problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 2 43 | problem_type stringclasses 8
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3. Find the number of natural numbers $k$, not exceeding 333300, such that $k^{2}-2 k$ is divisible by 303. Answer: 4400. | Solution. Factoring the dividend and divisor, we get the condition $k(k-2):(3 \cdot 101)$. This means that one of the numbers $k$ or $(k-2)$ is divisible by 101. Let's consider two cases.
a) $k: 101$, i.e., $k=101 p, p \in \mathrm{Z}$. Then we get $101 p(101 p-2):(3 \cdot 101) \Leftrightarrow p(101 p-2) \vdots 3$. The... | 4400 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 454500, such that $k^{2}-k$ is divisible by 505. | Answer: 3600.
Solution. By factoring the dividend and divisor, we get the condition $k(k-1):(5 \cdot 101)$. This means that one of the numbers $k$ or $(k-1)$ is divisible by 101. Let's consider two cases.
a) $k \vdots: 101$, i.e., $k=101 p, p \in \mathbb{Z}$. Then we get $101 p(101 p-1):(5 \cdot 101) \Leftrightarrow ... | 3600 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 700. The answer should be presented as an integer. | Answer: 2520.
Solution. Since $700=7 \cdot 2^{2} \cdot 5^{2}$, the sought numbers can consist of the following digits: (a) two twos, two fives, one seven, and three ones, or (b) one four, two fives, one seven, and four ones. We will calculate the number of variants in each case.
(a) First, we choose two places out of... | 2520 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 4900. The answer should be presented as an integer. | Answer: 4200.
Solution. Since $4900=7^{2} \cdot 2^{2} \cdot 5^{2}$, the sought numbers can consist of the following digits: (a) two twos, two fives, two sevens, and two ones, or (b) a four, two fives, two sevens, and three ones. We will calculate the number of variants in each case.
(a) First, we choose two places ou... | 4200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] A magician has a set of $12^{2}$ different cards. Each card has one side red and the other side blue; on each card, there is a natural number from 1 to 12 written on both sides. We will call a card a duplicate if the numbers on both sides of the card are the same. The magician wants to draw two cards such... | Answer: 1386.
Solution. Since the magician has a set of $12^{2}$ cards, all possible card variants exist (for each pair of numbers $(i ; j)$, where $1 \leqslant i \leqslant 12,1 \leqslant j \leqslant 12$, there is a card with the number $i$ written on the red side and $j$ on the blue side). Let's consider two types of... | 1386 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] A magician has a set of $15^{2}$ different cards. Each card has one side red and the other side blue; on each card, there is a natural number from 1 to 15 written on both sides. We will call a card a duplicate if the numbers on both sides of the card are the same. The magician wants to draw two cards such... | Answer: 2835.
Solution. Since the magician has a set of $15^{2}$ cards, all possible card variants exist (for each pair of numbers $(i ; j)$, where $1 \leqslant i \leqslant 15,1 \leqslant j \leqslant 15$, there is a card with the number $i$ written on the red side and $j$ on the blue side). Let's consider two types of... | 2835 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] A magician has a set of $20^{2}$ different cards. Each card has one side red and the other side blue; on each card, there is a natural number from 1 to 20 written on both sides. We will call a card a duplicate if the numbers on both sides of the card are the same. The magician wants to draw two cards such... | Answer: 7030.
Solution. Since the set contains $20^{2}$ cards, the magician has all possible card variants (for each pair of numbers $(i ; j)$, where $1 \leqslant i \leqslant 20,1 \leqslant j \leqslant 20$, there will be a card with the number $i$ written on the red side and $j$ on the blue side). Let's consider two t... | 7030 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] A magician has a set of $16^{2}$ different cards. Each card has one side red and the other side blue; on each card, there is a natural number from 1 to 16 written on both sides. We will call a card a duplicate if the numbers on both sides of the card are the same. The magician wants to draw two cards such... | Answer: 3480.
Solution. Since the magician has a set of $16^{2}$ cards, there are all possible card variants (for each pair of numbers $(i ; j)$, where $1 \leqslant i \leqslant 16,1 \leqslant j \leqslant 16$ there will be a card with the number $i$ written on the red side and $j$ on the blue side). Let's consider two ... | 3480 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On the coordinate plane, squares are considered, all vertices of which have integer non-negative coordinates, and the center is located at the point ( $60 ; 45$ ). Find the number of such squares. | Answer: 2070.
Solution. Draw through the given point $(60 ; 45)$ vertical and horizontal lines $(x=60$ and $y=45)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "bottom" vertex of the square can be located in 45 ways: ... | 2070 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. On the coordinate plane, squares are considered, all vertices of which have natural coordinates, and the center is located at the point $(55 ; 40)$. Find the number of such squares. | Answer: 1560.
Solution. Draw through the given point $(55 ; 40)$ vertical and horizontal lines $(x=55$ and $y=40)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "lower" vertex of the square can be located in 39 ways: $... | 1560 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. On the coordinate plane, squares are considered, all vertices of which have integer non-negative coordinates, and the center is located at the point ( $25 ; 60$ ). Find the number of such squares. | Answer: 650.
Solution. Draw through the given point $(25 ; 60)$ vertical and horizontal lines $(x=25$ and $y=60)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "left" vertex of the square can be located in 25 ways: $(0... | 650 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. On the coordinate plane, squares are considered, all vertices of which have natural coordinates, and the center is located at the point $(35 ; 65)$. Find the number of such squares. | Answer: 1190.
Solution. Draw through the given point $(35 ; 65)$ vertical and horizontal lines $(x=35$ and $y=65)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "left" vertex of the square can be located in 34 ways: $(... | 1190 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Given the number $5300 \ldots 0035$ (100 zeros). It is required to replace some two zeros with non-zero digits so that after the replacement, the resulting number is divisible by 495. In how many ways can this be done? | Answer: 22100.
Solution. $495=5 \cdot 9 \cdot 11$. Divisibility by 5 is always satisfied since the number ends in five. For investigating divisibility by 11, it is significant which positions the replaceable digits occupy.
First case. We replace two zeros in positions of the same parity (both even or both odd). For d... | 22100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given the number $800 \ldots 008$ (80 zeros). It is required to replace some two zeros with non-zero digits so that after the replacement, the resulting number is divisible by 198. In how many ways can this be done? | Answer: 14080.
Solution. $198=2 \cdot 9 \cdot 11$. Divisibility by 2 is always satisfied since the number ends in eight. For investigating divisibility by 11, it is significant on which positions the replaceable digits stand.
First case. We replace two zeros in positions of the same parity (both even or both odd). Fo... | 14080 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given the number $500 \ldots 005$ (80 zeros). It is required to replace some two zeros with non-zero digits so that after the replacement, the resulting number is divisible by 165. In how many ways can this be done? | Answer: 17280.
Solution. $165=3 \cdot 5 \cdot 11$. Divisibility by 5 is always satisfied since the number ends in a five. For divisibility by 11, it is significant on which positions the replaceable digits stand.
First case. We replace two zeros in positions of the same parity (both even or both odd). For divisibilit... | 17280 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given the number $200 \ldots 002$ (100 zeros). It is required to replace some two zeros with non-zero digits so that after the replacement, the resulting number is divisible by 66. In how many ways can this be done? | Answer: 27100.
Solution. $66=3 \cdot 2 \cdot 11$. Divisibility by 2 is always satisfied since the number ends in two. To investigate divisibility by 11, it is significant which positions the replaceable digits occupy.
First case. We replace two zeros in positions of the same parity (both even or both odd). For divisi... | 27100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. On each of the lines $x=0$ and $x=2$, there are 62 points with ordinates $1, 2, 3, \ldots, 62$. In how many ways can three points be chosen from the marked 124 points so that they form the vertices of a right triangle? | Answer: 7908.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle is on the second line. Let $ABC$ be the given triangle with a right angle at vertex $C$, and $CH$ be its height dropped to the hypotenuse. From the proportionality of the ... | 7908 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On each of the lines $y=0$ and $y=2$, there are 64 points with abscissas $1,2,3, \ldots, 64$. In how many ways can three points be chosen from the marked 128 points so that they form the vertices of a right triangle? | Answer: 8420.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle is on the second line. Let $ABC$ be the given triangle with a right angle at vertex $C$, and $CH$ be its height dropped to the hypotenuse. From the proportionality of the ... | 8420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Given 2117 cards, on which natural numbers from 1 to 2117 are written (each card has exactly one number, and the numbers do not repeat). It is required to choose two cards such that the sum of the numbers written on them is divisible by 100. In how many ways can this be done? | Answer: 22386.
Solution. We will take the cards in turn. There are several cases depending on the number written on the first card.
1) The number on the card ends in 00 (there are 21 such cards). For the sum to be divisible by 100, the second card must also end in 00. We get $C_{2} 1^{2}=\frac{21 \cdot 20}{2}=210$ op... | 22386 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Find the number of distinct quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have two distinct roots, which are powers of the number 7 with integer non-negative exponents, and at the same time, their coefficients in absolute value do not exceed $343^{36}$. | Answer: 2969.
Solution. Such quadratic trinomials can be represented in the form $\left(x-7^{a}\right)\left(x-7^{b}\right)$, where $a \geqslant 0$, $b \geqslant 0$ are integers. To avoid repetitions, we assume that $a>b$. Expanding the brackets, we get $x^{2}-\left(7^{a}+7^{b}\right) x+7^{a+b}$. By the condition
$$
\... | 2969 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Given 2414 cards, on which natural numbers from 1 to 2414 are written (each card has exactly one number, and the numbers do not repeat). It is required to choose two cards such that the sum of the numbers written on them is divisible by 100. In how many ways can this be done? | Answer: 29112.
Solution. We will take cards in turn. There are several cases depending on the number written on the first card.
1) The number on the card ends in 00 (there are 24 such cards). For the sum to be divisible by 100, the second card must also end in 00. We get a total of $C_{2} 4^{2}=\frac{24 \cdot 23}{2}=... | 29112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Find the number of distinct quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have two distinct roots, which are powers of the number 5 with integer non-negative exponents, and at the same time, their coefficients in absolute value do not exceed $122^{20}$. | Answer: 5699.
Solution. Such quadratic trinomials can be represented in the form $\left(x-5^{a}\right)\left(x-5^{b}\right)$, where $a \geqslant 0$, $b \geqslant 0$ are integers. To avoid repetitions, we assume that $a>b$. Expanding the brackets, we get $x^{2}-\left(5^{a}+5^{b}\right) x+5^{a+b}$. According to the condi... | 5699 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The numbers $x$ and $y$ are solutions to the system of equations param 1, where $a$ is a parameter. What param 2 value does the expression param 3 take?
| param1 | param2 | param3 | |
| :---: | :---: | :---: | :---: |
| $\left\{\begin{array}{l}a x+y=a+1 \\ x+4 a y=3\end{array}\right.$ | maximum | $x^{2}-6 y^{2}$ |... | 5. The numbers $x$ and $y$ are solutions to the system of equations param 1, where $a$ is a parameter. What param 2 value does the expression param 3 take?
| param1 | param2 | param3 | Answer |
| :---: | :---: | :---: | :---: |
| $\left\{\begin{array}{l}a x+y=a+1 \\ x+4 a y=3\end{array}\right.$ | maximum | $x^{2}-6 y^... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Given a regular param1. Find the number of its vertex quadruples that are vertices of a convex quadrilateral with exactly two angles equal to $90^{\circ}$. (Two quadruples of vertices that differ in the order of vertices are considered the same.)
| param1 | |
| :---: | :---: |
| 16-gon | |
| 18-gon | |
| 20-gon ... | 9. Given a regular param1. Find the number of its vertex quadruples that are vertices of a convex quadrilateral with exactly two angles equal to $90^{\circ}$. (Two quadruples of vertices that differ in the order of vertices are considered the same.)
| param1 | Answer |
| :---: | :---: |
| 16-gon | 336 |
| 18-gon | 504... | 504 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. Find the sum of the roots of the equation param1 that lie in the interval param2. Write the answer in degrees.
| param 1 | param2 | |
| :---: | :---: | :---: |
| $\sin x+\sin ^{2} x+\cos ^{3} x=0$ | $\left[360^{\circ} ; 720^{\circ}\right]$ | |
| $\cos x-\cos ^{2} x-\sin ^{3} x=0$ | $\left[180^{\circ} ; 540^{\cir... | 10. Find the sum of the roots of the equation param1 lying in the interval param2. Write the answer in degrees.
| param 1 | param2 | Answer |
| :---: | :---: | :---: |
| $\sin x+\sin ^{2} x+\cos ^{3} x=0$ | $\left[360^{\circ} ; 720^{\circ}\right]$ | 1800 |
| $\cos x-\cos ^{2} x-\sin ^{3} x=0$ | $\left[180^{\circ} ; 54... | -360 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=6 \\
\operatorname{LCM}(a ; b ; c)=2^{15} \cdot 3^{16}
\end{array}\right.
$$ | Answer: 7560.
Solution. Let $a=2^{\alpha_{1}} \cdot 3^{\alpha_{2}}, b=2^{\beta_{1}} \cdot 3^{\beta_{2}}, c=2^{\gamma_{1}} \cdot 3^{\gamma_{2}}$ (the numbers $a$, $b$, $c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=2^{... | 7560 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=15 \\
\operatorname{LCM}(a ; b ; c)=3^{15} \cdot 5^{18}
\end{array}\right.
$$ | Answer: 8568.
Solution. Let $a=3^{\alpha_{1}} \cdot 5^{\alpha_{2}}, b=3^{\beta_{1}} \cdot 5^{\beta_{2}}, c=3^{\gamma_{1}} \cdot 5^{\gamma_{2}}$ (the numbers $a$, $b$, $c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=3^{... | 8568 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=21 \\
\operatorname{LCM}(a ; b ; c)=3^{17} \cdot 7^{15}
\end{array}\right.
$$ | Answer: 8064.
Solution. Let $a=3^{\alpha_{1}} \cdot 7^{\alpha_{2}}, b=3^{\beta_{1}} \cdot 7^{\beta_{2}}, c=3^{\gamma_{1}} \cdot 7^{\gamma_{2}}$ (the numbers $a$, $b, c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=3^{\m... | 8064 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=10 \\
\operatorname{LCM}(a ; b ; c)=2^{17} \cdot 5^{16}
\end{array}\right.
$$ | Answer: 8640.
Solution. Let $a=2^{\alpha_{1}} \cdot 5^{\alpha_{2}}, b=2^{\beta_{1}} \cdot 5^{\beta_{2}}, c=2^{\gamma_{1}} \cdot 5^{\gamma_{2}}$ (the numbers $a$, $b$, $c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=2^{... | 8640 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=35 \\
\operatorname{LCM}(a ; b ; c)=5^{18} \cdot 7^{16}
\end{array}\right.
$$ | Answer: 9180.
Solution. Let $a=5^{\alpha_{1}} \cdot 7^{\alpha_{2}}, b=5^{\beta_{1}} \cdot 7^{\beta_{2}}, c=5^{\gamma_{1}} \cdot 7^{\gamma_{2}}$ (the numbers $a$, $b, c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=5^{\m... | 9180 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$, satisfying the system of equations
$$
\left\{\begin{array}{l}
\text { GCD }(a ; b ; c)=14 \\
\text { LCM }(a ; b ; c)=2^{17} \cdot 7^{18}
\end{array}\right.
$$ | Answer: 9792.
Solution. Let $a=2^{\alpha_{1}} \cdot 7^{\alpha_{2}}, b=2^{\beta_{1}} \cdot 7^{\beta_{2}}, c=2^{\gamma_{1}} \cdot 7^{\gamma_{2}}$ (the numbers $a$, $b, c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=2^{\m... | 9792 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\operatorname{GCD}(a ; b ; c)=22 \\
\operatorname{LCM}(a ; b ; c)=2^{16} \cdot 11^{19}
\end{array}\right.
$$ | Answer: 9720.
Solution. Let $a=2^{\alpha_{1}} \cdot 11^{\alpha_{2}}, b=2^{\beta_{1}} \cdot 11^{\beta_{2}}, c=2^{\gamma_{1}} \cdot 11^{\gamma_{2}}$ (the numbers $a, b, c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=2^{\... | 9720 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. [5 points] Find the number of triples of natural numbers $(a ; b ; c)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
\text { GCD }(a ; b ; c)=33, \\
\text { LCM }(a ; b ; c)=3^{19} \cdot 11^{15} .
\end{array}\right.
$$ | Answer: 9072.
Solution. Let $a=3^{\alpha_{1}} \cdot 11^{\alpha_{2}}, b=3^{\beta_{1}} \cdot 11^{\beta_{2}}, c=3^{\gamma_{1}} \cdot 11^{\gamma_{2}}$ (the numbers $a, b, c$ cannot contain any other prime factors - otherwise the second condition of the system is violated). From this,
$$
\operatorname{LCM}(a ; b ; c)=3^{\... | 9072 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Which whole numbers from 1 to 60000 (inclusive) are there more of, and by how many: those containing only even digits or those containing only odd digits? | Answer: There are 780 more numbers containing only odd digits.
Solution. Consider $k$-digit numbers ( $1 \leqslant k \leqslant 4$ ). The number of numbers consisting only of odd digits is $5^{k}$ (for each of the $k$ positions, any of the digits $1,3,5,7$, 9 can be chosen); the number of numbers consisting only of eve... | 780 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. On each of the lines $y=3$ and $y=4$, there are 73 points with abscissas $1,2,3, \ldots, 73$. In how many ways can three points be chosen from the marked 146 so that they form the vertices of a right triangle? | Answer: 10654.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle is on the second line. Let $ABC$ be the given triangle with a right angle at vertex $C$, and $CH$ be its height dropped to the hypotenuse. From the proportionality of the... | 10654 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On the extension of side $A C$ of triangle $A B C$ beyond point $A$, point $T$ is marked such that $\angle B A C=2 \angle B T C$. Find the area of triangle $A B C$, given that $A B=A C, B T=70$, $A T=37$. | Answer: 420.
Solution. Let $\angle B T A=\alpha$, then by the condition $\angle B A C=2 \alpha$. Triangle $A B C$ is isosceles with base $B C$, so $\angle A B C=\angle A C B=\frac{1}{2}\left(180^{\circ}-2 \alpha\right)=90^{\circ}-\alpha$. By the sum of angles in triangle $T B C$, we get that $\angle T B C=180^{\circ}-... | 420 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Which whole numbers from 1 to 80000 (inclusive) are there more of, and by how many: those containing only even digits or those containing only odd digits? | Answer: There are 780 more numbers containing only odd digits.
Solution. Consider $k$-digit numbers ( $1 \leqslant k \leqslant 4$ ). The number of numbers consisting only of odd digits is $5^{k}$ (for each of the $k$ positions, any of the digits $1,3,5,7$, 9 can be chosen); the number of numbers consisting only of eve... | 780 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. On each of the lines $x=5$ and $x=6$, there are 58 points with ordinates $1, 2, 3, \ldots, 58$. In how many ways can three points be chosen from the marked 116 so that they form the vertices of a right triangle? | Answer: 6724.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle is on the second line. Let $ABC$ be the given triangle with a right angle at vertex $C$, and $CH$ be its height dropped to the hypotenuse. From the proportionality of the ... | 6724 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On the extension of side $A C$ of triangle $A B C$ beyond point $A$, point $T$ is marked such that $\angle B A C=2 \angle B T C$. Find the area of triangle $A B C$, given that $A B=A C, B T=42$, $A T=29$. | Answer: 420.
Solution. Let $\angle B T A=\alpha$, then by the condition $\angle B A C=2 \alpha$. Triangle $A B C$ is isosceles with base $B C$, so $\angle A B C=\angle A C B=\frac{1}{2}\left(180^{\circ}-2 \alpha\right)=90^{\circ}-\alpha$. By the sum of angles in triangle $T B C$, we get that $\angle T B C=180^{\circ}-... | 420 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Given 6000 cards, on which natural numbers from 1 to 6000 are written (each card has exactly one number, and the numbers do not repeat). It is required to choose two cards such that the sum of the numbers written on them is divisible by 100. In how many ways can this be done? | Answer: 179940.
Solution. We will take the cards in turn. There are several cases depending on the number written on the first card.
1) The number on the card ends in 00 (there are 60 such cards). For the sum to be divisible by 100, the second card must also have a number ending in 00. We get a total of $C_{60}^{2}=\... | 179940 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Find the number of distinct quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have two distinct roots, which are powers of the number 3 with natural exponents, and at the same time, their coefficients in absolute value do not exceed \(27^{45}\). | Answer: 4489.
Solution. Such quadratic trinomials can be represented as $\left(x-3^{a}\right)\left(x-3^{b}\right)$, where $a, b$ are natural numbers. To avoid repetitions, we assume that $a>b$. Expanding the brackets, we get $x^{2}-\left(3^{a}+3^{b}\right) x+3^{a+b}$. According to the condition
$$
\left\{\begin{array... | 4489 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Given 5000 cards, on which natural numbers from 1 to 5000 are written (each card has exactly one number, and the numbers do not repeat). It is required to choose two cards such that the sum of the numbers written on them is divisible by 100. In how many ways can this be done? | Answer: 124950.
Solution. We will take the cards in turn. There are several cases depending on the number written on the first card.
1) The number on the card ends in 00 (there are 50 such cards). For the sum to be divisible by 100, the second card must be chosen so that the number on it also ends in 00. We get a tot... | 124950 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Find the number of distinct quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have two distinct roots, which are powers of the number 5 with natural exponents, and at the same time, their coefficients in absolute value do not exceed $125^{48}$. | Answer: 5112.
Solution. Such quadratic trinomials can be represented in the form $\left(x-5^{a}\right)\left(x-5^{b}\right)$, where $a, b$ are natural numbers. To avoid repetitions, we assume that $a>b$. Expanding the brackets, we get $x^{2}-\left(5^{a}+5^{b}\right) x+5^{a+b}$. By the condition
$$
\left\{\begin{array}... | 5112 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. On the table, there are 130 different cards with numbers $502, 504, 506, \ldots, 758, 760$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the sum of the numbers on the selected cards is divisible by 3? | Answer: 119282
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a difference of 2. Therefore, the remainders of these numbers when divided by 3 alternate. Indeed, if one of these numbers is divisible by 3, i.e., has the form $3k$, where $k \in \mathbb{N}$, then the next num... | 119282 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On the table, there are 140 different cards with numbers $4, 8, 12, \ldots, 556, 560$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the sum of the numbers on the selected cards is divisible by 3? | Answer: 149224.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a difference of 4. Therefore, the remainders of these numbers when divided by 3 alternate. Indeed, if one of these numbers is divisible by 3, i.e., has the form $3k$, where $k \in \mathbb{N}$, then the next nu... | 149224 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On the table, there are 200 different cards with numbers $201, 203, 205, \ldots, 597, 599$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the sum of the numbers on the selected cards is divisible by 3? | Answer: 437844.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a difference of 2. Therefore, the remainders of these numbers when divided by 3 alternate. Indeed, if one of these numbers is divisible by 3, i.e., has the form $3k$, where $k \in \mathbb{N}$, then the next nu... | 437844 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On the table, there are 160 different cards with numbers $5, 10, 15, \ldots, 795, 800$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the sum of the numbers on the selected cards is divisible by 3? | Answer: 223342.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a common difference of 5. Therefore, the remainders when these numbers are divided by 3 alternate. Indeed, if one of these numbers is divisible by 3, i.e., has the form $3k$, where $k \in \mathbb{N}$, then the... | 223342 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. There are 207 different cards with numbers $1,2,3,2^{2}, 3^{2}, \ldots, 2^{103}, 3^{103}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer divisible by 6? | Answer: 267903.
Solution. Consider the case when one of the selected cards has a one written on it. Then, on the other two cards, even powers of two and three must be recorded. There are 51 ways to choose an even power of two and 51 ways to choose an even power of three, and since this choice is made independently, th... | 267903 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. There are 183 different cards with numbers $1,2,11,2^{2}, 11^{2}, \ldots, 2^{91}, 11^{91}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the product of the numbers on the selected cards is a square of an integer divisible by 22? | Answer: 184275.
Solution. Consider the case when one of the selected cards has the number one written on it. Then, on the other two cards, even powers of the numbers two and eleven must be recorded. There are 45 ways to choose an even power of two and 45 ways to choose an even power of eleven, and since this choice is... | 184275 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. There are 195 different cards with numbers $1, 5, 7, 5^{2}, 7^{2}, \ldots, 5^{97}, 7^{97}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer divisible by 35? | Answer: 223488.
Solution. Consider the case when one of the selected cards has a one written on it. Then, on the other two cards, even powers of fives and sevens must be recorded. There are 48 ways to choose an even power of five and 48 ways to choose an even power of seven, and since this choice is made independently... | 223488 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. There are 167 different cards with numbers $1, 3, 11, 3^{2}, 11^{2}, \ldots, 3^{83}, 11^{83}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer divisible by 33? | Answer: 139523.
Solution. Consider the case when one of the selected cards has the number one written on it. Then, on the other two cards, even powers of the numbers three and eleven must be recorded. There are 41 ways to choose an even power of three and 41 ways to choose an even power of eleven, and since this choic... | 139523 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. On each of the lines $y=1$ and $y=6$, there are 200 points with abscissas $1,2,3, \ldots, 200$. In how many ways can three points be chosen from the 400 marked points so that they form the vertices of a right triangle? | Answer: $C_{200}^{2} \cdot 4 + 190 \cdot 2 + 174 \cdot 4 = 80676$.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle lies on the other line. Let $A B C$ be the given triangle with a right angle at vertex $C$, and $C H$ be its height dr... | 80676 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On each of the lines $x=2$ and $x=9$, there are 400 points with ordinates $1,2,3, \ldots, 400$. In how many ways can three points be chosen from the 800 marked points so that they form the vertices of a right triangle? | Answer: 321372.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle is on the second line. Let $ABC$ be the given triangle with a right angle at vertex $C$, and $CH$ be its height dropped to the hypotenuse. From the proportionality of th... | 321372 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On each of the lines $y=1$ and $y=12$, there are 200 points with abscissas $1,2,3, \ldots, 200$. In how many ways can three points be chosen from the 400 marked points so that they form the vertices of a right triangle? | Answer: 80268.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle is on the second line. Let $ABC$ be the given triangle with a right angle at vertex $C$, and $CH$ be its height dropped to the hypotenuse. From the proportionality of the... | 80268 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On each of the lines $x=2$ and $x=15$, there are 400 points with ordinates $1,2,3, \ldots, 400$. In how many ways can three points be chosen from the 800 marked points so that they form the vertices of a right triangle? | Answer: 320868.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle is on the second line. Let $ABC$ be the given triangle with a right angle at vertex $C$, and $CH$ be its height dropped to the hypotenuse. From the proportionality of se... | 320868 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of distinct reduced quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have at least one root, all their roots are powers of the number 3 with integer non-negative exponents, and their coefficients in absolute value do not exceed \(27^{47}\). | Answer: 5111.
Solution. Such quadratic trinomials can be represented in the form $\left(x-3^{a}\right)\left(x-3^{b}\right)$, where $a \geqslant 0$, $b \geqslant 0$ are integers. To avoid repetitions, we assume that $a \geqslant b$. Expanding the brackets, we get $x^{2}-\left(3^{a}+3^{b}\right) x+3^{a+b}$. According to... | 5111 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of distinct reduced quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have at least one root, all their roots are powers of the number 5 with integer non-negative exponents, and their coefficients in absolute value do not exceed $122^{85}$. | Answer: 16511.
Solution. Such quadratic trinomials can be represented in the form $\left(x-5^{a}\right)\left(x-5^{b}\right)$, where $a \geqslant 0$, $b \geqslant 0$ are integers. To avoid repetitions, we assume that $a \geqslant b$. Expanding the brackets, we get $x^{2}-\left(5^{a}+5^{b}\right) x+5^{a+b}$. According t... | 16511 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of distinct reduced quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have at least one root, all their roots are powers of the number 7 with integer non-negative exponents, and their coefficients in absolute value do not exceed \(49^{68}\). | Answer: 4760.
Solution. Such quadratic trinomials can be represented in the form $\left(x-7^{a}\right)\left(x-7^{b}\right)$, where $a \geqslant 0$, $b \geqslant 0$ are integers. To avoid repetitions, we assume that $a \geqslant b$. Expanding the brackets, we get $x^{2}-\left(7^{a}+7^{b}\right) x+7^{a+b}$. According to... | 4760 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of distinct reduced quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have at least one root,
all their roots are powers of the number 11 with integer non-negative exponents, and their coefficients in absolute value do not exceed $1331^{38}$... | Answer: 3363.
Solution. Such quadratic trinomials can be represented in the form $\left(x-11^{a}\right)\left(x-11^{b}\right)$, where $a \geqslant 0, b \geqslant 0$ are integers. To avoid repetitions, we assume that $a \geqslant b$. Expanding the brackets, we get $x^{2}-\left(11^{a}+11^{b}\right) x+11^{a+b}$. According... | 3363 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5. Does there exist thirteen consecutive natural numbers such that their sum is a 2021st power of a natural number? | Answer. They exist.
Solution. Let's denote 13 consecutive numbers as $N-6, N-5$, ..., $N+5, N+6$. Then their sum is $13N$. If $N=13^{2020}$, then the sum will be $13N = 13 \cdot 13^{2020} = 13^{2021}$.
Comment. A correct answer without justification - 0 points.
9.6-1. On a given circle $\omega$, points $A, B$, and $... | 90 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. [3 points] Find the number of eight-digit numbers, the product of the digits of each of which is equal to 3375. The answer should be presented as an integer. | Answer: 1680.
Solution. Since $3375=3^{3} \cdot 5^{3}$, the sought numbers can consist of the following digits: (a) three threes, three fives, and two ones, or (b) one three, one nine, three fives, and three ones. We will calculate the number of variants in each case.
(a) First, we choose three places out of eight fo... | 1680 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. [3 points] Find the number of eight-digit numbers, the product of the digits of each of which is equal to 16875. The answer should be presented as an integer. | Answer: 1120.
Solution. Since $16875=3^{3} \cdot 5^{4}$, the sought numbers can consist of the following digits: (a) three threes, four fives, and one one, or (b) one three, one nine, four fives, and two ones. We will calculate the number of variants in each case.
(a) First, we choose three places out of eight for th... | 1120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. [3 points] Find the number of eight-digit numbers, the product of the digits of each of which is equal to 9261. The answer should be presented as an integer. | Answer: 1680.
Solution. Since $9261=3^{3} \cdot 7^{3}$, the sought numbers can consist of the following digits: (a) three threes, three sevens, and two ones, or (b) one three, one nine, three sevens, and three ones. We will calculate the number of variants in each case.
(a) First, we choose three places out of eight ... | 1680 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. [3 points] Find the number of eight-digit numbers, the product of the digits of each of which is equal to 64827. The answer should be presented as an integer. | Answer: 1120.
Solution. Since $64827=3^{3} \cdot 7^{4}$, the sought numbers can consist of the following digits: (a) three threes, four sevens, and one one, or (b) one three, one nine, four sevens, and two ones. We will calculate the number of variants in each case.
(a) First, we choose three places out of eight for ... | 1120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, each of the 6 asterisks needs to be replaced with any of the digits $0,1,2,3,4,5,6,7,8$ (digits can be repeated) so that the resulting 12-digit number is divisible by 45. In how many ways can this be done | Answer: 13122.
Solution. For a number to be divisible by 45, it is necessary and sufficient that it is divisible by 5 and by 9. To ensure divisibility by 5, we can choose 0 or 5 as the last digit from the available options (2 ways).
To ensure divisibility by nine, we proceed as follows. We will choose four digits arb... | 13122 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In the number $2 * 0 * 1 * 6 * 0 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,1,2,3,4,5,6,7,8$ (digits can repeat) so that the resulting 10-digit number is divisible by 18. In how many ways can this be done? | Answer: 3645.
Solution. For a number to be divisible by 18, it is necessary and sufficient that it is divisible by 2 and by 9. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 6$ or 8 (5 ways).
To ensure divisibility by nine, we proceed as follows. Choose three digits ... | 3645 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the number $2 * 0 * 1 * 6 * 0 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,1,2,3,4,5,6,7,8$ (digits can repeat) so that the resulting 10-digit number is divisible by 45. In how many ways can this be done? | Answer: 1458.
Solution. For a number to be divisible by 45, it is necessary and sufficient that it is divisible by 5 and by 9. To ensure divisibility by 5, we can choose 0 or 5 as the last digit (2 ways).
To ensure divisibility by nine, we proceed as follows. We select three digits arbitrarily (this can be done in $9... | 1458 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, each of the 6 asterisks needs to be replaced with any of the digits $1,2,3,4,5,6,7,8,9$ (digits can repeat) so that the resulting 12-digit number is divisible by 18. In how many ways can this be done | Answer: 26244.
Solution. For a number to be divisible by 18, it is necessary and sufficient that it is divisible by 2 and by 9. To ensure divisibility by 2, we can choose the last digit from the available options as $2, 4, 6$ or 8 (4 ways).
To ensure divisibility by nine, we proceed as follows. We select four digits ... | 26244 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In the number $2016^{* * * *} 02 * *$, each of the 6 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 12-digit number is divisible by 15. In how many ways can this be done? | Answer: 5184.
Solution. For a number to be divisible by 15, it is necessary and sufficient that it is divisible by 5 and by 3. To ensure divisibility by 5, we can choose 0 or 5 as the last digit (2 ways).
To ensure divisibility by three, we proceed as follows. Choose four digits arbitrarily (this can be done in $6 \c... | 5184 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Given a right triangular prism $A B C A_{1} B_{1} C_{1}$. A sphere with diameter $B C$ intersects the edges $A C$ and $A B$ at points $P$ and $Q$, respectively, different from the vertices of the prism. Segments $B_{1} P$ and $C_{1} Q$ intersect at point $T$, and $B_{1} P=5, T Q=2$.
a) Find the angle $T P A$.
b) F... | Answer: a) $90^{\circ}$, b) $2: 1$, c) $V=15$.
Solution. a) Points $P$ and $Q$ lie on the circle with diameter $BC$; therefore, $\angle BPC=90^{\circ}, \angle BQC=90^{\circ}$ (i.e., $BP$ and $CQ$ are altitudes of triangle $ABC$). The line $BP$ is the projection of the line $TP$ onto the base plane, and since $BP \perp... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,6,7,8$ (digits can be repeated) so that the resulting 11-digit number is divisible by 6. In how many ways can this be done? | Answer: 2160.
Solution. For a number to be divisible by 6, it is necessary and sufficient that it is divisible by 2 and by 3. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 6, 8$ (5 ways).
To ensure divisibility by three, we proceed as follows. Choose three digits ar... | 2160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 11-digit number is divisible by 15. In how many ways can this be done? | Answer: 864.
Solution. For a number to be divisible by 15, it is necessary and sufficient that it is divisible by 5 and by 3. To ensure divisibility by 5, we can choose 0 or 5 as the last digit from the available options (2 ways).
To ensure divisibility by three, we proceed as follows. We will choose three digits arb... | 864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,7,8,9$ (digits can be repeated) so that the resulting 11-digit number is divisible by 6. In how many ways can this be done? | Answer: 1728.
Solution. For a number to be divisible by 6, it is necessary and sufficient that it is divisible by 2 and by 3. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 8$ (4 ways).
To ensure divisibility by three, we proceed as follows. We will choose three digi... | 1728 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, each of the 6 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 12-digit number is divisible by 75. In how many ways can this be done? | Answer: 2592.
Solution. For a number to be divisible by 75, it is necessary and sufficient that it is divisible by 25 and by 3. To ensure divisibility by 25, we can choose 5 as the last digit (1 way).
To ensure divisibility by three, we proceed as follows. Select four digits arbitrarily (this can be done in $6 \cdot ... | 2592 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In the number $2 * 0 * 1 * 6 * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,7,8,9$ (digits can repeat) so that the resulting 11-digit number is divisible by 12. In how many ways can this be done? | Answer: 1296.
Solution. For a number to be divisible by 12, it is necessary and sufficient that it is divisible by 4 and by 3. To ensure divisibility by 4, we can choose 0, 4, or 8 as the last digit (3 ways).
To ensure divisibility by 3, we proceed as follows. We will choose three digits arbitrarily (this can be done... | 1296 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the number $2 * 0 * 1 * 6 * 07 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,5,6,7$ (digits can repeat) so that the resulting 11-digit number is divisible by 75. In how many ways can this be done? | Answer: 432.
Solution. For a number to be divisible by 75, it is necessary and sufficient that it is divisible by 25 and by 3. To ensure divisibility by 25, we can choose 5 as the last digit (1 way).
To ensure divisibility by three, we proceed as follows. Choose three digits arbitrarily (this can be done in $6 \cdot ... | 432 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, each of the 6 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 12-digit number is divisible by 12. In how many ways can this be done? | Answer: 5184.
Solution. For a number to be divisible by 12, it is necessary and sufficient that it is divisible by 4 and by 3. To ensure divisibility by 4, we can choose 0 or 4 as the last digit (2 ways).
To ensure divisibility by 3, we proceed as follows. Choose four digits arbitrarily (this can be done in $6 \cdot ... | 5184 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
30. On the board, there are param1 natural numbers. It is known that the sum of any five of them is not less than param2. Find the smallest possible value of the sum of all the numbers written on the board.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| 20 | 117 | |
| 18 | 97 | |
| 19 | 107 | |
| 26 | 153... | 30. On the board, ragat1 natural numbers are written. It is known that the sum of any five of them is not less than param2. Find the smallest possible value of the sum of all numbers written on the board.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| 20 | 117 | 477 |
| 18 | 97 | 357 |
| 19 | 107 | 415 |
| 2... | 477 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. On the table, there are 100 different cards with numbers $3, 6, 9, \ldots 297, 300$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the sum of the numbers on the selected cards is divisible by $5?$ | Answer: 990.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a common difference of 3. Therefore, the remainders of these numbers when divided by 5 alternate. Indeed, if one of these numbers is divisible by 5, i.e., has the form $5k$, where $k \in \mathbb{N}$, then the nex... | 990 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In a right triangle $ABC (\angle B=90^{\circ})$, a circle $\Gamma$ with center $I$ is inscribed, touching sides $AB$ and $BC$ at points $K$ and $L$ respectively. A line passing through point $I$ intersects sides $AB$ and $BC$ at points $M$ and $N$ respectively. Find the radius of the circle $\Gamma$ if $MK=225$, $NL... | Answer: $R=120, AC=680$.
Solution. Angles $KIM$ and $LNI$ are equal as corresponding angles when lines $BC$ and $KI$ are parallel, so right triangles $KIM$ and $LNI$ are similar. Therefore, $\frac{MK}{KI}=\frac{IL}{LN}$, or (if we denote the radius of the circle by $r$) $\frac{225}{r}=\frac{r}{64}$, from which $r=120$... | 680 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. On the table, there are 150 different cards with numbers $2, 4, 6, \ldots 298, 300$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the sum of the numbers on the selected cards is divisible by $5?$ | Answer: 2235.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a difference of 2. Therefore, the remainders of these numbers when divided by 5 alternate. Indeed, if one of these numbers is divisible by 5, i.e., has the form $5k$, where $k \in \mathbb{N}$, then the next numb... | 2235 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. There are 306 different cards with numbers $3,19,3^{2}, 19^{2}, \ldots, 3^{153}, 19^{153}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer? | Answer: 17328.
Solution. To obtain the square of a natural number, it is necessary and sufficient that each factor enters the prime factorization of the number in an even power.
Suppose two cards with powers of three are chosen. We have 76 even exponents $(2,4,6, \ldots, 152)$ and 77 odd exponents $(1,3,5, \ldots, 15... | 17328 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. There are 294 different cards with numbers $7, 11, 7^{2}, 11^{2}, \ldots, 7^{147}, 11^{147}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer? | Answer: 15987.
Solution. To obtain the square of a natural number, it is necessary and sufficient that each factor enters the prime factorization of the number in an even power.
Suppose two cards with powers of seven are chosen. We have 73 even exponents $(2,4,6, \ldots, 146)$ and 74 odd exponents $(1,3,5, \ldots, 14... | 15987 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In the number $2016 * * * * 02 * *$, each of the 6 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 12-digit number is divisible by 15. In how many ways can this be done? | Answer: 5184.
Solution. For a number to be divisible by 15, it is necessary and sufficient that it is divisible by 5 and by 3. To ensure divisibility by 5, we can choose 0 or 5 as the last digit from the available options (2 ways).
To ensure divisibility by three, we proceed as follows. We will choose four digits arb... | 5184 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In the number $2016^{* * * *} 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,6,7,8$ (digits can be repeated) so that the resulting 11-digit number is divisible by 6. In how many ways can this be done | Answer: 2160.
Solution. For a number to be divisible by 6, it is necessary and sufficient that it is divisible by 2 and by 3. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 6, 8$ (5 ways).
To ensure divisibility by three, we proceed as follows. Choose three digits ar... | 2160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 11-digit number is divisible by 15. In how many ways can this be done? | # Answer: 864.
Solution. For a number to be divisible by 15, it is necessary and sufficient that it is divisible by 5 and by 3. To ensure divisibility by 5, we can choose 0 or 5 as the last digit from the available options (2 ways).
To ensure divisibility by three, we proceed as follows. We will choose three digits a... | 864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,7,8,9$ (digits can be repeated) so that the resulting 11-digit number is divisible by 6. In how many ways can this be done? | Answer: 1728.
Solution. For a number to be divisible by 6, it is necessary and sufficient that it is divisible by 2 and by 3. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 8$ (4 ways).
To ensure divisibility by 3, we proceed as follows. Select three digits arbitrari... | 1728 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On the plane $(x ; y)$, plot the set of points satisfying the equation $|3 x|+|4 y|+|48-3 x-4 y|=48$, and find the area of the resulting figure.
# | # Answer: 96.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ are non-negative (since if at least one of them is negative, the left side is greater than the right). Therefore, the first equation is equivalent to the system of inequalities
$$
\left\{\begin{array} {... | 96 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. On the plane $(x ; y)$, plot the set of points satisfying the equation $|5 x|+|12 y|+|60-5 x-12 y|=60$, and find the area of the resulting figure.
# | # Answer: 30.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ are non-negative (since if at least one of them is negative, the left side is greater than the right). Therefore, the first equation is equivalent to the system of inequalities
$$
\left\{\begin{array} {... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 1400. The answer should be presented as an integer. | Answer: 5880.
Solution. Since $1400=7 \cdot 2^{3} \cdot 5^{2}$, the sought numbers can consist of the following digits: (a) three twos, two fives, one seven, and two ones, (b) one four, one two, two fives, one seven, and three ones, or (c) one eight, two fives, one seven, and four ones. We will calculate the number of... | 5880 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. [5 points] Two circles of the same radius 9 intersect at points $A$ and $B$. On the first circle, a point $C$ is chosen, and on the second circle, a point $D$ is chosen. It turns out that point $B$ lies on the segment $C D$, and $\angle C A D=90^{\circ}$. On the perpendicular to $C D$ passing through point $B$, a po... | Answer: 18.
Solution. Let $R=9$ be the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$. Then $\angle B A C=\frac{\pi}{2}-\alpha$, and by the sine rule $B D=2 R \sin \alpha, B C=2 R \sin \left(\frac{\pi}{2}-\alpha\right)=2 R \cos \alpha$. Therefore, $C F^{2}=B C^{2}+B D^{2}=4 R^{2... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 7000. The answer should be presented as an integer. | Answer: 5600.
Solution. Since $7000=7 \cdot 2^{3} \cdot 5^{3}$, the sought numbers can consist of the following digits: (a) three twos, three fives, one seven, and one one, (b) four, two, three fives, one seven, and two ones, or (c) eight, three fives, one seven, and three ones. We will calculate the number of variant... | 5600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. [5 points] Two circles of the same radius 7 intersect at points $A$ and $B$. On the first circle, a point $C$ is chosen, and on the second circle, a point $D$ is chosen. It turns out that point $B$ lies on the segment $C D$, and $\angle C A D=90^{\circ}$. On the perpendicular to $C D$ passing through point $B$, a po... | Answer: 14.
Solution. Let $R=7$ be the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$. Then $\angle B A C=\frac{\pi}{2}-\alpha$, and by the sine rule $B D=2 R \sin \alpha, B C=2 R \sin \left(\frac{\pi}{2}-\alpha\right)=2 R \cos \alpha$. Therefore, $C F^{2}=B C^{2}+B D^{2}=4 R^{2... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] On a plane with a given rectangular Cartesian coordinate system, a square is drawn with vertices at points $(0 ; 0),(0 ; 59),(59 ; 59)$, and $(59 ; 0)$. Find the number of ways to choose two grid nodes inside this square (not including its boundary) such that at least one of these nodes lies on one of the... | Answer: 370330
Solution. There are two possible cases.
1) Both selected nodes lie on the lines specified in the condition. Each of them contains 58 points inside the square, and there are no repetitions among them (the intersection point of the lines has non-integer coordinates). There are 116 ways to choose the firs... | 370330 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] On a plane with a given rectangular Cartesian coordinate system, a square is drawn with vertices at points $(0 ; 0),(0 ; 69),(69 ; 69)$, and ( $69 ; 0)$. Find the number of ways to choose two grid nodes inside this square (not including its boundary) such that at least one of these nodes lies on one of th... | Answer: 601460.
Solution. There are two possible cases.
1) Both selected nodes lie on the specified lines. There are 68 points on each of them inside the square, and there are no duplicates among them (the intersection point of the lines has non-integer coordinates). There are 136 ways to choose the first point, and ... | 601460 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] On a plane with a given rectangular Cartesian coordinate system, a square is drawn with vertices at points $(0 ; 0),(0 ; 65),(65 ; 65)$ and ( $65 ; 0)$. Find the number of ways to choose two grid nodes inside this square (not including its boundary) such that at least one of these nodes lies on one of the... | # Answer: 500032.
Solution. There are two possible cases.
1) Both selected nodes lie on the lines specified in the condition. There are 64 points on each of them inside the square, and there are no repetitions among them (the intersection point of the lines has non-integer coordinates). There are 128 ways to choose t... | 500032 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] On a plane with a given rectangular Cartesian coordinate system, a square is drawn with vertices at points $(0 ; 0),(0 ; 63),(63 ; 63)$, and $(63 ; 0)$. Find the number of ways to choose two grid nodes inside this square (not including its boundary) such that at least one of these nodes lies on one of the... | Answer: 453902.
Solution. There are two possible cases.
1) Both selected nodes lie on the specified lines. There are 62 points on each of them inside the square, and there are no repetitions among them (the intersection point of the lines has non-integer coordinates). There are 124 ways to choose the first point, and... | 453902 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On the sides of triangle $A B C$, points were marked: 10 - on side $A B$, 11 - on side $B C$, 12 - on side $A C$. At the same time, none of the vertices of the triangle were marked. How many triangles exist with vertices at the marked points? | Answer: 4951.
Solution. Three points out of the 33 given can be chosen in $C_{33}^{3}=5456$ ways. In this case, a triangle is formed in all cases except when all three points lie on one side of the triangle. Thus, $C_{12}^{3}+C_{11}^{3}+C_{10}^{3}=220+165+120=505$ ways do not work. Therefore, there are $5456-505=4951$... | 4951 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On the coordinate plane, squares are considered, all vertices of which have integer non-negative coordinates, and the center is located at the point $(50 ; 30)$. Find the number of such squares. | Answer: 930.
Solution. Draw through the given point $(50 ; 30)$ vertical and horizontal lines $(x=50$ and $y=30)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "lower" vertex of the square can be located in 30 ways: $(... | 930 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. On the coordinate plane, squares are considered, all vertices of which have natural coordinates, and the center is located at the point $(55 ; 25)$. Find the number of such squares. | Answer: 600.
Solution. Draw through the given point $(55 ; 25)$ vertical and horizontal lines $(x=55$ and $y=25)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "lower" vertex of the square can be located in 24 ways: $(... | 600 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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