problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
21.2. In triangle $A B C$, the bisectors of angles $B A C$ and $A B C$ intersect at point $O$. Find angle $A C B$ if angle $A O B$ is $125^{\circ}$.
$(7-8$ grade $)$ | 21.2. In triangle $A B C$ (Fig. 15), we have: $\angle O A B + \angle O B A = 180^{\circ} - 125^{\circ} = 55^{\circ}$. Since $A O$ and $B O$ are the bisectors of angles $C A B$ and $C B A$, the sum of these angles is $2 \cdot 55^{\circ} = 110^{\circ}$. Then $\angle A C B = 180^{\circ} - (\angle C A B + \angle C B A) = 1... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
26.2. Find the smallest natural number divisible by 11, which after being increased by 1, would be divisible by 13.
$$
\text { (7-8 grades) }
$$ | 26.2. Let the desired number be denoted by $n$. According to the problem, we have: $n=11 k$ and $n+1=13 m$, where $k$ and $m$ are natural numbers. Then $11 k=13 m-1$ or $(13-2) k=13 m-1$, from which we get $13(k-m)=2 k-1$. In the obtained equation, the left side is divisible by 13. Therefore, the right side must also b... | 77 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
28.2. Among all triangles for which the sum of the medians is 3, find the triangle with the greatest sum of altitudes.
$$
(7-9 \text { grades })
$$ | 28.2. In any triangle, the height dropped to any side is not greater than the median drawn to the same side. Therefore, the sum of the heights of the triangle is not greater than the sum of its medians. In the conditions of this problem, the greatest sum of heights is 3, which will be the case for an equilateral triang... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
29.2. Find the smallest natural number that ends in 56, is divisible by 56, and has the sum of its digits equal to 56.
$$
(7-10 \text { cl.) }
$$ | 29.2. The desired number can be represented as $100k + 56$. Since $100k$ must be divisible by 56, the number $k$ must be divisible by 14, i.e., it is even, divisible by 7, and has a sum of digits equal to $56 - 5 - 6 = 45$. The smallest even number with a sum of digits 45 is 199998, but it is not divisible by 7. The ne... | 29899856 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
30.2. Find the largest natural number that is divisible by 37 and in which each subsequent digit is less than the previous one.
$$
\text { (7-10 grades) }
$$ | 30.2. The largest natural number with decreasing digits is the ten-digit number 9876543210. We will divide it by 37. During the division, we notice that it is not divisible by 37, whereas the eight-digit number 98765432 is divisible by 37. Therefore, the nine-digit number 987654320 is also divisible by 37. It is eviden... | 987654320 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
35.2. Two motorcyclists set out simultaneously from points $A$ and $B$ towards each other and met 50 km from $B$. Upon arriving at points $A$ and $B$, the motorcyclists immediately turned around and met again 25 km from $A$. How many kilometers is it from $A$ to $B$? | 35.2. Let the motorcyclists meet for the first time at point $C$, and for the second time at point $D$ (Fig. 19). Then $B C=50$ km and $A D=25$ km.
Fig. 19
Let $A B=x$ km. Clearly, before ... | 125 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
44.2. The sum of the first $n$ terms of an arithmetic progression is equal to $n^{2}+5 n$. Find the tenth term of the progression.
$$
\text { (9-10 grades) }
$$ | 44.2. The tenth term of the given progression can be found as the difference between the sums of its first ten and first nine terms. These sums can be found from the expression $n^{2}+5 n$, by substituting $n=10$ and $n=9$ respectively. Thus, the desired tenth term is
$$
\left(10^{2}+5 \cdot 10\right)-\left(9^{2}+5 \c... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
45.2. Find the natural number that is 40 times less than the sum of all preceding natural numbers.
$$
(9-10 \text { th } )
$$
10 | 45.2. We have the equation:
$$
\begin{gathered}
1+2+3+\ldots+(n-1)=40 n \text {, i.e., } \frac{1}{2} n(n-1)=40 n, \\
\text { hence } n=81 .
\end{gathered}
$$ | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.3. How to weigh 1 kg of cereal on balance scales using two weights of 300 g and 650 g?
$$
(4-5 \text { grade })
$$ | 1.3. This can be done in two weighings: first, weigh out 650 g, and then $650-300=350$ g. | 350 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3.3. Write the number 100 using four fives and arithmetic signs.
$(4-5$ grade.)
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3.3. The two solutions to this problem are given below:
1) $(5+5)(5+5)=100$
2) $(5 \cdot 5-5) \cdot 5=100$ | 100 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
5.3. In three piles, there are 22, 14, and 12 nuts. It is required to equalize the number of nuts in all piles by making three moves, while adhering to the following condition: from one pile to another, only as many nuts can be moved as there are in the pile to which the nuts are being moved.
$$
\text { (4-6 grades) }... | 5.3. Move 14 nuts from the first pile to the second. After this, the piles will contain 8, 28, and 12 nuts, respectively. Next, move 12 nuts from the second pile to the third. Now the piles will contain 8, 16, and 24 nuts, respectively. Finally, move 8 nuts from the third pile to the first, after which each of the thre... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.3. An old problem. On a hot summer day, 6 mowers drank a barrel of kvass in 8 hours. How many mowers will drink the same barrel of kvass in 3 hours?
$$
\text { (4-7 grades) }
$$ | 6.3. If 6 mowers drank a barrel of kvass in 8 hours, then one mower would drink this barrel in 48 working hours. Therefore, 16 mowers would drink the barrel of kvass in three hours. | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3. Add a digit to the left and right of the number 10 so that the resulting number is divisible by 72.
$$
(5-6 \text { grade) }
$$ | 9.3. The desired number must be divisible by 8 and 9, i.e., it must be even and the sum of its digits must be divisible by 9. Only 5 numbers need to be checked: $8100, 6102, 4104, 2106, 9108$. Among these numbers, only 4104 is divisible by 72. Answer: 4104. | 4104 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.3. Can two digits be appended to the number 277 so that the resulting number is divisible by any number from 2 to 12? | 10.3. For a number to be divisible by 10, it must end in zero. Let's choose the second-to-last digit so that the resulting number is divisible by 9. In this case, the sum of the digits of this number should be equal to 18. Therefore, we get the number 27720. The problem has a solution only if the found number is divisi... | 27720 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14.3. A 30% hydrochloric acid solution was mixed with a 10% solution, resulting in 600 g of a 15% solution. How many grams of each solution were taken?
$$
\text { (6-7 grades) }
$$ | 14.3. Let $x$ be the number of grams of the 30% solution. Then the 10% solution was taken in the amount of ( $600-x$ ) g. We have the equation: $0.3 x + 0.1(600-x) = 0.15 \cdot 600$. From this, $x=150$. Answer: 150 g of the 30% solution and 450 g of the 10% solution were taken.
36 | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15.3. Boys are dividing nuts. The first one took 3 nuts and a fifth of the remainder; the second took twice 3 nuts and a fifth of the new remainder; the third took thrice 3 nuts and a fifth of the next remainder, and so on. How many boys were there if it turned out that as a result of such division, each received an eq... | 15.3. Let the number of all boys be denoted by $n$. Then the last boy received $3 n$ nuts with nothing left over. The second-to-last boy received $3(n-1)$ nuts and one fifth of the remainder. Since the second-to-last boy received as many nuts as the last one, i.e., $3 n$ nuts, the one fifth of the remainder amounts to ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
19.3. Can the number of diagonals of a polygon be exactly twice the number of its sides?
$(6-8$ grades $)$ | 19.3. From one vertex of an $n$-sided polygon, $n-3$ diagonals emanate. Then the product $n(n-3)$ expresses twice the number of all diagonals, since in this counting each diagonal was counted twice. It remains to check whether the equation $0.5 n(n-3)=2 n$ has a solution in natural numbers. It is easy to see that such ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
20.3. The collective farm placed a certain amount of money in the savings bank. If the number of hundreds is added to the number formed by the last two digits, then the result is the annual income from this amount, calculated at $2 \%$ per annum. What is the amount of the deposit placed in the savings bank?
$(7-8$ gra... | 20.3. Let the number of hundreds in the deposited amount be denoted by $x$, and the number formed by the last two digits of this amount be denoted by $y$. We have the equation: $x+y=(100 x+y) \cdot 0.02$. From this, we find: $50 x=49 y$. The obtained equality allows us to conclude that $y$ is divisible by 50. Since $y$... | 4950 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
23.3. We have 10 natural numbers whose sum is 1001. What is the greatest value that the greatest common divisor of these numbers can take?
$$
\text { (7-8 grades) }
$$ | 23.3. It is clear that the greatest common divisor of these ten numbers will also be a divisor of the number 1001. Let's consider the divisors of this number. We have: $1001=7 \cdot 11 \cdot 13$. The sought greatest common divisor cannot be equal to $11 \cdot 13=143$, since $143 \cdot 10>1001$. However, it can be equal... | 91 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
28.3. On the sides $B C$ and $C D$ of the square $A B C D$, points $K$ and $M$ are chosen such that the perimeter of triangle $K M C$ is equal to twice the side length of the square $A B C D$. Find the angle $K A M$.
$$
(7-10 \text { grades })
$$ | 28.3. Rotate triangle $A M D$ around point $A$ by an angle of $90^{\circ}$ until segment $A D$ coincides with segment $A B$ (Fig. 25). Let
Fig. 25
point $M$ move to point $P$. We obtain tr... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
38.3. Find a four-digit number that is 4 times less than the number written with the same digits but in reverse order.
$$
\text { (8-10 grades) }
$$ | 38.3. It is required to find the number $\overline{a b c e}$ such that $4 \cdot \overline{a b c e}=\overline{e c b a}$. Since the product $4 \cdot \overline{a b c e}$ is even, the number $\overline{e c b a}$ is also even, i.e., the digit $a$ is even. Then it is easy to see that $a=2$ (since for $a>2$ the product $4 \cd... | 2178 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
20. Using arithmetic operation signs, write the largest natural number with two twos. | $\triangle$ Let's write down all natural numbers using two twos. There are not many of them:
$$
2+2=4, \quad 2 \cdot 2=4, \quad 2^{2}=4,22
$$
The largest of these is the number 22. Interestingly, for its representation, no arithmetic operation symbols were needed at all.
Answer: 22. | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
40. What is the smallest natural number by which the number 7 must be multiplied to obtain a number that consists only of nines? | $\triangle$ Let's find this number using division
$$
\text { 9999... } 7
$$
However, the quotient here is unknown, and the number of nines in the dividend is also unknown. We will perform the division by 7, appending nines to the dividend until the division is performed without a remainder for the first time. We get:... | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
114*. What is the maximum number of natural numbers that can be written in a row so that the sum of any three consecutive numbers is even, and the sum of any four consecutive numbers is odd? | Let's denote the consecutive natural numbers of the row as $a_{1}, a_{2}, a_{3}$, and so on.
By the condition, the sums
$$
a_{1}+a_{2}+a_{3}, \quad a_{2}+a_{3}+a_{4}, \quad a_{3}+a_{4}+a_{5}, \quad a_{4}+a_{5}+a_{6}
$$
and others are even. By subtracting each sum, starting from the second, from the previous one, we ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
121. Find all values of the digit $a$, if the number $\overline{875 a}$ is divisible by 6. | $\triangle$ Since this number is divisible by 6, it is divisible by 2 and 3, and vice versa.
Let's apply the divisibility rule for 3. For this, we will find the sum of the digits of the number:
$$
8+7+5+a=20+a
$$
From the divisibility of $(20+a): 3$ it follows that the digit $a$ is 1, 4, or 7. But according to the d... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
141. Find all values of the digit $a$, if the number $\overline{a 719}$ is divisible by 11. | $\triangle$ Divisibility of this number by 11 is equivalent to the divisibility by 11 of the sum $a-7+1-9=$ $=a-15$. The difference $a-15$ is divisible by 11 only when $a=4$.
Answer: 4. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
146. From a natural number, the sum of its digits was subtracted, and then one digit was erased from the resulting difference. The sum of the remaining digits of the difference is 131. Which digit was erased? | $\triangle$ The difference between a natural number and the sum of its digits is divisible by 9 (see problem 130). Let the erased digit be denoted by $x$. Then the sum $131+x$ is divisible by 9, from which $x=4$.
Answer: 4 | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
152. Find the smallest natural number that is written with identical digits and is divisible by 18. | $\triangle$ Let the desired number be denoted as
$$
\overline{a a a \ldots a} \text { ( } n \text { digits). }
$$
Since it is divisible by 18, and therefore by 2, $a$ must be an even digit.
Applying the divisibility rule for 9:
$$
(a+a+a+\ldots+a): 9, \quad \text { na:9 }
$$
Since $a$ is an even digit, the number ... | 666 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
158*. Using the digits from 1 to 9 once each, form the smallest nine-digit number that is divisible by 11. | $\triangle$ The sum of all digits of the desired number is equal to
$$
1+2+3+\ldots+9=45
$$
Let the sum of the digits in odd positions be denoted by $S_{1}$, and the sum of the digits in even positions by $S_{2}$. Then $S_{1}+S_{2}=45$.
The difference $S_{1}-S_{2}$, according to the divisibility rule by 11, is divis... | 123475869 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
274. How many natural numbers divisible by 3 or 7 are there among the natural numbers from 1 to 500? | $\triangle$ First, let's calculate how many natural numbers from 1 to 500 are divisible by 3. Let's list all such numbers:
$$
3, \quad 6=3 \cdot 2, \quad 9=3 \cdot 3, \quad 12=3 \cdot 4, \ldots, \quad 498=3 \cdot 166
$$
From this, we can see that the number of such numbers is 166.
Now let's find how many natural num... | 214 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
381. The numbers $p$ and $p+15$ are prime. Find all such $p$. | $\triangle$ When $p=2$, we get that the number $p+15=17$ is prime.
Let $p>2$. Then $p$ is odd. Consequently, the number $p+15$ is even. Hence, this number is composite. Therefore, no prime $p>2$ works.
Answer: 2.
$382^{\circ}$. Prove that the numbers $p, p+2$, and $p+4$ are all prime only in the case when they form ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
425*. A two-digit number was divided by a number written with the same digits but in reverse order, and the quotient and remainder were equal. Find all such two-digit numbers. | $\triangle$ Let the desired number be denoted by $\overline{x y}$. Then
$$
10 x+y=(10 y+x) r+r, \quad 10 x+y=(10 y+x+1) r
$$
where $r-$ is the remainder (and the quotient at the same time). Obviously, $1 \leqslant r < 10$. But this is impossible.
Answer: 52. | 52 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
427. If the numbers 826 and 4373 are divided by the same natural number, the remainders will be 7 and 8, respectively. Find all values of the divisor. | $\triangle$ Let's write the corresponding equalities:
$$
826=b q_{1}+7, \quad 4373=b q_{2}+8
$$
where $b-$ is the unknown divisor, $q_{1}$ and $q_{2}$ are the incomplete quotients. Then
$$
b q_{1}=819, \quad b q_{2}=4365
$$
Let's factorize the numbers 819 and 4365 into prime factors:
$$
819=3^{2} \cdot 7 \cdot 13,... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
431. When dividing a natural number $a$ by 2, the remainder is 1, and when dividing by 3, the remainder is 2. What remainder will be obtained when dividing $a$ by 6? | $\triangle$ Let
$$
a=6 q+r
$$
where the remainder $r$ satisfies the inequality $0 \leqslant r \leqslant 5$. We will consider all possible values of $r$.
The case $r=0$ is impossible, otherwise the number $a$ would be divisible by 2 and 3.
The case $r=1$ is also impossible, because then $a$ when divided by 3 would l... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
438*. What remainder does $46^{925}$ give when divided by $21?$ | $\triangle$ The solution method we applied in the last problems is too cumbersome here due to the large base of the exponent and the large divisor. Let's try to modify it.
Divide 46 by 21 with a remainder and transform the exponent:
$$
46^{925}=(21 \cdot 2+4)(21 \cdot 2+4) \ldots(21 \cdot 2+4)
$$
If we multiply 925 ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
447. Find the greatest common divisor of the numbers 111111 and 111111111. | $\triangle$ Obviously, each of the given numbers is divisible by 111, i.e., the number 111 is their common divisor. But will 111 be their greatest common divisor?
To answer this question, we will divide the given numbers by 111 and determine whether the resulting quotients are coprime numbers. We will have:
$$
111111... | 111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
449. Find the greatest common divisor of all nine-digit numbers, in the representation of which each of the digits $1,2,3, \ldots, 9$ appears exactly once. | $\triangle$ Let's denote this greatest common divisor by $d$.
From all nine-digit numbers of the specified form, we will take only two - 123456798 and 123456789.
Since these numbers are divisible by $d$, their difference, which is 9, is also divisible by $d$: $9: d$. Therefore, $d=1, d=3$ or $d=9$.
Which of these ca... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
459. Find all values of the greatest common divisor of the numbers $8 a+3$ and $5 a+2$, where $a$ is a natural number. | $\triangle$ Let's denote the greatest common divisor of these numbers by $d$. Then
$$
(8 a+3): d, \quad(5 a+2): d
$$
Multiply the sum $8 a+3$ by 5, and the sum $5 a+2$ by 8. We get:
$$
(40 a+15): d, \quad(40 a+16): d
$$
But two consecutive natural numbers $40 a+15$ and $40 a+16$ are coprime (see the statement of pr... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
472. For which number and for which natural $a$ are the fractions reducible:
a) $\frac{a^{4}+a^{3}-a^{2}+a+1}{a^{2}-1}$;
b) $\frac{a^{4}+6 a^{3}+15 a^{2}+18 a+8}{a^{4}+6 a^{3}+13 a^{2}+12 a+3}$?
473 $3^{\circ}$. Let $a$ and $b$ be natural numbers, where $a > b$, and the number $a$ is divided by the number $b$ with a r... | $\triangle$ Let's introduce the notation:
$$
\text { GCD }(a, b)=d_{1}, \quad \text { GCD }(b, r)=d_{2} .
$$
Since the numbers $a$ and $b$ are divisible by $d_{1}$, from the initial equation we get that the remainder $r$ is also divisible by $d_{1}$. Therefore, $d_{1}$ is a common divisor of the numbers $b$ and $r$. ... | 168 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
477*. What is the greatest common divisor of all numbers $7^{n+2}+8^{2 n+1}(n \in N)$? | $\triangle$ Let's denote the greatest common divisor of all such numbers by $d$. Replacing $n$ with $n+1$ in the given sum, we get the sum $7^{n+3} + 8^{2n+3}$, which must also be divisible by $d$. Additionally, let's multiply the sum $7^{n+2} + 8^{2n+1}$ by 7:
$$
\left(7^{n+2} + 8^{2n+1}\right) \cdot 7 = 7^{n+3} + 7 ... | 57 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
479*. For what least natural $n$ are each of the fractions
irreducible
$$
\frac{5}{n+6}, \quad \frac{6}{n+7}, \frac{7}{n+8}, \ldots, \quad \frac{24}{n+25}
$$ | $\triangle$ Transform the given fractions as follows:
$$
\frac{5}{n+6}=\frac{5}{(n+1)+5}, \quad \frac{6}{n+7}=\frac{6}{(n+1)+6}, \ldots, \frac{24}{n+25}=\frac{24}{(n+1)+24}
$$
From this, it is clear that each of these fractions is irreducible if and only if the number $n+1$ is coprime with each of the numbers $5,6,7,... | 28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
485. A father and son were walking one behind the other on a snow-covered road. The father's step length is $-80 \mathrm{~cm}$, the son's $-60 \mathrm{~cm}$. Their steps coincided 601 times, including at the beginning and the end of the journey. What distance did they cover? | $\triangle$ Let's find the distance that the father and son walked from one coincidence of steps to the next. It is equal to the least common multiple of the numbers 80 and 60, i.e., $240 \mathrm{~cm}=2.4$ m. Therefore, the entire distance they walked is
$$
600 \cdot 2.4 \mathrm{M}=1440 \mathrm{~m}=1 \mathrm{Km} 440 \... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
487. A customer wanted to buy all the eggs the seller had and asked how many eggs she had. She replied that she didn't remember, but she knew that if the eggs were divided into groups of $2, 3, 4, 5$ or 6, there would always be one egg left over. What is the smallest number of eggs the seller could have had? | $\triangle$ Let's temporarily set aside one egg. The number of remaining eggs is equal to the least common multiple of the numbers $2,3,4,5$ and 6, i.e., 60. We need to add one more egg to this number.
Answer: 61. | 61 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
526. A six-digit number ends with the digit 7. If this digit is moved to the beginning of the number, the number increases by 5 times. What is this number? | The problem can be solved in two ways.
1) We will solve it as a problem of restoring the record.
where the five-digit numbers denoted by asterisks in the first multiplier and in the produc... | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
532. A six-digit number starts with the digit 2. If this digit is moved to the last place, the number increases by 3 times. What is this number? | $\triangle$ Let's solve the problem in two ways.
1) Restore the record:
where the five-digit numbers denoted by asterisks in the first multiplier and in the product coincide. We sequentia... | 285714 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
543. Does there exist a six-digit number that increases by 6 times when the first three digits, without changing their order, are moved to the end of the number? | $\triangle$ Let the desired number be denoted as $\overline{a b c d e f}$. Then
$$
\overline{a b c d e f} \cdot 6=\overline{d e f a b c}
$$
Let
$$
\overline{a b c}=x, \quad \overline{d e f}=y
$$
The equation becomes
$$
(1000 x+y) \cdot 6=1000 y+x
$$
Simplify it:
$$
6000 x+6 y=1000 y+x, \quad 5999 x=994 y, \quad ... | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
552. Find all four-digit numbers that are 9 times greater than their reversals. | $\triangle$ Let the desired number be denoted as $\overline{a b c d}$. Then
$$
\overline{a b c d}=9 \cdot \overline{d c b a}
$$
Two digits, $d$ and $a$, can be found immediately. Since the product $9 \cdot \overline{d c b a}$ is a four-digit number, the digit $d$ can only be 1. Since the product $9a$ ends with the di... | 9801 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
560. How many zeros are at the end of the product of all natural numbers from 1 to 60? | $\triangle$ Since zeros in the product of natural numbers are formed by multiplying twos by fives, it is desirable to know how many twos and fives are in the prime factorization of the given product. However, since there are more twos than fives in this factorization, it is sufficient to count the number of fives.
Fir... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
582. The number $n^{2}+2 n(n \in N)$ ends with the digit 4. Find all possible values of its second-to-last digit. | $\triangle$ Add 1 to both sides of the equality
$$
n^{2}+2 n=10 a+4
$$
where $a-$ is the number of tens in the number $n^{2}+2 n$, to get:
$$
n^{2}+2 n+1=10 a+5, \quad(n+1)^{2}=10 a+5
$$
From this, it is clear that the number $n+1$ ends in 5, and therefore, its square ends in 25 (see the statement of problem 563). ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
585. Find the last non-zero digit of the product of all natural numbers from 1 to 40. | $\triangle$ First, let's find the last non-zero digit of the product $1 \cdot 2 \cdot 3 \cdot \ldots \cdot 10$. For this, we will remove the factors 10, 2, and 5 from the number. The remaining product is
$$
3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9
$$
which, as is not difficult to calculate, ends with the digit 8.
F... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
590. The last two digits of the decimal representation of the square of some natural number are equal and different from zero. What are these digits? Find all solutions. | $\triangle$ The square of a natural number can end in the digits $0,1,4,5,6$ or 9. The digit 0 is not suitable in this case. Then the square can end in
$$
11,44,55,66,99
$$
However, cases where the square ends in 11, 55, or 99 are impossible, as if the last digit of the square of a natural number is odd, then the sec... | 44 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
595*. Does there exist a three-digit number whose cube ends in three sevens? | Let's denote this three-digit number as $x$. Then
$$
x^{3}=1000 a+777
$$
where $a$ is the number of thousands in $x^{3}$.
Since $x^{3}$ ends with the digit 7, the number $x$ ends with the digit 3: $x=10 k+3$, and the number $k$ is two-digit. We obtain:
$$
\begin{gathered}
(10 k+3)^{3}=1000 a+777, \quad 1000 k^{3}+9... | 753 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
620. Find the smallest natural number that, when multiplied by 2, becomes a perfect square, and when multiplied by 3, becomes a perfect cube. | $\triangle$ The desired number $x$ is divisible by 2 or 3. Since it is the smallest of all natural numbers with the given properties, its prime factorization consists only of twos and threes:
$$
x=2^{k} \cdot 3^{l} \quad(k \in N, l \in N)
$$
After multiplying the number $x$ by 2, the result is a perfect square, and a... | 72 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
629. Prove that the number
$$
1998^{2}+1998^{2} \cdot 1999^{2}+1999^{2}
$$
is a perfect square. | $\triangle$ Let's replace 1998 with $a$ and transform the resulting expression:
$$
\begin{gathered}
a^{2}+a^{2}(a+1)^{2}+(a+1)^{2}=a^{2}+a^{4}+2 a^{3}+a^{2}+a^{2}+2 a+1= \\
=a^{4}+2 a^{3}+3 a^{2}+2 a+1=\left(a^{2}+a+1\right)^{2}
\end{gathered}
$$
Therefore, the given number is the square of the number $1998^{2}+1998+... | 3994003 | Algebra | proof | Yes | Yes | olympiads | false |
635. What exact square is equal to the product of four consecutive odd numbers? | Let's denote the smallest of the odd numbers by $n$. Then
$$
n(n+2)(n+4)(n+6)=a^{2}
$$
where $a$ is an integer, and also odd. We transform the left side of this equation by multiplying the first factor by the fourth, and the second by the third:
$$
\left(n^{2}+6 n\right)\left(n^{2}+6 n+8\right)=a^{2}, \quad\left(n^{... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
650. Find all natural $k$ for which the number
$$
2^{k}+8 k+5
$$
is a perfect square. | $\triangle$ Checking shows that the value $k=1$ does not work, the value $k=2$ works, and the values $k=3,4,5,6$ do not work. It seems that no $k>2$ satisfies the condition of the problem. Let's prove our assumption.
Assume that there exists such a $k>2$ that the number $2^{k}+8 k+5$ is a perfect square. Since this nu... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
663. Find all natural $n$ for which the number $n^{2}+3 n$ is a perfect square. | $\triangle$ For a natural $n$, the inequality holds:
$$
(n+1)^{2} \leqslant n^{2}+3 n<(n+2)^{2}
$$
(check it!). But since the numbers $(n+1)^{2}$ and $(n+2)^{2}$ are the two closest perfect squares to each other, the number $n^{2}+3 n$ will be a perfect square only when the left inequality turns into an equality:
$$... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
678. The number 43 is written in the septenary (base-7) numeral system. In which system is it written with the same digits but in reverse order? | $\triangle$ First, let's convert the number 43 to the decimal system:
$$
43_{7}=4 \cdot 7+3=28+3=31
$$
139
Now, let's determine the base $d$ of the system in which the equality $31=34_{d}$ holds. We will have:
$$
31=3 d+4, \quad 3 d=27, \quad d=9
$$
Answer: in base 9. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
700*. In what number system is the number $11111_{d}$ a perfect square? | $\triangle$ Let's represent this number as a sum, decomposing it by powers of $d$:
$$
11111_{d}=d^{4}+d^{3}+d^{2}+d+1
$$
We need to find a natural number $d>1$ such that the resulting sum is a square of a quadratic trinomial with argument $d$ and integer coefficients.
We will bound this sum between two squares:
$$
... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
719. Find all numbers of the form $222 \ldots 2$ that can be represented as the sum of two perfect squares. | $\triangle$ Let
$$
222 \ldots 2=a^{2}+b^{2}
$$
where $a$ and $b$ are integers. Then the numbers $a$ and $b$ can only be odd:
$$
a=2 k+1, b=2 l+1 \quad(k \in Z, l \in Z)
$$
Therefore, the sum $a^{2}+b^{2}$ when divided by 8 gives a remainder of 2:
$$
a^{2}+b^{2}=(2 k+1)^{2}+(2 l+1)^{2}=4 k(k+1)+4 l(l+1)+2
$$
On th... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
746. There are containers of two types: 130 kg and 160 kg. How many containers of the first type and how many of the second type are there if together they weigh 3 tons? List all solutions. | Let's denote the number of containers of the first type by $x$, and the second type by $y$. We get the equation
$$
130 x + 160 y = 3000, \quad 13 x + 16 y = 300
$$
Let's try to use divisibility by 13. For this, we represent $16 y$ as $13 y + 3 y$, and divide 300 by 13 with a remainder:
$$
13 x + 13 y + 3 y = 13 \cdo... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
752. Find the smallest natural number that is divisible by 28, and when divided by 15, gives a remainder of 4. | $\triangle$ The desired number is, on the one hand, $28 x$, and on the other $-15 y+4$, where $x$ and $y$ are natural numbers. We obtain the equation
$$
28 x=15 y+4
$$
We do not need to find all solutions to this equation in natural numbers, but only one solution - the one for which the values of $x$ and $y$ are mini... | 364 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
755. "Forty mice were walking, carrying forty groats,
Two poorer mice carried two groats each, Many mice - with no groats at all. The bigger ones carried seven each. And the rest carried four each.
How many mice walked without groats?" | $\triangle$ Let the number of mice that walked without groats be denoted by $x$, the number of large mice by $y$, and the number of those that carried four groats by $z$. We can set up a system of equations with the unknowns $x, y$, and $z$:
$$
\left\{\begin{array} { l }
{ 2 \cdot 2 + 7 y + 4 z = 4 0 , } \\
{ 2 + x +... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
757. Find all three-digit numbers that are 11 times the sum of their digits. | $\triangle$ Let the desired number be $\overline{a b c}$. We get:
$$
\overline{a b c}=11(a+b+c), \quad 100 a+10 b+c=11 a+11 b+11 c, \quad 89 a=b+10 c
$$
Let's try using divisibility by 10 (but we can also use 11):
$$
90 a-a=b+10 c, \quad 90 a-10 c=a+b .
$$
Therefore, the sum $a+b$ is divisible by 10, i.e., $a+b=10$... | 198 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
778*. Find the smallest natural number that can be represented in exactly two ways as $3 x+4 y$, where $x$ and $y$ are natural numbers. | $\triangle$ Let
$$
3 x+4 y=n
$$
where $n-$ is a natural number. Let's solve this equation first in integers:
$$
\begin{array}{cl}
3 x=n-4 y, & x=\frac{n-4 y}{3}=-y+\frac{n-y}{3} \\
\frac{n-y}{3}=t \in Z ; \quad n-y=3 t, \quad y=n-3 t ; \quad x=-(n-3 t)+t=4 t-n
\end{array}
$$
Next, let's find all $t$ for which $x$ a... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
841*. A natural number is equal to the cube of the number of its thousands. Find all such numbers. | $\triangle$ Let's denote the desired number as $1000 x+y$, where $x-$ is the number of its thousands, and $y<1000$. Estimate the number $k$ from above:
$$
k=\frac{y}{x}<\frac{1000}{31}<33
$$
However, the sum $1000+k$ must be a square of a natural number. Only $k=24$ fits:
$$
1000+k=1024=32^{2}
$$
Therefore, $x=32$.... | 32768 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
925. What is the smallest number of members in a math club if the number of girls in it is more than $40 \%$, but less than $50 \%$? | Let in the club there be $a$ girls, and the total number of club members is $b$. Then the ratio of the number of girls to the total number of club members is $\frac{a}{b}$.
Since $40 \%$ is $\frac{2}{5}$, and $50 \% - \frac{1}{2}$ of the club members, then by the condition
$$
\frac{2}{5}<\frac{a}{b}<\frac{1}{2}
$$
W... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
930. How many solutions in integers $x$ and $y$ does the inequality
$$
|x|+|y|<10 ?
$$
have? | $\triangle$ From the condition, it is clear that
$$
|x|<10, \text { i.e. }-10<x<10
$$
Let's consider all cases.
If $x= \pm 9$, then $y=0$. We have two solutions: $(9 ; 0)$ and $(-9 ; 0)$.
If $x= \pm 8$, then $y=0, \pm 1$. This gives us six solutions: $(8 ; 0),(-8 ; 0),(8 ; 1),(-8 ; 1)$, $(8 ;-1)$ and $(-8 ;-1)$.
W... | 181 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
998. Find a natural number that is divisible by 9 and 5 and has 14 different divisors. | $\triangle$ Let the desired number be denoted by
$$
a=3^{\alpha_{1}} \cdot 5^{\alpha_{2}} \cdot p_{3}^{\alpha_{3}} \ldots p_{n}^{\alpha_{n}}
$$
where $p_{3}, p_{4}, \ldots, p_{n}$ are distinct prime numbers (different from 3 and 5), $\alpha_{1}, \alpha_{2}$ are natural numbers, and $\alpha_{3}, \alpha_{4}, \ldots, \a... | 3645 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1000. A certain natural number has only two prime divisors (in some powers), and its square has 35 different divisors. How many different divisors does the cube of this number have? | $\triangle$ Let the original number have the form:
$$
a=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}}
$$
where $p_{1}$ and $p_{2}$ are distinct prime numbers, and $\alpha_{1}$ and $\alpha_{2}$ are natural numbers. Then
$$
a^{2}=p_{1}^{2 \alpha_{1}} \cdot p_{2}^{2 \alpha_{2}}, \quad a^{3}=p_{1}^{3 \alpha_{1}} \cdot p_{... | 70 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1007. From a natural number, the sum of its digits was subtracted, and from the resulting difference, the sum of its digits was subtracted again. If this process continues, with what number will the calculations end? | $\triangle$ The difference between a natural number and the sum of its digits is divisible by 9 (see supporting problem 120 from § 5), and this property is preserved at each step of the computation. But such a difference is non-negative and decreases with further continuation of the computations, so its last value is z... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1008. Does there exist a natural number $n$ such that
$$
n+S(n)=125
$$
where $S(n)$ is the sum of the digits of $n ?$ | $\triangle$ The number $n$ can only be a three-digit number, and its first digit is equal to 1. Let $n=\overline{1 a b}$. We get:
$$
\overline{1 a b}+1+a+b=125, \quad 100+10 a+b+1+a+b=125, \quad 11 a+2 b=24
$$
The last equation has a unique solution in digits: $a=2, b=1$.
Answer: It exists and is unique - 121. | 121 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1038. A piece of the book fell out. The first page of the piece has the number 163, and the number of the last page consists of the same digits but in a different order. How many sheets fell out of the book | $\triangle$ The last page of the piece has the number 316, as it must end with an even digit. Now we find the number of pages, and then the number of sheets in the piece:
$$
316-162=154, \quad 154: 2=77
$$
Answer: 77. | 77 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1048. Which is greater and by how much: the sum of all even two-digit numbers or the sum of all odd two-digit numbers? | $\triangle$ First, let's calculate the total number of two-digit numbers. There are 99 numbers in total from 1 to 99, including both single-digit and two-digit numbers. Among them, there are 9 single-digit numbers. Therefore, the number of two-digit numbers is $99-9=90$.
Out of the 90 two-digit numbers, 45 are even an... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1053. A vendor was selling eggs. To the first customer, she sold half of all the eggs she had and another half-egg, to the second - half of the remainder and another half-egg, to the third - half of the new remainder and another half-egg, to the fourth - half of the last remainder and another half-egg. After this, she ... | $\triangle$ Let's solve the problem from the beginning.
After the third customer, the seller had one egg left.
After the second $-2 \cdot 1+1=3$. Let's check: she sold half of the previous remainder plus half an egg to the third customer, i.e., $3 / 2+1 / 2=2$, and she was left with $3-2=1$ egg. After the first custo... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1061. There are candies in three boxes. In the first box, there are 8 candies less than in the other two together, and in the second box, there are 12 less than in the first and third together. How many candies are in the third box? | $\triangle$ Let the number of candies in the first, second, and third boxes be denoted by $a, b$, and $c$ respectively. We obtain the system of equations:
$$
\left\{\begin{array}{l}
a=b+c-8 \\
b=a+c-12
\end{array}\right.
$$
Adding these equations term by term:
$$
a+b=(a+b)+2 c-20
$$
Then
$$
2 c-20=0, \quad c=10
$$... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1067. Kolya had to square a certain natural number for his homework. Instead, he mistakenly doubled the number and got a two-digit number, written with the same digits as the square of the number, but in reverse order. What should the correct answer be? | $\triangle$ Let's denote the original number by $x$, its square - by $\overline{a b}$. Then
$$
\left\{\begin{array}{l}
x^{2}=10 a+b \\
2 x=10 b+a
\end{array}\right.
$$
From here, it is clear that the number $x$ is a single-digit number.
Let's add the equations of this system:
$$
x(x+2)=11(a+b)
$$
Then either $x$, ... | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1082. How many terms of the sum
$$
1+2+3+\ldots+n
$$
are needed to get a three-digit number that is written with identical digits? | $\triangle$ Let's denote this three-digit number as $\overline{a a a}$. Applying the formula for the sum of $n$ terms of an arithmetic progression, we get:
$$
\frac{n(n+1)}{2}=\overline{a a a}=a \cdot 111, \quad n(n+1)=2 \cdot a \cdot 111=2 \cdot 3 \cdot 37 \cdot a
$$
From this, either $n$ or $n+1$ is divisible by 37... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## PROBLEM 4. FERRIES
Two ferries shuttle across a river at constant speeds, turning at the banks without losing time. They depart simultaneously from opposite banks and meet for the first time 700 feet from one of the banks, continue to the banks, return, and meet for the second time 400 feet from the opposite bank. ... | Solution. The total distance covered by the ferries at the time of their first meeting is exactly equal to the width of the river (Fig. 5). However, it may be surprising that by the time they meet again, the total distance they have covered is three times the width of the river. Since the speeds are constant, the secon... | 1700 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## PROBLEM 13. TIC-TAC-TOE
Consider the game of "tic-tac-toe" on a three-dimensional cube $8 \times 8 \times 8$. How many straight lines can be indicated on which 8 symbols lie in a row? | Solution. This is not a difficult problem; it can be solved by simple counting. However, there is a brilliant solution proposed by Leo Moser. It involves considering a $10 \times 10 \times 10$ cube, obtained by covering the given $8 \times 8 \times 8$ cube with a layer of unit thickness. A winning line in the inner $8 ... | 244 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## PROBLEM 20. ROLLING DICE
A regular die has the numbers $1,2,3,4,5,6$ on its faces. It is rolled randomly until the sum of the points that come up exceeds the number 12.
What is the most likely total sum of points? | Solution. Consider the penultimate roll. After it, the total sum should be either $12,11,10,9,8$,
or 7. If it is 12, the overall result will with equal probability take the values $13,14,15$, $16,17,18$. Similarly, with a sum of 11, the final result will with equal probability take the values 13, 14, $15,16,17$, and so... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## PROBLEM 27. MATHEMATICAL JOKE
A person bought several one-cent stamps at the post office, as many two-cent stamps so that $3 / 4$ of their quantity equals the number of one-cent stamps, as many five-cent stamps so that $3 / 4$ of their quantity equals the number of two-cent stamps, and five eight-cent stamps. He pa... | Solution. Suppose a person bought $y$ one-cent stamps. Then he bought $3 y / 4$ two-cent stamps and $9 y / 16$ five-cent stamps. Since $9 y / 16$ is an integer, $y$ is divisible by 16. Then for some integer $16 x=y$ and the purchase consisted of 16 one-cent, $12 x$ two-cent, $9 x$ five-cent, and 5 eight-cent stamps. Su... | 18816 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## PROBLEM 33. SNOWBALLS
A boy made two snowballs, one of which has a diameter twice as large as the other. He brought them into a warm room and let them melt. Since only the surface of the snowball is exposed to the warm air, assume that the melting rate is proportional to the surface area. When half of the volume of... | Solution. We will prove that the assumption that the melting rate of a snowball is proportional to the area of its surface leads to a surprising result, namely that the radius decreases at a constant rate regardless of the size of the snowball. Therefore, both radii will decrease by the same amount.
The volume and sur... | 40500000001 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## PROBLEM 40. PERFECT NUMBERS
The ancient Greeks discovered that some natural numbers $n$ have a remarkable property: the sum of the divisors of the number $n$ equals the number $n$ itself (the number itself is not considered a divisor). For example, $n=28$ gives
$$
1+2+4+7+14=28
$$
Such numbers were called "perfec... | Solution. First, let us assume that $n$ is an odd perfect number. Then $\sigma(n)=2n$, where 2 and $n$ are coprime, and we get
$$
\sigma[\sigma(n)]=\sigma(2n)=\sigma(2)\sigma(n)=3\sigma(n)=3 \cdot 2n=6n
$$
Clearly, $6n$ is even, and if it is also perfect, then for some prime number $p$ we must have
$$
6n=2^{p-1}(2^{... | 6 | Number Theory | proof | Yes | Yes | olympiads | false |
## PROBLEM 44. COWS AND SHEEP
Two men jointly owned $x$ cows, which they sold for $x$ dollars a head. With the money they received, they bought sheep at 12 dollars a head. Since the proceeds from the sale of the cows did not divide by 12, they used the remaining money to buy a lamb. Then they divided the flock so that... | Solution. The revenue from sales equals $x^{2}$ dollars. If $x$ were divisible by 6, then $x^{2}$ would be divisible by 36, and consequently, by 12. Since this is not the case, $x$ is not divisible by 6. In this case, $x=12 k+r$, where $|r|=1,2,3,4$ or 5. Accordingly,
$$
\frac{x^{2}}{12}=\frac{(12 k+r)^{2}}{12}=\frac{... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## PROBLEM 47. RED AND BLUE POINTS
Consider points arranged in a square of 20 columns and 20 rows, each of which is colored either red or blue. Whenever two points of the same color are adjacent in some row or column, they are connected by a segment of the same color. Adjacent points of different colors are connected ... | Solution. Each row contains 19 segments, so we get $19 \cdot 20=380$ horizontal segments. There are as many vertical segments, so their total number is 760. Since 237 of them are black, the remaining 523 are blue or red.
Let $k$ be the number of red segments and let's count how many times red points are endpoints of s... | 223 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n^{3}-(n-1)^{3}}{(n+1)^{4}-n^{4}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n^{3}-(n-1)^{3}}{(n+1)^{4}-n^{4}}=\lim _{n \rightarrow \infty} \frac{n^{3}-n^{3}+3 n^{2}-3 n+1}{\left((n+1)^{2}-n^{2}\right) \cdot\left((n+1)^{2}+n^{2}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{3 n^{2}-3 n+1}{\left(n^{2}+2 n+1-n^{2}\right)\lef... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2}}-\sqrt{n^{2}+5}}{\sqrt[5]{n^{7}}-\sqrt{n+1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2}}-\sqrt{n^{2}+5}}{\sqrt[5]{n^{7}}-\sqrt{n+1}}=\lim _{n \rightarrow \infty} \frac{n^{-\frac{7}{5}}\left(\sqrt[3]{n^{2}}-\sqrt{n^{2}+5}\right)}{n^{-\frac{7}{5}}\left(\sqrt[5]{n^{7}}-\sqrt{n+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \f... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{n+3}{n+1}\right)^{-n^{2}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{n+3}{n+1}\right)^{-n^{2}}=\lim _{n \rightarrow \infty}\left(\frac{n+1+2}{n+1}\right)^{-n^{2}}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{2}{n+1}\right)^{-n^{2}}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{n+1}{2}\right)}\right)... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[7]{x}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[7]{x}}=\lim _{x \rightarrow 0} \frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{\sqrt[7]{x}(\sqrt{1+x}+\sqrt{1-x})}= \\
& =\lim _{x \rightarrow 0} \frac{1+x-(1-x)}{\sqrt[7]{x}(\sqrt{1+x}+\sqrt{1-x})}=\lim _{x \rightarrow 0... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{2 x}}{\sin 3 x-\tan 2 x}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{3 x}-e^{2 x}}{\sin 3 x-\tan 2 x}=\lim _{x \rightarrow 0} \frac{\left(e^{3 x}-1\right)-\left(e^{2 x}-1\right)}{\sin 3 x-\tan 2 x}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{3 x}-1\right)-\left(e^{2 x}-1\right)\right)}{\frac{1}{x}(\sin... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}=\lim _{x \rightarrow 0} \frac{(1-\sqrt{\cos x})(1+\sqrt{\cos x})}{(1-\cos \sqrt{x})(1+\sqrt{\cos x})}= \\
& =\lim _{x \rightarrow 0} \frac{1-\cos x}{(1-\cos \sqrt{x})(1+\sqrt{\cos x})}=
\end{aligned}
$$
Using the substitu... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\cos \left(\frac{x}{\pi}\right)\right)^{1+x}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\cos \left(\frac{x}{\pi}\right)\right)^{1+x}=\left(\cos \left(\frac{0}{\pi}\right)\right)^{1+0}=1^{1}=1$
## Problem Kuznetsov Limits 18-22 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1}(\sqrt{x}+1)^{\frac{\pi}{\operatorname{arctg} x}}$ | ## Solution
$\lim _{x \rightarrow 1}(\sqrt{x}+1)^{\frac{\pi}{\operatorname{arctg} x}}=(\sqrt{1}+1)^{\frac{\pi}{\operatorname{arctg} 1}}=2^{\frac{\pi}{\left(\frac{\pi}{4}\right)}}=2^{4}=16$
## Problem Kuznetsov Limits 20-22 | 16 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-1 ; 2 ;-3), B(3 ; 4 ;-6), C(1 ; 1 ;-1)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(3-(-1) ; 4-2 ;-6-(-3))=(4 ; 2 ;-3)$
$\overrightarrow{A C}=(1-(-1) ; 1-2 ;-1-(-3))=(2 ;-1 ; 2)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begi... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=3 p-2 q$
$b=p+5 q$
$|p|=4$
$|q|=\frac{1}{2}$
$(\widehat{p, q})=\frac{5 \pi}{6}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(3 p-2 q) \times(p+5 q)=3 \cdot p \times p+3 \cdot 5 \cdot p \times q-2 \cdot q \... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(2 ; 1 ;-1), B(6 ;-1 ;-4), C(4 ; 2 ; 1)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(6-2 ;-1-1 ;-4-(-1))=(4 ;-2 ;-3)$
$\overrightarrow{A C}=(4-2 ; 2-1 ; 1-(-1))=(2 ; 1 ; 2)$
We find the cosine of the angle $\phi_{\text{between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrig... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Calculate the area of the parallelogram constructed on vectors $a_{\text {and }} b$.
$a=3 p+2 q$
$$
\begin{aligned}
& b=p-q \\
& |p|=10 \\
& |q|=1 \\
& (\widehat{p, q})=\frac{\pi}{2}
\end{aligned}
$$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(3 p+2 q) \times(p-q)=3 \cdot p \times p-3 \cdot p \times q+2 \cdot q \times p-2 ... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-2 ; 3 ;-3)$
$a: 3 x+2 y-z-2=0$
$k=\frac{3}{2}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x+2 y-z-3=0$
Substitute the coordinates of point $A$ into the equat... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!}=\lim _{n \rightarrow \infty}\left(\frac{(2 n+1)!}{(2 n+3)!}+\frac{(2 n+2)!}{(2 n+3)!}\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{1}{(2 n+2)(2 n+3)}+\frac{1}{2 n+3}\right)=0+0=0
\end{aligned}
$$
## Problem Kuzne... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.