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Problem 77. Find the sum of the squares of the sides of an inscribed quadrilateral with perpendicular diagonals in a circle of radius $R$.
---
Translated as requested, maintaining the original text's line breaks and format. | Solution. For the problem $76 m^{2}+n^{2}+k^{2}+t^{2}=8 R^{2}$.
Answer. $8 R^{2}$. | 8R^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,811 |
Problem 78. Given a circle and a point $K$ inside it. Prove that the sum of the squares of the lengths of two mutually perpendicular chords of the given circle, intersecting at point $K$, is a constant.
.
\[
\begin{aligned}
& A B^{2}+C D^{2}=(A K+K B)^{2}+(C K+K D)^{2}= \\
& =A K^{2}+K B^{2}+C K^{2}+K D^{2}+ \\
& +2 A K \cdot K B+2 C K \cdot K D \\
& A B^{2}+C D^{2}=a^{2}+b^{2}+c^{2}+d^{2}+ \\
& +2(A K \cdot K B+C K \cdot ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,812 |
Problem 80. A quadrilateral ACBD with perpendicular diagonals is inscribed in a circle of radius $R$ (Fig. 81). The distance KO from the point of intersection of the diagonals to the center of the circle is $t$. Prove that the midpoints of the sides of the quadrilateral and the feet of the perpendiculars dropped from p... | Solution. Let $E, F, N, T$ be the midpoints of the sides of quadrilateral $A C B D$. Since $A B \perp C D$, it is obvious that $E F N T$ is a rectangle. Then $E N$ is the diameter of the circumscribed circle around $EFNT$.
$$
\begin{gathered}
E N^{2}=E T^{2}+T N^{2}=\frac{C D^{2}}{4}+\frac{A B^{2}}{4}= \\
\frac{C D^{2... | R_{1}=\frac{1}{2}\sqrt{2R^{2}-^{2}} | Geometry | proof | Yes | Yes | olympiads | false | 39,813 |
Problem 82. Prove that $A C=B C$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Proof. Tangent segments drawn from a given point to a given circle are equal.
$$
A C = C E \text{ and } C B = C E \text{, hence } A C = B C \text{. }
$$ | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,814 |
Problem 83. Prove that $\angle A E B=90^{\circ}$. | Proof. Since $A C=C E=C B$, point $C$ is the center of the circumcircle of triangle $A E B$, with diameter $A B$. Therefore, $\angle A E B=90^{\circ}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,815 |
Problem 86. Find the radius of the circle that touches two given circles of radii $R_{1}$ and $R_{2}$ and their common external tangent. | Solution. In the notation of Fig. 87, where $x$ is the radius of the desired circle, we write: $A C + C B = A B$, or, according to Archimedes' problem:
66
$$
2 \sqrt{R_{1} x} + 2 \sqrt{x R_{2}} = 2 \sqrt{R_{1} R_{2}}
$$
From this, it is not difficult to find $x$.
$$
x = \frac{R_{1} R_{2}}{\left(\sqrt{R_{1}} + \sqrt... | \frac{R_{1}R_{2}}{(\sqrt{R_{1}}\\sqrt{R_{2}})^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,818 |
Problem 89. $K$ to two externally tangent circles at point $E$ with radii $R_{1}$ and $R_{2}$, external tangents $A B$ and $C D$ are drawn. Prove that a circle can be inscribed in $A B C D$. Find its radius. | Solution. Obviously, $ABCD$ is a trapezoid. Draw a common internal tangent $MN$ to the given circles. Then:
$2 MN = AB + DC$ (Fig. 91). On the other hand, $MN$ is the midline of trapezoid $ABCD$, so $2 MN = AD + BC$. Therefore,
$$
AB + DC = AD + BC
$$
.
Solution 1. The area of a trapezoid is equal to the product of the height of the trapezoid and its midline. The height of the given trapezoid is equal to the diameter of the inscribed circle (see Problem 69), i.e., $\frac{4 R_{1} R_{2}}{R_{1}+R_{2}}$. The midlin... | Solution 2. It is obvious that the area of rectangle $C N E K$ (see Fig. 90) is half the area of triangle $A E B$, and the area of triangle $A E B$ (see Fig. 91) is half the area of trapezoid $A B P K$ (show this on your own!). It is further clear that the area of trapezoid $A B P K$ is half the area of the desired tra... | \frac{8R_{1}R_{2}\sqrt{R_{1}R_{2}}}{R_{1}+R_{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,822 |
Problem 91. Given a quadrilateral $ABCD$ in which a circle can be inscribed. Four circles are inscribed in it such that each touches the sides of the corresponding angle and two adjacent circles. Prove that at least two of the four adjacent circles are equal. | Solution. Let's denote the points of tangency of the circles with the sides of the quadrilateral as shown in Fig. 92. Considering that $ABCD$ is a circumscribed quadrilateral, we can write:
$$
AB + DC = AD + BC
$$
or
$AF + FT + TB + DQ + QP + PC = $
$= AN + MN + MD + BK + KL + LC$.
Next, taking into account that
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,823 |
Problem 92. A tangent $A B$ is drawn to two touching circles. Parallel to $A B$, a tangent to the smaller circle is drawn, intersecting the larger circle at points $D$ and $C$ (Fig. 93). Given that the radii of these circles are $R_{1}$ and $R_{2}$, find the radius of the circle circumscribed around triangle $BDC$.
$$
\frac{A I}{A L_{1}}=\frac{b+c}{a+b+c}
$$
From the sim... | \frac{(b+)}{+b+} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,829 |
Problem 100. The bisectors $A L_{1}$ and $B L_{2}$ of triangle $A B C$ intersect at point I. It is known that $A I: I L_{1}=3 ; B I: I L_{2}=2$. Find the ratio of the sides in triangle $A B C$. | Solution. According to the property of the incenter $\frac{b+c}{a}=3 ; \frac{a+c}{b}=2$, or $b+c=3a ; a+c=2b$. Solving this system of equations, we find the sides of the triangle: $\frac{3}{5} c ; \frac{4}{5} c ; c$. That is, the ratio of the lengths of the sides of the triangle is: $3: 4: 5^{1}$. | 3:4:5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,830 |
Problem 101. The center of the inscribed circle divides the height of an isosceles triangle, drawn to the base, into segments of 5 cm and 3 cm, measured from the vertex. Find the sides of the triangle.
. Then $\frac{b+c}{a}=\frac{2 b}{a}=\frac{5}{3}$, or $a=\frac{6}{5} b$. Then $\quad C L_{1}=B L_{1}=\frac{a}{2}=\frac{3}{5} b . \quad$ By the Pythagorean theorem for triangle $A L_{1} C \quad A L_{1}=\frac{4}{5} b$. Thus:
76
$... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,831 |
Problem 102. Prove that if in triangle $ABC$ the equality $\operatorname{ctg} \frac{A}{2}=\frac{b+c}{a}$ holds, then the given triangle is a right triangle. | Proof. By the property of the incenter: $\frac{A I}{I L_{1}}=\frac{b+c}{a}$. But by the property of the angle bisector for triangle $A C L_{1} \quad \frac{A I}{I L_{1}}=\frac{A C}{C L_{1}}$ (Fig. 103). Therefore:
$$
\frac{A C}{C L_{1}}=\frac{b+c}{a}=\operatorname{ctg} \frac{A}{2}
$$
 | Proof. The given inequality is equivalent to the following:
$$
\frac{1}{\sin \frac{A}{2}} \geq \frac{b+c}{a}
$$
From triangle $A T T \sin \frac{A}{2}=\frac{r}{A I}$ (Fig. 104). We will prove that:
$$
\frac{A I}{r} \geq \frac{b+c}{a}, \text{ or } \frac{A I}{r} \geq \frac{A I}{I L_{1}}, \text{ or }
$$
or $r \leq I L_... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 39,833 |
Problem 104. Prove Mollweide's formula:
$$
\frac{\cos \frac{|B-C|}{2}}{\sin \frac{A}{2}}=\frac{b+c}{a}
$$ | Proof. It is known that the angle between the altitude and the bisector drawn from the vertex of angle $A$ is $\frac{|B-C|}{2}$ (prove this fact independently).
Then, in triangle $I K L_{1}$, $I L_{1}=\frac{r}{\cos \frac{|B-C|}{2}}$ (Fig. 105).
. Therefore, $A I \geq \frac{b+c}{a} \cdot r$. Similarly, $B I \geq \frac{a+c}{b} \cdot r, C I \geq \frac{a+b}{c} \cdot r$. Then
$$
\begin{gathered}
A I+B I+C I \geq r\left(\f... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 39,835 |
Problem 106. Prove the validity of the inequality in an arbitrary triangle:
$$
\sqrt{\frac{I L_{1}}{A I}}+\sqrt{\frac{I L_{2}}{B I}}+\sqrt{\frac{I L_{3}}{C I}} \geq 2
$$ | Proof. According to the property of the incenter, the original inequality reduces to the inequality
$$
\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}} \geq 2 .
$$
Consider the expression $\sqrt{\frac{b+c}{a}}$. By the Cauchy-Schwarz inequality for the geometric and arithmetic means:
$$
\sqrt{\frac{b+c... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 39,836 |
Problem 107. Prove the validity of the inequality in any triangle \(ABC\):
\[
\frac{AI}{AL_1} \cdot \frac{BI}{BL_2} \cdot \frac{CI}{CL_3} \leq \frac{8}{27}
\]
(International Mathematical Olympiad, 1991) | ## Proof. We have
$$
\begin{gathered}
\frac{A I}{A L_{1}}=\frac{b+c}{a+b+c}, \frac{B I}{B L_{2}}=\frac{a+c}{a+b+c}, \frac{C I}{C L_{3}}=\frac{a+b}{a+b+c} \\
\frac{A I}{A L_{1}}+\frac{B I}{B L_{2}}+\frac{C I}{C L_{3}}=\frac{b+c+a+c+a+b}{a+b+c}=2
\end{gathered}
$$
According to the Cauchy inequality for three numbers,
... | \frac{AI}{AL_{1}}\cdot\frac{BI}{BL_{2}}\cdot\frac{CI}{CL_{3}}\leq\frac{8}{27} | Inequalities | proof | Yes | Yes | olympiads | false | 39,837 |
Problem 117. The 4 medians of a tetrahedron (segments connecting each vertex to the center of mass of the opposite face) intersect at one point and are divided in the ratio $3:1$, counting from the vertex. Prove it! | Proof. We find the center of gravity of three balls of unit mass located at three vertices of a tetrahedron. According to Archimedes' theorem on medians - this is the ball $m(3)$. Then we find the center of mass of the system consisting of the ball $m(3)$ and the ball $m(1)$, located at the fourth vertex of the tetrahe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,839 |
Problem 118. The medians of tetrahedron DABC intersect at point $M$. Prove that $\overline{M A}+\overline{M B}+\overline{M C}+\overline{M D}=\overline{0}$. | Proof. The equality is obvious, since, according to the method of masses, point $M$ is the center of mass of the tetrahedron. At this point, the tetrahedron is in equilibrium, which means that the resultant of all forces at this point is zero, which is equivalent to the given formula. | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,840 |
Problem 119. Prove that for any point $T$ outside the tetrahedron DABC, the following formula holds:
$$
T A^{2}+T B^{2}+T C^{2}+T D^{2}=M A^{2}+M B^{2}+M C^{2}+M D^{2}+4 T M^{2}
$$
(where $M$ is the center of mass of the tetrahedron). | Proof. It is obvious that $\overline{T A}=\overline{T M}+\overline{M A}$ (Fig. 110). Similarly, $\overline{T B}=\overline{T M}+\overline{M B}, \overline{T C}=\overline{T M}+\overline{M C}$, $\overline{T D}=\overline{T M}+\overline{M D}$. Then
$$
\begin{gathered}
T A^{2}+T B^{2}+T C^{2}+T D^{2}= \\
=4 T M^{2}+M A^{2}+M... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,841 |
Problem 120. Find the point for which the sum of the squares of the distances to the four vertices of a tetrahedron is minimal. | Solution. Since the center of mass of the given tetrahedron is fixed, according to problem 119, the expression $T A^{2}+T B^{2}+T C^{2}+T D^{2}$ will be minimal when $T M=0$, or $T \equiv M$.
Answer. This point is the center of mass of the tetrahedron.
We will also demonstrate the application of the method of masses ... | This\point\is\the\center\of\mass\of\the\tetrahedron | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,842 |
Problem 121. The midpoints of the sides of an arbitrary hexagon are connected every other one. Prove that the points of intersection of the medians of the two resulting triangles coincide. | Proof. Let $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ be the given hexagon (Fig. 111). Let also $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}$ be the midpoints of its sides. Place unit masses at the vertices of the hexagon. Then the system of points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ can be replaced by two systems of po... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,843 |
Problem 122. A line passing through vertex C and the midpoint of median $m_{a}$ cuts off one third of AB. Prove it!
 | Proof. Place unit mass balls at points $B$ and $C$ (Fig. 112). Obviously, a ball of mass 2 must be located at point $M_{1}$ $\left(B M_{1}=C M_{1}\right)$. We require that point $E$ be the center of mass of the system of points $A, B, C$. For this, a ball of mass 2 must be placed at point $A$ (since $A E=E M_{1} \quad-... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,844 |
Problem 123. Prove that the point of intersection of the diagonals of a trapezoid, the point of intersection of the extensions of its non-parallel sides, and the midpoints of the bases lie on the same line. | Proof. Let $O$ be the point of intersection of the diagonals of the trapezoid; $E$ be the point of intersection of the extensions of its lateral sides (Fig. 113). We require that the point $O$ be the center of mass of triangle $A E D$. Place a unit mass $m(1)$ at point $E$. And place masses $m(k)$ and $m(t)$ at points ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,845 |
Problem 126. Given angle $A O B$. From point $O$ as the center, a circle with radius $A O = B O = R$ is drawn (Fig. 118). Suppose in the segment $A B$ we managed to "insert" $F H$ - perpendicular to $A B$ - such that $A H = B H + B F$. Prove that in this case the trisection of angle $A O B$ is achieved. | Proof. Let $H K = B H$. Then, considering the condition, $F B = A K$. But $F B = F K$ (triangle $B F K$ is isosceles). Therefore, $A K = F K$. Let $\angle F B K = \alpha$. Then $\cup A F = 2 \alpha$. For the isosceles triangle $A K F$, $\angle F K B = \alpha$ is the exterior angle. Then $\angle F A K = \frac{\alpha}{2}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,847 |
Problem 129. Construct an isosceles triangle given the height and the bisector drawn to the lateral side, or
$$
(b=c) ; h_{b} ; l_{b}
$$ | Solution. The analysis shows that $\Delta B H_{2} L_{2}$ is not difficult to construct using the leg $h_{b}$ and the hypotenuse $I_{b}$ (Fig. 121). But then... Let $\angle A B L_{2}=\alpha$. Then $\angle A C B=2 \alpha$. By summing the angles, we find that $\angle B L_{2} H_{2}=180^{\circ}-3 \alpha$. But this angle is ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,848 |
Problem 130. Given an angle of $54^{\circ}$. Divide it into three equal parts using a compass and a straightedge. | Solution. We will use the fact that $\frac{1}{3} \cdot 54^{\circ}=18^{\circ}$. By supplementing the angle $54^{\circ}$ to a right angle, we obtain the angle $36^{\circ}$ (Fig. 122). The bisector of the angle $36^{\circ}$ will give us the desired angle $18^{\circ}$. | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,849 |
Problem 133. A perpendicular is erected from point $D$ to the line $A C$ until it intersects the outer circle at point $B$. Prove that the area of the arbelos is equal to the area of the circle with diameter $D B$.
---
Note: The term "арbelos" in the original text is a transliteration of the Greek word for a geometri... | Proof. The area of the arbelos is equal to the difference between the area of the semicircle $ABC$ (Fig. 125) and the semicircles $AMD$ and $DNC$. That is:
$$
S_{a}=\frac{1}{2}\left(\pi\left(r_{1}+r_{2}\right)^{2}-\pi r_{1}^{2}-\pi r_{2}^{2}\right)=\pi r_{1} r_{2}
$$
Since angle $ABC$ is a right angle, $BD$ is the al... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,851 |
Problem 134. A circle with diameter $D B$ intersects the small arcs of the arbelos at points $M$ and $N$ (Fig. 126). Prove that point $M$ lies on the line $A B$, and point $N$ lies on the line $B C$. | Proof. Angle $A M D$ is a right angle, as it subtends the diameter $A D$. Angle $B M D$ is also a right angle, as it subtends the diameter $B D$. Therefore, angle $A M B$ is a straight angle, i.e., point $M$ lies on the line $A B$. Similarly, it can be proven that point $N$ lies on the line $B C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,852 |
Problem 138. $B$ semicircles are inscribed on opposite sides of the perpendicular $D B$, as shown in Fig. 129. Prove that their radii are equal.
 | Solution. Draw the diameter $H N$ of the inscribed circle parallel to $A C$. Let points $F$ and $G$ be the points of tangency of the inscribed circle with the semicircles of the arbelos (Fig. 129). Then, by the lemma on parallel diameters, points $A, H$ and $F$, as well as points $F, N$ and $C$; $H, G$ and $D$; $A, G$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,855 |
Problem 139. A circle is inscribed in an arbelos (Fig. 130). Prove that the distance from its center to the line AC is equal to its diameter. | Solution. Draw the diameter $E F$ of the inscribed circle parallel to $A C$. Using the lemma about parallel diameters, construct the connecting lines as shown in Fig. 130. Then, from the parallelism of the lines $B M$ and $C F$ and the lines $E K$ and $F L$,
=2 n \cdot 2(n+1)=4 n(n+1)$. Of two consecutive integers $n$ and $n+1$, one is always even, and therefore $n(n+1)$ is always divisible by 2; consequently, $4 n(n+1)$ is divisible by 8. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,866 |
14. Prove that an even number, not divisible by 4, cannot equal the sum of two consecutive odd numbers. | 14. $(2 n+1)+(2 n+3)=4 n+4=4(n+1)$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,867 |
15. Prove that if $n$ is a prime number, different from 2 and 3, then $n^{2}-1$ is divisible by 24. | 15. $n^{2}-1=(n-1)(n+1)$. Among three consecutive integers $n-1, n, n+1$, one is divisible by 3, but since $n$ is the original number, either $n-1$ or $n+1$ is divisible by 3. On the other hand, these numbers are two consecutive even numbers, so one is divisible by 2 and the other by 4. Therefore, their product is divi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,868 |
16. Prove that the product of a whole perfect square and the whole number immediately preceding this square is always divisible by 12. | 16. $n^{2}\left(n^{2}-1\right)=n^{2}(n-1)(n+1)$. Among three consecutive numbers $n-1, n, n+1$, one is divisible by 3; if $n$ is even, then $n^{2}$ is divisible by 4; if $n$ is odd, then $n-1$ and $n+1$ are even, and therefore their product is divisible by 4. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,869 |
17. Every natural number can be represented in one of three forms: $3 n, 3 n+1, 3 n+2$, where $n$ stands for a natural number or zero. | 17. When dividing an integer by 3, there are three (and only three) cases: either the number is divisible by 3 with no remainder, or the remainder is 1, or the remainder is 2. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,870 |
18. If both of two natural numbers give a remainder of 1 when divided by 3, then their product gives the same remainder when divided by 3. | 18. One of the given numbers can be expressed in the form $3 m+1$, the other $3 n+1$ ( $m$ and $n$ are the integer quotients of division by 3 ); their product $(3 m+1)(3 n+1)=3(3 m n+m+n)+1$, and therefore it also gives a remainder of 1 when divided by 3. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,871 |
19. Two natural numbers both give a remainder of 2 when divided by 3. What remainder does their product give when divided by 3? | 19. $(3 m+2)(3 n+2)=3(m n+m+n+1)+1$.
Translating the above text into English, while preserving the original text's line breaks and format, results in:
19. $(3 m+2)(3 n+2)=3(m n+m+n+1)+1$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,872 |
21. A number that represents a perfect square is divisible by 3 or gives a remainder of 1 when divided by 3. | 21. Each integer has one of three forms: $3 m, 3 m+1$, $3 m-1$; the corresponding squares have the form: $9 m^{2}, 9 m^{2}+6 m+1$, $9 m^{2}-6 m+1$. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,874 |
22. If a number is not divisible by 5, then its square, increased or decreased by 1, is divisible by 5. | 22. A number not divisible by 5 ends with one of the digits: $1,2,3,4$; the square of this number ends with one of the digits: $1,4,9,6$. This is what leads to the given statement. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,875 |
23. Given two arbitrary natural numbers; determine the probability that their product will be an even number; that it will be an odd number. (Definition of probability see question 375.)
Determine the probability that their product will be an even number; that it will be an odd number. | 23. There are four equally possible cases: $and and, and not$, and and, not and; in three of them the product is an even number, in one only an odd number. Therefore, the corresponding probabilities will be: $\frac{3}{4}, \frac{1}{4}$. | \frac{3}{4},\frac{1}{4} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,876 |
24. Two natural numbers are given at random; determine the probability that their product is divisible by 3 without a remainder; that when divided by 3, the remainder is 1; the remainder is 2. | 24. A number can be of one of three types: $3 m, 3 m+1$, $3 m+2$, and another can be of one of the types: $3 n, 3 n-1-1, 3 n+2$; from the nine equally possible cases (each type of the first number can be combined with each type of the second), in five cases the product gives a remainder of 0, in two cases a remainder o... | \frac{5}{9};\frac{2}{9};\frac{2}{9} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,877 |
25. Two integers are given at random; determine the probability that their product will give remainders of $0 ; 1 ; 2 ; 3$ when divided by 4. | 25. Sixteen equally possible cases. Corresponding probabilities: $\frac{8}{16} ; \frac{2}{16} ; \frac{4}{16} ; \frac{2}{16}$. | \frac{8}{16};\frac{2}{16};\frac{4}{16};\frac{2}{16} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,878 |
30. The number of diagonals in a polygon with $\boldsymbol{n}$ angles is given by the formula $\frac{n(n-3)}{2}$. This number, which is essentially always an integer, appears in the form of a fraction. Explain this apparent contradiction. | 30. The fraction $\frac{n(n-3)}{2}$ is simplified only by $n$ and not by the fraction, because if $\boldsymbol{n}$ is an even number, it cancels out with the denominator; if $n$ is odd, then $n-3$ is even. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,881 |
32. If $n$ is an integer, then $\frac{n(n+1)(n+2)}{2 \cdot 3}$ is also an integer. Why? | 32. Of the three factors of the numerator $n(n+1)(n+2)$, at least one is an even number and certainly one is a multiple of 3. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,882 |
33. Prove that for an integer $n$ the expression
$$
\frac{n(n+1)(n+2)(n+3)}{1.2 .3 .4}
$$
is an integer. Check for several larger values of $n$ and $k$ that, generally, the expression
$$
\frac{n(n+1)(n+2) \ldots(n+k-2)(n+k-1)}{1.2 .3 \ldots(k-1) k}
$$
is an integer. | 33. The number $n$ gives a remainder of 0 or 1, or 2, or 3 when divided by 4; in the first case, $n$ is divisible by 4, in the second case $n+3$ is divisible by 4, in the third case $n+2$, and in the fourth case $n+1$. Since two of the factors in the numerator are even, the one that is not divisible by 4 is divisible b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,883 |
34. Show that every odd number is the difference of the squares of two consecutive natural numbers. | 34. $2 n+1=(n+1)^{2}-n^{2}$.
Translating the above text into English, while preserving the original text's line breaks and format, results in:
34. $2 n+1=(n+1)^{2}-n^{2}$. | 2n+1=(n+1)^{2}-n^{2} | Number Theory | proof | Yes | Yes | olympiads | false | 39,884 |
35. Show that every initial number, excluding the number 2, can be expressed as the difference of two squares in one and only one way. | 35. Every original number, except for the number 2, is an odd number of the form $2 n+1$ (where $n$ is an integer). Let us assume that $2 n+1=x^{2}-y^{2}$ (where $x$ and $y$ are integers) or: $2 n+1=$ $=(x-y)(x+y)$. The left side of the equation, by definition, is a prime number, and therefore $x-y=1, x+y=2 n+1$, from ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,885 |
37. Is the following statement true: every odd number can be expressed as the difference of two squares in one and only one way? | 37. No. For example, $15=4^{2}-1^{2}=8^{2}-7^{2} ; 105=19^{2}-16^{2}=$ $=13^{2}-8^{2}=11^{2}-4^{2}=53^{2}-52^{2}$. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,887 |
38. How to find all possible expressions of a given odd number as the difference of squares of two integers? How to determine in advance the number of these expressions? | 38. Let the number 63 be given. We will decompose it into two factors in all possible ways: $63=1\cdot63=3\cdot21=7\cdot9$. It should be:
$$
63=x^{2}-y^{2}=(x-y)(x+y)
$$
Let's set sequentially:
$$
\begin{array}{lll}
x-y=1, & x-y=3, & x-y=7 \\
x+y=63, & x+y=21, & x+y=9
\end{array}
$$
from which we get:
$$
\begin{ar... | 63=32^{2}-31^{2}=12^{2}-9^{2}=8^{2}-1^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,888 |
39. There are numbers that can be expressed as the semi-product of two consecutive natural numbers (for example, 10; 15; and others). If such an expression is possible for a given number, is it possible in only one way? | 39. Such a representation, since it is possible (for example, numbers $4,5,7,8 \ldots$ cannot be represented in this way), can only be achieved in one way. For if a natural number could simultaneously satisfy two equations of the form Elem. Math.
$x=\frac{n(n+1)}{2}$ and $x=\frac{k(k+1)}{2}$, where $k \neq n$, then it ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,889 |
40. Explain with examples what the expression "in one and only one way" means. Explain what "in one way" and "only in one way" mean separately and together. | 40. Each number can be represented as the final result of known operations. A requirement can be set (see questions $34,35,37,38$) to represent the given number as the result of a known type of calculation (it is clear that the "calculation" can consist of a long series of actions). For example: represent the given num... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,890 |
41. Any natural number that is not divisible by 2 or 3 can be represented in one of two forms: $6n+1$ or $6n+5$, where $n$ denotes a natural number or zero. The product of two numbers of the form $6n+1$ is also a number of the form $6n+1$, the product of two numbers of the form $6n+5$ is a number of the form $6n+1$, an... | 41. A natural number that is not divisible by 2, nor by 3, is also not divisible by 6, and therefore it can only have one of the following forms: $6k+1, 6k+2, 6k+3, 6k+4, 6k+5$. But since it is not divisible by 2, the forms $6k+2$ and $6k+4$ are excluded; the form $6k+3$ is also excluded, as it is not divisible by 3. T... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,891 |
42. Show that any prime number $N$ can be represented in the form $30 n + p$, where $n$ is an integer, and $p$ is a prime number less than 30. | 42. Let $n$ be the quotient of the division of $N$ by $30$, and $p$ the remainder of this division, so we have $N=30 n+p$, where $05$, then it would be $p \geqslant 49$, which contradicts the condition; hence, it must be divisible by 2, or by 3, or by 5. But then the number $N$ would not be a prime number, since 30 is ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,892 |
43. When is the least common multiple of several numbers simply their product? | 43. If these numbers are coprime. | If\these\\\coprime | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,893 |
44. How many common multiples do the numbers 180 and 300 have? Which one is the smallest? If the smallest is determined, how can the others be obtained? How many common divisors do these numbers have and what are they? | 44. These numbers (like any pair of numbers) have an infinite number of common multiples; the smallest of them is 900; all other common multiples are obtained by multiplying 900 by any integer; the common divisors of the given numbers are: $2,3,5 ; 4,6,10$, $15 ; 12,20,30 ; 60$. (The number of common divisors is always... | 900 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,894 |
45. The least common multiple of five numbers was determined, and it turned out that in finding it, two of the given numbers could be completely disregarded, i.e., that the problem simply reduced to determining the least common multiple of the remaining three numbers. What special properties must the given five numbers... | 45. The three remaining numbers included all the prime factors entering into those two numbers, and moreover in powers not
less. For example, when determining the least common multiple of five numbers $2^{3} \cdot 3^{2} \cdot 5 \cdot 7 ; \quad 2^{2} \cdot 3 \cdot 7 ; \quad 2^{4} \cdot 3^{2} \cdot 7 \cdot 11 ; 2 \cdot 3... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,895 |
46. When finding the least common multiple of several numbers, it turned out to be one of these numbers. Why? The greatest common divisor of several numbers turned out to be one of these numbers. Why? | 46. 1) Because one of the given numbers contained all the prime factors present in the other numbers, and in powers no less than those in the other numbers; 2) because one of the given numbers did not contain any prime factor in a higher power than all the other numbers. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,896 |
47. The greatest common divisor of four numbers has been found; how can we quickly select three more numbers so that the found greatest common divisor remains the greatest common divisor of all seven numbers? | 47. It is enough to multiply the greatest common divisor of the given numbers by any three numbers and add the resulting products to the given numbers. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,897 |
49. How many divisors do the following integers have: 32 ; $2^{10} ; 2^{m} ; 2^{10} \cdot 3 ; 2^{10} \cdot 3^{2} ; 2^{m} \cdot 3^{n} ; 2^{m} \cdot 3^{n} \cdot 5^{p}$ ? | 49. The number $2^{10} \cdot 3^{2}$, for example, has the following divisors:
$1, 2, 2^{2}, 2^{3}, 2^{4}, 2^{3}, 2^{6}, 2^{7}, 2^{8}, 2^{9}, 2^{10}, 3, 2 \cdot 3, 2^{2} \cdot 3, 2^{3} \cdot 3, 2^{4} \cdot 3, 2^{3} \cdot 3, 2^{6} \cdot 3, 2^{7} \cdot 3, 2^{8} \cdot 3, 2^{9} \cdot 3, 2^{10} \cdot 3, 3^{2}, 2 \cdot 3^{2}... | (+1)(n+1)(p+1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,899 |
50. How many divisors does the integer
$$
N=a^{\alpha} \cdot b^{\beta} \cdot c^{\gamma} \ldots l^{\lambda}
$$
have, where $a, b, c, \ldots l$ are prime numbers and $\alpha, \beta, \gamma, \ldots \lambda$ are any integers? | 50. Generalizing the method used in the answer to the previous question, we get for the total number of divisors the expression:
$$
(\alpha+1)(\beta+1)(\gamma+1) \cdots(\lambda+1)
$$ | (\alpha+1)(\beta+1)(\gamma+1)\cdots(\lambda+1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,900 |
51. In algebra, it is proven (see the previous question) that if $\alpha$, $\beta, \gamma, \ldots \lambda$ are the exponents of the prime factors of a number $N$, then the number $n$ of divisors of this number is given by the formula: $n=(\alpha+1)(\beta+1)(\gamma+1) \ldots(\lambda+1)$. Prove that the necessary and suf... | 51. If $n$ is an odd number, then all numbers $\alpha+1, \beta+1$, $\gamma+1, \ldots \lambda+1$ are odd numbers, and thus $\alpha, \beta, \gamma, \ldots \lambda$ are even; therefore, $N$ is a perfect square. Conversely, if $N$ is a perfect square, then $\alpha, \beta, \gamma \ldots \lambda$ are even numbers, $\alpha+1,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,901 |
56. If all sides of a right-angled triangle are expressed as integers, then one of them or all three numbers are even; there can only be two odd numbers. Why? (Pythagorean theorem.) | 56. Let $a$ and $b$ be the lengths of the legs, and $c$ the length of the hypotenuse. According to the Pythagorean theorem, we have $a^{2}+b^{2}=c^{2}$. Note that the square of an even number is even, and the square of an odd number is odd. If $a$ and $b$ are both even or both odd, then in both cases $c$ is an even num... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,904 |
57. Given an integer $a$, representing a leg of a right triangle, all sides of which are measured in integers; determine the length of the other leg $b$ and the hypotenuse $c$. How many solutions are there? Solve the problem, for example, for $a=15$ and for $a=12$. | 57. $c^{2}-b^{2}=a^{2}$ or $(c+b)(c-b)=a^{2}$.
$a=15 \quad(c+b)(c-b)=15^{2}=3 \cdot 3 \cdot 5 \cdot 5$.
1) $(c+b)(c-b)=225 \cdot 1$
$c+b=225$
$c-b=1$
$c=113, b=112$
3) $(c+b)(c-b)=45 \cdot 5$
$c+b=45$
$c-b=5$
$c=25, b=20$
2) $(c+b)(c-b)=75 \cdot 3$
$c+b=75$
$\frac{c-b=3}{c=39, b=36}$
4) $(c+b)(c-b)=25 \cdot 9$... | =113,b=112;=25,b=20;=39,b=36;=17,b=8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,905 |
58. If in the formulas $a=m^{2}-n^{2}, b=2 m n, c=m^{2}+n^{2}$ we assign consecutive integer values to the letters $m$ and $n$, then we will obtain triplets of integer values for $a, b$, and $c$, and, as is easy to verify, $a^{2}+b^{2}=c^{2}$ [perform the operations on both sides of the identity $\left(m^{2}-n^{2}\righ... | 58. Indeed: one of the legs, $b=2 m n$, is an even number; therefore, the other, $c^{2}=m^{2}-n^{2}$, must be an odd number, for otherwise the hypotenuse would also be even, i.e., all three sides would be even numbers, which contradicts the condition. Thus, the hypotenuse and the other leg are odd numbers. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,906 |
59. Why, if calculating according to the formulas $a=m^{2}-n^{2}, b=2 m n$, $c=m^{2}+n^{2}$ "Pythagorean triples" (i.e., triples of integers expressing the sides of right triangles) and limiting oneself to triangles that are essentially different in shape (i.e., not similar), must the letters $m$ and $n$ be given only ... | 59. 1) $m \neq n$, for if $m=n$ one of the legs would be $=0$; 2) if $m$ and $n$ had common divisors, then $a, b$ and $c$ would also have common divisors and similar triangles would result, which is unnecessary; 3) if $m$ and $n$ were both odd, then $m^{2}-n^{2}$ and $m^{2}+n^{2}$ would be even; hence, all sides would ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,907 |
61. In a right-angled triangle, all sides of which are expressed as integers, 1) one of the legs is a multiple of 4, 2) one of the legs (the same or another) is a multiple of 3, 3) one of the sides is a multiple of five. | 61. 1) The cathetus $2 m n$ is a multiple of 4, since $m$ or $n$ is an even number; 2) or one of the numbers $m$ and $\boldsymbol{n}$ is a multiple of three, then the cathetus $2 m n$ satisfies the condition, or the squares of $m$ and $n$ have the form $3 k+1,3 l+1$ (see questions 18 and 19), then the second cathetus h... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,908 |
62. Given a four-digit number; it is required to check whether this number is prime or composite. For this purpose, we sequentially check whether it is divisible without a remainder by $2 ; 3 ; 5 ; 7 ; 11$; 13 ; and so on. At which prime divisor can we stop the check? | 62. You can stop the check at the first of consecutive initial numbers, when dividing by which the quotient is less than this divisor. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,909 |
65. What whole number is equal to the sum of all the whole numbers preceding it? Is there only one such number? | 65. $1+2=3$. No other number $>3$ has the required property: indeed, if we move forward in the sequence of natural numbers, then with each step taken, the sum of natural numbers increases by $3, 4, 5$, etc., while each subsequent natural number increases only by 1, and therefore the difference between the sum of natura... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,910 |
66. Do there exist two consecutive natural numbers, the quotient of which is an integer? | 66. Only 1 and 2 ; indeed: $\frac{n+1}{n}=1+\frac{1}{n} ; \frac{1}{n}$ will be an integer only for $n=1$. | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,911 |
67. How can you check the correctness of the addition of a long column of integers? | 67. a) If the addition is performed from top to bottom, then perform it from bottom to top; b) break the column into several groups, perform the addition of separate groups and add the partial sums. | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 39,912 |
68. Verification of subtraction: a) by addition, b) by subtraction; verification of multiplication: a) by division, b) by multiplication. | 68. a) Difference + subtrahend - minuend; b) minuend - difference = subtrahend; c) product: multiplier = multiplicand; d) multiplicand $\times$ multiplier $=$ product. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,913 |
69. What is the basis for checking performed arithmetic operations by dividing by 9? What errors can this check not detect? | 69. On the following properties of numbers: 1) the remainder of dividing any number by 9 is the same as the remainder of dividing the sum of the digits of this number by 9; 2) the remainder of dividing the sum of several numbers by 9 is equal to the sum of the remainders of dividing each of the addends by 9; the remain... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,914 |
71. What conveniences does dividing a degree (hour) into 60 parts offer? What conveniences does counting by dozens offer? | 71. Dividing an hour into 60 minutes is convenient because $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$, $\frac{1}{5}, \frac{1}{6}, \frac{1}{10}, \frac{1}{12}, \frac{1}{15}, \frac{1}{20}, \frac{1}{30}$ of an hour can be expressed as whole numbers of minutes. A dozen is divisible by $2,3,4,6$ parts. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,915 |
72. What is the main difference between our and the Roman numeral systems?
## II. Arithmetic operations and the expansion of the concept of number.
## The main difference between our numeral system (the Hindu-Arabic numeral system) and the Roman numeral system lies in their structure and the way they represent number... | 72. One of the significant differences between our and the Roman numeral systems is that in our system, a small number of symbols (ten: $0,1,2,3,4,5,6$, $7,8,9$) are sufficient to write any arbitrarily large number, whereas in the Roman system, a small number of symbols (e.g., $\boldsymbol{I}, \boldsymbol{V}, \boldsymb... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,916 |
74. Multiplication has the distributive property, or distributivity, with respect to addition
$[a(b+c)=a b+a c]$. Does exponentiation have the distributive property with respect to multiplication? Does multiplication have the distributive property with respect to exponentiation? | 74. Let there be any two operations on numbers, which we will symbolically denote by the signs $\bigcirc$ and $\cup$ (to give our reasoning a general character, we deliberately chose signs not used in arithmetic, where each of the signs $\bigcirc$ and $\cup$ can represent any operation: addition, multiplication, subtra... | ()^{}=^{}b^{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,917 |
75. What actions are always executable in the domain of natural numbers (i.e., positive integers)? Which ones force us to go beyond this domain? | 75. Addition, multiplication, and exponentiation are always possible within the domain of natural numbers; subtraction, division, root extraction, and logarithmization - not always. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,918 |
78. Before the introduction of negative numbers, at the very beginning of arithmetic, the following statement is made:
$$
\begin{array}{ll}
\text { if } & a-c=b-d \\
\text { then } & a+d=b+c
\end{array}
$$
Can the reverse statement be considered valid before the introduction of negative numbers? | 78. No, for before the introduction of negative numbers, expressions $a-c$ and $b-d$ might lack meaning if the minuend is less than the subtrahend. For example, at the indicated stage of our arithmetic knowledge, from the equality $3+9=5+7$ we cannot conclude that $3-7=5-9$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,921 |
79. Why are negative numbers introduced in mathematics? | 79. In addition to practical purposes (see question 80), then, to form a domain of numbers within which subtraction is always possible. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,922 |
85. From three relative (positive and negative) numbers, all possible products of two are formed. What can be the signs of these three products? What can be said about the number of positive products in each case? | 85. Or all three numbers have the same sign: then all three products of these numbers, taken in pairs, are positive; or one number has a sign opposite to the other two: then one product is positive, two are negative. Thus, the number of positive products can be either 3 or 1. (Construct the corresponding table). | 3or1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,925 |
89. If $a<\frac{1}{b}$, what can be concluded about the relative magnitude of $\frac{1}{a}$ and $b$? | 89. If $a$ and $b$ have the same sign (shorter: if $\boldsymbol{a} b 0$), then $\frac{1}{a}>b$; if $a b<0$, then $\frac{1}{a}<b$. | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 39,927 |
92. When can two inequalities of the same sense be multiplied term by term? What sense will the resulting inequality have? | 92. When both sides of both inequalities are positive or when both sides of both inequalities are negative; in the first case, an inequality of the same sense is obtained, in the second case, an inequality of the opposite sense. | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 39,929 |
94. Is it always useful to reduce a fraction? In what cases is it better not to rush into reducing it? | 94. Sometimes you shouldn't rush to reduce a fraction, especially when you will need to bring it to a common denominator later; for example, for addition, subtraction, or clearing denominators from an equation. Provide examples. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,931 |
95. Given two fractions; when can one say at a glance which is greater and which is less? | 95. If the fractions have the same denominators, the larger one is the one with the larger numerator; if the numerators are the same, the larger fraction is the one with the smaller denominator; finally, one fraction is clearly larger than the other if the numerator of the first is larger than the numerator of the othe... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,932 |
96. The sum of two irreducible fractions with different denominators is never a whole number. | 96. $\frac{a}{b}+\frac{c}{d}=\frac{a d+c b}{b d}$. For this fraction to become a whole number, the numerator must be divisible by both $b$ and $d$ simultaneously. For it to be divisible by $b$, $a d$ must be divisible by $b$, which means that $d$ must be divisible by $b$ (since $a$ and $b$ are coprime). For the same re... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,933 |
98. In what sense is arithmetic on integers a special case of operations on polynomials arranged in descending powers of a known letter? | 98. Each integer written in the decimal system can be represented as a polynomial arranged in descending powers of the number 10. For example,
$$
30275=3.10^{4}+2.10^{2}+7.10+5
$$ | 30275=3.10^{4}+2.10^{2}+7.10+5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,934 |
99. Write a polynomial arranged in descending powers of the number five, with coefficients being integers, non-negative, and less than 5. How can this polynomial be symbolically recorded in a manner similar to the notation of numbers in our decimal system? (See the previous question). | 99. For example, the polynomial $3.5^{6}+1.5^{3}+2.5^{2}+4.5+1$ can be briefly written as 3001241 or, to avoid misunderstandings, as (3001241), where the 5 outside the parentheses indicates that the base of the numeral system is 5. Conversely, the symbol (432) represents the polynomial $4.5^{2}+3.5+2=117$ (in the decim... | 432 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,935 |
100. Write a polynomial arranged in descending powers of 2, with coefficients equal to 1 or 0. How can this polynomial be symbolically represented? (See previous questions). | 100. For example, $(110101101)_{2}=2^{8}+2^{7}+2^{5}+2^{3}+2^{2}+1=429$. Conversely, the number $15=2^{3}+2^{2}+2+1=(1111)_{\mathbf{2}}$. | 429 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,936 |
101. If a number is divisible by 3, then, as is known, the sum of its digits is divisible by 3 (and vice versa); this is true in our decimal system of numeration; is this rule of divisibility by 3 applicable to numbers written in another, for example, in a duodecimal system of numeration? | 101. No; recall the derivation of this divisibility rule, which is based on the fact that the base of the decimal system of numeration $10=3\cdot3+1, \quad 10^{2}=3\cdot33+1$ and so on. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,937 |
102. A certain number is written in the duodecimal system; for which divisor $m$ is the following divisibility rule valid: if the sum of the digits of the number is divisible by $m$, then the number is also divisible by $m$? | 102. $m=11$, because $12=11\cdot1+1, \quad 12^{2}=11\cdot13+1$ and so on. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,938 |
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