problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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23. Prove that the semi-sum of two consecutive odd prime numbers is a composite number. | 23. Since odd numbers are considered, their half-sum will be an integer. It is known that the half-sum of two unequal numbers is always between these numbers. But all natural numbers between two consecutive prime numbers are composite. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,725 |
24. Do there exist two consecutive natural numbers such that the sum of the digits of each of them is divisible by 13? | 24. If the last three digits of a number are 9, and the fourth digit from the end is not 9, then adding one to such a number will decrease the sum of its digits by \(9+9+9-1=26\). Thus, if a number ends with exactly three nines and the sum of its digits is divisible by 13, then the sum of the digits of the next number ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,726 |
25. Prove that there exists an odd number, starting from which every odd number can be represented as the sum of two composite numbers.
32 | 25. The smallest even composite number is 4, and the smallest odd composite number is 9. Therefore, the smallest odd number that can be represented as the sum of two composite numbers is 13 = 4 + 9. Obviously, each subsequent odd number can be represented as the sum of an even number greater than 4 with the number 9. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,727 |
26. Does there exist a three-digit number that is divisible by each of eight consecutive natural numbers? | 26. The only such number is 840. It is divisible by $1,2,3,4,5,6,7,8$. | 840 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,728 |
27. Find the smallest number that can be represented as the sum of five, six, and seven consecutive natural numbers. | 27. The sum of five consecutive natural numbers has the form:
$$
n+(n+1)+(n+2)+(n+3)+(n+\dot{4})=5(n+2)
$$
i.e., this sum is divisible by 5. Similarly, we prove that the sum of seven consecutive natural numbers is divisible by 7, and the sum of six consecutive natural numbers has the form $3(2 n+5)$, i.e., it is an o... | 105 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,729 |
28. Find the smallest prime number that can be represented as the sum of two, three, four, five, and six different prime numbers. | 28. It is clear that the sought prime number is not less than $2+3+5+7+11+13=41$. The number 41 is prime, but it cannot be represented as the sum of two prime numbers. Indeed, since one of these two numbers must be the even prime number, i.e., 2, the second number would need to be 39, but it is not prime.
The next pri... | 61 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,730 |
29. Prove that for no natural number $n$ the number $7n+3$ is a perfect square. | 29. If there exist such natural $n$ and $k$ that $7 n + 3 = k^{2}$, then $k^{2} - 3 = 7 n$, i.e., $k^{2} - 3$ is divisible by 7. However, it is not difficult to prove that such a $k$ does not exist. The further solution of the problem is analogous to the solution of problem 20. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,731 |
30. Prove that if $(n-1)!+1$ is divisible by $n$, then $n$ is a prime number. | 30. Suppose that $n$ is a composite number, i.e., $n$ has a natural divisor $q$ different from one and less than $n$. But $(n-1)!+1$ is not divisible by any of the numbers $2,3,4, \ldots$, $n-1$, i.e., it is not divisible by $q$, and therefore not by $n$. This leads to a contradiction. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,732 |
31. Find all prime numbers $p$ such that $p^{2}+14$ is also a prime number. | 31. It is easy to notice that the prime number $p=3$ satisfies the condition of the problem. If $p \neq 3$, then $p^{2}$ has the form $3 \mathrm{k}+1$ (see problem 5 a). Then $p^{2}+14=3 k+15=3(k+5)$. The obtained number is composite. Therefore, the problem has a unique solution $p=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,733 |
32. Prove that any number of the form $999 \ldots 91$, containing an odd number of nines, is composite. | 32. Let the number $999 \ldots 91$ contain $2 k-1$ nines, where $k \in N$. Then $999 \ldots 91=10^{2 k}-9=\left(10^{k}\right)^{2}-3^{2}=\left(10^{k}-3\right)\left(10^{k}+3\right)$. The obtained number is composite for any natural $k$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,734 |
33. Find the smallest number \( n > 1980 \) such that the number
\[
\frac{x_{1}+x_{2}+x_{3}+\ldots+x_{n}}{5}
\]
is an integer for any assignment of integers \( x_{1}, x_{2}, x_{3}, \ldots, x_{n} \), none of which are divisible by 5. | 33. In problem 12, it was shown that for any integer $n$ not divisible by 5, the number $n^{4}-1$ is divisible by 5. Therefore, the fourth power of an integer not divisible by 5, when divided by 5, leaves a remainder of one. Hence, the given expression in the problem will be an integer only if the number of terms in th... | 1985 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,735 |
34. Prove that when any prime number is divided by 30, the remainder cannot be a composite number. | 34. Let $p$ be a prime number and $r$ the remainder of its division by 30. Then $p=30k+r$, where $0<r<30$. Suppose that $r$ is a composite number. Let $q$ be one of its prime divisors. Clearly, $q$ cannot coincide with any prime divisor of the number 30. Therefore, $q \geqslant 7$. Since the composite number $r$ has at... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,736 |
35. Prove that if for some integers $k$ and $n$ the number $k+4n$ is divisible by 13, then the number $10k+n$ is also divisible by 13. | 35. If the number $k+4n$ is divisible by 13, then the number $3(k+4n)$ is also divisible by 13. But then the number $13(k+n)-3(k+4n)=10k+n$ is also divisible by 13. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,737 |
36. Find the prime number $p$, given that the number $13 p+1$ is a perfect cube. | 36. If $13 p+1=n^{3}$, then $13 p=(n-1)\left(n^{2}+n+1\right)$. Since on the left side of the last equation we have the product of two prime numbers, there are only three possible cases:
$$
\left\{\begin{array} { l }
{ n - 1 = 1 } \\
{ n ^ { 2 } + n + 1 = 1 3 p ; }
\end{array} \quad \left\{\begin{array} { l }
{ n - ... | 211 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,738 |
37. Prove that for any natural $n$ the number $10^{n}+$ $+18 n-1$ is divisible by 27. | 37. We have: $10^{n}+18 n-1=\underbrace{999 \ldots 9}_{n}+18 n=9(\underbrace{(111 \ldots 1}_{n}+2 n)$.
It remains to prove that the expression in parentheses is divisible by 3. For this, we consider three cases: $n=3 k, n=3 k+1$ and $n=3 k+2$. In the first case, both terms in parentheses are divisible by 3. In the sec... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,739 |
38. Prove that the number 1439831 is composite. | 38. We have: $1439831=1440000-169=1200^{2}-13^{2}=$ $=1187 \cdot 1213$ | 1439831=1187\cdot1213 | Number Theory | proof | Yes | Yes | olympiads | false | 44,740 |
39. Prove that the sum of the squares of three integers cannot give a remainder of 7 when divided by 8. | 39. The square of an odd number, when divided by 8, leaves a remainder of 1 (see problem 3b), whereas the square of an even number either divides by 8 without a remainder or leaves a remainder of 4. It is easy to verify that from three remainders, each of which can only take the values $1,0,4$, it is impossible to get ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,741 |
40. Let $a, b, c, d$ be integers, and $a \neq b$. Prove that if $a c + b d$ is divisible by $a - b$, then $a d + b c$ is also divisible by $a - b$. | 40. The truth of the statement of the problem follows from the equality:
$$
(a-b)(d-c)+(a c+b d)=a d+b c
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 44,742 |
41. All natural numbers less than $p^{3}$, where $p$ is a prime number, are considered. How many of these numbers are coprime with the number $p^{3}$? | 41. Among the first $p^{3}$ natural numbers, there are $\mathrm{p}^{3}: \mathrm{p}=\mathrm{p}^{2}$ numbers that are multiples of $p$ (including $p^{3}$). Since $p$ is a prime number, the remaining $p^{3}-p^{2}$ numbers will be less than $p^{3}$ and coprime with $p$. | p^{3}-p^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,743 |
42. Prove that the smallest natural number that is coprime with each of the numbers $2, 3, 4, \ldots, n$ is a prime number. | 42. Let $a$ be the smallest number that is coprime with each of the numbers $2,3,4, \ldots, n$. Suppose that $a$ is a composite number. Then it has a prime divisor $p<a$. Since $a$ is coprime with the numbers $2,3,4, \ldots, n$, $p$ is also coprime with these numbers. But $p$ cannot be less than $n$, otherwise it would... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,744 |
44. Prove that the number 244141649 is composite. | 44. Extracting the square root of the given number, we get:
$$
\begin{aligned}
& 244141649=15625^{2}+1024=5^{12}+2^{10}=5^{12}+2^{6} \cdot 5^{6}+2^{10}- \\
& -2^{6} \cdot 5^{6}=\left(5^{6}+2^{5}\right)^{2}-\left(2^{3} \cdot 5^{3}\right)^{2}= \\
& =\left(5^{6}+2^{5}-1000\right)\left(5^{6}+2^{5}+1000\right)=14657 \cdot ... | 14657\cdot16657 | Number Theory | proof | Yes | Yes | olympiads | false | 44,746 |
45. Prove that for no natural $k$ the number $2^{2 k-1}+2^{k}+1$ is divisible by 7. | 45. Since the numbers 2 and 7 are coprime, it is sufficient to prove that the doubled number, i.e., the number of the form
$$
2\left(2^{2 k-1}+2^{k}+1\right)=2^{2 k}+2 \cdot 2^{k}+1+1=\left(2^{k}+1\right)^{2}+1
$$
does not divide by 7 for any natural $k$. Setting $n=2^{k}+1$, we reduce the problem to problem 20. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,747 |
46. Prove that for no natural numbers $n$ and $k$ can the number $2 n^{3 k} + 4 n^{k} + 10$ be the product of two or more consecutive natural numbers. | 46. In problem 6a, it was shown that the product of two consecutive integers is either divisible by 6 (and hence by 3), or when divided by 18 (and hence by 3) leaves a remainder of 2. Let \( n^{k} = q \). Now the given number will take the form:
$$
\begin{aligned}
& 2 n^{3 k} + 4 n^{k} + 10 = 2 q^{3} + 4 q + 10 = \lef... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,748 |
47. Prove that for no natural number $n$ the number $2 \cdot 13^{n}+5 \cdot 7^{n}+26$ can be a perfect square. | 47. Let $A(n)=2 \cdot 13^{n}+5 \cdot 7^{n}+20$. It is not difficult to prove by mathematical induction that $A(n)$ is divisible by 9 for any natural number $n$. The number given in the problem is $A(n)+6$. It is divisible by 3, but not by 9. It is obvious that such a number cannot be a perfect square. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,749 |
48. Prove that there exist arbitrarily long sequences of consecutive natural numbers, each of which is composite. | 48. Let $n$ be any large natural number. We will show that there exist $n$ consecutive natural numbers, each of which is composite. Consider the following $n$ consecutive natural numbers:
\[
\begin{aligned}
& (n+1)!+2 \\
& (n+1)!+3 \\
& (n+1)!+(n+1)
\end{aligned}
\]
Clearly, the first of these numbers is divisible by... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,750 |
49. Prove that the set of prime numbers is infinite. | 49. Suppose there are only a finite number of prime numbers. Then among the prime numbers there must be a greatest one. Let it be the number $p$. Consider the number $A$, which is the product of all prime numbers, plus one, i.e., $A=2 \cdot 3 \cdot 5 \cdot 7 \ldots p+1$. Clearly, the number $A$ is not divisible by any ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,751 |
50. Prove that there are infinitely many prime numbers that give a remainder of 2 when divided by 3.
## Solutions and hints
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 50. Suppose that there is only a finite set of prime numbers of the form $3k+2$: $p_{1}, p_{2}, p_{3}, \ldots, p_{s}$. Then the number $B=3 p_{1} \cdot p_{2} \cdot p_{3} \ldots p_{s}+2$ is not divisible by any of these numbers. It is also not divisible by 3. Let us factorize the number $B$ into prime factors: $B=q_{1} ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,752 |
1. Prove that any triangle can be cut into two or more right triangles. | 1. The proof is based on the fact that any triangle can be cut into two right triangles. For this, it is sufficient to draw a perpendicular from the vertex of the triangle to the opposite side. In an acute triangle, the perpendicular can be drawn from the vertex of any angle, whereas in a right triangle, it can only be... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,753 |
2. Prove that any obtuse triangle can be cut into any number of obtuse triangles. | 2. The proof is based on the fact that any obtuse triangle can be cut into two obtuse triangles. For this, it is sufficient to connect the vertex of the acute angle with any internal point of the opposite side. Figure 30 shows the cutting of an obtuse triangle into 3 obtuse triangles.
## 54 | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,754 |
13. Cut a square into 11 smaller squares. | 13. The problem allows for several solutions. Two of them are shown in Figure 41.
56
## Fig. 41
 | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,765 |
14. Prove that if a square can be cut into $k$ smaller squares, then it can be cut into $k+3$ smaller squares. | 14. Suppose a square is cut into $k$ smaller squares. If we cut one of them into 4 smaller squares, we will get $k+3$ smaller squares. In Figure 39, this method is used to obtain seven squares from four smaller squares. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,766 |
18. This square should be cut by a broken line into two parts of equal area so that each segment of the broken line is parallel to the side or diagonal of the square, and the sum of the lengths of the segments parallel to the sides equals the length of the side, while the sum of the lengths of the segments parallel to ... | 18. It is obvious that each segment of the broken line must be shorter than the side or diagonal of the square parallel to it, since otherwise, this segment alone would already divide the square into two parts. From this, it follows that the number of segments of the broken line cannot be less than four. We will now sh... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,770 |
20. Can a triangle be cut into two acute triangles?
50 | 20. It is impossible, since when a triangle is cut into two triangles, adjacent angles are formed, which cannot both be acute. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,772 |
22. Cut an equilateral triangle into three equal triangles. | 22. Cutting should be done along the bisectors of the triangle to the point of their intersection. The solution to the problem is illustrated in Figure 47. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,774 |
23. Prove that any triangle can be cut into four equal triangles. | 23. It is necessary to draw all three midlines in the triangle. The solution to the problem is shown in Figure 48.
Fig. 48
 be given (Fig. 50). It is easy to prove that the angle bisectors of the triangle intersect at one point. 58
Denote it by \(O\). We will prove that triangles \(OAB\), \(OBC\), and \(OCA\) are obtuse. Indeed, since \(\angle BAC + \angle BCA < 90^\circ\), i.e., triangle \(OAC\) is obtuse. Simil... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,777 |
29. Does there exist a pentagon (hexagon, heptagon) that can be cut into two triangles? | 29. The required pentagon and hexagon exist. Examples of them are shown in Figure 54. As for the heptagon, it does not exist, since two triangles cannot, under any application to each other, result in more than six vertices. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,781 |
31. Prove that any right triangle can be cut into two isosceles triangles. | 31. It is known that in a right-angled triangle, the midpoint of the hypotenuse is the center of the circumscribed circle. Therefore, the median of the hypotenuse is the radius of this circle and is equal to half the hypotenuse. Thus, the median of the hypotenuse divides the right-angled triangle into two isosceles tri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,783 |
32. Prove that any right triangle can be cut into three isosceles triangles. | 32. Let there be a triangle $A B C$, in which $\angle A B C=90^{\circ}$ (Fig. 56). If triangle $A B C$ is isosceles, then the solution to problem 60 is obvious. If, however, $|A C|>|B C|$, we construct an isosceles triangle $A B D$ ($D \in [A C]$), and then divide the right triangle $B C D$ into two isosceles triangles... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,784 |
33. Prove that any acute triangle can be cut into three isosceles triangles. | 33. Let there be an acute-angled triangle $A B C$ (Fig. 57). Describe a circle around it. In an acute-angled triangle, the center $O$ of the circumscribed circle lies inside the triangle. By connecting point $O$ to the vertices of the triangle, we obtain three isosceles triangles: $O A B, O B C$, and $O C A$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,785 |
34. Prove that any triangle can be cut into 4 or more isosceles triangles. | 34. Let's consider separately the dissection of a triangle into an even and an odd number of isosceles triangles.
Suppose the given triangle needs to be dissected into \(2n (n \geqslant 2)\) isosceles triangles. For this, we first dissect it into \(n\) right triangles (see problem 1), and then each of the right triang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,786 |
35. One of the angles of an acute triangle is twice another. Prove that this triangle can be cut into two isosceles triangles. | 35. Let an acute-angled triangle $ABC$ be such that $\angle BAC = \alpha$ and $\angle BCA = 2\alpha$ (Fig. 58). On the side $AC$, it is always possible to construct a point $D$ such that $|BD| = |BC|$. Since $\angle BDC = 2\alpha$ and $\angle BAD = \alpha$, then $\angle ABD = \alpha$. Thus, triangles $BCD$ and $ABD$ wi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,787 |
36. One of the angles of a triangle is three times another. Prove that this triangle can be cut into two isosceles triangles. | 36. Let triangle $ABC$ be such that $\angle BAC = \alpha$ and $\angle BCA = 3\alpha$ (Fig. 59). On side $AB$, it is always possible to construct a point $D$ such that $|AD| = |CD|$. Then $\angle ACD = \alpha$ and $\angle BCD = \angle BDC = 2\alpha$, i.e., triangles $ACD$ and $BCD$ will be isosceles.
. From their similarity, we get: $|A K|:|K M|=|K M|:|M C|$. Let: $|A D|=a,|A B|=b,|A K|=x$. Now the proportion will take the form: $x: b=b:(a-x)$. From this, $x^{2}-a x+b^{2}=0$. Taking into account that $0 ... | 2b | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,801 |
53. Prove that any triangle can be cut into three trapezoids. | 53. From a point taken inside a triangle, make cuts along rays parallel to the sides of the triangle (Fig. 75).
Fig. 75
Fig. 81
. The center of the circle inscribed in triangle $A B C$ is denoted by $O$. From point $O$ as the center, construct a circle of radius $|O C|$. Let this circle intersect side $A B$ at points $K$ and $M$, and sides $B C$ and $C A$ at points $P$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,818 |
1. Form all possible subsets for a set consisting of: a) two elements; b) three elements. | 1. As a two-element and three-element set, we will take the following sets: $\{A ; B\}$ and $\{A ; B ; C\}$. Let's list their subsets:
a) $\varnothing,\{A\},\{B\},\{A ; B\}$
b) $\varnothing,\{A\},\{B\},\{C\},\{A ; B\},\{A ; C\},\{B ; C\},\{A ; B ; C\}$.
It is useful to draw students' attention to the fact that a two... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,822 |
2. Three girls, Anya, Sanya, and Tanya, lined up one after another in a queue at the theater box office. List the possible cases of the girls' arrangement in the queue.
Write out the possible cases of the girls' arrangement in the queue. | 2. Let's denote the girls by the first letters of their names. We get the following 6 cases of the girls' arrangement in the queue:
ACT, ATC, CAT, CTA, TAC, TCA. | ACT,ATC,CAT,CTA,TAC,TCA | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,823 |
5. List all three-digit numbers that can be formed using the digits 4 and 5. | 5. $444,445,454,544,455,545,554,555$ (8 numbers). | 444,445,454,544,455,545,554,555 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,826 |
6. List all four-digit numbers that can be formed using the digits 0 and 9. | 6. $9000,9009,9090,9900,9099,9909,9990,9999$ (8 numbers). | 9000,9009,9090,9900,9099,9909,9990,9999 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,827 |
7. List all two-element subsets of a five-element set. | 7. Let's take the five-element set $\{A ; B ; C ; D ; E\}$. We will list its two-element subsets:
$\{A ; B\},\{A ; C\},\{A ; D\},\{A ; E\},\{B ; C\},\{B ; D\},\{B ; E\},\{C ; D\}$, $\{C ; E\},\{D ; E\}$. In total, we have 10 subsets. | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,828 |
8. Given 5 points, no three of which lie on the same line. How many segments need to be drawn to connect each pair of points? | 8. The solution to the problem is shown in Figure 93. We will obtain 10 segments. It is useful to compare this problem with problem 7.
Fig. 93
 | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,829 |
10. In how many ways can two identical pencils be distributed among five people? | 10. If we give one pencil at a time, we will have 10 ways of distributing the pencils (equal to the number of two-element subsets of a five-element set). We will get another 5 ways if we give both pencils to one person. Ultimately, we will have 15 ways. | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,831 |
11. Write down all the letter combinations that can be obtained by rearranging the letters in the word MAMA. | 11. We will get the following 6 letter combinations: MAMA, MMAA, MAAM, AMAM, AAMM, AMMA. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,832 |
13. In how many ways can four people be seated on four chairs? | 13. To solve this problem, it is advisable to use the previous problem as a model. We will get the same number of ways, i.e., 24. | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,834 |
14. There are 5 different books: three on mathematics and two on physics. In how many ways can two books be selected: one on mathematics and one on physics? | 14. Let's denote the mathematics books as $M_{1}, M_{2}, M_{3}$, and the physics books as $\Phi_{1}$, and $\Phi_{2}$. We obtain the following 6 ways: $M_{1} \Phi_{1}, M_{2} \Phi_{1}, M_{3} \Phi_{1}, M_{1} \Phi_{2}, M_{2} \Phi_{2}, M_{3} \Phi_{2}$. We can also reason as follows: the selection of a mathematics book can b... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,835 |
15. In how many ways can two letters be chosen from the word УЧЕБНИК so that one of the letters is a consonant and the other is a vowel? | 15. The number of ways can be visually represented using the graph shown in Figure 94. In total, we get \(4 \cdot 3 = 12\) ways.
Fig. 94
After solving problems 14 and 15, it is appropriate... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,836 |
16. From city $A$ to city $B$, one can travel by one of three types of transport, and from city $B$ to city $C$ - by one of four types of transport. In how many ways can one travel from city $A$ to city $C$, visiting city $B$ along the way? | 16. From city $A$ to city $B$, there are three ways to travel, after which from city $B$ to city $C$, there are four ways. By the rule of product, from city $A$ to city $C$, there are $3 \cdot 4=12$ ways to travel. | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,837 |
17. In the class, there are 16 boys and 14 girls. In how many ways can two students be assigned duty in the class: one boy and one girl? | 17. By the rule of product, we get $16 \cdot 14=224$ ways. | 224 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,838 |
18. How many different two-digit numbers can be formed from the digits: a) $1,2,3,4,5,6$; b) $0,1,2,3,4,5,6$? | 18. a) The choice of both the first and second digit can be made in six ways. In total, we will have $6 \cdot 6=36$ two-digit numbers.
b) The choice of the first digit can be made in six ways (zero cannot be taken), and the choice of the second digit can be made in seven ways. Thus, we will have $6 \cdot 7=42$ two-dig... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,839 |
20. In how many ways can two people be elected from ten for two different positions? | 20. There are ten ways to choose for the first position, after which there are only nine ways to choose for the second position. In total, we get $10 \cdot 9=90$ ways. | 90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,841 |
21. Ten people exchanged photographs. How many photographs were needed? | 21. The choice of one person can be made in ten ways, after which this person can give a photograph to nine remaining people. In total, $10 \cdot 9=90$ photographs are required. | 90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,842 |
22. How many different positions can appear on a chessboard if both players, starting from the initial position, make only one move each? | 22. In the initial position, each pawn and each knight can move in two ways, while the other pieces cannot make a move. Therefore, from the initial position, 20 different moves can be made by both the white and black pieces. Thus, after each player makes one move, $20 \cdot 20=400$ different positions can result. | 400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,843 |
23. In the canteen for lunch, 4 first-course dishes, 5 second-course dishes, and 3 desserts are prepared for selection. In how many ways can one lunch be composed? | 23. Applying the multiplication rule for the case of three choices, we get: $4 \cdot 5 \cdot 3=60$. | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,844 |
24. The teacher prepared three problems for the class to solve. In how many ways can he offer these problems to the students, if there are 30 students in the class? | 24. Tasks can be proposed to students $30 \cdot 29 \cdot 28=$ $=24360$ ways. | 24360 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,845 |
25. How many even four-digit numbers can be formed from the digits $0,1,2,3,4,5$? | 25. Since zero cannot be the first digit, the first digit can be chosen in five ways. The second and third digits can each be chosen in six ways. Finally, the fourth digit, which must be even, can be chosen in three ways. Then, by the rule of product, we will have $5 \cdot 6 \cdot 6 \cdot 3=540$ four-digit even numbers... | 540 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,846 |
26. How many four-digit numbers can be formed from the digits $0,1,2,3,4,5,6$, if the repetition of digits in the number is not allowed? | 26. $6 \cdot 6 \cdot 5 \cdot 4=720$ numbers. | 720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,847 |
27. In how many ways can 6 people be seated on six chairs? | 27. The first person can choose a chair in six ways, the second in five ways, and so on. Finally, only one chair will remain for the sixth person. In total, we get $6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=720$ ways.
Note. The product of the first $n$ natural numbers is denoted by $n$! (read: n factorial). Thus, $6 \... | 720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,848 |
28. At the meeting, there are 25 people present. In how many ways can four people be elected from them to fill 4 different positions | 28. There are $25 \cdot 24 \cdot 23 \cdot 22=303600$ different possibilities for the election to the positions. | 303600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,849 |
29. In how many ways can 5 different items be distributed among three people? | 29. Each item can be distributed in three ways. By the rule of product, we get: $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3=3^{5}=243$. | 243 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,850 |
30. In how many ways can 3 different items be distributed among five people? | 30. Each item can be distributed in five ways. In total, we will have $5 \cdot 5 \cdot 5=5^{3}=125$ ways. | 125 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,851 |
31. In how many ways can 5 different items be distributed into eight boxes if no more than one item can be placed in one box? | 31. The choice of the box for the first item can be made in eight ways, for the second in seven ways, and so on. In total, we get $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4=6720$ ways. | 6720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,852 |
32. How many five-digit numbers are there that read the same from left to right and right to left? | 32. The first digit can be chosen in nine ways, and the second and third digits can be chosen in ten ways each. Since the fourth digit must match the second, and the fifth must match the first, each of the last two digits can be chosen in only one way. Thus, the number of such numbers will be $9 \cdot 10 \cdot 10 \cdot... | 900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,853 |
33. How many even five-digit numbers are there that read the same from left to right and from right to left? | 33. Since the last digit must be even, the first digit will also be even. But the first digit cannot be zero. Therefore, there are only 4 ways to choose the first digit. In total, we get $4 \cdot 10 \cdot 10 \cdot 1 \cdot 1=400$ required even numbers. | 400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,854 |
34. How many five-digit numbers are divisible by 5? | 34. Since the last digit can be chosen in only two ways (0 or 5), we will get a total of $9 \cdot 10 \cdot 10 \cdot 10 \cdot 2=18000$ required numbers. The same result can be obtained by dividing the total number of five-digit numbers by $5: 9 \cdot 10^{4}: 5=18000$. | 18000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,855 |
35. How many five-digit numbers are there that do not contain the digit 5? | 35. When composing the required five-digit numbers, we will have only 9 digits at our disposal. Therefore, we will have a total of $8 \cdot 9 \cdot 9 \cdot 9 \cdot 9=8 \cdot 9^{4}$ numbers. | 8\cdot9^4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,856 |
36. How many five-digit numbers are there in which at least the first two digits are the same? | 36. $9 \cdot 1 \cdot 10 \cdot 10 \cdot 10 \cdot=9000$ numbers. | 9000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,857 |
37. How many five-digit numbers are there in which at least the last two digits are different? | 37. $9 \cdot 10 \cdot 10 \cdot 10 \cdot 9=81000$ numbers. | 81000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,858 |
38. How many six-digit numbers are there in which only the middle two digits are the same? | 38. $9 \cdot 9 \cdot 8 \cdot 1 \cdot 7 \cdot 6=27216$ numbers. | 27216 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,859 |
39. How many six-digit numbers are there with the second-to-last digit being 1, which are divisible by 4? | 39. It is clear that the numbers divisible by 4 are those and only those numbers whose last two digits form a number divisible by 4. Then, if the second-to-last digit is 1, the last digit can only be 2 or 6. In total, we get $9 \cdot 10 \cdot 10 \cdot 10 \cdot 1 \cdot 2=18000$ required numbers. | 18000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,860 |
40. How many even four-digit numbers are there that do not contain the digits 5 and 6? | 40. The first digit can be chosen in seven ways (we cannot take 0, 5, and 6), the second and third digits can each be chosen in eight ways, and the last digit, which must be even, can be chosen in four ways (we cannot take 6). In total, we get $7 \cdot 8 \cdot 8 \cdot 4 = 1792$ numbers. | 1792 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,861 |
41. In how many ways can 5 people be seated in a row so that Kolya and Olya sit next to each other? | 41. First, let's seat Kolya and Olya. There are 4 ways to seat them next to each other with Olya sitting to the right of Kolya, and 4 ways with Olya sitting to the left of Kolya. Thus, for 5 free seats, there are 8 ways to seat Kolya and Olya next to each other. After seating Kolya and Olya, 3 seats will remain free. T... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,862 |
42. In how many ways can all ten digits be arranged in a row so that the digits $2,5,8$ stand together? | 42. We will consider the three digits $2,5,8$ as one number 258. Together with the other seven digits, we will have 8 numbers: $258,0,1,3,4,6,7,9$. There are $8!=40320$ ways to arrange them in a row. In this case, the digits $2,5,8$ will always be together and in the specified order. | 40320 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,863 |
43. On a bookshelf, there are ten different books, three of which are on mathematics. In how many ways can all these books be arranged in a row so that the mathematics books are together? | 43. Let's bind three mathematics books together so that they form a single book. Then we will have 8 books. They can be arranged in a row on a bookshelf in $8!=40320$ ways. Obviously, in each such arrangement, the mathematics books will continue to remain together if they are rearranged among themselves. This can be do... | 241920 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,864 |
45. Five girls are sitting on five chairs in a row, and opposite them, on five chairs, are sitting five boys. It was decided that the boys would swap places with the girls. In how many ways can this be done? | 45. Boys can be seated on the girls' chairs in $5!=$ $=120$ ways. The girls can also be seated on the boys' chairs in the same number of ways. Ultimately, we will have $120 \cdot 120=14400$ ways. | 14400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,866 |
48. How many subsets does an $n$-element set have? | 48. Let there be a set with $n$ elements. For each element, there are two possibilities: to be included in a subset or not to be included. Then for all $n$ elements, we get $\underbrace{2 \cdot 2 \cdot 2 \ldots 2}_{n}=$ $=2^{n}$ such possibilities, i.e., $2^{n}$ different subsets. | 2^n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,869 |
49. In the room, there are 6 light bulbs, each with its own switch. In how many ways can the room be illuminated if at least one light bulb must be turned on? | 49. The number of ways to light a room will be equal to the number of subsets of a set of six elements, except for the case when the subset is the empty set. Thus, $2^{6}-1=63$. | 63 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,870 |
50. How many five-digit numbers are there that are divisible by 5 and do not have any repeated digits in their representation? | 50. The number of numbers that end in 0 and do not have identical digits in their representation is $9 \cdot 8 \cdot 7 \cdot 6 \cdot 1$. The number of numbers that end in 5 and do not have identical digits in their representation is $8 \cdot 8 \cdot 7 \cdot 6 \cdot 1$. Therefore, the total number of numbers of interest... | 5712 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,871 |
51. How many four-digit numbers contain only one even digit? | 51. Four-digit numbers, in which only the first digit is even, amount to $4 \cdot 5 \cdot 5 \cdot 5=500$. Four-digit numbers, in which only the second (third, fourth) digit is even, amount to $5 \cdot 5 \cdot 5 \cdot 5$. Then, the total number of four-digit numbers, in which only one digit is even, will be $4 \cdot 5 \... | 2375 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,872 |
53. How many four-digit numbers contain at least one even digit? | 53. From four-digit numbers, we need to discard all those numbers that do not have a single even digit. We will get: $9 \cdot 10 \cdot 10 \cdot 10-5 \cdot 5 \cdot 5 \cdot 5 \cdot 5=8375$ numbers. | 8375 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,874 |
54. How many five-digit numbers are there in which
a) the digit 5 appears exactly once?
b) the digit 5 appears no more than once?
c) the digit 5 appears at least once? | 54. a) The number of five-digit numbers where only the first digit is 5 is $1 \cdot 9 \cdot 9 \cdot 9 \cdot 9=9^{4}$. The number of five-digit numbers where only the second (third, fourth, fifth) digit is 5 is $8 \cdot 9^{3}$. Therefore, the total number of required five-digit numbers is $9^{4}+4 \cdot 8 \cdot 9^{3}=41... | 37512 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,875 |
55. How many different letter combinations can be obtained by rearranging the letters in the word a) NONNA; b) MATHEMATICA? | 55. a) The word NONNA contains 5 letters. If all of them were different, we could obtain $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120$ different letter combinations from this word. In reality, in each such combination, the letter N is repeated three times. Permuting these three letters among themselves does not change the co... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,876 |
56. In the class, 12 subjects are studied. In how many ways can 5 different subjects be scheduled for Monday? | 56. For the first lesson, the subject can be chosen in twelve ways, for the second - only eleven, and so on. Thus, using the multiplication rule, we get that a schedule of five different subjects, if there are 12 subjects in total, can be created $12 \cdot 11 \cdot 10 \cdot 9 \cdot 8=9504$ ways. | 9504 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,877 |
59. In a city with $n$ cinemas, $n$ tourists have arrived. In how many ways can they disperse among the cinemas so that each cinema has one tourist? | 59. A n s w e r. $n$ ! ways. | n! | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,880 |
60. In a city with $n$ cinemas, $k$ tourists have arrived. In how many ways can they disperse into the cinemas? | 60. For each tourist, there are $n$ ways to visit the cinema. For all $k$ tourists, we get $n \cdot n \cdot n \ldots n=n^{k}$ ways for them to disperse among the cinemas.
86 | n^k | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,881 |
61. A $k$-member sequence, the members of which are elements of a given $n$-element set, is called a $k$-member replacement arrangement from $n$ elements. The number of all replacement arrangements from $n$ elements by $k$ is denoted by $\tilde{A}_{n}^{k}$. Prove that $\tilde{A}_{n}^{k}=n^{k}$. | 61. Each member of a $k$-member sequence can be chosen in $n$ ways. Therefore, there exist $n^{k}$ different $k$-member sequences whose members are elements of the given $n$-element set. This is the number of arrangements with repetitions of $n$ elements taken $k$ at a time. It is useful to compare this with problem 60... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 44,882 |
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