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 |
|---|---|---|---|---|---|---|---|---|
4.48. Find the remainder when the number
$$
10^{10}+10^{\left(10^{2}\right)}+10^{\left(10^{3}\right)}+\ldots+10^{\left(10^{10}\right)}
$$
is divided by 7.
|
4.48. Answer: 5. Note that $10^{6} \equiv 1(\bmod 7)$, since $10^{3}+1$ is divisible by 7, and $10^{k} \equiv 4(\bmod 6)$ for $k \geqslant 1$, since the number $99 \ldots 96$ is even and divisible by 3. Therefore, $10^{10^{k}} \equiv 10^{4}(\bmod 7)$ for $k \geqslant 1$. Hence, the required remainder is the remainder of the division of the number $10 \cdot 10^{4}=10^{5}$ by 7. This remainder is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.12. Prove that the number $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ is rational.
|
6.12. The identity $(2 \pm \sqrt{2})^{3}=20 \pm 14 \sqrt{2}$ shows that $\sqrt[3]{20 \pm 14 \sqrt{2}}=2 \pm \sqrt{2}$. Therefore, the considered number is 4.
|
4
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
6.23. Find the first digit after the decimal point of the number $(2+\sqrt{3})^{1000}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
6.23. Answer: 9. Let $(2+\sqrt{3})^{n}=A_{n}+B_{n} \sqrt{3}$, where $A_{n}$ and $B_{n}$ are natural numbers. According to problem 6.22, $(2-\sqrt{3})^{n}=A_{n}-B_{n} \sqrt{3}$. Therefore, $(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}$ is a natural number. But $2-\sqrt{3} \approx$ $\approx 0.2679 < 0.3$, so $(2-\sqrt{3})^{1000} \approx 0.0 \ldots$ (and then several more zeros follow).
|
9
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. A pedestrian left point $A$ for point $B$, which are 5 km apart. At the same time, a cyclist left point $B$ towards the pedestrian, with a speed twice that of the pedestrian. Upon meeting the pedestrian, he turned around and rode back to $B$. Upon reaching $B$, the cyclist turned again and rode towards the pedestrian, and so on. What distance will the cyclist have traveled by the time the pedestrian arrives at $B$?
## 7.2. Calculations
|
7.2. A n s w e r: 10 km. We can assume that the cyclist is always moving in one direction (the length of the path does not change from this).
His speed is twice the speed of the pedestrian, so in the time it takes the pedestrian to walk 5 km, the cyclist will travel 10 km.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3. A flock of geese was flying over the lakes. On each lake, half of the geese and another half-goose landed, while the rest continued flying. All the geese landed on seven lakes. How many geese were in the flock?
|
7.3. Answer: 127. Let's mentally add a bird, for example, a duck, that flies with the flock all the way to the last lake. Then, at each lake, exactly half of the birds land, and on the seventh lake, two birds land - the goose and the duck. This means there were $2^{7}$ birds in the flock, of which $2^{7}-1=127$ were geese.
|
127
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. Two cyclists set off simultaneously towards each other from points $A$ and $B$ and met 70 km from $A$. Continuing to move at the same speeds, they reached $A$ and $B$ and turned back. They met for the second time 90 km from $B$. Find the distance from $A$ to $B$.
|
7.4. Answer: 120 km. Let $x$ be the distance from $A$ to $B$. By the first meeting, both cyclists together had traveled $x$ km, and by the second meeting, $-3x$ km. Therefore, the cyclist who started from $A$ had traveled $3 \cdot 70 = 210$ km by the second meeting. On the other hand, he had traveled $x + 90$ km. Thus, $x = 210 - 90 = 120$ km.
|
120
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. A steamship from Gorky to Astrakhan takes 5 days, while from Astrakhan to Gorky it takes 7 days. How many days will it take for driftwood to float from Gorky to Astrakhan?
|
7.5. Answer: 35. Let $v$ be the speed of the river current, $u$ be the speed of the steamship, and $l$ be the distance from Gorky to Astrakhan. According to the problem, $\frac{l}{u+v}=5$ and $\frac{l}{u-v}=7$; we need to find $l / v$. From the system of equations $l=5 u+5 v, l=7 u-7 v$, we find $l / v=35$.
|
35
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.13. In the report on the ski race, it is stated that $96 \%$ of its participants met the standard. It is known that this figure is given with an accuracy of $0.5 \%$. What is the smallest number of participants in this race?
|
7.13. Let $n$ be the number of skiers in the race, and $k$ be the number of them who did not meet the standard. According to the condition, $3.5 \leqslant 100 k / n \leqslant 4.5$. Therefore, $k \geqslant 1$ and $n \geqslant \frac{100 k}{4.5}>22.2 k \geqslant 22.2$. Hence, $n \geqslant 23$. If there were 23 skiers in the race, and only one did not meet the standard, then the required condition is satisfied.
|
23
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.14. In a quarry, 120 granite slabs weighing 7 tons each and 80 slabs weighing 9 tons each are prepared. Up to 40 tons can be loaded onto a railway platform. What is the minimum number of platforms required to transport all the slabs?
|
7.14. It is impossible to load 6 slabs, each weighing 7 tons, onto one platform. Therefore, at least $200 / 5=40$ platforms are needed. Forty platforms are sufficient: onto each platform, 3 slabs weighing 7 tons each and 2 slabs weighing 9 tons each can be loaded.
|
40
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.16. Ten workers need to assemble 50 items from parts. First, the parts of each item need to be painted; it takes one worker 10 minutes to do this. After painting, the parts dry for 5 minutes. Assembling an item takes one worker 20 minutes. How many workers should be assigned as painters and how many as assemblers to complete the work in the shortest time? (Two workers cannot paint or assemble the same item simultaneously.)
## 7.5. Correspondences
|
7.16. Answer: 3 painters and 6 assemblers (one worker is extra, i.e., you can assign 4 painters and 6 assemblers or 3 painters and 7 assemblers). It is easy to verify that 3 painters and 6 assemblers can complete the work in 195 minutes. Indeed, after 15 minutes from the start of the painters' work, 3 items are ready for assembly, and 3 assemblers will assemble them in 20 minutes. After that, 6 items will be ready for assembly, and the assemblers will never have to wait for the painters to finish painting the next item. After \(15 + 20 + 140 = 175\) minutes from the start of the work, \(3 + 7 \cdot 6 = 45\) items will be assembled. To assemble the remaining 5 items, another 20 minutes are needed (only 5 assemblers can work on this).
If there are fewer than three painters, painting will take no less than 250 minutes, and if there are fewer than six assemblers, assembly will take no less than 200 minutes.
|
3
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.11. The numbers $a_{1}, a_{2}, \ldots, a_{n}$ are such that the sum of any seven consecutive numbers is negative, and the sum of any eleven consecutive numbers is positive. For what largest $n$ is this possible?
118 Chapter 9. Computation of Sums and Products
$$
\text { 9.3. Sums } S_{k}(n)=1^{k}+2^{k}+\ldots+n^{k}
$$
The sum $1+2+3+\ldots+n$ can be computed as follows. Add the equations $(k+1)^{2}=k^{2}+2 k+1$ for $k=1,2, \ldots, n$. After simplification, we get $(n+1)^{2}=1+2 S_{1}(n)+n$, where $S_{1}(n)$ is the desired sum. Therefore, $S_{1}(n)=\frac{n(n+1)}{2}$.
|
9.11. According to problem $9.10\left(a_{1}+a_{2}+\ldots+a_{7}\right)+\left(a_{2}+\ldots+a_{8}\right)+\ldots$ $\ldots+\left(a_{11}+\ldots+a_{17}\right)=\left(a_{1}+a_{2}+\ldots+a_{11}\right)+\left(a_{2}+\ldots+a_{12}\right)+\ldots+$ $+\left(a_{7}+\ldots+a_{17}\right)$. Therefore, $n<17$.
For $n=16$, such a sequence exists: $5,5,-13,5$, $5,5,-13,5,5,-13,5,5,5,-13,5,5$.
|
16
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.7. Determine the coefficients that will stand by \(x^{17}\) and \(x^{18}\) after expanding the brackets and combining like terms in the expression
\[
\left(1+x^{5}+x^{7}\right)^{20}
\]
|
10.7. The number 18 cannot be represented as the sum of several numbers 5 and 7, so the coefficient of $x^{18}$ will be zero.
The number 17 can be represented as the sum of several numbers 5 and 7 as follows: $17=7+5+5$; this representation is unique up to the permutation of the addends. In one of the 20 expressions $1+x^{5}+x^{7}$, we must choose $x^{7}$, and in two of the 19 remaining such expressions, we must choose $x^{5}$. Therefore, the coefficient of $x^{17}$ is $20 \cdot \frac{19 \cdot 18}{2}=3420$.
|
3420
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.9. When dividing the polynomial $x^{1951}-1$ by $x^{4}+x^{3}+2 x^{2}+$ $+x+1$, a quotient and a remainder are obtained. Find the coefficient of $x^{14}$ in the quotient.
## 10.4. Vieta's Theorem
|
10.9. Answer: -1. The equalities $x^{4}+x^{3}+2 x^{2}+x+1=\left(x^{2}+1\right) \times$ $\times\left(x^{2}+x+1\right)$ and $x^{12}-1=(x-1)\left(x^{2}+x+1\right)\left(x^{3}+1\right)\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)$ show that
$$
\begin{aligned}
& x^{4}+x^{3}+2 x^{2}+x+1=\frac{x^{12}-1}{(x-1)\left(x^{3}+1\right)\left(x^{4}-x^{2}+1\right)}= \\
& =\frac{x^{12}-1}{x^{8}-x^{7}-x^{6}+2 x^{5}-2 x^{3}+x^{2}+x-1}
\end{aligned}
$$
Therefore, dividing the polynomial $x^{1951}-1$ by $x^{4}+x^{3}+2 x^{2}+x+1$ is the same as first dividing it by $x^{12}-1$, and then multiplying by $x^{8}-x^{7}-x^{6}+2 x^{5}-2 x^{3}+x^{2}+x-1$. But
$$
\frac{x^{1951}-1}{x^{12}-1}=x^{1939}+x^{1927}+x^{1915}+\ldots+x^{19}+x^{7}+\frac{x^{7}-1}{x^{12}-1}
$$
so the desired coefficient is the coefficient of $x^{14}$ in the product
$$
\left(x^{1939}+\ldots+x^{19}+x^{7}+\frac{x^{7}-1}{x^{12}-1}\right)\left(x^{8}-x^{7}-x^{6}+2 x^{5}-2 x^{3}+x^{2}+x-1\right)
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.31. What remainder does $x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}$ give when divided by $(x-1) ?$
|
10.31. Answer: 6. Let $x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}=P(x) \times$ $\times(x-1)+r$. By setting $x=1$, we get $r=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.17. How many roots does the equation $\sin x=\frac{x}{100}$ have?
## 11.4. Sums of Sines and Cosines Related to Regular Polygons
In solving the problems of this section, the following geometric problem is useful.
|
11.17. Answer: 63. First, note that the number of positive roots is equal to the number of negative roots, and there is also a root 0. Therefore, it is sufficient to verify that the number of positive roots is 31. If $\sin x = x / 100$, then $|x| = 100|\sin x| \leqslant 100$. Consider the graphs of the functions $y = x / 100$ and $y = \sin x$. The segment of the $O x$ axis from 0 to 100 contains 15 segments of length $2 \pi$ and one segment of length less than $2 \pi$. By examining the specified graphs, it is easy to see that on the first segment of length $2 \pi$ there is one root of the given equation, and on each of the other 14 segments of length $2 \pi$ there are two roots. Calculations show that the length of the last segment is greater than $\pi$, so there are also two roots on it. In total, we get 31 positive roots.
|
63
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12.6. Solve the equation $x+y=x y$ in natural numbers. 12.7. Solve the equation $2 x y+3 x+y=0$ in integers.
|
12.6. It is clear that $(x-1)(y-1)=x y-x-y+1=1$, therefore $x-1=y-1=1$, i.e., $x=y=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12.15. Solve the equation in integers
$$
x^{3}-2 y^{3}-4 z^{3}=0
$$
|
12.15. Answer: $x=y=z=0$.
Let $x^{3}-2 y^{3}-4 z^{3}=0$, where $x, y, z$ are integers. Then the number $x$ is even. After substituting $x=2 x_{1}$, we get the equation $8 x_{1}^{3}-2 y^{3}-4 z^{3}=0$. Dividing by 2: $4 x_{1}^{3}-y^{3}-2 z^{3}=0$. Therefore, the number $y$ is even. After substituting $y=2 y_{1}$, we get the equation $4 x_{1}^{3}-8 y_{1}^{3}-2 z^{3}=0$. Dividing by 2 again: $2 x_{1}^{3}-4 y_{1}^{3}-z^{3}=0$. Therefore, the number $z$ is even. After substituting $z=2 z_{1}$, we get the equation $x_{1}^{3}-2 y_{1}^{3}-4 z_{1}^{3}=0$, which has the same form as the original equation. Therefore, we can again prove that the numbers $x_{1}, y_{1}, z_{1}$ are even, and so on. But this is only possible if $x=y=z=0$.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14.4. How many necklaces can be made from five white beads and two black ones?
|
14.4. Answer: 3. Black beads divide the white beads into two groups (one of these groups may contain 0 beads). The appearance of the necklace is completely determined by the number of beads in the smaller group. This number can be 0, 1, or 2.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14.5. a) Seven girls are forming a circle. In how many different ways can they stand in a circle?
b) How many necklaces can be made from seven different beads?
|
14.5. a) Answer: 720. Seven girls can be arranged in a circle in $7!=5040$ ways. However, in a circle, arrangements that can be obtained by rotating the circle are not considered different. By rotating the circle, one arrangement can result in 7 other arrangements. Therefore, the total number of ways is $7!/ 7=6!=720$.
b) Answer: 360. Unlike a circle, a necklace can not only be rotated but also flipped. As a result, we get $720 / 2=360$ different necklaces.
|
360
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14.21. How many four-digit numbers (from 0001 to 9999) exist where the sum of the first two digits is equal to the sum of the last two digits?
|
14.21. Answer: 669. Let the sum of the first two digits be $n$, and the sum of the last two digits also be $n$. The value of $n$ ranges from 1 to 18. If the number of two-digit numbers whose digits sum to $n$ is $a_{n}$, then the desired number is $a_{1}^{2}+a_{2}^{2}+\ldots+a_{18}^{2}$. A two-digit number whose digits sum to $n$ consists of digits $a$ and $n-a$, where $0 \leqslant a \leqslant 9$ and $0 \leqslant n-a \leqslant 9$. Thus, $0 \leqslant a \leqslant 9$ and $n-9 \leqslant a \leqslant n$. If $n \leqslant 9$, the inequality remains $0 \leqslant a \leqslant n$, and if $n \geqslant 9$, the inequality remains $n-9 \leqslant a \leqslant 9$. In the end, we get $a_{1}=2, a_{2}=3, \ldots, a_{8}=9, a_{9}=10, a_{10}=9, \ldots, a_{17}=2, a_{18}=1$.
|
669
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14.22. How many pairs of integers $x, y$, lying between 1 and 1000, are there such that $x^{2}+y^{2}$ is divisible by $7 ?$
|
14.22. Answer: $142^{2}=20164$. The number $x^{2}+y^{2}$ is divisible by 7 if and only if both numbers $x$ and $y$ are divisible by 7. Indeed, the square of an integer when divided by 7 gives remainders of $0,1,2$, and 4. The number of integers between 1 and 1000 that are divisible by 7 is 142. Therefore, the desired number is $142^{2}=20164$.
|
20164
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14.23. How many natural numbers less than a thousand are there that are not divisible by 5 or 7?
|
14.23. Answer: 686 numbers. First, we will cross out from the set of numbers $1,2, \ldots, 999$ the numbers that are multiples of 5; their quantity is $\left[\frac{999}{5}\right]=199$. Then, from the same set of numbers $1,2, \ldots, 999$, we will cross out the numbers that are multiples of 7; their quantity is $\left[\frac{999}{7}\right]=142$. In this process, numbers that are multiples of 35 will be crossed out twice. Their quantity is $\left[\frac{999}{35}\right]=28$. Therefore, we have crossed out a total of $199+142-28=313$ numbers, and $999-313=686$ numbers remain.
|
686
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14.24. How many different integer solutions does the inequality $|x|+|y|<100$ have?
|
14.24. Answer: 338350. The equations $|x|+|y|=0,|x|+|y|=1$, $|x|+|y|=2, \ldots,|x|+|y|=99$ have, respectively, $1,2^{2}, 3^{2}, \ldots$ $\ldots, 100^{2}$ integer solutions. Therefore, the desired number is $1^{2}+2^{2}+3^{2}+\ldots+100^{2}=\frac{100 \cdot 101 \cdot 201}{6}$ (see problem 9.12).
|
338350
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14.25. How many natural numbers $x$, less than 10000, exist for which $2^{x}-x^{2}$ is divisible by $7 ?$
|
14.25. Answer: 2857. The remainders of $2^{x}$ and $x^{2}$ when divided by 7 repeat with periods of 3 and 7, respectively, so the remainders of $2^{x}-x^{2}$ when divided by 7 repeat with a period of 21. Among the numbers $x$ from 1 to 21, the remainders of $2^{x}$ and $x^{2}$ when divided by 7 are equal for exactly 6 numbers. Therefore, among the numbers from 1 to $9996=21 \cdot 476$, there are $476 \cdot 6=2856$ required numbers. Direct verification using the obtained sequence of remainders shows that among the remaining numbers 9997, 9998, and 9999, only the number 9998 has the required property.
|
2857
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14.27. Given 6 digits: $0,1,2,3,4,5$. Find the sum of all four-digit even numbers that can be written using these digits (the same digit can be repeated in a number).
|
14.27. Answer: 1769580. We will separately calculate the sum of thousands, hundreds, tens, and units for the considered numbers. The first digit can be any of the five digits $1,2,3,4,5$. The number of all numbers with a fixed first digit is $6 \cdot 6 \cdot 3=108$, since the second and third places can be any of the six digits, and the fourth place can be any of the three digits $0,2,4$ (we are considering only even numbers). Therefore, the sum of thousands is $(1+2+3+4+5) \cdot 108 \cdot 1000=1620000$. The number of numbers with a fixed second digit is $5 \cdot 6 \cdot 3=90$ (the first place can be any of the five digits). Therefore, the sum of hundreds is $(1+2+3+4+5) \cdot 90 \cdot 100=135000$. Similarly, the sum of tens is $(1+2+3+4+5) \cdot 90 \cdot 10=13500$, and the sum of units is $(2+4) \cdot 5 \cdot 6 \cdot 6=1080$.
|
1769580
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14.28. In how many different ways can 1000000 be represented as a product of three natural numbers? Products that differ only in the order of the factors are considered the same.
## 14.5. Inequalities for binomial coefficients
|
14.28. Answer: 139. Let the factors have the form $2^{a_{1}} 5^{b_{1}}, 2^{a_{2}} 5^{b_{2}}$, and $2^{a_{3}} 5^{b_{3}}$. Then $a_{1}+a_{2}+a_{3}=6$ and $b_{1}+b_{2}+b_{3}=6$. Here, the numbers $a_{i}$ and $b_{i}$ can be zero. If $a_{1}=k$, then for the decomposition $a_{2}+a_{3}=6-k$ we get $7-k$ options. Therefore, for the decomposition $a_{1}+a_{2}+a_{3}=6$ we get $7+6+5+4+3+2+1=28$ options. In total, we get $28^{2}=784$ ways.
However, we have not yet accounted for the identical decompositions that differ only in the order of the factors. There is exactly one decomposition that does not depend on the order of the factors, in which all factors are equal to 100. The decompositions in which there are two equal factors, we have counted three times. In each of the equal factors, 2 can enter in degrees $0,1,2$ or 3, i.e., in four different ways; 5 can also enter in four different ways. In total, we get 16 decompositions of this kind, but one of them is the decomposition considered above with three equal factors. This leaves 15 decompositions, each of which we have counted three times. The number of decompositions with pairwise distinct factors is $784-1-45=738$. Each of them we have counted 6 times, so among them there will be $738 / 6=123$ distinct decompositions. In total, we get $1+15+123=139$ decompositions.
|
139
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14.48. Which sum of the numbers rolled is more likely when throwing two dice: 9 or $10?$
|
14.48. A n s w e r: 9. The numbers 9 and 10 can be obtained in two different ways: $9=3+6=4+5$ and $10=4+6=5+5$. However, we need to consider the order in which the numbers fall on the dice. Therefore, the number 9 can be obtained in four different ways, while the number 10 can only be obtained in three: $9=3+6=6+3=4+5=5+4,10=4+6=6+4=5+5$.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14.51. In a drawer, there are red and black socks. If two socks are randomly pulled out of the drawer, the probability that both are red is $1 / 2$.
a) What is the smallest number of socks that can be in the drawer?
b) What is the smallest number of socks that can be in the drawer if it is known that the number of black socks is even?
|
14.51. a) Answer: 4. Let there be $m$ red socks and $n$ black socks in the box. The probability that the first selected sock is red is $\frac{m}{n+m}$. Given that the first selected sock is red, the probability that the second selected sock is also red is $\frac{m-1}{n+m-1}$. Therefore, the probability that both socks are red is $\frac{m}{n+m} \cdot \frac{m-1}{n+m-1}$. Thus, we need the equality
$$
\frac{m}{n+m} \cdot \frac{m-1}{n+m-1}=\frac{1}{2}
$$
For $n=1$, we get $m=3$. This set of socks suits us.
b) Answer: 21. Clearly, $n>0$. Then, as it is easy to check, $\frac{m}{n+m}>\frac{m-1}{n+m-1}$. Therefore, the inequalities must hold
$$
\left(\frac{m}{n+m}\right)^{2}>\frac{1}{2}>\left(\frac{m-1}{n+m-1}\right)^{2}
$$
i.e., $(\sqrt{2}+1) n<m<(\sqrt{2}+1) n+1$. By the condition, the number $n$ is even. For $n=2,4,6$ we get $m=5,10,15$. In the first two cases, the equality (1) does not hold, but in the third case, it does. In this case, $n+m=6+15=21$.
Remark. Let $x=m-n$ and $y=n$. Then the equality (1) can be rewritten as $(2 x-1)^{2}-2(2 y)^{2}=1$. Therefore, in the general case, we are dealing with a Pell's equation.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
16.13. a) There is a piece of chain consisting of 60 links, each weighing 1 g. What is the smallest number of links that need to be unbuckled so that from the resulting parts, all weights of 1 g, 2 g, 3 g, ..., 60 g can be formed (an unbuckled link also weighs 1 g)?
b) The same question for a chain consisting of 150 links.
|
16.13. a) Answer: 3 links. Let's determine the largest $n$ for which it is sufficient to break $k$ links of an $n$-link chain so that all integer weights from 1 to $n$ can be formed from the resulting parts. If $k$ links are broken, any number of links from 1 to $k$ can be formed from them. But we cannot form $k+1$ links if there is no part with $k+1$ or fewer links (we do not consider the broken links here). The most advantageous situation is to have a part with exactly $k+1$ links. Then we can form any number of links from 1 to $2k+1$. (Otherwise, we can form only the number of links from 1 to $l_1 + k$, where $l_1 \leq k$.) Next, the most advantageous situation is to have a part with $2(k+1)$ links, then with $4(k+1)$ links, and so on. Thus, if we break $k$ links, the most advantageous situation is when the $k+1$ parts obtained consist of $k+1$, $2(k+1)$, $4(k+1)$, $8(k+1)$, ..., $2^k(k+1)$ links (we do not consider the broken links here). In this case, any number of links from 1 to $n = 2^{k+1}(k+1) - 1$ can be formed. Therefore, if $2^k k \leq n \leq 2^{k+1}(k+1) - 1$, then $k$ breaks are sufficient and $k-1$ breaks are not sufficient. In particular, if $24 \leq n \leq 63$, the smallest number of broken links is 3. The four parts of the chain obtained after breaking should consist of 4, 8, 16, and 29 links.
b) Answer: 4 links. According to the solution of part a), for a chain consisting of $n$ links, where $64 \leq n \leq 159$, it is sufficient to break 4 links.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
17.21. We took three numbers $x, y, z$ and computed the absolute values of their pairwise differences $x_{1}=|x-y|, y_{1}=|y-z|$, $z_{1}=|z-x|$. In the same way, from the numbers $x_{1}, y_{1}, z_{1}$ we constructed the numbers $x_{2}, y_{2}, z_{2}$ and so on. It turned out that for some $n$ we got $x_{n}=x, y_{n}=y, z_{n}=z$. Knowing that $x=1$, find $y$ and $z$.
|
17.21. Answer: $y=z=0$. The numbers $x_{n}, y_{n}, z_{n}$ are non-negative, so the numbers $x, y, z$ are also non-negative. If all the numbers $x$, $y, z$ were positive, then the largest of the numbers $x_{1}, y_{1}, z_{1}$ would be strictly less than the largest of the numbers $x, y, z$, and then the largest of the numbers $x_{n}, y_{n}, z_{n}$ would be strictly less than the largest of the numbers $x, y, z$. Therefore, among the numbers $x, y, z$ there is a 0. Similarly, it is proved that among the numbers $x_{1}, y_{1}, z_{1}$ there is a 0 (for $n=1$ there is nothing to prove, because then $x_{1}=x, y_{1}=y, z_{1}=z$). This means that two of the numbers $x, y, z$ are equal. In the end, we get that the unordered set of numbers $x, y, z$ can be either $0,0,1$ or $0,1,1$. It is easy to check that the second set does not have the required property.
228 Chapter 17. The Pigeonhole Principle. The Rule of the Extreme
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
19.6. In the institute, there are truth-lovers, who always tell the truth, and liars, who always lie. One day, each employee made two statements.
1) There are not even ten people in the institute who work more than I do.
2) At least a hundred people in the institute earn more than I do.
It is known that the workload of all employees is different, and so are their salaries. How many people work in the institute?
|
19.6. Answer: 110. From the first statement, for the truth-teller with the least workload, it follows that there are no more than 10 truth-tellers in the institute, and for the liar with the greatest workload, it follows that there are no fewer than 10 truth-tellers in the institute. From the second statement, for the truth-teller with the highest salary, it follows that there are no fewer than 100 liars in the institute, and for the liar with the lowest salary, it follows that there are no more than 100 liars in the institute. Thus, there are 10 truth-tellers and 100 liars working in the institute.
|
110
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
20.3. The family approached the bridge at night. Dad can cross it in 1 minute, Mom in 2, the son in 5, and grandma in 10. They have one flashlight. The bridge can only hold two people at a time. How can they cross the bridge in 17 minutes? (If two people are crossing the bridge, they move at the slower of their speeds. It is not possible to cross the bridge without a flashlight. Shining the light from a distance or throwing the flashlight across the river is not allowed.)
|
20.3. The main idea is that the grandmother should cross the bridge together with the grandson. First, the father and mother go, then the father returns with the flashlight, after that the grandmother and grandson go, then the mother returns with the flashlight, finally, the father and mother cross the bridge. In total, this will take $2+1+10+2+2=17$ minutes.
|
17
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
20.16. Some of the 20 metal cubes, identical in size and appearance, are aluminum, the rest* are duralumin (heavier). How can you determine the number of duralumin cubes using 11 weighings on a balance with two pans and no weights?
|
20.16. Let's put one cube on each pan of the balance. There are two possible cases.
Case 1. One of the cubes turned out to be heavier in the first weighing.
In this case, one of the selected cubes is aluminum, and the other is duralumin. Place the selected cubes on one pan and compare them with the remaining cubes. Specifically, divide the remaining 18 cubes into 9 pairs and place them on the other pan one by one. Each time, we can determine how many duralumin cubes are in the pair. Indeed, if the reference pair is lighter, then we placed two duralumin cubes; if the reference pair has the same weight, then we placed one aluminum and one duralumin cube; if the reference pair is heavier, then we placed two aluminum cubes. Thus, in the first case, 10 weighings are sufficient.
Case 2. The cubes turned out to be of equal weight in the first weighing.
In this case, either both selected cubes are aluminum or both are duralumin. Place the selected cubes on one pan and sequentially compare them with the remaining cubes. Suppose the first $k$ pairs have the same weight, and the $(k+1)$-th pair has a different weight. (If $k=9$, then all cubes have the same weight, so there are no duralumin cubes.) Let's assume for definiteness that the $(k+1)$-th pair is heavier. Then the first two cubes and the cubes of the first $k$ pairs are aluminum. Place one cube from the $(k+1)$-th pair on each pan of the balance. If these cubes have the same weight, then they are both duralumin. If the cubes have different weights, then one is aluminum and the other is duralumin. In both cases, we can form a pair of cubes, one of which is aluminum and the other is duralumin. We can compare the remaining pairs of cubes with this pair, as in the first case. The total number of weighings in the second case is 11.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
20.22. In a table of size $1987 \times 1987$, real numbers are written, the absolute value of each of which does not exceed 1. It is known that the sum of the four numbers at the intersection of any two rows and two columns is zero. Prove that the sum of all numbers does not exceed 1987.
|
20.22. Consider a $3 \times 3$ table and cut out its top-left corner. We will show that if such a figure is placed in this table, the sum of the 8 numbers in it does not exceed 2. Indeed, consider two $2 \times 2$ squares located in the bottom-left and top-right corners of the $3 \times 3$ table, and take the sum of the numbers in one square and add it to the sum of the numbers in the other square. On the one hand, the result will be zero, since by condition the sum of the numbers in each of these squares is zero. On the other hand, the result will be the sum of 7 numbers in these squares minus the number in the center of the table. Therefore, the sum of these 7 numbers is equal to the number in the center, so the sum of the 8 numbers is equal to the sum of two numbers (standing in the center of the table and in the bottom-right corner). It remains to note that the absolute value of each number does not exceed 1.
Place 993 such figures along the diagonal running from the top-left corner to the bottom-right corner of the original table, so that they do not overlap. The sum of the numbers in the top-left corner and in all these figures does not exceed $1 + 2 \cdot 993 = 1987$. The rest of the table is divided into $2 \times 2$ square tables, in each of which the sum of the numbers is zero.
|
1987
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
21.3. The numbers $2^{n}$ and $5^{n}$ start with the digit $a$. What is $a$?
|
21.3. Answer: 3. By the condition $a \cdot 10^{p}<2^{n}<(a+1) 10^{p}$ and $a \cdot 10^{q}<5^{n}<$ $<(a+1) 10^{q}$. Therefore, $a^{2} 10^{p+q}<10^{n}<(a+1)^{2} 10^{p+q}$, i.e., $a^{2}<10^{n-p-q}<$ $<(a+1)^{2}$. At the same time, $(a+1)^{2} \leqslant 100$. Thus, $a^{2}<10<(a+1)^{2}$, i.e., $a=3$. The number 3 is the leading digit of the numbers $2^{5}$ and $5^{5}$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
21.12. Find a four-digit number that is a perfect square and has the first two digits the same and the last two digits also the same.
|
21.12. Let $a$ be the first and second digits, $b$ be the third and fourth. Then the given number is equal to $11(b+100 a)$, so $b+100 a=11 x^{2}$ for some natural number $x$. Moreover, $100 \leqslant b+100 a \leqslant$ $\leqslant 908$, hence $4 \leqslant x \leqslant 9$. By calculating the squares of the numbers $44, \ldots, 99$, we find that exactly one of them has the required form: $88^{2}=7744$.
|
7744
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
21.15. Find all three-digit numbers that are equal to the sum of the factorials of their digits.
|
21.15. Answer: 145. Let \( N = 100x + 10y + z \) be the desired number, for which \( N = x! + y! + z! \). The number \( 7! = 5040 \) is a four-digit number, so no digit of \( N \) exceeds 6. Therefore, the number \( N \) is less than 700. But then no digit of \( N \) exceeds 5, since \( 6! = 720 \). The inequality \( 3 \cdot 4! = 72 < 100 \) shows that at least one digit of \( N \) is equal to 5. At the same time, \( x \neq 5 \), since \( 3 \cdot 5! = 360 < 500 \). The equality \( 3 \cdot 5! = 360 \) also shows that \( x \leq 3 \). Moreover, \( x \leq 2 \), since \( 3! + 2 \cdot 5! = 246 < 300 \). The number 255 does not satisfy the condition of the problem, and if only one digit of the desired number is equal to 5, then \( x \leq 1 \), since \( 2! + 5! + 4! = 146 < 200 \). Since \( 1! + 5! + 4! = 145 < 150 \), we get \( y \leq 4 \). Therefore, \( z = 5 \). Considering that \( x = 1 \) and \( 0 \leq y \leq 4 \), we find the unique solution \( N = 145 \).
|
145
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
21.16. All integers are written in a row, starting from one. What digit stands at the 206,788th place?
## 21.6. Periods of Decimal Fractions and Repunits
| Let $p$ be a prime number different from 2 and 5. The length of the period of the number $p$ is the number of digits in the period of the decimal representation of the fraction $1 / p$.
|
21.16. Answer: the digit 7. There are exactly 9 single-digit numbers, 99-9=90 two-digit numbers, 999-99-9=900 three-digit numbers, 9000 four-digit numbers, and so on. Single-digit numbers occupy the first 9 positions in the written sequence, two-digit numbers occupy $90 \cdot 2=180$ positions, three-digit numbers occupy $900 \cdot 3=2700$ positions, four-digit numbers occupy $9000 \cdot 4=36000$ positions, and five-digit numbers occupy $90000 \cdot 5=450000$ positions. Therefore, the digit we are interested in belongs to a five-digit number.
Digits belonging to numbers with no more than four digits have numbers from 1 to $9+180+2700+36000=38889$. The difference $206788-38889=167899$ needs to be divided by 5 with a remainder: $167899=5 \cdot 33579+4$. The digit we are interested in belongs to the 33580-th five-digit number, i.e., the number 43579 (the first five-digit number is 10000). In this number, the digit we are interested in is in the $4-th$ position.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
21.33. If we consider digits in different positions as different, then in a $d$-ary numeral system, $n d$ digits allow us to write $d^{n}$ numbers (from 0 to $d^{n}-1$). Which numeral system is the most economical in this respect, i.e., allows recording the largest number of numbers using a given number of digits? (When comparing numeral systems with bases $d_{1}$ and $d_{2}$, we consider only sets of $m$ digits, where $m$ is divisible by $d_{1}$ and $d_{2}$.)
See also problems $20.15,20.17$.
### 21.10. Other representations of numbers
|
21.33. Answer: system with base 3. Using $m=d n$ digits, we can write $d^{m / d}$ numbers. Therefore, we need to prove that $3^{m / 3} \geqslant d^{m / d}$, i.e., $3^{d} \geqslant d^{3}$ for any natural $d$. This inequality is proven in the solution to problem 13.11.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
23.21. For which natural numbers $n$ is the expression $a^{n}(b-c)+$ $+b^{n}(c-a)+c^{n}(a-b)$ divisible by $a^{2}+b^{2}+c^{2}+a b+b c+c a$?
|
23.21. Answer: only for $n=4$. It is easy to verify that for $n=4$ the result of the division is $(a-b)(b-c)(a-c)$. We will show that for all other natural $n$ the expression $a^{n}(b-c)+b^{n}(c-a)+c^{n}(a-b)$ does not divide $a^{2}+b^{2}+c^{2}+ab+bc+ca$. It is sufficient to check that the first expression does not divide the second when $b=2$ and $c=1$, i.e., $a^{n}-(2^{n}-1)a+(2^{n}-2)$ does not divide $a^{2}+3a+7$. For $n=2$ and 3, this is verified directly. Now let $n \geqslant 5$. The quadratic equation $a^{2}+3a+7=0$ has roots $\frac{-3 \pm i \sqrt{19}}{2}$; the modulus of each root is $\sqrt{7}$. Therefore, it is sufficient to check that if $z$ is a complex number and $|z|=\sqrt{7}$, then $z^{n}-(2^{n}-1)z+(2^{n}-2) \neq 0$. Clearly, $\left|z^{n}-(2^{n}-1)z+(2^{n}-2)\right| \geqslant \left|z^{n}\right|-(2^{n}-1)|z|-(2^{n}-2)=7^{n/2}-(2^{n}-1)\sqrt{7}-2^{n}+2>7^{n/2}-2^{n}(1+\sqrt{7})>7^{n/2}-4 \cdot 2^{n}$, since $\sqrt{7}>2$, it is sufficient to consider the case $n=5$. In this case, we need to check that $7^{5}>(4 \cdot 2^{5})^{2}$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
25.15. Calculate $\lim _{n \rightarrow \infty}(\sqrt{n+1}-\sqrt{n})$.
|
25.15. Answer: 0. It is clear that
$$
0<\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{\sqrt{n}}
$$
Therefore, $\lim _{n \rightarrow \infty}(\sqrt{n+1}-\sqrt{n})=0$.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
25.35. Find the first digit of the number $2^{400}$.
|
25.35. Answer: 2. The first digit of the number $2^{400}=\left(2^{10}\right)^{40}=1024^{40}$ coincides with the first digit of the number $1,024^{40}$. On the one hand, according to problem 25.34, we get $1,024^{40}1+40 \cdot 0,024+\frac{40 \cdot 39}{2} 0,024^{2}=2,40928>2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
25.40. Does the sequence $a_{n}=\sin (2 \pi n!e)$ converge?
|
25.40. A n s w e r: yes, converges to 0. According to problem 25.39, the fractional part of the number $n!e$ is between 0 and $1 / n$. Therefore, $0<\sin (2 \pi n!e)<$ $<\sin (2 \pi / n)$ for $n \geqslant 4$.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
27.13. Solve the equation $5^{2 x-1}+5^{x+1}=250$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
|
27.13. Answer: $x=2$. The function $f(x)=5^{2 x-1}+5^{x+1}$ is monotonically increasing, so it takes the value 250 at only one value of $x$. It is also clear that $f(2)=5^{3}+5^{3}=250$.
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
27.15. How many digits does the number $2^{100}$ have?
## 27.3. Identities for logarithms
|
27.15. Answer: 31 digits. It is clear that $2^{100}=1024^{10}>1000^{10}$, so the number $2^{100}$ has no fewer than 31 digits. On the other hand,
$$
\frac{1024^{10}}{1000^{10}}<\left(\frac{1025}{1000}\right)^{10}=\left(\frac{41}{40}\right)^{10}<\frac{41}{40} \cdot \frac{40}{39} \cdot \frac{39}{38} \cdot \ldots \cdot \frac{32}{31}=\frac{41}{31}<10
$$
Thus, $2^{100}=1024^{10}<10 \cdot 1000^{10}$, so the number $2^{100}$ has fewer than 32 digits.
|
31
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
27.26. Provide an example of positive irrational numbers $a$ and $b$ such that the number $a^{b}$ is an integer.
## 27.6. Some Notable Limits
|
27.26. Let $a=\sqrt{2}$ and $b=\log _{\sqrt{2}} 3$. The numbers $a$ and $b$ are irrational (problems 6.16 and 27.25). In this case, $a^{b}=(\sqrt{2})^{\log _{\sqrt{2}} 3}=3$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
28.62. Calculate the limit $\lim _{x \rightarrow 0} \frac{\tan x - x}{x - \sin x}$.
### 28.10. The number of roots of the equation
|
28.62. Answer: 2. Transform the ratio of derivatives:
$$
\frac{\frac{1}{\cos ^{2} x}-1}{1-\cos x}=\frac{1}{\cos ^{2} x} \cdot \frac{1-\cos x^{2}}{1-\cos x}=\frac{1+\cos x}{\cos ^{2} x}
$$
The last expression tends to 2 as $x \rightarrow 0$.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
29.27. Calculate the area under the graph of the function $y=$ $=\sin x$ on the interval from 0 to $\pi$.
## 29.5. Calculation of Volumes
Let a body be located in space with rectangular coordinates $O x y z$, such that the projection of this body onto the $O x$ axis is the segment $[a, b]$. Suppose that a plane passing through
a point $x$ of the segment $[a, b]$ perpendicular to the $O x$ axis cuts out a figure of area $S(x)$ on this body. Then the volume of this body is given by $\int_{a}^{b} S(x) d x$.
|
29.27. The desired area is equal to $\int_{0}^{\pi} \sin x d x=-\left.\cos x\right|_{0} ^{\pi}=2$.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
31.11. a) Find the remainder of the division of $171^{2147}$ by 52.
b) Find the remainder of the division of $126^{1020}$ by 138.
|
31.11. a) Answer: 7. The numbers 171 and 52 are coprime, so $171^{\varphi(52)} \equiv 1(\bmod 52)$. Further, $\varphi(52)=\varphi(4) \varphi(13)=24$. Therefore, $171^{2147} \equiv 15^{24 \cdot 89+11} \equiv 15^{11} \equiv 7(\bmod 52)$.
b) Answer: 54. The numbers 126 and 138 are not coprime: their GCD is 6. We will use the fact that if $a \equiv b(\bmod 23)$, then $6 a \equiv 6 b(\bmod 138)$. Let $a=21 \cdot 126^{1019}$. Then $a \equiv 21 \cdot 11^{22 \cdot 46+7} \equiv$ $\equiv-2 \cdot 11^{7} \equiv 9(\bmod 23)$. Therefore, $126^{1020}=6 a \equiv 54(\bmod 138)$.
|
54
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
32.15. Prove that if $a+b+c+d=2$ and $1 / a+1 / b+1 / c+$ $+1 / d=2$, then
$$
\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+\frac{1}{1-d}=2
$$
$$
\% * *
$$
|
32.15. Let $\sigma_{k}$ be the $k$-th elementary symmetric function of $a, b, c, d$. Given that $\sigma_{1}=2$ and $\sigma_{3}=2 \sigma_{4}$. Therefore,
$$
\begin{aligned}
\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+\frac{1}{1-d}=\frac{4-3 \sigma_{1}+2 \sigma_{2}-\sigma_{3}}{1-\sigma_{1}+\sigma_{2}-\sigma_{3}+\sigma_{4}} & = \\
& =\frac{4-6+2 \sigma_{2}-2 \sigma_{4}}{1-2+\sigma_{2}-2 \sigma_{4}+\sigma_{4}}=2
\end{aligned}
$$
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
1. Write a million: a) using three hundreds and operation signs; b) using six tens and operation signs.
|
1. a) $100 \cdot 100 \cdot 100=1000000$; б) $10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 \cdot 10=1000000$.
|
1000000
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Write down and read all seven-digit numbers, the sum of the digits in each of which is equal to 2. How many such numbers are there?
|
2. $2000000,1100000,1010000,1001000,1000100,1000010,1000001$. Bcero 7 numbers.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. How many two-digit numbers are there in which: a) there is at least one digit 3; b) the number of tens is less than the number of units?
|
7. a) The digit 3 starts 10 numbers (from 30 to 39 inclusive), and ends in this digit in a total of 9 numbers, but the number 33 has already been counted, so there are 8 such numbers left. Therefore, the total number of such numbers is $10+8=18$.
b) From 10 to 20, there are 8 such numbers (12, 13, ..., 19), from 21 to 30 there are 7, from 31 to 40 there are 6, and so on, from 81 to 90 there is one number 89. Therefore, the total number of such numbers is $8+7+6+\ldots+2+1=9 \cdot 4=36$.
|
18
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. How many natural numbers not exceeding 1000 are there such that each subsequent digit is greater than the previous one?
|
8. Among two-digit numbers, there are 36 (see problem 7). Let's count how many such numbers there are among three-digit numbers. From 123 to $129-7$ numbers, from 134 to $139 \cdots 6$ and so on; finally, one number 189. Thus, among three-digit numbers, where the hundreds digit is 1, there are $7+6+\ldots+2+1=28$ such numbers. Among three-digit numbers starting with the digit 2, there are $6+5+\ldots+2+1=21$ such numbers; starting with the digit 3, there are $5+4+3+2+1=15$ such numbers; starting with the digit 4, there are $4+3+2+1=10$ such numbers; starting with the digit 5, there are $3+2+1=6$ such numbers; starting with the digit 6, there are $2+1=3$ such numbers; starting with the digit 7, there is only one number 789. Thus, in total, there are $28+21+15+10+6+$ $+3+1=84$ such three-digit numbers, and together with the two-digit numbers, there will be $36+84=120$.
|
120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
13. In the number 513879406, cross out 4 digits so that the remaining digits in the same order form: a) the largest number; b) the smallest number.
|
13. a) $89406 ;$ б) 13406 .
|
89406
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14. All numbers from 1 to 40 are written in a row. Without changing the order of writing, erase 60 digits so that the remaining digits express: a) the smallest number; b) the largest number.
|
14. a) In writing numbers from 1 to 40 inclusive, 71 digits are used. To make the remaining 11 digits express the smallest number, one must erase 60 larger digits, and in the remaining number, the smallest digits should be on the left. First, there are three zeros (remaining from the numbers 10, 20, 30). Then come 1, 2, 3, 3, 3, 3, 3, and 0. Therefore, the smallest number is $00012333330=12333330$. b) The largest number should start with the largest number of nines (there can be no more than three): these are the nines from the first three tens, the fourth digit should be the largest of those to the right of which there are still 7 digits. This digit is 6. Therefore, the largest number is 99967383940.
|
12333330
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
17. Students were given the task to write several different three-digit numbers that do not contain the digit 7 in their notation. How many such numbers can be written in total?
|
17. First, let's calculate the number of three-digit numbers in which there is at least one digit 7. Firstly, we notice that from 700 to 800, there are 100 such numbers; secondly, within each other hundred, there are 19 such numbers. (For example, from 100 to 200, the numbers with the tens digit 7 will be 10. These are the numbers 170, $171, \ldots, 179$. The numbers with the units digit 7 are also 10. These are the numbers 107, $117, \ldots$, 197. In total, $10+10=20$. But the number 177 is counted twice, so $20-1=19$.) The total number of three-digit numbers with the digit 7 will be $19 \cdot 8+100=252$. Considering that there are 900 three-digit numbers in total, the number of three-digit numbers in which there is no digit 7 at all is $900-252=648$.
|
648
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
19. The lock code is a seven-digit number, the first three digits of which are the same, and the last four digits are also the same. The sum of all the digits of this number is a two-digit number, the first digit of which matches the first digit of the code, and the last digit matches the last digit of the code. Find this code.
|
19. Let the cipher be $\overline{a a b b b b b}$, then by the condition $3 a+4 b=10 a+b$, from which $7 a=3 b$. Since $a$ and $b$ are digits, the equality is possible only when $a=3$ and $b=7$. Therefore, the cipher is 3337777. (The notation $\bar{a} a b b b b$ represents a seven-digit number, where $b$ is the unit of the first place, $b$ is the unit of the second place, ..., $a$ is the unit of the fifth place, ..., $a$ is the unit of the seventh place.)
|
3337777
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
20. Andrei was asked to name the apartment number that his family received in the new building. He replied that this number is 17 times greater than the number in the units place of the apartment number. What is the apartment number?
|
20. If the desired number is $10 a+b$, then $10 a+b=17 b$, from which $5 a=8 b$. The equality is possible only when $a=8$ and $b=5$. Answer: 85.
|
85
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
21. Observant Yura noticed that if in a two-digit number expressing the distance in kilometers they traveled today, a zero is inserted between the tens and units digits, the resulting number is 9 times greater than the original. What distance did they travel?
|
21. Let the distance be expressed by the number $10 a+b$. If we insert a zero between the tens and units digits, we get $100 a+b$. According to the condition, we have $100 a+b=9(10 a+b)$, so $10 a=8 b$, which means $5 a=4 b, a=4, b=5$. Answer: 45 km.
|
45
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
23. The distance between two cities in kilometers is expressed as a two-digit number, where the left digit is equal to the difference between this number and the number written with the same digits but in reverse order. What is this distance?
|
23. Let the desired number be $10 a+b$. Then
$$
a=(10 a+b)-(10 b+a), a=9 a-9 b, 9 b=8 a
$$
from which $a=9, b=8$. Answer: 98 km.
|
98
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
24. Find the number whose sum of digits equals the difference between 328 and the sought number.
|
24. Obviously, the desired number is a three-digit number. Let it be $100a + 10b + c$, then $328 - 100a - 10b - c = a + b + c$, which simplifies to $328 = 101a + 11b + 2c$, from which $a = 3$. Therefore, $11b + 2c = 328 - 303 = 25$, from which $b = 1$ and $c = 7$. Thus, the desired number is 317.
|
317
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
27. Find the digit that has the property that if it is appended to the end of any natural number, the resulting number is equal to the sum of three terms, the first of which is the original number, the second is the number represented by the sought digit, and the third is the product of the first two terms.
|
27. Let the natural number be $a$, and the appended digit be $x$. Then according to the condition $a \cdot 10 + x = a + x + a x$, from which $9 a = a x$, and therefore, $x = 9$.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
28. Write down a four-digit number where each subsequent digit is 1 greater than the previous one, then write the number with the same digits but in reverse order and subtract the smaller number from the larger one. Repeat this several times with different numbers and compare the results. Solve the problem in general terms. What do you notice?
|
28. Let the digits of the number $a, a+1, a+2$ and $a+3$, then one number is $1000 a + 100(a+1) + 10(a+2) + (a+3)$, and the number written with the same digits but in reverse order is $1000(a+3) + 100(a+2) + 10(a+1) + a$. Let's find the difference between these numbers: $1000 a + 3000 + 100 a + 200 + 10 a + 10 + a - 1000 a - 100 a - 100 - 10 a - 20 - a - 3 = 3000 + 200 + 10 - 100 - 20 - 3 = 3087$. As we can see, the difference under the given condition of the problem is always 3087.
|
3087
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
29. All natural numbers from 1 to 100 inclusive are divided into 2 groups - even and odd. Determine in which of these groups the sum of all digits used to write the numbers is greater and by how much.
|
29. Odd numbers $1,3,5,7, \ldots, 99$; even numbers $2,4,6,8, \ldots, 100$. Let's compare the sums of the single-digit numbers in these groups: $1+3+5+7+9=25,2+$ $+4+6+8=20,25-20=5$. Consider the numbers with the tens digit 1. The sum of the digits will be greater in the group that has a greater sum of the units digits. As shown above, such a sum is greater in the group of odd numbers by 5. Similarly, for numbers with the tens digit $2,3, \ldots, 9$. Thus, the sum of all digits of odd numbers from 1 to 99 is greater than the sum of the digits of even numbers from 2 to 98 by $5 \cdot 10=50$. But the number 100 belongs to the group of even numbers, and the sum of its digits is 1, so $50-1=49$. Therefore, the sum of all digits used to write the odd numbers from 1 to 100 is 49 greater than the sum of the digits used to write the even numbers from 1 to 100.
|
49
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
30. The number of students studying in the V-VI grades of the school is expressed as a three-digit number. From the digits of this number (without repeating them), 6 different two-digit numbers can be formed, the sum of which is twice the number of students in the V-VI grades. How many students are there in these grades?
|
30. Let the desired number be $100a + 10b + c$, then we can form the following two-digit numbers: $10a + b, 10b + a, 10a + c, 10c + a, 10b + c, 10c + b$. According to the condition, we have
$(10a + b) + (10b + a) + (10a + c) + (10c + a) + (10b + c) + (10c + b) = 2(100a + 10b + c), 22a + 22b + 22c = 2(100a + 10b + c), 10c + b = 89a$. The number $10c + b$ is a two-digit number, so $89 \cdot a$ is also a two-digit number, which is possible only when $a = 1$. Then $c = 8$ and $b = 9$. Therefore, in grades V-VI of the school, there are 198 students.
|
198
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
31. The number of poplars planted by the students is expressed as a three-digit number, and the number of lindens as a two-digit number. In total, they planted 144 trees. If the extreme digits in these two numbers are swapped and the resulting numbers are added, the sum is 603. How many poplars and how many lindens were planted?
|
31. Since the sum of a three-digit and a two-digit number is 144, the hundreds digit of the three-digit number is 1. The sum of this digit and the tens digit of the two-digit number is 3, as the sum of the numbers written in reverse order is 603; therefore, the tens digit of the two-digit number is $2(3-1=2)$. The units digit of the desired three-digit number is $5(6-1=5)$, since the sum of the reversed numbers is 603, which means the units digit of the two-digit number is $9(144-1 * 5=29)$. Therefore, 115 poplars and 29 lindens were planted.
|
115
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
32. Two identical two-digit numbers were written on the board. To one of them, 100 was added to the left, and to the other, 1 was added to the right, as a result of which the first number became 37 times larger than the second. What numbers were written on the board?
|
32. When multiplying the second number, ending with the digit 1, by 37, we get a number ending with the digit 7. This means the original numbers ended with the digit 7, i.e., $\overline{a 7}$. According to the condition, $\overline{100 a 7}=\overline{a 71} \cdot 37$, $\overline{a 71} \cdot 37=(100 a+71) \cdot 37=3700 a+2627$. It is easy to notice that $a=2$. Indeed, $3700 \cdot 2=7400,7400+2627=10027$. Therefore, the number written on the board was $27,27$.
|
27
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
33. How many six-digit numbers are there in which all digits are odd?
|
33. The total number of different six-digit numbers is 900000. (How to determine this?) Of these, half end with an odd digit, i.e., 450000. Of these numbers, those with the second digit from the right being odd are $450000: 2=225000$, of which those with the third digit from the right being odd are $225000: 2=112500$, of which those with the fourth digit from the right being odd are 56250, of which those with the fifth digit from the right being odd are 28125, of which those with the sixth digit from the right being odd are $\frac{28125}{9} \cdot 5=15625$ (the sixth digit from the right, i.e., the first from the left, cannot be 0). Therefore, there are 15625 six-digit numbers where all digits are odd.
|
15625
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
36. The number of students in grades V-VI of one of the schools is expressed by a three-digit number. If you find the product of the digits of this number, and then the product of the digits of the resulting number, then all these three numbers can be written as:
$$
\triangle \bigcirc O ; \triangle \square ; \square
$$
(Identical shapes correspond to identical digits.) How many students are there in grades V-VI in the school?
|
36. By comparing the numbers $\Delta \square$ and $\square$, we get that $\triangle=1$, because only by multiplying the number $\square$ by 1 do we get the same number $\square$. By comparing 100 and $1 \square$, we get that $\bigcirc=4$, because only by multiplying one by 4 and then again by 4 do we get a number with the tens digit 1. We have $144 ; 1 \cdot 4 \cdot 4=16 ; 1 \cdot 6=6$. In grades V-VI, there are 144 students.
|
144
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
37. In the hundred-digit number $12345678901234567890 \ldots 1234567890$, all digits in odd positions were erased. In the resulting fifty-digit number, digits in odd positions were erased again. The erasing continued until nothing was left. What was the last digit to be erased?
|
37. The first time, 50 odd digits were crossed out, leaving 50 even ones. The second time, 25 digits were crossed out, including 5 twos, 5 fours, 5 sixes, 5 eights, and 5 zeros. 25 remained in the following sequence: $4,8,2,6,0,4$, $8,2,6,0, \ldots, 2,6,0$. (The group of digits 4, 8, 2, 6,0 is written in sequence 5 times.) During the third crossing out, 13 digits will be crossed out, leaving 12 such digits: $8,6,4,2,0,8,6,4,2,0,8,6$. After the fourth crossing out, $6,2,8,4$, 0,6 will remain, after the fifth $-2,4,6$ and after the sixth - 4. Therefore, the last (seventh) time, the digit 4 will be crossed out.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
38. "How many students are there in our city?" the children asked. "And you will find out yourself if you know that the number representing the number of students is the largest of all such numbers, where any two adjacent digits always form a number that is divisible by 23," the teacher replied. Using this information, the children solved the problem. Solve it too.
|
38. Let the desired number be $\overline{x_{1} x_{2} x_{3} \ldots x_{n}}$. It is known that $\overline{x_{i} x_{i+1}}: 23$. For $\overline{x_{i} x_{i+1}}$, the possible values are: $00,23,46,69,92$. Considering these 5 possible options, we get 46923. This number will be the largest of the possible such numbers, since none of the five numbers divisible by 23 end in 4 and do not start with 3, and therefore no digit can be written to the left or right.
|
46923
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In a school from I to VIII grade, there is only one class per grade. In grades I-IV, there are a total of 130 students. In grade V, there are 7 more students than in grade II, in grade VI, there are 5 fewer students than in grade I, in grade VII, there are 10 more students than in grade IV, and in grade VIII, there are 4 fewer students than in grade I. How many students are there in grades I-VIII in total?
|
3. Since there are 8 more students in grades V-VIII than in grades I-IV $(7-5+10-4=8)$, there are a total of 138 students in grades V-VIII ( $130+8=138$ ), and a total of 268 students in grades I-VIII $(130+138=268)$.
|
268
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Vitya and Vova collected 27 kg of waste paper together. If the number of kilograms of waste paper collected by Vitya were increased by 5 times, and that collected by Vova by 3 times, they would have 111 kg together. How many kilograms did each boy collect?
|
4. If the number of kilograms of waste paper collected by the two boys is tripled, we would get $81 (27 \cdot 3=81)$. Then $111-81=30$ (kg) is the doubled number of kilograms of waste paper collected by Vitya, which means Vitya collected 15 kg $(30: 2=15)$, then Vova - 12 kg $(27-15=12)$.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Over three days, juice was sold in the cafeteria. On the first day, 1 large and 4 medium cans were sold, on the second day - 2 large and 6 liter cans, and on the third day - 1 large, 3 medium, and 3 liter cans. How many liters of juice were sold over 3 days, if the same amount of juice was sold each day?
|
5. Comparing what was sold on the first and third days (I - 1 large jar and 4 medium; III - 1 large jar, 3 medium, and 3 liter jars), we get that the medium jar has a capacity of 3 liters. Therefore, for:
I day - 1 large jar and $3 \cdot 4=12$ (L);
II day - 2 large jars and 6 L. From this, the capacity of the large jar is 6 L. Thus, for one day, $6+12=18$ (L) were sold, and for three days - 54 L.
|
54
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. How many different products divisible by ten can be formed from the numbers $2,3,5,7,9 ?$
|
8. For the product of numbers to be divisible by 10, the factors must include the numbers 2 and 5. Taking this into account, the following products satisfy the condition of the problem:
1) $2 \cdot 5$
2) $2 \cdot 5 \cdot 3$
3) $2 \cdot 5 \cdot 7$
4) $2 \cdot 5 \cdot 9$;
5) $2 \cdot 5 \cdot 3 \cdot 7$
6) $2 \cdot 5 \cdot 3 \cdot 9$
7) $2 \cdot 5 \cdot 7 \cdot 9$
8) $2 \cdot 5 \cdot 3 \cdot 7 \cdot 9$.
In total, there are 8 products.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. A store needed to receive 185 kg of candies in closed boxes from the warehouse. The warehouse has boxes of candies weighing 16 kg, 17 kg, and 21 kg. What boxes and how many could the store receive?
|
9. We notice that $185=37 \cdot 5$ and $16+21=37$. Therefore, the store could have received 5 boxes weighing 21 kg each and 5 boxes weighing 16 kg each. Since the number 185 has been factored into prime factors, the problem has a unique solution.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. A kilogram of one type of candy is 80 k more expensive than another. Andrey and Yura bought 150 g of candies, which included both types of candies, with Andrey having twice as much of the first type as the second, and Yura having an equal amount of each. Who paid more for their purchase and by how much?
14
|
10. Andrei has 100 g of one type of candy and 50 g of another, while Yura has 75 g of each type. Andrei's candies are more expensive than Yura's by the same amount that 25 g of one type is more expensive than 25 g of the other. Since 1 kg is more expensive by 80 k., then 25 g is more expensive by 2 k. $(80: 40=2)$. Therefore, Andrei paid 2 k. more for his candies than Yura.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. How many strikes will a clock make in a day if it strikes the whole number of hours and also marks the midpoint of each hour with one strike?
|
11. $(1+2+3+\ldots+12) \cdot 2+24=13 \cdot 6 \cdot 2+24=180$.
Translating the above text into English, while preserving the original text's line breaks and format:
11. $(1+2+3+\ldots+12) \cdot 2+24=13 \cdot 6 \cdot 2+24=180$.
|
180
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. The Zmey Gorynych has 2000 heads. A legendary hero can cut off 33, 21, 17, or 1 head with one strike of his sword, but in return, the Zmey grows 48, 0, 14, or 349 heads respectively. If all heads are cut off, no new ones grow. Can the hero defeat the Zmey? How should he act?
|
12. $2000=94 \cdot 21+26$. This means that after cutting off 21 heads 94 times, 26 heads will remain. We notice that if 17 heads are cut off, 14 will grow back, i.e., the number of heads will decrease by 3. Then, we perform this three times, and $26-3 \cdot 3=$ $=17$ will remain, which can be cut off with the final blow.
|
17
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
16. The store released two batches of notebooks. If the second batch had 5 more notebooks than it actually did, the product of the numbers expressing the quantities of notebooks in each batch would have increased by 1250, and if the first batch had 7 more notebooks, then this product would have increased by 3150. How many notebooks in total did the store release?
|
16. Since increasing the second factor by 5 increases the product by 1250, the first factor is $1250: 5=250$, i.e., in the first batch there were 250 notebooks. When the first factor is increased by 7, the product increases by 3150, so the second factor is: $3150: 7=450$, i.e., in the second batch there were 450 notebooks. Therefore, the store released a total of $250+$ $+450=700$ (notebooks).
|
700
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
17. Find the largest number that, when divided by 31, gives a quotient of 30.
|
17. The number will be the largest if when divided by 31 it yields the largest possible remainder. This remainder will be 30. Therefore, the desired number is: $30 \cdot 31 + 30 = 32 \cdot 30 = 960$.
|
960
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
20. A magazine consists of 16 nested double sheets. On which double sheet will the sum of the numbers indicating the page numbers be the greatest?
|
20. The page numbers range from 1 to 64. Pages on double sheets are arranged as follows:
$$
\begin{aligned}
& 1-\bar{h}-1,2,63,64 ; \\
& 2-\bar{и}-3,4,61,62 ;
\end{aligned}
$$
16th-31, 32, 33, 34. Since in the sequence the sums of numbers equidistant from the ends are the same, i.e., $1+64=2+63=\ldots=32+33$, the sums of the numbers indicating the page numbers of the double sheets are the same, namely $65+65=130$.
|
130
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
24. L. N. Tolstoy's Problem. Five brothers divided their father's inheritance equally. The inheritance included three houses. Since the three houses could not be divided into 5 parts, the three older brothers took them, and in return, the younger brothers were given money. Each of the three brothers paid 800 rubles, and the younger brothers divided this money among themselves, so that everyone ended up with an equal share. How much is one house worth?
|
24. Two younger brothers received 2400 rubles, and each received 1200 rubles. $(2400: 2=$ $=1200)$. Therefore, the total value of the inheritance, i.e., three houses, is 6000 rubles. $(1200 \times$ $\times 5=6000)$. Then one house costs 2000 rubles. $(6000: 3=2000)$.
|
2000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
28. How many standard sheets of paper will be required to print all numbers from 1 to 1000000 inclusive, if printing on one side of the sheet? Each sheet has 30 lines with 60 characters each, and there are intervals between numbers, each equivalent to two characters.
|
28. For printing single-digit numbers, 9 characters are required; for two-digit numbers - $180(2 \cdot 90=180)$; for three-digit numbers $-2700(900 \cdot 3=2700)$; for four-digit numbers $36000(4 \cdot 9000=36000)$; for five-digit numbers $-450000(5 \cdot 90000=450000)$; for six-digit numbers $-5400000(6 \cdot 900000=5400000)$ and one seven-digit number (1000000) 7 characters, and in total $9+180+2700+36000+450000+5400000+7=5888896$. The intervals between numbers will take up: $2.999999=1999998$ characters. Therefore, a total of 7887894 characters are needed $(5888896+1999998=7887894)$. On one sheet (page), 1800 characters can fit $(60 \cdot 30=1800)$. Since
$7887894: 1800=4382$ (remainder 294), a total of 4383 sheets of paper are required.
|
4383
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
30. Think of a single-digit number: double it, add 3, multiply by 5, add 7, using the last digit of the resulting number, write down a single-digit number, add 18 to it, and divide the result by 5. What number did you get? No matter what number you thought of, you will always get the same number in the final result. Explain why.
|
30. Let the number thought of be $x$. Then we sequentially obtain:
1) $2 x ; 2) 2 x+3 ; 3) 10 x+15=10(x+1)+5$
2) $10(x+1)+12 ; 5) 2 ; 6) 2+18=20$; 7) $20: 5=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
31. The quotient of dividing one number by another is an integer that is two times smaller than one of them and six times larger than the other. Find this quotient.
|
31. Let the dividend be $a$, and the divisor be $b$, then according to the condition, the quotient is $a: 2$ or $6 b$, i.e., $a: 2 = 6 b$, from which $a = 12 b$ and $a: b = 12$. Therefore, the quotient $a: b$ is 12.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
32. If mom gives each of her children 13 notebooks, she will have 8 notebooks left; if she gives them 15 notebooks each, all the notebooks will be distributed. How many notebooks did mom have?
|
32. If the number of children was $a$, then the total number of notebooks was $15a$. When dividing $15a$ by 13, the quotient is $a$ and the remainder is 8. Therefore, $15a = 13a + 8$, from which $2a = 8$ and $a = 4$. Thus, the total number of notebooks was $60 (15 \cdot 4 = 60)$.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
34. In the book of records, a whole number expressing the number of meters of fabric sold was blotted out with ink. The total revenue could not be read either, but the end of this entry was visible: 7 r. 28 k. - and it is known that this amount does not exceed 500 r. The price of 1 m of fabric is 4 r. 36 k. Help the inspector restore this entry.
|
34. Since one factor ends with the digit 6 (436), and the product ends with the digit 8 (...728), the second factor must end with the digit 8 or 3. If it ends with the digit 8, then \(436 \cdot 8 = 3488\). We have
Since \((8 + x)\) ends with the digit 2, then \(x = 4\). Therefore, \(6 \cdot y\) must end with the digit 4, so \(y = 4\) or \(y = 9\). If \(y = 4\), then the amount of sold material is 48 meters or 148 meters, but neither of these numbers satisfies the condition, as we get 208 rubles 28 kopecks and 1081 rubles 28 kopecks. The only remaining option is \(y = 9\). Then, 4 rubles 36 kopecks \(\cdot 98 = 427\) rubles 28 kopecks. If we assume that the second factor (the number of meters) ends with the digit 3, then a similar analysis shows that none of the possible numbers \(23, 123, 223, 73, 173\) satisfy the condition of the problem. Therefore, 98 meters were sold at 4 rubles 36 kopecks per meter for a total of 427 rubles 28 kopecks.
|
98
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
35. Think of a number written in one column:
| 10 | 23 | 16 | 29 | 32 |
| ---: | ---: | ---: | ---: | ---: |
| 27 | 15 | 28 | 31 | 9 |
| 14 | 32 | 30 | 8 | 26 |
| 36 | 24 | 12 | 25 | 13 |
| 23 | 16 | 24 | 17 | 30 |
How to guess the thought number by the sum of the numbers (excluding this number) in the row or column where the number is written? (Propose this trick to your friends.)
|
35. The sum of the numbers in each row or column is 110. Therefore, if, for example, the number 30 is chosen, then the sum of the numbers in the row without it: $14+32+8+$ $+26=80$. If you are given this sum, then you find $110-80=30$.
|
30
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
38. At a math club meeting, the students were offered the following math trick:
Write down any number,
I'll take a look at it
And then on a piece of paper
I'll put it in an envelope in front of you.
Under your number, any
Write down a number again,
And I will allow myself
To write down one number.
We will draw a line and add up
Our neat row of numbers.
No mistakes? Be more careful?!
Everything in order? I'm very glad!
The sum - I ask, check! -
I have had it in the envelope for a long time!
What's the trick here? What's the secret?
Give me your answer.
Try to uncover the secret of this trick.
|
38. Let's say, for example, the numbers 58431 and 61388 were recorded. The guesser added such a third addend that, when added to the second number recorded by the children, it would result in a number consisting only of nines. For example, $61388 + 38611 = 99999$. The sum of the three addends can be easily obtained using the number 58431, by placing the digit 1 at the far left, reducing the units digit by 1, and leaving all other digits unchanged. The guesser placed the number 158430 in the envelope.
|
158430
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
40. In a football tournament, each of the participating teams played against each other once. The teams scored $16,14,10,10,8,6,5,3$ points. How many teams participated in the tournament and how many points did the teams that played and finished in the top 4 positions lose? (A team gets 2 points for a win and 1 point for a draw.)
|
40. If the number of teams is $n$, then the total number of games is $\frac{n(n-1)}{2}$, and the total number of points scored by them is $(n-1) n$. In our example, we have $16+14+10+10+8+$ $+6+5+3=72$. Therefore, $(n-1) \cdot n=72$, or $(n-1) \cdot n=8 \cdot 9$, so $n=9$, i.e., there were 9 teams. The maximum number of points a team can score is: $2 \cdot 8=16$. This means the first team did not lose any points, the second team lost 2 points, the third and fourth teams lost 6 points each, and in total they lost 14 points $(2+6 \cdot 2=14)$. (One team did not score any points.)
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
42. In a regular set of dominoes, there are 28 tiles. How many tiles would a domino set have if the values indicated on the tiles ranged not from 0 to 6, but from 0 to $12?$
|
42. If the values on the dominoes changed from 0 to 12, then the dominoes would be:
This means there would be 13 dominoes with zero, 12 dominoes with one (but without zero), 11 dominoes with two (but without zero and one), and so on. Therefore, the total number of dominoes would be:
$13+12+11+\ldots+3+2+1=(13+1) \cdot 6+7=91$.
|
91
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
44. Find the sum of all possible different three-digit numbers, all digits of which are odd.
|
44. Let's write the sum in ascending order of the addends: $111+113+\ldots+$ $+119+131+\ldots+311+313+\ldots+319+\ldots+911+913+\ldots+993+995+$ $+997+999$. The number of addends starting with the digits $1,3,5,7,9$ will be 25 each, so the total number of addends is $25 \cdot 5=125$. The sum of two addends equally distant from the ends is 1110. There are 62 such sums, plus the middle addend is 555, so the desired sum will be $1110 \cdot 62+555=69375$.
|
69375
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
47. Over the past five years, 27 collective farmers have been awarded. Moreover, in each subsequent year, more were awarded than in the previous year. In the last year, three times as many were awarded as in the first year. How many collective farmers were awarded in the third year?
|
47. If in the first year $x$ collective farm workers are awarded, then in the last year it is $3x$, and for these two years it is $4x$. The values $x=1$ and $x=2$ do not satisfy the condition, as in the second, third, and fourth years, 23 (27-4=23) and 19 (27-8=19) would be awarded, which does not meet the condition that "in each subsequent year, more were awarded than in the previous year." Let's try $x=3$. In this case, in the first year, 3 collective farm workers were awarded, and in the fifth year, 9. In the remaining three years, 15 collective farm workers were awarded. Since $15=4+5+6$, in the third year, 5 collective farm workers were awarded ( $3<4<5<6<9$ ).
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
49. Sergei wrote down a certain five-digit number and multiplied it by 9. To his surprise, he got a number consisting of the same digits but in reverse order. What number did Sergei write down?
|
49. Since multiplying a five-digit number by 9 results in a five-digit number, the leftmost digit of the original number is 1.
Since the number obtained after multiplying this number by 9 ends with the digit 1, the original number ends with the digit $9(9 \cdot 9=81)$
Since the second digit after 1 can only be 0, then
Since the second digit in the original number is 8, then
$$
\begin{array}{r}
10 * 89 \\
\times \quad 9 \\
\hline 98 * 01
\end{array}
$$
Since the third digit in the original number is 9, then
10989
$$
\times \frac{9}{98901}
$$
|
10989
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
51. If in a certain six-digit number the leftmost digit 7 is moved to the end of the number, the resulting number is 5 times smaller than the original. Find the original number.
18
|
51. From the condition, it follows that when dividing the original six-digit number, starting with the digit 7, by 5, we get a six-digit number ending with the digit 7. Therefore, in the resulting number, the leftmost digit will be 1, and in the original number, the rightmost digit will be 5 (since the number is divisible by 5). We have $1 * * * * 7 \cdot 5=7 * * * * 5$. The second rightmost digit in the number $1 * * * * 7$ will be 5, as this is the rightmost digit of the original number. Then $1 * * * 57 \cdot 5=7 * * * 85$. We get the third rightmost digit 8, and then $1 * * 857 \cdot 5=7 * * 285$. We get the fourth rightmost digit 2, and then $1 * 2857 \cdot 5=7 * 4285$. Finally, we get the digit between 1 and 2 as 4: $142857 \cdot 5=714285$. Thus, the original number is 714285.
|
714285
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
52. When the number POTOП was taken as an addend 99999 times, the resulting number had the last three digits 285. What number is denoted by the word POTOП? (Identical letters represent identical digits.)
|
52. Taking a number as an addend 99999 times means multiplying it by 99999 or by $(100000-1)$, i.e., appending 5 zeros to the given number and subtracting the given number from the result. In this case, we have
$$
\overline{\text { ПОТОП }} 00000-\overline{\text { ПОТОП }}=\ldots 285 .
$$
From this, we get $\Pi=5, \mathrm{O}=1, \mathrm{~T}=7(10-5=5,9-8=1,9-2=7)$. Therefore, the desired number is 51715.
|
51715
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
53. The Merry Clown Nibumbum
Today is gloomy and sullen.
What is troubling Nibumbum?
He solved an example eight times,
And each time a different sum!
A sad case! (And you?)
When solving, don't forget
(That's the subtlety of the matter!)
Identical letters represent identical digits!
$$
\begin{array}{r}
\text { K O S H K A } \\
+ \text { K O S H K A } \\
\text { K O S H K A } \\
\hline \text { S O B A K A }
\end{array}
$$
|
53. The sum of three $A$ ends in $A$, so $A=0$ or $A=5$. But, if $A=5$, then $(K+K+K+1)$ cannot end in $K$. Therefore, $\mathbf{A}=0, \mathrm{~K}=5$. Since (Ш + Ш + ІІ +1 ) ends in $\mathbf{A}=0$, then $Ш=3$. Since $\mathrm{K}+\mathrm{K}+\mathrm{K}=15$, then $\mathrm{C}=1$. We have
$$
\begin{aligned}
& 5 * 350 \quad 56350 \quad 57350 \\
& +5 * 350 \quad 56350 \text { or } 57350 \\
& \frac{5 * 350}{1 * * 050} \quad \frac{56350}{169050} \quad \frac{57350}{172050}
\end{aligned}
$$
|
172050
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
54. The task is very difficult-
Not everyone can find:
What does the star,
Bicycle and hedgehog equal?
| BICYCLE | HEDGEHOG | 7 |
| :---: | :---: | :---: |
| + STAR | HEDGEHOG | 4 |
| 6 | BICYCLE HEDGEHOG | |
| 1 BICYCLE | 0 | STAR |
Decode the rebuses:
|
54. (HEDGEHOG + HEDGEHOG + BICYCLE + 1) ends with the digit 0. Therefore, (HEDGEHOG + HEDGEHOG + BICYCLE) = 9 (or 19). The equation HEDGEHOG + HEDGEHOG + BICYCLE = 19 is impossible. Therefore, the sum 9 is possible, and from the cases 1+1+7=9, 2+2+5=9, 3+3+3=9, and 4+4+1=9, only 2+2+5=9 fits. As a result, HEDGEHOG = 2, STAR = 3, BICYCLE = 5:
$$
\begin{array}{r}
527 \\
+324 \\
652 \\
\hline 1503
\end{array}
$$
128
|
1503
|
Logic and Puzzles
|
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.