus \{a, b\} $ such that $ r(c) = 2 $, then by setting $ d = f(c) $ and denoting by $ e $ the unique element of the set $ S \setminus \{a, b, c, d\} $, we have
Such a function $ f $ is determined by the choice of the number $ e $ (which can be done in $ 5 $ ways) and the partition of the set $ S \setminus \{e\} $ into two two-element subsets $ \{a, b\} $ and $ \{c, d\} $ (there are $ 3 $ such partitions). We thus get $ 15 $ functions of the form (1).
In total, there are $ 50 $ functions satisfying the conditions of the problem. | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXII OM - III - Task 3
How many locks at least need to be placed on the treasury so that with a certain distribution of keys among the 11-member committee authorized to open the treasury, any 6 members can open it, but no 5 can? Determine the distribution of keys among the committee members with the minimum number of locks. | Suppose that for some natural number $ n $ there exists a key distribution to $ n $ locks among an 11-member committee such that the conditions of the problem are satisfied. Let $ A_i $ denote the set of locks that the $ i $-th member of the committee can open, where $ i = 1, 2, \ldots, 11 $, and let $ A $ denote the set of all locks. Then from the conditions of the problem, we have
for any five-element subset $ \{ i_1, \ldots, i_5\} $ of the set $ \{1, 2, \ldots, 11\} $ and
for any six-element subset $ \{j_1, \ldots, j_6\} $ of the set $ \{1,2,\ldots, 11\} $.
From (1), it follows that the set $ A - (A_{i_1} \cup \ldots \cup A_{i_5}) $ is non-empty. Let $ x_{i_1}, \ldots, x_{i_5} $ be one of its elements, i.e., a lock that the group of committee members numbered $ i_1, \ldots, i_5 $ cannot open. From (2), it follows that for every $ j \not \in \{i_1, \ldots, i_5 \} $ we have $ x_{i_1 , \ldots, i_5} \in A_j $.
Suppose that $ x_{i_1,\ldots,i_5} = x_{k_1, \ldots, k_5} $ for some subsets $ \{i_1, \ldots, i_5\} $ and $ \{k_1, \ldots, k_5\} $. If these subsets were different, then, for example, $ i_t \not \in \{ k_1, \ldots, k_5 \} $. Therefore, $ x_{k_1, \ldots, k_5} \in A_{i_t} $; but on the other hand, this leads to a contradiction. The obtained contradiction proves that $ \{i_1, \ldots,i_5\} = \{ k_1, \ldots, k_5 \} $.
In other words, different five-element subsets $ \{i_1, \ldots, i_5\} $ correspond to different locks. Therefore, the number of locks is not less than the number of five-element subsets of an 11-element set, i.e., $ n \geq \binom{11}{5} = 462 $.
We will now prove that if we install $ \binom{11}{5} $ locks on the treasury, then we can distribute the keys to them among the members of the 11-member committee in such a way that the conditions of the problem are satisfied.
Let us associate each of the $ \binom{11}{5} $ locks with a five-element subset of the set $ \{1, 2, \ldots, 11\} $ in a one-to-one manner. If a lock corresponds to the subset $ \{i_1, \ldots, i_5\} $, then the keys to it are given to all members of the committee whose numbers are different from $ i_1, \ldots, i_5 $.
We will show that no five members of the committee can open a certain lock, and therefore the treasury. Indeed, the members of the committee numbered $ i_1, \ldots, i_5 $ do not have the key to the lock corresponding to the subset $ \{i_1, \ldots, i_5\} $.
We will show that any six members of the committee can open any lock, and therefore the treasury. If the members of the committee have numbers $ j_1, \ldots, j_6 $ and want to open a lock corresponding to the subset $ \{i_1, \ldots, i_5\} $, then one of the six numbers $ j_1, \ldots, j_6 $ does not belong to this five-element subset, say $ j_t \not \in \{i_1, \ldots, i_5 \} $. Therefore, the member of the committee numbered $ j_t $ has the key to the lock corresponding to the subset $ \{i_1, \ldots, i_5\} $.
Thus, the smallest number satisfying the conditions of the problem is $ \binom{11}{5} = 462 $. | 462 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Consider the set $M$ of integers $n \in[-100 ; 500]$, for which the expression $A=n^{3}+2 n^{2}-5 n-6$ is divisible by 11. How many integers are contained in $M$? Find the largest and smallest of them? | Answer: 1) 164 numbers; 2) $n_{\text {min }}=-100, n_{\text {max }}=497$. | 164 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Solution. According to the problem, the sum of the original numbers is represented by the expression:
$$
\begin{aligned}
& \left(a_{1}+2\right)^{2}+\left(a_{2}+2\right)^{2}+\ldots+\left(a_{50}+2\right)^{2}=a_{1}^{2}+a_{2}^{2}+\ldots+a_{50}^{2} \rightarrow \\
& {\left[\left(a_{1}+2\right)^{2}-a_{1}^{2}\right]+\left[\left(a_{2}+2\right)^{2}-a_{2}^{2}\right]+\ldots\left[\left(a_{50}+2\right)^{2}-a_{50}^{2}\right]=0 \rightarrow} \\
& \rightarrow 4\left(a_{1}+1\right)+4\left(a_{2}+1\right)+\ldots+4\left(a_{50}+1\right)=0 \rightarrow a_{1}+a_{2}+\ldots+a_{50}=-50
\end{aligned}
$$
Then, if we add 3, we get:
$$
\begin{aligned}
& \left(a_{1}+3\right)^{2}+\left(a_{2}+3\right)^{2}+\ldots+\left(a_{50}+3\right)^{2}-\left(a_{1}^{2}+a_{2}^{2}+\ldots+a_{50}^{2}\right)= \\
& =\left[\left(a_{1}+3\right)^{2}-a_{1}^{2}\right]+\left[\left(a_{2}+3\right)^{2}-a_{2}^{2}\right]+\ldots\left[\left(a_{50}+3\right)^{2}-a_{50}^{2}\right]= \\
& =3\left(2 a_{1}+3\right)+3\left(2 a_{2}+3\right)+\ldots+3\left(2 a_{50}+3\right)=6\left(a_{1}+a_{2}+\ldots+a_{50}\right)+9 \cdot 50= \\
& =-300+450=150
\end{aligned}
$$ | Answer: will increase by 150. | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A set of 60 numbers is such that adding 3 to each of them does not change the value of the sum of their squares. By how much will the sum of the squares of these numbers change if 4 is added to each number? | Answer: will increase by 240. | 240 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A set of 70 numbers is such that adding 4 to each of them does not change the magnitude of the sum of their squares. By how much will the sum of the squares of these numbers change if 5 is added to each number? | Answer: will increase by 350. | 350 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A set of 80 numbers is such that adding 5 to each of them does not change the magnitude of the sum of their squares. By how much will the sum of the squares of these numbers change if 6 is added to each number? | Answer: will increase by 480. | 480 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the fraction $\frac{p}{q}$ with the smallest possible natural denominator, for which $\frac{1}{2014}<\frac{p}{q}<\frac{1}{2013}$. Enter the denominator of this fraction in the provided field | 5. Find the fraction $\frac{p}{q}$ with the smallest possible natural denominator, for which
$\frac{1}{2014}<\frac{p}{q}<\frac{1}{2013}$. Enter the denominator of this fraction in the provided field
Answer: 4027 | 4027 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. If $\quad a=\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}}, \quad$ then $\quad P(a)=\overline{a_{6} a_{1} a_{2} a_{3} a_{4} a_{5}}$, $P(P(a))=\overline{a_{5} a_{6} a_{1} a_{2} a_{3} a_{4}} \quad$ with $\quad a_{5} \neq 0, a_{6} \neq 0, a_{1} \neq 0 . \quad$ From the equality $P(P(a))=a$ it follows that $a_{1}=a_{5}, a_{2}=a_{6}, a_{3}=a_{1}$, $a_{4}=a_{2}, a_{5}=a_{3}, a_{6}=a_{4}$, that is, $a_{1}=a_{3}=a_{5}=t, t=1,2, \ldots, 9$ and $a_{2}=a_{4}=a_{6}=u, u=1,2, \ldots, 9$. Thus, the sought $a=\overline{\text { tututu }}$ and there are 81 such different numbers ( $t$ and $u$ can take any values of the decimal system digits from 1 to 9).
Let $n>2-$ be a prime number, $a=\overline{a_{1} a_{2} a_{3} a_{4} \ldots a_{n-3} a_{n-2} a_{n-1} a_{n}}$. Then
$$
\begin{gathered}
P(a)=\overline{a_{n} a_{1} a_{2} a_{3} a_{4} \ldots a_{n-3} a_{n-2} a_{n-1}} \\
P(P(a))=\overline{a_{n-1} a_{n} a_{1} a_{2} a_{3} a_{4} \ldots a_{n-3} a_{n-2}}
\end{gathered}
$$
The property $P(P(a))=a$ gives the relations $a_{1}=a_{n-1}=a_{n-3}=\ldots=a_{1}$. For a prime $n>2$, all the digits of the number $a$ are involved in the chain, so they are all equal to each other. | Answer: 1) 81 is the number; 2) $a=\overline{t u t u t u}, t, u$, where $t, u$ - are any digits, not equal to zero. | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Square the numbers $a=10001, b=100010001$. Extract the square root of the number $c=1000200030004000300020001$. | 1) $a^{2}=100020001$; 2) $b^{2}=10002000300020001$; 3) $\sqrt{c}=1000100010001$. | 1000100010001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. By what natural number can the numerator and denominator of the ordinary fraction of the form $\frac{5 n+3}{7 n+8}$ be reduced? For which integers $n$ can this occur? | Answer: it can be reduced by 19 when $n=19k+7, k \in Z$. | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The angle at vertex $B$ of triangle $A B C$ is $130^{\circ}$. Through points $A$ and $C$, lines perpendicular to line $A C$ are drawn and intersect the circumcircle of triangle $A B C$ at points $E$ and $D$. Find the acute angle between the diagonals of the quadrilateral with vertices at points $A, C, D$ and $E$.
Problem 1 Answer: 12 students. | Solution. Let $a$ be the number of students in the first category, $c$ be the number of students in the third category, and $b$ be the part of students from the second category who will definitely lie in response to the first question (and say "YES" to all three questions), while the rest of the students from this category will answer "NO" to all three questions. Then, "YES" to the first question will be given by $a+b+c=25$ students. "YES" to the second question will be given by $b+c=21$ students. "YES" to the third question will be given by $b=6$ students. Solving the system, we get: $a=4, b=6, c=15$. Then, to the second category, we should assign $31-a-c=12$ students.
Problem 2 Answer: $a= \pm 5-2 \sqrt{6}$.
Solution. By Vieta's theorem, the roots of the equation are $m=(a+2 \sqrt{6})$ and $n=\left(\frac{1}{a}-2 \sqrt{6}\right)$.
Then we have:
$$
\left[\begin{array}{l}
a=m-2 \sqrt{6} \\
\frac{1}{a}=n+2 \sqrt{6}
\end{array} \rightarrow 1=m n+2 \sqrt{6}(m-n)-24 \rightarrow 2 \sqrt{6}(m-n)=25-m n\right.
$$
If $m, n \in Z, m \neq n$, then $2 \sqrt{6}=\frac{25-m n}{m-n}$ is a rational number, which is incorrect. Therefore, $m=n$ and then $m^{2}=25 \rightarrow m= \pm 5 \rightarrow a= \pm 5-2 \sqrt{6}$.
Problem 3 Answer: 5400 numbers.
Solution. Write $496125=3^{4} \cdot 5^{3} \cdot 7^{2}$. There are a total of $3^{4} \cdot 5^{3}=10125$ multiples of 49. Among them, $3^{3} \cdot 5^{3}=3375$ numbers are divisible by 3, $3^{4} \cdot 5^{2}=2025$ numbers are divisible by 5, and $3^{3} \cdot 5^{2}=675$ numbers are divisible by 15. Then, among the numbers that are multiples of 49, there are $3375+2025-675=4725$ numbers that are divisible by either 3 or 5. Therefore, the numbers that are not divisible by either 3 or 5 will be $10125-4775=$ 5400 desired numbers.
Problem 4 Answer: $n=210$.
Solution. According to the problem, $n=7 k$ and the expression
$$
n^{2}+25 n+100=(n+5)(n+20)=(7 k+5)(7 k+20)
$$
must be divisible by 5 and 23. Note that if one of the factors is divisible by 5, then the other is also divisible by 5 and vice versa, so:
$$
7 k+5=5 m \rightarrow 7 k=5(m-1) \rightarrow\left\{\begin{array}{c}
k=5 t \\
m=7 t+1
\end{array} \rightarrow n=35 t, t \in Z\right.
$$
Case 1. $7 k+5$ is divisible by 23:
$$
35 t+5=23 u \rightarrow\left\{\begin{array}{l}
t=23 v-10 \\
u=35 v-15
\end{array} \rightarrow n=35 t=35(23 v-10) \rightarrow n_{\min }=455\right.
$$
Case 2. $7 k+20$ is divisible by 23:
$$
35 t+20=23 u \rightarrow\left\{\begin{array}{c}
t=23 v+6 \\
u=35 v+10
\end{array} \rightarrow n=35 t=35(23 v+6) \rightarrow n_{\min }=210\right.
$$
Choosing the smallest of the found $n_{\min }$, we get $n_{\min }=210$.
Problem 5 Answer: $80^{\circ}$.
Solution. Figures 1 and 2 show the possible geometric configurations.
Fig 1
Fig 2
In the figures, point $O$ is the center of the circle $K$ circumscribed around triangle $A B C$.
Case 1. $\alpha>90^{\circ}$ (Fig. 1). Quadrilateral $A B C D$ is inscribed in circle $K$ by construction, so point $O$ is equidistant from points $C$ and $D$ at a distance equal to the radius of circle $K$. Similarly, point $O$ is equidistant from points $A$ and $E$ at the same distance. Therefore, point $O$ is the intersection point of the diagonals of rectangle $A C D E$. The angle $C D O$, as the opposite angle to $\alpha$ in the inscribed quadrilateral $A B C D$, is equal to $180^{\circ}-\alpha$. The desired angle between the diagonals of rectangle $A C D E$ is: $180^{\circ}-2\left(180^{\circ}-\alpha\right)=2\left(\alpha-90^{\circ}\right)$.
Case 2. $\alpha<90^{\circ}$ (Fig. 2). Point $O$ is equidistant from points $A, C, D$ and $E$, so it is the intersection point of the diagonals of rectangle $A C D E$ with these vertices. The angle $A D C$ is equal to $\alpha$, as it is inscribed and subtends the same arc of circle $K$ as angle $A B C$. Then the desired angle $D O C$ is: $180^{\circ}-2 \alpha=2\left(90^{\circ}-\alpha\right)$.
In case 1, $\alpha=130^{\circ}$, so case 1 applies, and $2\left(\alpha-90^{\circ}\right)=80^{\circ}$. | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In a convex quadrilateral $A B C D$, the lengths of sides $B C$ and $A D$ are 2 and $2 \sqrt{2}$ respectively. The distance between the midpoints of diagonals $B D$ and $A C$ is 1. Find the angle between the lines $B C$ and $A D$. | Answer: $\alpha=45^{\circ}$. | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Let's introduce the notation: $A B=2 c, A C=2 b, \measuredangle B A C=\alpha$. The feet of the perpendicular bisectors are denoted by points $P$ and $Q$. Then, in the right triangle $\triangle A M Q$, the hypotenuse $A M=\frac{b}{\cos \alpha}$. And in the right triangle $\triangle A N P$, the hypotenuse $A N=\frac{c}{\cos \alpha}$. By the cosine theorem for triangles $A M N$ and $A B C$ respectively, we have
$$
\begin{aligned}
& N M^{2}=A M^{2}+A N^{2}-2 A M \cdot A N \cdot \cos \alpha=\frac{c^{2}+b^{2}-2 b c \cos \alpha}{\cos ^{2} \alpha} \\
& B C^{2}=A B^{2}+A C^{2}-2 A B \cdot A C \cdot \cos \alpha=4\left(c^{2}+b^{2}-2 b c \cos \alpha\right)
\end{aligned}
$$
By the condition, $M N=B C$, therefore $\cos ^{2} \alpha=\frac{1}{4} \Leftrightarrow \cos \alpha= \pm \frac{1}{2}$, from which $\alpha=60^{\circ}$ or $\alpha=120^{\circ}$. We will show that both cases are possible, that is, if $\alpha=60^{\circ}$ or $\alpha=120^{\circ}$, then $M N=B C$.
Case 1. If $\alpha=60^{\circ}$, then $\measuredangle P N A=30^{\circ}$, so $A N=2 c=A B$, and $\measuredangle A M Q=30^{\circ}$, so $A M=2 b=A C$. Therefore, $\triangle A N M=\triangle A B C$ by two sides and the angle $\alpha$ between them. Consequently, $M N=B C$.
Case 2. If $\alpha=120^{\circ}$, then $\measuredangle B A N=60^{\circ}$. Further, $A N=N B$, so $\triangle N A B$ is isosceles, hence $\measuredangle A B N=60^{\circ}$, thus $\triangle N A B$ is equilateral. Therefore, $A N=A B$. Similarly, $\triangle M A C$ is equilateral. Therefore, $A M=A C$. Hence,
$\triangle A N M=\triangle A B C$ by two sides and the angle $\alpha$ between them. Consequently, $M N=B C$. | Answer: $60^{\circ}$ or $120^{\circ}$. | 60 | Geometry | proof | Yes | Yes | olympiads | false |
5. In triangle $A B C$, the perpendicular bisectors of sides $A B$ and $A C$ intersect lines $A C$ and $A B$ at points $N$ and $M$ respectively. The length of segment $N M$ is equal to the length of side $B C$ of the triangle. The angle at vertex $C$ of the triangle is $40^{\circ}$. Find the angle at vertex $B$ of the triangle. | Answer: $80^{\circ}$ or $20^{\circ}$.
## Final round of the "Rosatom" Olympiad, 9th grade, CIS, February 2020
# | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The sum $b_{6}+b_{7}+\ldots+b_{2018}$ of the terms of the geometric progression $\left\{b_{n}\right\}, b_{n}>0$ is 6. The sum of the same terms taken with alternating signs $b_{6}-b_{7}+b_{8}-\ldots-b_{2017}+b_{2018}$ is 3. Find the sum of the squares of the same terms $b_{6}^{2}+b_{7}^{2}+\ldots+b_{2018}^{2}$. | Answer: $b_{6}^{2}+b_{7}^{2}+\ldots+b_{2018}^{2}=18$. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Integers, the decimal representation of which reads the same from left to right and from right to left, we will call symmetric. For example, the number 513315 is symmetric, while 513325 is not. How many six-digit symmetric numbers exist such that adding 110 to them leaves them symmetric? | Answer: 81 numbers of the form $\overline{a b 99 b a}$, where $a=1,2, \ldots, 9, b=0,1,2, \ldots, 8$. | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In city "N", there are 12 horizontal and 16 vertical streets, of which a pair of horizontal and a pair of vertical streets form the rectangular boundary of the city, while the rest divide it into blocks that are squares with a side length of 100m. Each block has an address consisting of two integers $(i ; j), i=1,2, . ., 11, j=1,2, \ldots, 15-$ the numbers of the streets that bound it from below and from the left. Taxis transport passengers from one block to another, adhering to the following rules: 1) pick-up and drop-off can be made at any point on the boundary of the block at the passenger's request; 2) it is forbidden to enter inside the block; 3) transportation is carried out along the shortest path; 4) a fee of 1 coin is charged for every 100m traveled (rounding the distance to the nearest 100m in favor of the driver). How many blocks are there in the city? What is the maximum and minimum fare that a driver can charge a passenger for a ride from block $(7,2)$ to block $(2 ; 1)$ without violating the rules? | Answer: 165 blocks; $c_{\min }=4$ coins, $c_{\max }=8$ coins. | 165 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Integers, the decimal representation of which reads the same from left to right and from right to left, we will call symmetric. For example, the number 5134315 is symmetric, while 5134415 is not. How many seven-digit symmetric numbers exist such that adding 1100 to them leaves them symmetric? | Answer: 810 numbers of the form $\overline{a b c 9 c b a}$, where $a=1,2, \ldots, 9$, $b=0,1,2, \ldots, 9, c=0,1,2, \ldots, 8$. | 810 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In city "N", there are 7 horizontal and 13 vertical streets, of which a pair of horizontal and a pair of vertical streets form the rectangular boundary of the city, while the rest divide it into blocks that are squares with a side length of 100 m. Each block has an address consisting of two integers $(i ; j), i=1,2, . ., 6, j=1,2, \ldots, 12$ - the numbers of the streets that bound it from below and from the left. Taxis transport passengers from one block to another, adhering to the following rules: 1) pick-up and drop-off can be made at any point on the boundary of the block at the passenger's request; 2) it is forbidden to enter inside the block; 3) transportation is carried out along the shortest path; 4) a fee of 1 coin is charged for every 100 m traveled (rounding the distance to the nearest 100 m in favor of the driver). How many blocks are there in the city? What is the maximum and minimum fare that the driver can charge the passenger for a ride from block $(4,2)$ to block $(1 ; 9)$ without violating the rules. | # Answer: 72 blocks; $c_{\min }=8$ coins, $c_{\max }=12$ coins. | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Integers, whose decimal notation reads the same from left to right and from right to left, we will call symmetric. For example, the number 513151315 is symmetric, while 513152315 is not. How many nine-digit symmetric numbers exist such that adding 11000 to them leaves them symmetric? | Answer: 8100 numbers of the form $\overline{a b c d 9 d c b a}$, where $a=1,2, \ldots, 9$, $b=0,1,2, \ldots, 9, c=0,1,2, \ldots, 9, d=0,1,2, \ldots, 8$. | 8100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In city "N", there are 10 horizontal and 12 vertical streets, of which a pair of horizontal and a pair of vertical streets form the rectangular boundary of the city, while the rest divide it into blocks that are squares with a side length of 100 meters. Each block has an address consisting of two integers $(i ; j), i=1,2, . ., 9, j=1,2, \ldots, 11-$ the numbers of the streets that bound it from below and from the left. Taxis transport passengers from one block to another, adhering to the following rules: 1) pick-up and drop-off can be made at any point on the boundary of the block at the passenger's request; 2) it is forbidden to enter inside the block; 3) transportation is carried out along the shortest path; 4) a fee of 1 coin is charged for every 100 meters traveled (rounding the distance to the nearest 100 meters in favor of the driver). How many blocks are there in the city? What is the maximum and minimum fare that the driver can charge the passenger for a ride from block $(7,1)$ to block $(2 ; 10)$ without violating the rules? | Answer: 99 blocks; $c_{\min }=10$ coins, $c_{\max }=14$ coins. | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Kuzya the flea can make a jump in any direction on a plane for exactly 19 mm. Her task is to get from point $A$ to point $B$ on the plane, the distance between which is 1812 cm. What is the minimum number of jumps she must make to do this? | Answer: $n_{\min }=954$ jumps. | 954 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. When purchasing goods for an amount of no less than 1000 rubles, the store provides a discount on subsequent purchases of $50 \%$. Having 1200 rubles in her pocket, Dasha wanted to buy 4 kg of strawberries and 6 kg of sugar. In the store, strawberries were sold at a price of 300 rubles per kg, and sugar - at a price of 30 rubles per kg. Realizing that she didn't have enough money for the purchase, Dasha still managed to buy what she intended. How did she do it? | First purchase: 3 kg of strawberries, 4 kg of sugar. Its cost is $300 \times 3 + 4 \times 30 = 1020$ rubles. Second purchase: 1 kg of strawberries, 2 kg of sugar. With a $50\%$ discount, its price is $(300 + 60) \cdot 0.5 = 180$ rubles. The total amount of both purchases is $1200$ rubles. | 1200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. The polynomial $p_{1}=x-a$ can have a root $x=a$ coinciding with one of the roots of the product $p(x)=p_{1}(x) \cdot p_{2}(x)$.
Case $1 \quad a=1$
Then the polynomial $p_{2}(x)=(x-1)^{r}(x-2)^{s}(x+3)^{t}$, where $r \geq 1, s \geq 1, t \geq 1-$ are integers, $r+s+t=4$, satisfies the condition of the problem. The polynomial sum $p_{1}(x)+p_{2}(x)$ has a constant term $u=(-1)^{r+s} \cdot 2^{s} \cdot 3^{t}-1=(-1)^{t} \cdot 2^{s} \cdot 3^{t}-1$ which has a maximum value $u_{\max }=17$ when $r=1, s=1, t=2$.
Case $2 \quad a=2$
Then the polynomial $p_{2}(x)=(x-1)^{r}(x-2)^{s}(x+3)^{t}$ and the polynomial $p_{1}(x)+p_{2}(x)$ has a constant term $u=(-1)^{r+s} \cdot 2^{s} \cdot 3^{t}-2=(-1)^{t} \cdot 2^{s} \cdot 3^{t}-2$, taking the maximum value $u_{\max }=16$ when $r=1, s=1, t=2$.
Case $3 \quad a=-3$
The constant term $u=(-1)^{r+s} \cdot 2^{s} \cdot 3^{t}+3$ takes the maximum value $u_{\max }=21$ when $s=r=1, t=2$
The required polynomials: $p_{2}(x)=(x-1)(x-2)(x+3)^{2}, p_{1}(x)=x+3$ | Answer: $p_{1}(x)=x+3, p_{2}(x)=(x-1)(x-2)(x+3)^{2} ; a_{0}=21$ | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A natural number $a$ is divisible by 21 and has 105 different divisors, including 1 and $a$. Find the smallest such $a$. | Answer: $a_{\min }=2^{6} \cdot 3^{4} \cdot 7^{2}=254016$ | 254016 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. A natural number $a$ is divisible by 35 and has 75 different divisors, including 1 and $a$. Find the smallest such $a$. | Answer: $a_{\text {min }}=2^{4} \cdot 5^{4} \cdot 7^{2}=490000$. | 490000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. A natural number $a$ is divisible by 55 and has 117 distinct divisors, including 1 and $a$. Find the smallest such $a$. | Answer: $a_{\min }=2^{12} \cdot 5^{2} \cdot 11^{2}=12390400$. | 12390400 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2.17. Final round of the "Rosatom" Olympiad, 7th grade
# Answers and solutions
Problem 1 Answer: 9
There exists a set of 8 buttons in which there are no three buttons of the same color: each color has two buttons. In any set of 9 buttons, there will be at least one triplet of buttons of the same color.
If we assume the opposite, that there are no more than 2 buttons of the same color, then there can be no more than 8 such buttons in total, which contradicts the condition. | Answer: 9 buttons.
Problem 2 Answer: 1261
\[
\left\{\begin{array}{l}
a=35 n+1 \\
a=45 m+1
\end{array} \rightarrow 35 n=45 m \rightarrow 7 n=9 m \rightarrow\left\{\begin{array}{c}
n=9 t \\
m=7 t, t \in Z
\end{array} \rightarrow\right.\right.
\]
\[
a=315 t+1 \geq 1000 \rightarrow t \geq 4 \rightarrow a_{\text {min }}=1261
\]
## Problem 3 Answer: 1) 4026 2) Sasha
At each step of the game, the sum of the numbers written on the cards in the bag does not change.
At the beginning of the game, it was \(2013 \times 2=4026\). Each step of the game changes the parity of the number of cards in the bag, as it reduces their number by one. Initially, the total number of cards in the bag is odd, so Sasha will make his move when the number of cards in the bag is odd, and Dasha when it is even. The last move occurs when there is one card in the bag, i.e., their number is odd, and Sasha will notice this.
Problem 4 Answer: 1) \(17 \quad\) 2) 90
\((12,18,21,24,27,30,36,42,45,48,54,60,63,72,81,84,90)\)
\(x, y\) - digits of the desired number
\[
a=10 x+y=3 k, \quad a=10 x+y=(x+y) m \rightarrow\left\{\begin{array}{c}
x+y=\frac{3 k}{m}, 2 \leq m \leq 10, \\
10 x+y=3 k, 4 \leq k \leq 33
\end{array}\right.
\]
\[
\left\{\begin{array}{l}
x=\frac{k}{3} \cdot \frac{m-1}{m} \\
y=\frac{k}{3} \cdot \frac{10-m}{m}
\end{array}\right.
\]
Case \(m=2 \rightarrow\left\{\begin{array}{l}
x=k / 6 \\
y=4 k / 3
\end{array} \rightarrow k=6 t \rightarrow\left\{\begin{array}{l}
x=t \\
y=8 t
\end{array} \rightarrow t=1 \rightarrow a=18\right.\right.
\]
Case \(m=3 \rightarrow\left\{\begin{array}{l}
x=2 k / 9 \\
y=7 k / 9
\end{array} \rightarrow k=9 t \rightarrow\left\{\begin{array}{l}
x=2 t \\
y=7 t
\end{array} \rightarrow a=27\right.\right.
\]
Case \(m=4 \rightarrow\left\{\begin{array}{l}
x=k / 4 \\
y=k / 2
\end{array} \rightarrow k=4 t \rightarrow\left\{\begin{array}{l}
x=t \\
y=2 t
\end{array} \rightarrow a=12,24,36,48\right.\right.
\]
Case \(m=5 \rightarrow\left\{\begin{array}{c}
x=4 k / 15 \\
y=k / 3
\end{array} \rightarrow k=15 t \rightarrow\left\{\begin{array}{l}
x=4 t \\
y=5 t
\end{array} \rightarrow a=45\right.\right.
\]
Case \(m=6 \rightarrow\left\{\begin{array}{l}
x=5 k / 18 \\
y=2 k / 9
\end{array} \rightarrow k=18 t \rightarrow\left\{\begin{array}{l}
x=5 t \\
y=4 t
\end{array} \rightarrow a=54\right.\right.
\]
Case \(m=7 \rightarrow\left\{\begin{array}{l}
x=2 k / 7 \\
y=k / 7
\end{array} \rightarrow k=7 t \rightarrow\left\{\begin{array}{l}
x=2 t \\
y=t
\end{array} \rightarrow a=21,42,63,84\right.\right.
\]
Case \(m=8 \rightarrow\left\{\begin{array}{c}
x=7 k / 24 \\
y=k / 12
\end{array} \rightarrow k=24 t \rightarrow\left\{\begin{array}{l}
x=7 t \\
y=2 t
\end{array} \rightarrow a=72\right.\right.
\]
Case \(m=9 \rightarrow\left\{\begin{array}{c}
x=8 k / 27 \\
y=k / 27
\end{array} \rightarrow k=27 t \rightarrow\left\{\begin{array}{c}
x=8 t \\
y=t
\end{array} \rightarrow a=81\right.\right.
\]
Case \(m=10 \rightarrow\left\{\begin{array}{c}
x=3 k / 10 \\
y=0
\end{array} \rightarrow k=10 t \rightarrow\left\{\begin{array}{l}
x=3 t \\
y=0
\end{array} \rightarrow a=30,60,90\right.\right.
\]
\[
a_{\max }=90, \text{ number of numbers } 17
\]
Problem 5 Answer: 2 weights
Indicate the minimum number of 3 kg weights needed to weigh loads of 1, 2, 3, and 4 kg.
Load of 1 kg - 2 weights of 3 kg are needed
Load of 2 kg or 3 kg - 1 weight of 3 kg is needed
Load of 4 kg - 2 weights of 3 kg are needed
The weight of any load when divided by 5 has a remainder of 0, 1, 2, 3, or 4. If the remainder is 0, the load can be weighed using only 5 kg weights. If the weight of the load is \(5 n+1\), the weighing picture is obtained from the picture of weighing 1 kg:
If the weight of the load is \(5 n+2\), the weighing picture can be used for a 2 kg load:
If the weight of the load is \(5 n+3\), the weighing picture can be used for a 3 kg load:
Finally, a load of \(5 n+4\) kg is weighed similarly to a 4 kg load:
 | 1261 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In the decimal representation of a six-digit number $a$, there are no zeros and the sum of its digits is 18. Find the sum of all different numbers obtained from the number $a$ by cyclic permutations of its digits. In a cyclic permutation, all digits of the number, except the last one, are shifted one place to the right, and the last one is moved to the first place. | 3. Solution. Case 1. The number $a=333333$. This number does not change under cyclic permutations, so it is the only one and the sum of the numbers is the number itself, that is, 333333.
Case 2. The number $a$ consists of three identical cycles of two digits each, for example, $a=242424$. Such numbers have two different cyclic permutations: 242424 and 424242, the sum of which is $666666=2 \cdot 333333$. For any other such number, we will get the same sum of its cyclic permutations: $151515+515151=666666$ and so on.
Case 3. The number $a$ consists of two identical cycles of three digits each, for example, $a=423423$. Such numbers have three different cyclic permutations: 423423, 342342, and 234234, the sum of which is $999999=3 \cdot 333333$. For any other such number, we will get the same sum of its cyclic permutations: $513513+351351+135135=999999$ and so on.
Case 4. All six cyclic permutations are different. Let $a=a_{5} \cdot 10^{5}+a_{4} \cdot 10^{4}+a_{3} \cdot 10^{3}+a_{2} \cdot 10^{2}+a_{1} \cdot 10+a_{0}$.
If we write out all its cyclic permutations in the same form, add the results, and regroup the terms, we get
$a_{5}\left(10^{5}+10^{4}+\ldots+1\right)+a_{4}\left(10^{5}+10^{4}+\ldots+1\right)+\ldots+a_{0}\left(10^{5}+10^{4}+\ldots+1\right)=$ $=\left(a_{5}+a_{4}+a_{3}+a_{2}+a_{1}+a_{0}\right)\left(10^{5}+10^{4}+\ldots+1\right)=18 \cdot 111111=1999998$ | 1999998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. On a sheet of paper, 14 parallel lines $L$ and 15 lines $P$ perpendicular to them are drawn. The distances between adjacent lines from $L$ from the first to the last are given: 2;4;6;2;4;6;2;4;6;2;4;6;2. The distances between adjacent lines from $P$ are also known: 3;1;2;6;3;1;2;6;3;1;2;6;3;1. Find the greatest length of the side of a square whose boundaries lie on the lines $L$ and $P$. | 5. Solution. We will prove that the maximum length of the side of the square is 40. Calculate the distance from the first to the last line in $P: 3+1+2+6+3+1+2+6+3+1+2+6+3+1=40$. Therefore, the side length of the square cannot be more than 40. On the other hand, in $L$ the distance from the second line to the third line from the end is $4+6+2+4+6+2+4+6+2+4=40$. This means that a square with a side length of 40 exists. | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. The polynomial $P(x)$ with integer coefficients satisfies the condition $P(29)=P(37)=2022$. Find the smallest possible value of $P(0)>0$ under these conditions. | Answer: $P(0)_{\min }=949$. | 949 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The polynomial $P(x)$ with integer coefficients satisfies the condition $P(11)=P(13)=2021$. Find the smallest possible value of $P(0)>0$ under these conditions. | Answer: $P(0)_{\text {min }}=19$. | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The polynomial $P(x)$ with integer coefficients satisfies the condition $P(19)=P(21)=2020$. Find the smallest possible value of $P(0)>0$ under these conditions. | Answer: $P(0)_{\min }=25$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The number $A$ in decimal form is represented as $A=\overline{7 a 631 b}$, where $a, b$ are non-zero digits. The number $B$ is obtained by summing all distinct six-digit numbers, including $A$, that are formed by cyclic permutations of the digits of $A$ (the first digit moves to the second position, the second to the third, and so on, with the last digit moving to the first position). How many numbers $A$ exist such that $B$ is divisible by 121? Find the largest such $A$. | 2. Solution. The sum of the digits of number $A$ and the numbers obtained from $A$ by cyclic permutations of its digits is $a+b+17$. After summing these numbers (there are 6 of them), in each digit place of number $B$ we get
$a+b+17$, so $B=(a+b+17)\left(10^{5}+10^{4}+10^{3}+10^{2}+10+1\right)=(a+b+17) \cdot 111111$. Since
$111111=3 \cdot 7 \cdot 11 \cdot 13 \cdot 37$, $B$ is divisible by 121 if and only if $a+b+17$ is divisible by 11. Since $a$ and $b$ are digits, there are only two possible cases: $a+b=5$ and $a+b=16$. From this, we find the possible pairs of $a$ and $b$: $a=1, b=4 ; a=2, b=3 ; a=3, b=2 ; a=4, b=1 ; a=7, b=9 ; a=8, b=8 ; a=9, b=7$. Thus, 7 numbers satisfy the condition of the problem. The largest number is obtained when $a=9, b=7$. It is 796317. | 796317 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. How many different pairs of integers $a$ and $b$ exist such that the equation $a x^{2}+b x+1944=0$ has positive integer roots? | Answer: $\quad 108+24=132$ pairs. | 132 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. How many different pairs of integers $a$ and $b$ exist such that the equation $a x^{2}+b x+432=0$ has positive integer roots | Answer: $\quad 78+20=98$ pairs. | 98 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. By the condition
$$
T(t)=\frac{270-s(t)}{s(t) / t}=\frac{t(270-s(t))}{s(t)}=C>1, t \in[0.5 ; 1]
$$
Then $s(t)=\frac{270 t}{t+C}$ on this interval. The speed of movement
$$
\begin{aligned}
& v(t)=s^{\prime}(t)=\frac{270 C}{(t+C)^{2}}=60 \text { when } t=1 \text {, i.e. } \\
& \qquad 2 c^{2}-5 c+2=0 \rightarrow C_{1}=2, C_{2}=\frac{1}{2}
\end{aligned}
$$
The second value of the constant is not allowed by the condition. Thus, $s(t)=\frac{270 t}{t+2}$ and $s(1)=90$.
$$
v(t)=\left.\frac{540}{(t+2)^{2}}\right|_{t=\frac{1}{2}}=\frac{432}{5}=86.4 \text { km } / \text { hour }
$$ | Answer: 1) 90 km; 2) 86.4 km/hour | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. If $m+2019 n$ and $n+2019 m$ are divisible by $d$, then the number
$$
2019(m+2019 n)-(n+2019 m)=(2019^2-1) n
$$
is also divisible by $d$. If $n$ is divisible by $d$, and $m+2019 n$ is divisible by $d$, then $m$ is divisible by $d$ and the numbers $m$ and $n$ are not coprime. Therefore, $d$ divides the number
$$
2019^2-1=2018 \cdot 2020=2^3 \cdot 5 \cdot 101 \cdot 1009
$$
Thus, the smallest possible prime number is $d=101$. It remains to find coprime $m$ and $n$ for which it is realized. For example, $m=102, n=1$. Then
$$
\begin{gathered}
m+2019 n=102+2019=2121=21 \cdot 101 \text { and } \\
n+2019 m=1+2019 \cdot 102=205939=2039 \cdot 101 .
\end{gathered}
$$ | Answer: $d_{\text {min }}=101$. | 101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Kostya is making a car trip from point A to point B, which are 320 km apart. The route of the trip is displayed on the computer screen. At any moment in time $t$ (hours), Kostya can receive information about the distance traveled $s(t)$ (km), the speed of movement $v(t)$ (km/hour), and the estimated time $T=T(t)$ (hours) until the end of the trip. The program for calculating $T(t)$ is based on the assumption that the remaining part of the journey will be traveled at a speed equal to the average speed of the vehicle's movement over the time interval $[0 ; t]$. One hour after the start of the trip, he looked at the speedometer - 60 km/h. On the time interval $[1 ; 2]$, Kostya noticed that $T>1$ and does not change. How far from point $A$ was the car two hours after the start of the trip? What was the speed of the car 2 hours after the start of the trip? | Answer: 1) 128 km; 2) 38.4 km/h. | 128 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that for some positive coprime numbers $m$ and $n$, the numbers $m+2024 n$ and $n+2024 m$ have a common prime divisor $d>7$. Find the smallest possible value of the number $d$ under these conditions. | Answer: $d_{\min }=17$.
For example,
$$
\begin{aligned}
& m=16, n=1 \rightarrow 2024 m+n=2024 \cdot 16+1=32385=17 \cdot 1905 \\
& m+2024 n=16+2024=17 \cdot 120
\end{aligned}
$$ | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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