problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
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 | 44,994 |
27.2. Prove that for no integer $n$ the number $n^{2}+1$ is divisible by 3.
$$
\text { (7-8 grades) }
$$ | 27.2. Any integer $n$ has one of the three forms:
$$
n=3 k, \quad n=3 k+1, \quad n=3 k+2
$$
Then for $n^{2}+1$ we will have respectively:
$$
9 k^{2}+1, \quad 9 k^{2}+6 k+2, \quad 9 k^{2}+12 k+5
$$
Obviously, none of these numbers is divisible by 3 for any integer $k$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,995 |
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 | 44,996 |
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 | 44,997 |
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 | 44,998 |
31.2. Inside the square $A B C D$, a point $M$ is taken such that $M A=1$, $M B=2$, $M C=3$. Find $M D$.
$$
\text { (8-9 grades) }
$$ | 31.2. Through point $M$ draw lines parallel to the sides of rectangle $A B C D$ (Fig. 16). Denote the points of intersection with sides $A B, \quad B C, C D, D A$ as $K, P, R, T$ respectively. Let:
$$
A K=R D=a, \quad K B=C R=b, \quad B P=A T=c, \quad P C=T D=d
$$
Then, by the Pythagorean theorem, we get:
$$
a^{2}+c... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,999 |
32.2. Find the smallest and largest values of the expression
$$
A=1.5 x+y-z
$$
if $x, y, z$ are non-negative numbers satisfying the system of equations:
$$
\left\{\begin{array}{l}
x-2 y-3 z=-10 \\
x+2 y+z=6
\end{array}\right.
$$
$$
\text { (8-9 grades) }
$$ | 32.2. Expressing $x$ and $y$ in terms of $z$ from the given system, we get:
$$
x=z-2, \quad y=4-z
$$
Considering that $x \geqslant 0$ and $y \geqslant 0$, i.e., $z-2 \geqslant 0$ and $4-z \geqslant 0$, we arrive at the following relation: $2 \leqslant z \leqslant 4$. Now, substituting the obtained expressions for $x$... | A_{}=-1,A_{max}=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,000 |
33.2. In an isosceles trapezoid $A B C D$, a perpendicular $C E$ is dropped from the vertex $C$ of the smaller base $B C$ to $A D$, and the midpoint $M$ of side $A B$ is connected to vertex $C$. Prove that if the segment $B E$ is equal to the midline of the trapezoid, then angle $M C D$ is a right angle.
$(8-9$ grades... | 33.2. Let $MK$ be the midline of the given trapezoid (Fig. 17). Connect point $K$ to points $B$ and $E$, and point $E$ to point $M$.
Fig. 17
Since $KM$ is the perpendicular bisector of seg... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,001 |
34.2. A line passing through the point $M(0 ; 1)$ intersects the parabola $y = x^2$ at points $A$ and $B$. Prove that the product of the distances from points $A$ and $B$ to the y-axis is equal to 1. | 34.2. The equation of the line passing through the point $M(0 ; 1)$ and intersecting the parabola $y=x^{2}$ (Fig. 18) is $y=k x+1$. The abscissas of points $A$ and $B$ will be the roots of the equation $x^{2}=k x+1$, which we can write as $x^{2}-k x-1=0$. By Vieta's theorem, the product of the roots of this equation is... | proof | Algebra | proof | Yes | Yes | olympiads | false | 45,002 |
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 | 45,003 |
36.2. Solve the system of equations:
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 \\
a x=b y=c z
\end{array}\right.
$$
$(8-9$ grades)
$\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1, \\ a x=b y=c z .\end{array}\right.$ | 36.2. Since none of the numbers $x, y, z$ can be equal to 0, multiplying all terms of the first equation by $x$, we get:
$$
1+\frac{x}{y}+\frac{x}{z}=x
$$
At the same time, from the second equation we have: $\frac{x}{y}=\frac{b}{a}, \frac{x}{z}=\frac{c}{a}$. Then $x=1+\frac{b}{a}+\frac{c}{a}=\frac{a+b+c}{a}$. Similar... | \frac{+b+}{},\quad\frac{+b+}{b},\quad\frac{+b+}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,004 |
37.2. It is known that $a>b>0$ and $a^{2}+b^{2}=6 a b$. Find $\frac{a+b}{a-b}$.
$$
(8-10 \text { th grade) }
$$ | 37.2. Given that $a>b>0$ and $a^{2}+b^{2}=6 a b$, we get:
$$
\frac{a+b}{a-b}=\sqrt{\left(\frac{a+b}{a-b}\right)^{2}}=\sqrt{\frac{a^{2}+2 a b+b^{2}}{a^{2}-2 a b+b^{2}}}=\sqrt{\frac{8 a b}{4 a b}}=\sqrt{2}
$$ | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,005 |
38.2. Solve the equation
$$
\left(x^{2}+x+1\right)\left(3-x-x^{2}\right)=3
$$
$$
\text { (8-10 grades) }
$$ | 38.2. Setting $t=x^{2}+x+1$, we get the equation:
$$
t(4-t)=3 \text { or } t^{2}-4 t+3=0 .
$$
From this, $t_{1}=1, t_{2}=3$. Now it remains to solve the equations:
$$
x^{2}+x+1=1 \quad \text { and } \quad x^{2}+x+1=3 \text {. }
$$
We obtain four roots of the given equation: $x_{1}=0, x_{2}=-1, x_{3}=1$, $x_{4}=-2$. | x_{1}=0,x_{2}=-1,x_{3}=1,x_{4}=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,006 |
39.2. Solve the system of equations in integers:
$$
\left\{\begin{array}{l}
x+y+z=6 \\
x+y z=7
\end{array}\right.
$$
$$
\text { (8-10 grades) }
$$ | 39.2. Expressing $x$ from the second equation and substituting into the first, we get:
$$
7-y z+y+z=6, \text { from which } y=\frac{z+1}{z-1}=1+\frac{2}{z-1}
$$
It is clear that $y$ will be an integer only if $z-1$ equals one of the numbers $-2,-1,1,2$, i.e., $z$ takes one of the values $-1,0,2,3$. Then for $y$ we wi... | \begin{aligned}&x_{1}=7,\quady_{1}=0,\quadz_{1}=-1;\\&x_{2}=7,\quady_{2}=-1,\quadz_{2}=0\text{;}\\&x_{3}=1,\quady_{3}=3,\quadz_{3}=2\\&x_{4}=1,\quady_{4}=2, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,007 |
40.2. Through the given point in the circle, draw a chord of a given length.
$$
\text { (8-10 grades) }
$$ | 40.2. Let a circle with center $O$, a point $M$ inside the circle, and a chord $A B$ (Fig. 20) be given. Take an arbitrary point $K$ on the circumference and construct a chord $K P = A B$. Obviously, if the segment $A B$ is greater than the diameter of the given circle, the problem has no solution.
 }
$$ | 41.2. We have: $\left(a^{2}-b^{2}\right)^{2} \geqslant 0$, from which $a^{4}+b^{4} \geqslant 2 a^{2} b^{2}$. It remains to prove that for $a>b>0$ the inequality $2 a^{2} b^{2}>2 a b^{3}$ holds. Indeed, $2 a^{2} b^{2}-2 a b^{3}=2 a b^{2}(a-b)>0$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 45,009 |
42.2. Prove that the numbers 23! and 26! give the same remainders when divided by 821.
$$
\text { (8-10 grades) }
$$
Note. $n!=1 \cdot 2 \cdot 3 \ldots n$. | 42.2. If the numbers 26! and 23! give the same remainders when divided by 821, then the difference between these numbers must be divisible by 821. Obviously, the converse statement is also true: if the difference 26! - 23! is divisible by 821, then the numbers 26! and 23! give the same remainders when divided by 821. T... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,010 |
43.2. Prove that the side of a regular nonagon is equal to the difference between its largest and smallest diagonals. | 43.2. In a regular nonagon \(A B C D E F G H I\), we draw the diagonals \(B D\) and \(A E\) (Fig. 21). It is clear that the first one is the smallest, and the second one is the largest. In the isosceles trapezoid \(A B D E\), from vertex \(B\) we draw a ray parallel to \(D E\) until it intersects the diagonal \(A E\) a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,011 |
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
$$
\left(10^{2}+5 \cdot 10\right)-\left(9^{2}+5 \c... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,012 |
45.2. Find the natural number that is 40 times less than the sum of all preceding natural numbers.
$$
(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}
$$ | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,013 |
46.2. In an isosceles triangle, the altitudes intersect on the inscribed circle. Find the angles of the triangle.
$$
\text { (9-10 grades) }
$$ | 46.2. Let in an isosceles triangle \(ABC (AB = BC)\) the altitudes \(AK, BD\), and \(CM\) intersect at point \(P\), which lies on the inscribed circle with center \(O\) (Fig. 22). Let \(\angle BAC = \alpha\).
=\lg x+\lg y
\end{array}\right.
$$
$$
\text { (9-10 grades) }
$$ | 47.2. Given the condition $x>0, y>0$, the given system is equivalent to the following:
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=80 \\
x y=x+y
\end{array}\right.
$$
Multiplying both sides of the second equation by 2 and adding it to the first equation, we arrive at the following equation:
$$
(x+y)^{2}-2(x+y)-80=0
$$
S... | \begin{pmatrix}x_{1}=5+\sqrt{15},&y_{1}=5-\sqrt{15}\\x_{2}=5-\sqrt{15},&y_{2}=5+\sqrt{15}\end{pmatrix} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,015 |
48.2. Find the range of the function
$$
f(x)=\sin ^{6} x+\cos ^{6} x
$$ | 48.2. We have: $f(x)=\sin ^{6} x+\cos ^{6} x=\left(\sin ^{2} x+\cos ^{2} x\right)^{3}-3 \sin ^{4} x$. - $\cos ^{2} x-3 \sin ^{2} x \cdot \cos ^{4} x=1-0.75 \sin ^{2} 2 x$. Since $0 \leqslant \sin ^{2} 2 x \leqslant 1$, then $0.25 \leqslant 1-0.75 \sin ^{2} 2 x \leqslant 1$. | 0.25\leqslant1-0.75\sin^{2}2x\leqslant1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,016 |
49.2. It is known that the parabola $y=x^{2}+p x+q$ touches the line $y=2 x-3$ at the point $M(2 ; 1)$. Find $p$ and $q$. | 49.2. If this parabola passes through the point $M(2 ; 1)$, then $1=2^{2}+2 p+q$, i.e., $2 p+q=-3$. Moreover, since the parabola is tangent to the line $y=2 x-3$ at the point $\mathrm{M}(2 ; 1)$, the derivative of the function $y=x^{2}+p x+q$ at $x=2$ will be equal to the slope of the line, i.e., 2. Thus: $y^{1}=2 x+p,... | x^{2}-2x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,017 |
50.2. Solve the equation
$$
\operatorname{tg}^{2} x=\frac{1-\cos x}{1-\sin x}
$$ | 50.2. We can rewrite the given equation as follows:
$$
\frac{\sin ^{2} x}{\cos ^{2} x}=\frac{1-\cos x}{1-\sin x}, \text { i.e., } \frac{(1-\cos x)(1+\cos x)}{(1-\sin x)(1+\sin x)}-\frac{1-\cos x}{1-\sin x}-=0
$$
From this, we have: $1-\cos x=0$ or $\cos x-\sin x=0$.
Finally, we get: $x_{1}=\frac{\pi}{2}+2 \pi k, x_{... | x_{1}=\frac{\pi}{2}+2\pik,x_{2}=\frac{\pi}{4}+\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,018 |
1.3. How to weigh 1 kg of cereal on balance scales using two weights of 300 g and 650 g?
$$
(4-5 \text { grade })
$$ | 1.3. This can be done in two weighings: first, weigh out 650 g, and then $650-300=350$ g. | 350 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,019 |
2.3. A wrong equation is depicted using matches:
$$
\mathrm{VI}-\mathrm{IV}=\mathrm{IX}
$$
Obtain a correct equation by moving just one match. Find two solutions. | 2.3. Here are these two solutions:
$$
\mathrm{V}+\mathrm{IV}=\mathrm{IX} \quad \text { and } \quad \mathrm{VI}+\mathrm{IV}=\mathrm{X} .
$$ | \mathrm{V}+\mathrm{IV}=\mathrm{IX}\quad\text{}\quad\mathrm{VI}+\mathrm{IV}=\mathrm{X} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,020 |
3.3. Write the number 100 using four fives and arithmetic signs.
$(4-5$ grade.)
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3.3. The two solutions to this problem are given below:
1) $(5+5)(5+5)=100$
2) $(5 \cdot 5-5) \cdot 5=100$ | 100 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 45,021 |
5.3. In three piles, there are 22, 14, and 12 nuts. It is required to equalize the number of nuts in all piles by making three moves, while adhering to the following condition: from one pile to another, only as many nuts can be moved as there are in the pile to which the nuts are being moved.
$$
\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 | 45,023 |
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 | 45,024 |
7.3. Petya and Vasya have a thin wooden stick. First, Petya breaks it into two pieces, and then Vasya breaks one of the resulting pieces into two pieces again. Prove that Petya can always do this in such a way that it is impossible to form a triangle from the resulting three pieces.
$$
\text { (5-6 grades) }
$$ | 7.3. If Petya breaks the rod into two equal parts, then no matter how Vasya breaks one of these parts, it is impossible to form a triangle from the resulting three pieces. | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,025 |
8.3. On the board, there are 7 natural numbers. Prove that one of them can be erased so that the sum of the remaining numbers is even.
$$
\text { (5-6 grades) }
$$ | 8.3. Among seven natural numbers, there is an odd number of either even or odd numbers. One of them needs to be erased. Then, both the number of even and odd numbers left will be even. Therefore, the sum of them will be an even number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,026 |
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 | 45,027 |
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 | 45,028 |
11.3. A cyclist and a motorist simultaneously set off from point $A$ to point $B$, with the motorist traveling 5 times faster than the cyclist. However, halfway through the journey, the car broke down, and the motorist had to walk the rest of the way at a speed that was half that of the cyclist. Who arrived at $B$ firs... | 11.3. While the motorist was driving the first half of the distance from $A$ to $B$, the cyclist covered one fifth of the half, i.e., 0.1 of the total distance from $A$ to $B$. Therefore, the cyclist had 0.9 of the distance $AB$ left to cover. While the motorist was walking the second half of the distance, the cyclist ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,029 |
12.3. Can we select such digits $a, b, c$, so that the numbers $\overline{a b}$, $\overline{b a}, \overline{b c}, \overline{c b}, \overline{c a}, \overline{a c}$ are prime?
$$
(6-7 \text { grade })
$$ | 12.3. It is clear that among the selected three digits, there should be neither even digits nor the digit 5. Moreover, since the number 39 is composite, the digits 3 and 9 cannot be taken together. This leaves us to check two triplets: $1,3,7$ and $1,7,9$. In the first case, it is easy to verify that the numbers $13,31... | 1,3,7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,030 |
13.3. Find the number if it is known that the product of the number and the number written in reverse order is 78445.
$$
\text { (6-7 grades) }
$$ | 13.3. It is clear that the sought multipliers are three-digit numbers. If the product of two three-digit numbers is 78445, then one of them ends in 5. Therefore, the second number starts with the digit 5. Then the first digit of the first number and the last digit of the second number must be one, otherwise the product... | 145\cdot541=78445 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,031 |
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 | 45,032 |
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 | 45,033 |
16.3. An old problem. The master hired a worker for a year and promised to give him 12 rubles and a coat. But after 7 months, the worker wanted to leave. Upon settlement, he received the coat and 5 rubles. How much did the coat cost
$$
\text { (6-8 grades) }
$$
12 | 16.3. Let the cost of the kaftan be $x$ rubles. Monthly, the worker's share was $(1+x: 12)$ rubles. Therefore, over 7 months, he should have received $(7+7 x: 12)$ rubles, but he received $(5+x)$ rubles. We have the equation: $7+7 x: 12=5+x$. From this, $x=4.8$. Answer: The kaftan cost 4 rubles and 80 kopecks. | 4.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,034 |
17.3. Prove that among 18 consecutive three-digit numbers, there is always at least one that is divisible by the sum of its digits.
$$
\text { (6-8 grades) }
$$ | 17.3. Among 18 consecutive natural numbers, there is always a number divisible by 18 (and therefore by 9). If this number is a three-digit number, the sum of its digits is either 9 or 18. Therefore, it is divisible by the sum of its digits. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,035 |
18.3. A motorist and a cyclist set off from $A$ to $B$ simultaneously. After covering one-third of the distance, the cyclist stopped and only resumed when the motorist had one-third of the distance left to $B$. The motorist, having reached $B$ without stopping, turned back towards $A$. Who will arrive earlier: the moto... | 18.3. Since the cyclist covered one third of the distance from $A$ to $B$ faster than the motorist covered two thirds of this distance, the doubled speed of the cyclist is greater than the speed of the motorist. At the moment the cyclist set off again, he had two thirds of the distance $A B$ left to travel to reach $B$... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,036 |
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 | 45,037 |
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 | 45,038 |
21.3. Find all two-digit numbers that are equal to three times the product of their digits.
$$
(7-8 \text {th grades })
$$ | 21.3. Let $\overline{a b}$ be the required two-digit number. According to the problem, we have the equation: $10 a+b=3 a b$, from which
$$
b=\frac{10 a}{3 a-1}
$$
Since the numbers $a$ and $3 a-1$ are coprime, 10 must be divisible by $3 a-1$. The following 4 cases are possible: $3 a-1=1, 3 a-1=2$, $3 a-1=5, 3 a-1=10$... | 1524 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,039 |
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 | 45,041 |
24.3. In a scalene triangle $ABC$, on the extension of its largest side $AC$, a segment $CM = BC$ is laid off from $C$. Prove that angle $ABM$ is obtuse.
$$
\text { (7-8 grades) }
$$ | 24.3. For triangle $C B M$, angle $A C B$ is an exterior angle (Fig. 23).
Fig. 23
Therefore, $\angle A C B = \angle C B M + \angle C M B$. Since $C M = B C$, then $\angle C M B = \angle C M... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,042 |
26.3. In an isosceles triangle \(ABC\), the angle at vertex \(A\) is \(100^\circ\), and the bisector \(BD\) is drawn. Prove that \(AD + BD = BC\).
$$
(7-9 \text{ grades })
$$ | 26.3. It is clear that $\angle A B C=\angle A C B=40^{\circ}$ (Fig. 24). Since $B D$ is the angle bisector, then $\angle A B D=\angle C B D=20^{\circ}$. On side $B C$, we mark segment $B E=B D$. Then $\angle B D E=\angle B E D=80^{\circ}$. It remains to prove that $E C=A D$. On side $B C$, we take point $F$ such that $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,044 |
27.3. A square is divided into 100 equal squares. In each of them, one of the numbers: $1,2,3$ is written. Then the sum of these numbers is calculated for each row, each column, and both diagonals. Prove that at least two of these sums will coincide. $(7-10$ grades $)$ | 27.3. We will get $10+10+2=22$ sums, then among them, there will be no more than 21 different ones (since the smallest among the sums is not less than 10, and the largest is not more than 30). Therefore, there will be at least two equal sums. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 45,045 |
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 | 45,046 |
29.3. On the board, natural numbers from 1 to 100 inclusive are written. It is allowed to erase any two numbers and write their difference in place of one of them, and zero in place of the other. Is it possible, by performing such operations one after another, to transform the initial set of 100 numbers into a set of 1... | 29.3. Let's divide the 100 numbers into 50 pairs: 1 and 2, 3 and 4, ..., 99 and 100. Obviously, we will replace each such pair with a pair of 1 and 0. Thus, we will have a set of 50 ones and 50 zeros. Now, we will divide the 50 ones into 25 pairs of two ones each. We will replace such a pair with a pair of two zeros. U... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,047 |
30.3. In a trapezoid with bases $x$ and 5, express the distance between the midpoints of the diagonals as a function of $x$.
$$
\text { (8-9 grades) }
$$ | 30.3. Let in trapezoid $ABCD$ points $K$ and $M$ be the midpoints of the lateral sides $AB$ and $CD$, respectively. Denote by $P$ and $Q$ the points of intersection of the midline $KM$ with the diagonals $AC$ and $BD$, respectively. Let $AD = x$, $BC = 5$, and consider two cases:
1) Suppose $x > 5$ (Fig. 26a). Then, t... | \frac{|x-5|}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,048 |
31.3. Prove the inequality
$$
a^{2}\left(1+b^{4}\right)+b^{2}\left(1+a^{4}\right) \leqslant\left(1+a^{4}\right)\left(1+b^{4}\right)
$$ | 31.3. Dividing both sides of the inequality by the expression
$$
\left(1+a^{4}\right)\left(1+b^{4}\right)
$$
which will be positive for any $a$ and $b$, we arrive at the following inequality:
$$
\frac{a^{2}}{1+a^{4}}-\frac{b^{2}}{1+b^{4}} \leqslant 1
$$
We will prove that both terms in the left part of the obtained... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 45,049 |
32.3. Prove that in a right triangle for the legs $a$ and $b$ and the hypotenuse $c$ the inequality
$$
a+b \leqslant c \sqrt{2}
$$
holds. | 32.3. For a right-angled triangle, we have: \(a^{2}+b^{2}=c^{2}\). At the same time, \(2ab \leqslant a^{2}+b^{2}=c^{2}\), i.e., \(2ab \leqslant c^{2}\). Therefore, \(a^{2}+2ab+b^{2} \leqslant 2c^{2}\) or \((a+b)^{2} \leqslant 2c^{2}\), from which \(a+b \leqslant c\sqrt{2}\).
33.3 The right-hand side of the equation, w... | +b\leqslant\sqrt{2} | Inequalities | proof | Yes | Yes | olympiads | false | 45,050 |
34.3. In a circle, a diameter $A B$ and a chord $C D$ are drawn, not intersecting the diameter. Perpendiculars $A E$ and $B F$ are dropped from points $A$ and $B$ to the line $C D$. Prove that $C E = D F$.
$(8-10$ grades $)$ | 34.3. From the center $O$ of the circle, drop a perpendicular $O M$ to the chord $C D$ (Fig. 27). Then $C M=M D$. Moreover, since $A O=O B$, then $E M=M F$. Therefore, $C E=D F$.
40 | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,051 |
35.3. Find all integers $n$ for which the number $\left|n^{2}-6 n-27\right|$ is prime | 35.3. We have: $\left|n^{2}-6 n-27\right|=|n+3| \cdot|n-9|$. The product of the absolute values will be a prime number if one of the absolute values is equal to one, and the other is a prime number at that time. The first absolute value is equal to one at $n=-4$ and $n=-2$, and the second at $n=8$ and $n=10$. It is eas... | -4,-2,8,10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,052 |
36.3. In triangle $A B C$, the bisectors $A A_{1}$ and $B B_{1}$ intersect at point $O$. Find the ratio $A A_{1}: O A_{1}$, if $A B=6, B C=5$, $C A=4$. | 36.3. First, we will prove that the bisector of an angle in a triangle divides the opposite side into segments proportional to the adjacent sides. We will use the bisector $B B_{1}$ (Fig. 28).
 }
$$ | 37.3. We will paint blue the arcs that are symmetric to the black arcs with respect to the center of the circle. Obviously, the total length of the painted arcs will be less than the length of the circumference. Therefore, there will be an unpainted point on the circumference. The diameter drawn from this point will be... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,054 |
38.3. Find a four-digit number that is 4 times less than the number written with the same digits but in reverse order.
$$
\text { (8-10 grades) }
$$ | 38.3. It is required to find the number $\overline{a b c e}$ such that $4 \cdot \overline{a b c e}=\overline{e c b a}$. Since the product $4 \cdot \overline{a b c e}$ is even, the number $\overline{e c b a}$ is also even, i.e., the digit $a$ is even. Then it is easy to see that $a=2$ (since for $a>2$ the product $4 \cd... | 2178 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,055 |
39.3. Three circles of the same radius intersect in pairs so that one of the intersection points is common to all three. Prove that the circle passing through the other three intersection points is equal to the three given circles.
$$
\text { (8-10 grades) }
$$ | 39.3. Let $O$ be the common intersection point of the three given circles, and let $K, M, P$ be the other points of pairwise intersection of these circles (Fig. 29). Draw a line through point $K$ perpendicular to $OK$. Let $A$ and $B$ be the points of intersection of this line with the circles intersecting at point $K$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,056 |
40.3. From point $A$ along the circumference, two bodies start moving simultaneously in opposite directions: the first with a constant speed $v$, and the second with a constant linear acceleration $a$ and an initial speed of 0. After what time did the bodies meet for the first time, if they met again for the second tim... | 40.3. Let's denote the length of the circumference by $l$. Suppose the time until the first meeting of the bodies is $t_{1}$, and the time until the second meeting is $t_{2}$. According to the problem, we have three equations:
$$
v t_{1}+\frac{a t_{1}^{2}}{2}=l, v t_{2}=l, \frac{a t_{2}^{2}}{2}=l
$$
By expressing $t_... | t_{1}=\frac{v}{}(\sqrt{5}-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,057 |
41.3. Construct two pentagons with respectively equal sides, the area of one of which is twice the area of the other.
$$
\text { (9-10 grades) }
$$ | 41.3. The problem allows for infinitely many solutions. One of them is shown in Figure 30. The pentagons $A B C D E_{1}$ and $A B C D E_{2}$ (vertices $E_{1}$ and $E_{2}$ are symmetric with respect to the line $A D$) satisfy the requirements of the problem. Using the given dimensions, it is easy to calculate that the a... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,058 |
42.3. Using only a compass, construct three points lying on the same straight line.
$$
(9-10 \text {th grade) }
$$ | 42.3. In solving the problem, the fact that the side of a regular hexagon is equal to the radius of the circumscribed circle around it is used. Take an arbitrary point $O$ and describe a circle of arbitrary radius with it as the center. On the obtained circle, take an arbitrary point $A$ and, using a compass with a rad... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,059 |
43.3. In a circle, mutually perpendicular diameters $A B$ and $C D$ are drawn, and an arbitrary point $M$ is taken on the arc $A C$. Find $M B + M D$, if $M A + M C = a$.
$$
\text { (9-10 grades) }
$$ | 43.3. Let the center of the circle be denoted by \( O \) and set \( AM = x \), \( CM = y \). Rotate triangle \( AMB \) around point \( O \) by an angle of \( 90^\circ \) so that point \( A \) moves to point \( C \) (see Fig. 31). Since angle \( AMB \) is a right angle, point \( M \) will move to some point \( E \) on t... | (1+\sqrt{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,060 |
44.3. Prove that any number of the form $10101 \ldots 101$, containing $k$ zeros, for $k \geqslant 2$ is composite.
$$
\text { (9-10 grades) }
$$ | 44.3. We have: $10101 \ldots 101=1+10^{2}+10^{4}+\ldots 10^{2 k}=$
$$
\frac{10^{2 k} \cdot 100-1}{99}=\frac{\left(10^{k+1}-1\right)\left(10^{k+1}+1\right)}{99}
$$
The obtained number is an integer, as it is the sum of the terms of an integer geometric progression. It is clear that for $k \geqslant 2$, each of the bra... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,061 |
46.3. Can the difference of the squares of two prime numbers be equal to 4048?
$$
\text { (9-10 grades) }
$$ | 46.3. Suppose there exist such primes \( p \) and \( q \) for which the equation \( p^{2} - q^{2} = 4048 \) holds. Clearly, \( q \neq 2 \) and \( q \neq 3 \), because in these cases we would have \( p^{2} = 4052 \) and \( p^{2} = 4057 \), respectively, which is impossible since the square of an integer cannot end in 2 ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,063 |
47.3. Form an equation with integer coefficients, one of the roots of which is $\sqrt[3]{2}+\sqrt[3]{3}$.
$$
\text { (9-10 grades) }
$$ | 47.3. Let $x=\sqrt[3]{2}+\sqrt[3]{3}$. Cubing, we get:
$$
x^{3}=2+3 \sqrt[3]{4} \cdot \sqrt[3]{3}+3 \sqrt[3]{2} \cdot \sqrt[3]{9+3}=5+3 \sqrt[3]{6}(\sqrt[3]{2}+\sqrt[3]{3})
$$
Thus, $x^{3}=5+3 \sqrt[3]{6} \cdot x$, i.e., $\left(x^{3}-5\right)^{3}=162 x^{3}$ or finally: $x^{9}-15 x^{6}-87 x^{3}-125=0$. | x^{9}-15x^{6}-87x^{3}-125=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,064 |
48.3. A string contains $n$ natural numbers. Prove that either one of them, or the sum of several consecutive numbers, is divisible by $n$.
$$
\text { (9-10 grades) }
$$ | 48.3. Let $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ be given natural numbers. Consider $n$ numbers:
$$
\begin{array}{r}
a_{1} \\
a_{1}+a_{2} \\
a_{1}+a_{2}+a_{3} \\
a_{1}+a_{2}+\ldots+a_{n}
\end{array}
$$
If any of these numbers is divisible by $n$, then the statement of the problem is true. Otherwise, when divided by $n$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,065 |
49.3. Triangles $A B C$ and $A_{1} B_{1} C_{1}$ are such that $\angle A+\angle A_{1}=180^{\circ}$ and $\angle B=\angle B_{1}$. Prove that $A B \cdot A_{1} B_{1}+A C \cdot A_{1} C_{1}=B C \cdot B_{1} C_{1}$.
$(9-10$ grades$)$ | 49.3. Let $\angle A \geqslant \angle A_{1}$. On the ray $B A$, take a point $D$ such that $\angle A C D=\angle A_{1} B_{1} C_{1}$ (Fig. 32). Then triangles $A C D$ and $A_{1} B_{1} C_{1}$ are similar, since $\angle C A D=180^{\circ}-\angle B A C=\angle B_{1} A_{1} C_{1}$ and $\angle A C D=$ $=\angle A_{1} B_{1} C_{1}$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,066 |
50.3. In a cylinder with an axial section of $2 \times 2$, there are 9 points. Prove that there will be two points whose distance does not exceed $\sqrt{3}$. | 50.3. Let's describe a cube with a side of 2 around a cylinder and cut it into 8 smaller cubes with a side of 1. Then at least one of the smaller cubes will contain no fewer than two points. The distance between them does not exceed the diagonal of the cube, i.e., $\sqrt{ } \overline{3}$. | \sqrt{3} | Geometry | proof | Yes | Yes | olympiads | false | 45,067 |
13. Place the addition and subtraction signs between the digits
$$
\begin{array}{lllllllll}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9
\end{array}
$$
so that the result is 100. | $\triangle$ For example, such an arrangement will work:
$$
123-4-5-6-7+8-9=100
$$
Answer: for example, $123-4-5-6-7+8-9=100$. | 123-4-5-6-7+8-9=100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,069 |
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 | 45,070 |
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 | 45,076 |
53. Solve the rebus:
$$
\begin{aligned}
& \text { COCA } \\
&+ \\
& \text { COLA } \\
& \hline \text { WATER. }
\end{aligned}
$$ | $\triangle$ Let's look at the last column: it contains the same digit A. What is it equal to? Only zero.
Now let's pay attention to the second column: there is a similar situation with the digit O. Hence, O is equal to zero or 9. But the first possibility is ruled out; thus, $\mathrm{O}=9$.
To find K, consider the fi... | 3930+3980=7910 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,081 |
69*. Solve the puzzle:
$$
\begin{array}{r}
\text { MOUNTAIN } \\
+\begin{array}{c}
\text { FIRE }
\end{array} \\
\hline \text { VOLCANO. }
\end{array}
$$ | $\triangle$ Three digits can be found immediately:
$$
\mathrm{B}=1, \quad \mathrm{y}=0, \quad \mathrm{O}=9
$$
Let's find the digit K. Since it results from adding 9 to 9 and is certainly different from 9, then $\mathrm{K}=8$.
Further solution reduces to restoring the record:
$$
\begin{array}{r}
\Gamma 9 \mathrm{PA}... | 6947+96925=103872 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,084 |
83. Even or odd sum of all natural numbers from 1 to 17? | $\triangle$ Out of 17 natural numbers, 8 are even:
$$
2,4,6,8,10,12,14,16 \text{, }
$$
the remaining 9 are odd. The sum of all these even numbers is even (property 3), the sum of the odd numbers is odd (property 5). Therefore, the sum of all 17 numbers is odd as it is the sum of an even and an odd number (property 4)... | odd | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,085 |
86. In a five-story building with four entrances, the number of residents on each floor and in each entrance was counted. Could all 9 obtained numbers be odd? | $\triangle$ Let's denote the number of residents on the floors by $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$, and the number of residents in the entrances by $b_{1}, b_{2}, b_{3}, b_{4}$. Then, the total number of residents in the building can be calculated in two ways - by floors and by entrances:
$$
a_{1}+a_{2}+a_{3}+a_{4}... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,086 |
90. Even or odd product
$$
(7 a+b-2 c+1)(3 a-5 b+4 c+10)
$$
where the numbers $a, b, c$ are integers | $\triangle$ We can consider cases related to the parity of numbers $a, b$, and $c$ (8 cases!), but it is simpler to proceed differently. Let's add the factors:
$$
(7 a+b-2 c+1)+(3 a-5 b+4 c+10)=10 a-4 b+2 c+11
$$
Since the resulting sum is odd, one of the factors of the given product is even, and the other is odd. Th... | even | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,087 |
92. Let $a_{1}, a_{2}, a_{3}, \ldots, a_{25}$ be integers, and $b_{1}, b_{2}, b_{3}, \ldots, b_{25}$ be the same numbers taken in a different order. Prove that the number
$$
\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right)\left(a_{3}-b_{3}\right) \ldots\left(a_{25}-b_{25}\right)
$$
is even.
$\mathbf{9 3}^{\circ}$. P... | $\triangle$ When proving, we can consider three cases: when the integers $a$ and $b$ are even, when they are odd, and when they have different parities. However, it is simpler to add the sum $a+b$ to the difference $a-b$:
$$
(a+b)+(a-b)=2a
$$
Since the resulting sum is even, the addends $a+b$ and $a-b$ of this sum mu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,088 |
94. Even or odd number
$$
1-2+3-4+5-6+\ldots+993 ?
$$ | $\triangle$ Based on the statement of problem 93, the difference $1-2$ has the same parity as the sum $1+2$, the difference $3-4$ has the same parity as the sum $3+4$, and so on. Therefore, the given sum has the same parity as the sum
$$
1+2+3+4+5+6+\ldots+993
$$
We can reason further in the same way as in the soluti... | odd | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,089 |
96. Seryozha wrote on the board:
$$
1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9=33
$$
where instead of each asterisk, he put either a plus or a minus. Kolya corrected several signs to their opposites and as a result, instead of the number 33, he got the number 32. Is it true that at least one of the boys made a mistake in thei... | $\triangle$ If all asterisks are replaced with plus signs, the resulting sum will be odd (check this!), and therefore, the given sum is also odd. Therefore, Kolya must have made at least one mistake.
Answer: correct. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,090 |
103*. On seven cards, the numbers $1,2,3,4,5,6$ and 7 were written. Then the cards were flipped, shuffled, and the same numbers $1,2,3,4,5,6$, 7 were written on the reverse sides. The numbers written on both sides of each card were added, and the resulting sums were multiplied. Is the resulting product even or odd? | $\triangle$ Suppose the product is odd. For this to be true, all 7 factors must be odd. But then, for the four cards with odd numbers 1, 3, 5, and 7 on one side, the numbers on the other side must be even. However, there are only three even numbers here.
Therefore, this case is impossible.
Answer: even. | even | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,091 |
111*. A circle is divided into 40 equal arcs, and a checker is placed at each of the 40 points of division. Among the checkers, 25 are white and 15 are black. Prove that there will be a white and a black checker that are at the ends of the same diameter. | $\triangle$ Suppose this is incorrect. Consider any white checker and the diameter on which it stands. Then, by our assumption, at the other end of this diameter, there should also be a white checker. This means that the total number of white checkers (as well as black ones) is even. But this contradicts the condition ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 45,092 |
113*. In some natural number, the digits were arbitrarily rearranged. Prove that the sum of the obtained number and the original one cannot be equal to $999...9$ (125 nines). | $\triangle$ Let's denote the original number by $a$, and the number obtained after rearranging the digits - by $b$.
Suppose the equality:
$$
a+b=999 \ldots 9 \text { (125 nines) }
$$
27
is possible. Then, when adding the numbers $a$ and $b$, there cannot be a carry to the next digit. Since the sums of the digits of ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,093 |
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 | 45,094 |
## 5.1.
$1_{16}{ }^{\circ}$. Prove that the divisibility of a natural number by 2 is equivalent to the last digit being even.
When in mathematics it is said that one statement is equivalent to another, in this case - one divisibility is equivalent to another, it means the following: from one divisibility follows the ... | $\triangle$ Let's take a natural number $a$. Denote the number of its tens by $b$ (not the digit, but the number of tens; for example, for the number 3258, the number of tens is $325$), and the last digit by $c$. Then
$$
a=10 b+c
$$
The number $10 b$ is always divisible by 2. Therefore, if $c$ is divisible by 2, then... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,095 |
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 | 45,096 |
124. To the number 43, add one digit to the left and one digit to the right so that the resulting four-digit number is divisible by 45. Find all solutions. | $\triangle$ Let's denote the unknown digits by $a$ and $b$. Then the four-digit number can be written as $\overline{a 43 b}$.
According to the divisibility rule for 5, $b=0$ or $b=5$. Let's consider both cases.
1) Suppose $b=0$.
The number $\overline{a 430}$ is divisible by 9 if and only if the sum of its digits, wh... | 2430or6435 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,097 |
134. To the number 41, append one digit to the left and one digit to the right so that the resulting four-digit number is divisible by 36. Find all solutions. | $\triangle$ Let's denote the unknown digits by $x$ and $y$. We apply the divisibility rule for 4 to the number $\overline{x 41 y}$:
$$
\overline{1 y}: 4
$$
Then $y=2$ or $y=6$. Let's consider both cases.
1) Suppose $y=2$.
We get the number $\overline{x 412}$. Based on the divisibility rule for 9,
$$
(x+4+1+2): 9, ... | 2412or7416 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,099 |
## 5.2.
$140^{\circ}$. Prove that the divisibility of a natural number by 11 is equivalent to the divisibility by 11 of the difference between the sum of its digits in the odd positions and the sum of its digits in the even positions (in other words, the divisibility by 11 of the alternating sum of all its digits). | $\triangle$ Let's take a five-digit number $\overline{a b c d e}$ and represent it in the following form:
$$
\begin{gathered}
\overline{a b c d e}=10000 a+1000 b+100 c+10 d+e= \\
=(9999+1) a+(1001-1) b+(99+1) c+(11-1) d+e= \\
=9999 a+a+1001 b-b+99 c+c+11 d-d+e= \\
=(9999 a+1001 b+99 c+11 d)+(a-b+c-d+e) .
\end{gathered... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,100 |
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 | 45,101 |
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 | 45,102 |
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 | 45,103 |
154. A five-digit number is divisible by 72, and three of its digits are ones. Find all such numbers. | Let's denote two unknown digits of a five-digit number by $x$ and $y$. One of them, for example, $y$, stands at the last place, since 1 cannot be the last digit of an even number.
We apply the divisibility rule for 9:
$$
(1+1+1+x+y): 9, \quad(3+x+y): 9
$$
From this, $x+y=6$ or $x+y=15$.
There is a problem with the ... | 41112,14112,11016,11160 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,104 |
156*. In a natural number, the digits were rearranged to obtain a number three times larger than the original. Prove that the resulting number is divisible by 27. | $\triangle$ Let the original number be $\overline{a_{1} a_{2} a_{3} \ldots a_{n}}$, and the number formed after rearranging its digits be $-\overline{b_{1} b_{2} b_{3} \ldots b_{n}}$. Then
$$
\begin{gathered}
\overline{b_{1} b_{2} b_{3} \ldots b_{n}}=3 \cdot \overline{a_{1} a_{2} a_{3} \ldots a_{n}} \\
a_{1}+a_{2}+a_{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,105 |
158*. Using the digits from 1 to 9 once each, form the smallest nine-digit number that is divisible by 11. | $\triangle$ The sum of all digits of the desired number is equal to
$$
1+2+3+\ldots+9=45
$$
Let the sum of the digits in odd positions be denoted by $S_{1}$, and the sum of the digits in even positions by $S_{2}$. Then $S_{1}+S_{2}=45$.
The difference $S_{1}-S_{2}$, according to the divisibility rule by 11, is divis... | 123475869 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,106 |
6.1.
$161^{\circ}$. Prove that for any integer $a$ the difference $a^{2}-a$ is divisible by 2. | $\triangle$ Let's factor this difference:
$$
a^{2}-a=a(a-1)
$$
Of two consecutive integers $a-1$ and $a$, one (and only one) is even.
162. Prove that for any integer $a$, the difference $a^{3} - a$ is divisible by 3.
$\triangle$ Let's factor the difference $a^{3}-a$ into linear factors:
$$
a^{3}-a=a\left(a^{2}-1\r... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,107 |
164. Is it true that the sum $a^{3}+2 a$ is divisible by 3 for any integer $a$? | $\triangle$ To use the divisibility $\left(a^{3}-a\right): 3$, subtract and add $a$:
$$
a^{3}+2 a=\left(a^{3}-a\right)+3 a
$$
Each term of the resulting sum is divisible by 3 for any integer $a$, so the entire sum is divisible by 3 for any integer $a$.
Answer: correct. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,108 |
169. Prove that if the sum $a+b+c$, where $a, b, c$ are integers, is divisible by 3, then the sum $a^{3}+b^{3}+c^{3}$ is also divisible by 3. | $\triangle$ Transform the last sum as follows:
$$
a^{3}+b^{3}+c^{3}=\left(a^{3}-a\right)+\left(b^{3}-b\right)+\left(c^{3}-c\right)+(a+b+c)
$$
Each of the four terms of the last sum is divisible by 3, therefore, the entire sum is divisible by 3. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,109 |
171. Prove that if an integer $a$ is not divisible by 3, then the number $5a^2 + 1$ is divisible by 3. | $\triangle$ The square of an integer, when divided by 3, gives a remainder of 0 or 1. But since $a$ is not divisible by 3, the second case remains: $a^{2}=3 k+1$ (where $k$ is a non-negative integer). Then
$$
5 a^{2}+1=5(3 k+1)+1=15 k+6
$$
The last sum is always divisible by 3. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,110 |
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