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Given that the area of triangle $\mathrm{ABC}$ is 1, and $\mathrm{BC}=1$, when the product of the three altitudes of this triangle is maximized, $\sin A =$ | Since the area of the triangle is fixed, the product of the three altitudes with the three sides is also fixed. To maximize the product of the three altitudes, we should minimize the product of the three sides. Since angle $\mathrm{A}$ is fixed with respect to side $\mathrm{BC}$, and the area of the triangle is given by $\frac{1}{2} bc \sin A$, which is constant, we should maximize $\sin A$, so that $bc$ is minimized. As shown in the diagram below, draw line $l$ parallel to $\mathrm{BC}$ at a distance of 2 units. Then, draw the perpendicular from $\mathrm{B}$ to $l$, intersecting $l$ at $\mathrm{A}$. Construct the circumcircle $\mathrm{O}$ of triangle $\mathrm{ABC}$. In this configuration, $\mathrm{A}$ is at a position that maximizes angle $\mathrm{A}$, as the angle subtended at the circumference of the circle is greater than the angle subtended at the center. Calculate $\mathrm{AB}=\mathrm{AC}=\frac{\sqrt{17}}{2}$ using the Pythagorean theorem, and then use the cosine rule to find the trigonometric value of angle $\mathrm{A}$. | AIChallenge_Geo17 | \frac{8}{17} |
If each face of a tetrahedron is not an isosceles triangle, then it has at least\_\_\_\_\_ distinct edge lengths. | Each pair of edges needs to be equal. | AIChallenge_Geo18 | 3 |
In tetrahedron $ABCD$, where $AC=15$, $BD=18$, $E$ is the trisect point of $AD$ closer to $A$, $F$ is the trisect point of $BC$ closer to $C$, and $EF=14$. Then, the cosine value of the angle between edges $AC$ and $BD$ is\_\_\_\_\_. | Taking the trisect point $G$ on $CD$ closer to $C$, we have $EG // AC$ and $FG // BD$, where $EG=10$ and $FG=6$.
Then, $\cos \angle EGF = \frac{6^2 + 10^2 - 14^2}{2 \times 6 \times 10} = -\frac{1}{2}$.
Since the angle between skew lines is acute, the result is $\frac{1}{2}$. | AIChallenge_Geo19 | \frac{1}{2} |
Given that each face of the tetrahedron has edges of lengths $\sqrt{2}$, $\sqrt{3}$, and $2$, the volume of this tetrahedron is \_\_\_\_\_. | Considering a rectangular parallelepiped with three face diagonals of lengths $\sqrt{2}, \sqrt{3}, 2$, the lengths of its three edges are $\frac{\sqrt{2}}{2}, \frac{\sqrt{6}}{2}, \frac{\sqrt{10}}{2}$.
Its volume $V^{\prime}=\frac{\sqrt{30}}{4}$.
Therefore, the volume of the tetrahedron is $V=\frac{V^{\prime}}{3}=\frac{\sqrt{30}}{12}$. | AIChallenge_Geo20 | \frac{\sqrt{30}}{12} . |
Given that the line $l$ intersects two parabolas $\Gamma_{1}: y^{2}=2px (p>0)$ and $\Gamma_{2}: y^{2}=4px$ at four distinct points $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, $D\left(x_{3}, y_{3}\right)$, and $E\left(x_{4}, y_{4}\right)$, where $y_{4}<y_{2}<y_{1}<y_{3}$.
Let $l$ intersect the $x$-axis at point $M$. Given that $AD = 6 BE$, then the value of $\frac{AM}{ME}$ is \_\_\_\_\_. | According to the given conditions, the slope of the line $l$ exists. Let $l: y=kx+m$. Then
\[\begin{array}{l}
\left\{\begin{array}{l}
y^{2}=2px, \\
y=kx+m
\end{array}\right.
\Rightarrow ky^{2}-2 py+2 pm
=0
\Rightarrow \frac{1}{y_{1}}+\frac{1}{y_{2}}=\frac{y_{1}+y_{2}}{y_{1} y_{2}}
=\frac{1}{m} .
\end{array}\]
Similarly, $\frac{1}{y_{3}}+\frac{1}{y_{4}}=\frac{1}{m}$.
Hence, $\frac{1}{y_{1}}+\frac{1}{y_{2}}=\frac{1}{y_{3}}+\frac{1}{y_{4}}
\Rightarrow \frac{1}{y_{1}}-\frac{1}{y_{3}}=\frac{1}{y_{4}}-\frac{1}{y_{2}}
\Rightarrow \frac{y_{3}-y_{1}}{y_{1} y_{3}}=\frac{y_{2}-y_{4}}{y_{2} y_{4}}
\Rightarrow\frac{AD}{BE}=\frac{\left|y_{3}-y_{1}\right|}{\left|y_{2}-y_{4}\right|} =\frac{\left|y_{1} y_{3}\right|}{\left|y_{2} y_{4}\right|} $.
By $y_{1} y_{2}=\frac{2 pm}{k}$ and $y_{3} y_{4}=\frac{4 pm}{k}$, $\Rightarrow \frac{y_{1} y_{3}}{y_{2} y_{4}}=\frac{y_{1} \cdot \frac{4 p m}{k y_{4}}}{\frac{2 p m}{k y_{1}} \cdot y_{4}}=\frac{2 y_{1}^{2}}{y_{4}^{2}}=6$.
Hence, $ \frac{A M}{M E}=\frac{\left|y_{1}\right|}{\left|y_{4}\right|}=\sqrt{3}$. | AIChallenge_Geo21 | \sqrt{3} |
Given that $\triangle ABC$ is an acute-angled triangle, with $A$, $B$, and $C$ as its internal angles, the minimum value of $2 \cot A+3 \cot B+4 \cot C$ is \_\_\_\_\_. | By the inequality $(2x-3y \cos \gamma-4z \cos \beta)^{2}+(3y \sin \gamma- 4z \sin \beta)^{2} \geqslant 0$,
we can rearrange it to obtain $(2 x+3 y+4 z)^{2} \geqslant 12(\cos \gamma+1) xy+24(1-\cos (\beta+\gamma)) y z+ \\
16(\cos \beta+1) zx$.
Let's assume the coefficients, and denote
$
12(\cos \gamma+1)=24(1-\cos (\beta+\gamma))=16(\cos \beta+1)=k .
$
This leads to the equation
$
\left(\frac{k}{12}-1\right)\left(\frac{k}{16}-1\right)-\sqrt{1-\left(\frac{k}{12}-1\right)^{2}} \sqrt{1-\left(\frac{k}{16}-1\right)^{2}} =1-\frac{k}{24} \Rightarrow k=23 $.
Therefore, $(2x+3y+4z)^{2} \geqslant 23(xy+yz+zx)$.
Also $\cot A \cdot \cot B+\cot B \cdot \cot C+\cot C \cdot \cot A=1 $,
thus, $(2 \cot A+3 \cot B+4 \cot C)^{2} \geqslant 23 $.
Therefore, the minimum value is $\sqrt{23}$. | AIChallenge_Geo22 | \sqrt{23} |
There are 10 points in the plane, with no three points lying on the same line. Using these 10 points as vertices of triangles, such that any two triangles have at most one common vertex, we can form at most \_\_\_\_\_ triangles. | Considering all triangles containing one of the points $A$, since there is at most one common point (which is $A$), the other two points of these triangles must be different. Therefore, there are $\left[\frac{10-1}{2}\right]=4$ triangles containing point $A$. Similarly, for each point, we can obtain 4 triangles. Hence, there are a total of $4 \times 10=40$ triangles. Since each triangle is counted 3 times, at most $\left[\frac{40}{3}\right]=13$ triangles can be formed. Here is a construction:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline & $\mathrm{A}$ & $\mathrm{B}$ & $\mathrm{C}$ & $\mathrm{D}$ & $\mathrm{E}$ & $\mathrm{F}$ & $\mathrm{G}$ & $\mathrm{H}$ & $\mathrm{I}$ & $\mathrm{J}$ \\
\hline First & $\sqrt{ }$ & $\sqrt{ }$ & $\sqrt{ }$ & & & & & & & \\
\hline Second & $\sqrt{ }$ & & & $\sqrt{ }$ & $\sqrt{ }$ & & & & & \\
\hline Third & $\sqrt{ }$ & & & & & $\sqrt{ }$ & & & & $\sqrt{ }$ \\
\hline Fourth & $\sqrt{ }$ & & & & & & & $\sqrt{ }$ & $\sqrt{ }$ & \\
\hline Fifth & & $\sqrt{ }$ & & & & & & $\sqrt{ }$ & & $\sqrt{ }$ \\
\hline Sixth & & $\sqrt{ }$ & & & $\sqrt{ }$ & & & & $\sqrt{ }$ & \\
\hline Seventh & & $\sqrt{ }$ & & $\sqrt{ }$ & & $\sqrt{ }$ & & & & \\
\hline Eighth & & & & & & $\sqrt{ }$ & $\sqrt{ }$ & $\sqrt{ }$ & & \\
\hline Nineth & & & $\sqrt{ }$ & & $\sqrt{ }$ & & & $\sqrt{ }$ & & \\
\hline Tenth & & & $\sqrt{ }$ & $\sqrt{ }$ & & & $\sqrt{ }$ & & & \\
\hline Eleventh & & & $\sqrt{ }$ & & & $\sqrt{ }$ & & & $\sqrt{ }$ & \\
\hline Twelfth & & & & $\sqrt{ }$ & & & & & $\sqrt{ }$ & $\sqrt{ }$ \\
\hline Thirteenth & & & & & $\sqrt{ }$ & & $\sqrt{ }$ & & & $\sqrt{ }$ \\
\hline
\end{tabular} | AIChallenge_Geo23 | 13 |
Given that quadrilateral $ABCD$ is a parallelogram, with the lengths of $AB$, $AC$, $AD$, and $BD$ being distinct integers, then the minimum perimeter of quadrilateral $ABCD$ is \_\_\_\_\_. | Let $\mathrm{AB} = \mathrm{a}$, $\mathrm{AD} = \mathrm{b}$, $\mathrm{AC} = \mathrm{p}$, $\mathrm{BD} = \mathrm{q}$. Obviously, $2\left(a^{2}+b^{2}\right)=p^{2}+q^{2}$. Without loss of generality, assume $\mathrm{a}>\mathrm{b}$ and $\mathrm{p}>\mathrm{q}$.
Let $r=a^{2}+b^{2}$.
If the equation $x^{2}+y^{2}=r(\mathrm{x}>\mathrm{y})$ has a unique non-negative integer solution $\left\{\begin{array}{l}x=a \\ y=b\end{array}\right.$, then $\mathrm{p}=\mathrm{a}+\mathrm{b}$ and $\mathrm{q}=\mathrm{a}-\mathrm{b}$, and in this case, a parallelogram cannot be formed.
Hence, if the equation $x^{2}+y^{2}=r(\mathrm{x}>\mathrm{y})$ has at least 2 sets of non-negative integer solutions, then $r$ is composite, and in its factorization, there are at least 2 primes of the form $4p+1$ multiplied together.
If $r=25=5 \times 5$, then $\mathrm{a}=4$, $\mathrm{~b}=3$, $\mathrm{p}=\mathrm{q}=5$, which contradicts the inequality of $AC$ and $BD$.
If $r=50=2 \times 5 \times 5$, then $\mathrm{a}=\mathrm{b}=5$, $\mathrm{p}=8$, $\mathrm{q}=6$, which contradicts the inequality of $AB$ and $AD$.
If $r=65=5 \times 13$, then $\mathrm{a}=7$, $\mathrm{~b}=4$, $\mathrm{p}=9$, $\mathrm{q}=7$, which contradicts the inequality of $AB$ and $BD$.
If $r=85=5 \times 17$, then $\mathrm{a}=7$, $\mathrm{~b}=6$, $\mathrm{p}=11$, $\mathrm{q}=7$, which contradicts the inequality of $AB$ and $BD$.
If $r=130=2 \times 5 \times 13$, then $a=9$, $b=7$, $p=14$, $q=8$, and in this case, $a+b=16$, so the perimeter is 32.
If $r=170=2 \times 5 \times 17$, then $a=11$, $b=7$, $p=14$, $q=12$, and in this case, $a+b=18$, so the perimeter is 36.
If $r=221=13 \times 17$, then $a=11$, $b=10$, $p=19$, $q=9$, and in this case, $a+b=21$, so the perimeter is 42.
For $r>226$, $a+b>16$, hence the minimum perimeter of $ABCD$ is 32. | AIChallenge_Geo24 | 32 |
When the two ends of a strip of paper are glued together, forming a loop, it is called a circular ring. Cutting along the bisector of the paper strip will result in two circular rings. When a strip of paper is twisted 180 degrees and then the ends are glued together again, forming a loop, it is called a Mobius strip. Cutting along the bisector of the Mobius strip will result in a longer loop-like structure. If cut along the bisector of this longer loop-like structure, it will yield \_\_\_\_\_ loop-like structure again. | Examine the topological structure.
As shown in the figure below, cutting along the quadrisection line of the Möbius strip is equivalent to cutting along the $\frac{1}{4}$-line. At this point, the second strip and the third strip are glued together to form a larger loop-like structure, while the first strip and the fourth strip are glued together to form the same loop-like structure. Therefore, there are a total of 2 loop-like structures. (It is also easy to obtain by cutting along the $\frac{1}{n}$-line of the Möbius strip using this method.) | AIChallenge_Geo25 | 2 |
In the Cartesian coordinate system xOy, let \(\Gamma_{1}\) be the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)\) and \(\Gamma_{2}\) be the parabola \(y^{2}=\frac{1}{2}ax\). They intersect at points A and B, and P is the rightmost point of \(\Gamma_{1}\). If points O, A, P, and B are concyclic, then the eccentricity of \(\Gamma_{1}\) is\_\_\_\_\_. | By symmetry, we know that $\angle O A P=\angle O B P=90^{\circ}$, so points A and B lie on the circle with OP as its diameter. From
\[
\left\{\begin{array}{l}
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, \\
x^{2}-a x+y^{2}=0
\end{array}\right. \]
$\Rightarrow \frac{c^{2}}{a^{2}} x^{2}-a x+b^{2}=0$ . Then $x_{A} x_{P}=\frac{a^{2} b^{2}}{c^{2}} \Rightarrow x_{A}=\frac{a b^{2}}{c^{2}}
\Rightarrow y_{A}^{2}=-x_{A}^{2}+a x_{A}=-\frac{a^{2} b^{4}}{c^{4}}+\frac{a^{2} b^{2}}{c^{2}}$.
Combining with $y_{A}^{2}=\frac{1}{2} a x_{A}$ gives
$-\frac{a^{2} b^{4}}{c^{4}}+\frac{a^{2} b^{2}}{c^{2}}=\frac{1}{2} \cdot \frac{a^{2} b^{2}}{c^{2}}
\Rightarrow c^{2}=2 b^{2} \Rightarrow 3 c^{2}=2 a^{2}$.
Therefore, the eccentricity $e$ of ellipse
$\Gamma_{1}$ is
$e=\frac{c}{a}=\frac{\sqrt{6}}{3}$. | AIChallenge_Geo26 | \frac{\sqrt{6}}{3} |
Given the circle $\Gamma: x^{2}+y^{2}=1$, points A and B are two points symmetric about the x-axis on the circle. M is any point on the circle $\Gamma$ distinct from A and B. If MA and MB intersect the x-axis at points P and Q respectively, then the product of the abscissas of P and Q is \_\_\_\_\_. | Let $A(m, n), B(m,-n), M\left(x_{0}, y_{0}\right)$.
Then
$l_{MA}: y-y_{0}=\frac{y_{0}-n}{x_{0}-m}\left(x-x_{0}\right)$ ,
$l_{MB}: y-y_{0}=\frac{y_{0}+n}{x_{0}-m}\left(x-x_{0}\right)$.
Setting y=0 yield
$x_{P}=x_{0}-\frac{y_{0}\left(x_{0}-m\right)}{y_{0}-n},
x_{Q}=x_{0}-\frac{y_{0}\left(x_{0}-m\right)}{y_{0}+n}$,
So $x_{P} x_{Q}$
\[
\begin{array}{l}
=x_{0}^{2}+\frac{y_{0}^{2}\left(x_{0}-m\right)^{2}-2x_{0} y_{0}^{2}\left(x_{0}-m\right)}{y_{0}^{2}-n^{2}} \\
=x_{0}^{2}-\frac{y_{0}^{2}\left(x_{0}^{2}-m^{2}\right)}{y_{0}^{2}-n^{2}} \\
=\frac{y_{0}^{2} m^{2}-x_{0}^{2} n^{2}}{y_{0}^{2}-n^{2}} .
\end{array}
\]
Substituting $x_{0}^{2}=1-y_{0}^{2}$, $m^{2}=1-n^{2} $ into the above expression, we get
$x_{P}x_{Q}=\frac{y_{0}^{2}\left(1-n^{2}\right)-\left(1-y_{0}^{2}\right) n^{2}}{y_{0}^{2}-n^{2}}=1 $. | AIChallenge_Geo27 | 1 |
If three points are randomly chosen from the vertices of a regular 17-sided polygon, what is the probability that the chosen points form an acute-angled triangle? | When selecting any three points, all forming triangles, there are a total of $C_17^3=680$ triangles, among which there are no right-angled triangles.
Classify obtuse-angled triangles based on the length of the longest side. The longest side corresponds to the diagonal of the regular 17-sided polygon, which has 7 different lengths (there are 1, 2, ..., 7 vertices between the two ends), with exactly 17 diagonals of each length. Thus, there are $17(1+2+...+7)=467$ obtuse-angled triangles. Therefore, the probability of forming an acute-angled triangle is $p=\frac{680-476}{680}=\frac{3}{10}$. | Combinary-1 | \frac{3}{10} |
A rook piece moves through each square of a $2023\times 2023 $ grid paper once, each time moving only one square (i.e., from the current square to an adjacent square). If the squares are numbered from 1 to $n^2$ according to the order in which the rook piece reaches them, let M denote the maximum difference in numbers between adjacent squares. Then the minimum possible value of $M$ is? | Firstly, it is explained that the minimum possible value of $M$ when operating on a $n \times n $ grid paper is $2n-1$.
Firstly, it is explained that $M$ can be equal to $2n-1$.
In fact, as long as the rook piece moves in a 'serpentine' manner on the chessboard: moving along the bottom row from the leftmost to the rightmost, then moving up one square, then moving along that row from the rightmost to the leftmost, then moving up one square, and so on.
Then it is proved that $M \geq 2n-1$.
By contradiction:
Assume $M < 2n-1$,
observe the numbers in the top row. Since the difference between any two adjacent numbers in this row is not greater than $2n-2$,
then when the rook piece moves from the smallest number in this row to the largest number, it cannot pass through the squares in the bottom row, because to reach the bottom row, it needs to take at least $n-1$ steps,
and to return, it needs another $n-1 $ steps, and it still needs to spend one step for horizontal movement. This indicates that the rook piece does not pass through the squares in the bottom row when traversing all the numbers in the top row. Similarly, when the rook piece traverses all the numbers in the bottom row, it does not pass through the squares in the top row. This indicates that all the numbers in the top row are either all greater than or all less than all the numbers in the bottom row. Similarly, all the numbers in the leftmost column are either all greater than or all less than all the numbers in the rightmost column.
Without loss of generality, assume that all numbers in the leftmost column are greater than those in the rightmost column, and all numbers in the bottom row are greater than those in the top row. Now observe the number $A$ at the top left corner and the number $B$ at the bottom right corner. On one hand, when viewed by column, we have $A > B$, at the bottom right corner. On one hand, when viewed by column, we have A<B. Contradiction.
Therefore, the answer to this problem is $2 \times 2023- 1 = 4045$. | Combinary-2 | 4045 |
Using the 24-hour clock, the probability of the sum of four digits at a certain moment being smaller than the sum of the four digits at 20:21 is \_\_\_\_\_. | Because the sum of the four digits at the time 20:21 is 5, the digits at other times satisfying the condition must sum up to 0, 1, 2, 3, or 4.
When the sum of the four digits is 0, there is 1 possibility.
When the sum is 1, there are 4 possibilities.
When the sum is 2, there are 4 + 6 = 10 possibilities.
When the sum is 3, there are 3 + 4 + 12 = 19 possibilities.
When the sum is 4, there are 1 + 3 + 9 + 6 + 12 = 31 possibilities.
Thus, there are a total of 65 possibilities satisfying the condition.
Therefore, the required probability is $\frac{65}{24\times 60}=\frac{13}{288}$. | Combinary-3 | \frac{13}{288} |
A six-digit number $N=\overline{a_1a_2...a_{6}}$ composed of non-repeating digits from 1 to 6 satisfies the condition $|a_{k+1}-a_{k}| \neq 1, (k\in \{1,2, \cdots, 5\})$. Then the number of such six-digit numbers is \_\_\_\_\_. | According to the problem, we know that $a_{k+1}$ and $a_k$ are not adjacent numbers, there are five cases where the digits are adjacent: (1,2), (2,3), (3,4), (4,5), and (5,6).
(1) If at least one pair is adjacent, there are $5A_2^2A_5^5=1200$ possibilities.
(2) If at least two pairs are adjacent, there are two cases:
\ (i) Three consecutive numbers are adjacent, resulting in $4\times 2A_4^4=192$ possibilities;
\ (ii) Two numbers are adjacent, but the two groups of numbers are not adjacent, resulting in $6\times 2^2 A_4^4=576$ possibilities.
Therefore, there are a total of 768 possibilities where at least two pairs are adjacent.
(3) If at least three pairs are adjacent, there are three cases:
\ (i) Four consecutive numbers are adjacent, resulting in $3\times 2A_3^3=36$ possibilities;
\ (ii) Three consecutive numbers and two adjacent numbers, but the two groups of numbers are not adjacent, resulting in $6\times 2\times 2A_3^3=144$ possibilities;
\ (iii) Each of the three pairs has only two adjacent numbers, resulting in $2^3A_3^3=48$ possibilities.
Therefore, there are a total of 228 possibilities where at least three pairs are adjacent.
(4) If at least four pairs are adjacent, there are two cases:
\ (i) Five consecutive numbers are adjacent, resulting in $2\times 2 A_2^2=8$ possibilities;
\ (ii) Four consecutive numbers and two adjacent numbers, but the two groups of numbers are not adjacent, or both groups consist of three consecutive numbers but are not adjacent, resulting in $3 \time 2^2A_2^2=24$ possibilities.
Therefore, there are a total of 32 possibilities where at least four pairs are adjacent.
(5) If all five pairs are adjacent, there are 2 possibilities.
By the principle of inclusion-exclusion, we know that the total number of permutations satisfying the condition is \(6! - 1200 + 768 - 228 + 32 - 2 = 90\). | Combinary-4 | 90 |
Given $M=\{1,2, \ldots, 8\}, A, B$ are two distincet subsets of set $M$, satisfying
(1) The number of elements in set \(A\) is fewer than the number of elements in set \(B\).
(2) The smallest element in set \(A\) is larger than the largest element in set \(B\).
Then the total number of ordered pairs \((A, B)\) satisfying these conditions is \_\_\_\_\_. | Based on the elements of \(A \cup B\), we can discuss:
If \(|A \cup B|=3\), then there is only one way to split these three elements, giving \(C_8^3 - 1 = 56\) possibilities.
If \(|A \cup B|=4\), then there is only one way to split these four elements, giving \(C_8^4 = 70\) possibilities.
If \(|A \cup B|=5\), then there are two ways to split these five elements, giving \(C_8^5 \times 2 = 112\) possibilities.
If \(|A \cup B|=6\), then there are two ways to split these six elements, giving \(C_8^6 \times 2 = 56\) possibilities.
If \(|A \cup B|=7\), then there are three ways to split these seven elements, giving \(C_8^7 \times 3 = 24\) possibilities.
If \(|A \cup B|=8\), then there are three ways to split these eight elements, giving \(C_8^8 \times 3 = 3\) possibilities.
In summary, the number of ordered pairs satisfying the conditions is \(56 + 70 + 112 + 56 + 24 + 3 = 321\). | Combinary-5 | 321 |
Using six different colors to color each edge of the regular tetrahedron \(ABCD\), each edge can only be colored with one color and edges sharing a vertex cannot have the same color. The probability that all edges have different colors is \_\_\_\_\_. | Classify according to whether the opposite edges of the tetrahedron have the same color.
If all three pairs of opposite edges have the same color, meaning three colors are used, then there are \(A_6^3\) different coloring schemes.
If only two pairs of opposite edges have the same color, meaning four colors are used, then there are \(C_3^2 A_6^4\) different coloring schemes.
If only one pair of opposite edges have the same color, meaning five colors are used, then there are \(C_3^1 A_6^5\) different coloring schemes.
If all edges have different colors, meaning six colors are used, then there are \(A_6^6\) different coloring schemes.
Therefore, the probability that all edges have different colors is \(\frac{A_6^6}{A_6^3 + C_3^2 A_6^4 + C_3^1 A_6^5 + A_6^6} = \frac{3}{17}\). | Combinary-6 | \frac{3}{17} |
The king summons two wizards into the palace. He demands Wizard A to write down 100 positive real numbers on cards (allowing duplicates) without revealing them to Wizard B. Then, Wizard B must accurately write down all of these 100 positive real numbers. Otherwise, both wizards will be beheaded. Wizard A is allowed to provide a sequence of numbers to Wizard B, where each number is either one of the 100 positive real numbers or a sum of some of them. However, he cannot tell Wizard B which are the numbers on the cards and which are the sums of numbers on the cards. Ultimately, the king decides to pull off the same number of beards from both wizards based on the count of these numbers. Without the ability to communicate beforehand, the question is: How many beard pulls does each wizard need to endure at least to ensure their own survival? | If only 100 hint numbers are given, it's impossible to distinguish whether all 100 numbers are on the cards or there are 99 numbers on the cards, and the largest hint number is the sum of the number on the 100th card and another number on the card. Therefore, at least 101 hint numbers are needed. For the 101 hint numbers, we can write down powers of 2 from $2^1$ to $2^{100}$ on the cards, and give hints for these 100 numbers as well as their sum. In this way, by using the number 2, we can determine that there must be a number on one of the cards that is not greater than 2. Then, using the number 4, we can determine that there must be another number on one of the cards that is not greater than 4, and so on. This process allows us to sequentially determine that the numbers on the 100 cards are not greater than 2 raised to the power of 1 to 100, and then we can infer the specific values of these 100 numbers based on their sum. | Combinary-7 | 101 |
Using $1 \times 1$, $2 \times 2$, and $3 \times 3$ tiles to cover a $23 \times 23$ floor (without overlapping or leaving gaps), what is the minimum number of $1 \times 1$ tiles needed? (Assuming each tile cannot be divided into smaller tiles). | Paint the $3k+1$ columns black, where $k=0,1,2, \ldots, 7$, and the rest white. Then, there are an odd number of white squares, but both $2 \times 2$ and $3 \times 3$ cover an even number of white squares. Therefore, at least one $1 \times 1$ tile is needed. Use $2 \times 2$ and $3 \times 3$ tiles to first form $2 \times 6$ and $3 \times 6$ sections, which can then be combined to form an $11 \times 12$ grid with $4$ copies of $11 \times 12$ and $1$ $1 \times 1$ tile, completing the construction of a $23 \times 23$ grid. | Combinary-8 | 1 |
In a number-guessing game, the host has prearranged a permutation of the numbers 1 to 100, and participants also need to provide a permutation of these 100 numbers. Interestingly, as long as the permutation provided by a participant has at least one number whose position matches that of the host's permutation, it is considered a successful guess. How many participants are needed to ensure that at least one person guesses correctly? | The first person guesses that the first 51 numbers are from 1 to 51, the second person guesses that the first 51 numbers are $2, 3, 4, \ldots, 51, 1$, and so on in a rotating fashion. If all 51 people fail to guess correctly, it implies that the numbers from 1 to 51 are all in the last 49 positions, which is clearly contradictory. For 50 people, the permutation arranged by the host can be constructed as follows: fill in numbers from 1 to 100 in the positions 1 to 100 successively, ensuring that each number differs from the guesses of the 50 participants (at least 50 different numbers). If, for any position $\mathrm{k}$, it's not possible to do so, it implies that at least 50 numbers that could have been placed in position $\mathrm{k}$ have already been placed elsewhere. Since any number not yet placed can be placed in at most 50 different positions (including position $\mathrm{k}$), there will always be a number that can replace the one initially placed at position $\mathrm{k}$. This process ensures that all 50 participants fail to guess correctly. | Combinary-9 | 51 |
There is a stack of 52 face-down playing cards on the table. Mim takes 7 cards from the top of this stack, flips them over, and puts them back at the bottom, calling it one operation. The question is: how many operations are needed at least to make all the playing cards face-down again? | Starting from the top of this stack of playing cards, color the 52 cards as follows: the top three blue, the next four red, the next three blue, the next four red, and so on until all cards are colored. Note that 52 mod 7 equals 3, meaning that each of Mim's operations does not change the distribution of colors in this stack of playing cards. If we only consider the blue cards, there are a total of eight groups of blue cards (each group consisting of three cards). After one operation, the top group of blue cards is flipped over and moved to the bottom. Therefore, flipping all the blue cards requires 8 operations, and flipping them back requires a total of 16 operations. Similarly, there are seven groups of red cards, and flipping all the red cards requires 14 operations. Thus, it takes $[14, 16] = 112$ operations to make all the playing cards face-down again. | Combinary-10 | 112 |
There are two segments of length $3n (0 \leq n \leq 1011)$. How many different shapes of triangles can be formed from these 2024 segments? (Congruent triangles are considered the same.) | From the triangle inequality, it's known that the triangle must be isosceles, and the length of the legs is not shorter than the length of the base. Then, classifying by the length of the legs, there are $1 + 2 + \ldots + 1011 = 511566$ types of triangles. | Combinary-11 | 511566 |
Let $A$ and $B$ be two subsets of the set $\{1,2, \ldots, 20\}$, where $A \cap B = \varnothing$, and if $n \in A$, then $2n + 2 \in B$. Let $M(A)$ denote the sum of the elements of $A$. The maximum value of $M(A)$ is: | From 2n+2 \leq 20 \Rightarrow n \leq 9.
According to the pigeonhole principle, $A$ can have at most 6 elements. When $A=\{9,8,7,6,5,4\}$, we get the maximum value of $M(A)$ as 39. | Combinary-12 | 39 |
Alice and Bob are playing a game. They write down four expressions on four cards: $x+y$, $x-y$, $x^2+xy+y^2$, and $x^2-xy+y^2$. They place these four cards face down on the table, then randomly choose one card to reveal its expression. Alice can pick two of the four cards and hand the other two to Bob, then all four cards are revealed. Alice can assign a value (real number) to one of the variables $x$ or $y$ and inform Bob of which variable she has assigned and what value. Afterwards, Bob assigns a value (real number) to the other variable. Finally, they each calculate the product of the values on their two cards, and the person with the larger product wins. Who has a winning strategy?
A. Alice
B. Bob
C. Neither of them has a winning strategy. | Alice has a winning strategy.
Firstly, let $A$ and $B$ represent the products of the two cards in Alice's and Bob's hands respectively.
If $x-y$ or $x+y$ is revealed, then Alice chooses any two hidden cards. Otherwise, she picks one hidden card and one revealed card, ensuring she doesn't get both $x-y$ and $x+y$.
If Bob gets both $x-y$ and $x+y$, then Alice can choose $y=1$. In this case,
$A = (x^2-xy+y^2)(x^2+xy+y^2) = x^4+x^2+1$,
$B = (x-y)(x+y) = x^2-y^2 = x^2-1$,
thus $A-B = x^2+2 > 0$, and Alice wins.
If $B = (x-y)(x^2+xy+y^2) = x^3-y^3$,
$A = (x+y)(x^2-xy+y^2) = x^3+y^3$,
then Alice chooses $y>0$.
If $A = x^3-y^3, B = x^3+y^3$, then Alice chooses $y<0$.
If $A = (x-y)(x^2-xy+y^2)$,
$B = (x+y)(x^2+xy+y^2)$,
then $A-B = -4x^2y-2y^3 = -2y(y^2+2x^2)$, in this case Alice chooses $y<0$.
If $A = (x+y)(x^2+xy+y^2)$,
$B = (x-y)(x^2-xy+y^2)$, then $A-B = 4x^2y+2y^3 = 2y(y^2+2x^2)$, and Alice chooses $y>0$.
In conclusion, Alice has a winning strategy. | Combinary-13 | A |
Find the smallest integer $k > 2$ such that any partition of $\{2,3,\ldots,k\}$ into two sets must contain at least one set containing $a$, $b$, and $c$ (which are allowed to be the same), satisfying $ab=c$. | Firstly, we provide a counterexample for $k=31$.
Let $A=\{2,3,16,17,\ldots,31\}$ and $B=\{4,5,\ldots,15\}$.
If $k<31$, then simply remove integers greater than $k$ from sets $A$ and $B$ respectively.
Next, we prove that the conclusion holds when $k=32$.
Conversely, if the conclusion does not hold, then $2$ and $4$, and $4$ and $16$ are not in the same set respectively. Thus, $2$ and $16$ are in the same set. Similarly, $4$ and $8$ are in the same set. In this case, $32$ cannot be in any set $(32=2\times16=4\times8)$.
Therefore, the minimum value of $k$ is $32$. | Combinary-14 | 32 |
On a plane, there are 2019 points. Drawing circles passing through these points, a drawing method is called "$k$-good" if and only if drawing $k$ circles divides the plane into several closed shapes, and there is no closed shape containing two points. Then, find the minimum value of $k$ , it ensures that no matter how these 2019 points are arranged, there always exists a $k$-good drawing method. | First, we prove that $k \geqslant 1010$.
Take 2019 points on the same line. Obviously, among these 2019 points on the line, at least 2018 of them need to be passed through by a circle.
Furthermore, since each circle can pass through the points on the line at most twice, if $k \leqslant 1009$, then the number of points passed through by these $k$ circles does not exceed 2018. Note that when equality holds, the first and last points are not contained within any circle, leading to a contradiction.
Next, we strengthen the proof that no matter how these 2019 points are arranged, there always exists a $1010$-good drawing method.
The strengthened proposition is: "Prove: there exists a $1010$-good drawing method such that these $1010$ circles pass through the same point."
Perform an inversion transformation with a point on the plane not coinciding with any of the 2019 points and not lying on the circumcircle of any three points as the center of inversion. Then, after inversion, to prove the strengthened proposition, it suffices to prove that for any 2019 non-collinear points on the plane, there exist 1010 lines such that each pair of points has at least one line passing through them.
A line is said to "bisect" a set of points if the difference between the number of points on each side of the line is either 1 or equal.
Lemma:
For any two sets of points $A$ and $B$, there exists a line $l$ that bisects $A$ and $B$.
Proof: The portion of a plane divided by a line is called a half-plane. Obviously, there exists a half-plane containing half of the points in both $A$ and $B$, i.e., $2 \times$ the number of points in $A$ in the half-plane $-|A| \leqslant 1$.
It is evident that there exists a half-plane containing fewer than half of the elements in $B$ and half of the elements in $A$; there exists a half-plane containing more than half of the elements in $B$ and half of the elements in $A$; there exists a way to rotate and translate the half-plane so that it can be transformed from any half-plane containing half of the elements in $A$ to another, and the route of translation and rotation through half-planes. Therefore, by the intermediate value theorem, the proposition is proved.
A region including a given point is called a figure formed by closed lines, and the modulus of a region is the number of points in the region. Start by connecting a sufficiently large polygon enclosing all the points. Obviously, there is only one region at the beginning.
Consider drawing 1010 lines in order. Consider making each line bisect the two regions with the largest moduli at the moment, defining this as one operation. Next, we prove that this operation can be performed 1010 times.
The first operation generates a new region, and thereafter each operation generates two new regions. By induction, it is easy to see that the modulus of the region with the maximum modulus is no more than twice the modulus of the region with the minimum modulus, so if there is a region with a modulus of 1, then the moduli of the other regions must be either 1 or 2.
Therefore, when the last operation stops, there is at most one region with a modulus of 2 left. Due to parity considerations, there are 2018 regions at this point, and 1009 lines have been drawn. Drawing one more line to separate the two points left in the same region is enough. Thus, it is proved that it is possible to sequentially draw 1010 closed shapes such that no closed shape contains two points. Thus, the proposition is proved. | Combinary-15 | 1010 |
Zheng flips an unfair coin 5 times. If the probability of getting exactly 1 head is equal to the probability of getting exactly 2 heads and is nonzero, then the probability of getting exactly 3 heads is \_\_\_\_. | Let the probability of getting a head be $p$.
According to the given conditions, we have $C_5^1 p(1-p)^4 = C_5^2 p^2(1-p)^3$.
Solving this equation, we find $p = \frac{1}{3}$.
Therefore, the probability of getting exactly 3 heads is $C_5^3 p^3(1-p)^2 = \frac{40}{243}$. | Combinary-16 | \frac{40}{243} |
Let $a_{1}, a_{2}, \ldots, a_{6}$ be any permutation of $\{1,2, \ldots, 6\}$. If the sum of any three consecutive numbers cannot be divided by 3, then the number of such permutations is \_\_\_\_\_. | Taking the numbers in the set modulo 3, we get two 0s, two 1s, and two 2s. Thus, permutations satisfying the condition that the sum of any three consecutive numbers cannot be divided by 3 are $002211$, $001122$, $112200$, $110022$, $220011$, $221100$, $011220$, $022110$, $100221$, $122001$, $211002$, and $200112$.
Therefore, the number of permutations satisfying the condition is $12 \times A_2^2 \times A_2^2 \times A_2^2 = 96$. | Combinary-17 | 96 |
Given an integer $n > 2$. Now, there are $n$ people playing a game of "Passing Numbers". It is known that some people are friends (friendship is mutual), and each person has at least one friend. The rules of the game are as follows: each person first writes down a positive real number, and the $n$ positive real numbers written by everyone are all different; then, for each person, if he has $k$ friends, he divides the number he wrote by $k$, and tells all his friends the result obtained; finally, each person writes down the sum of all the numbers he hears. The question is: what is the minimum number of times that someone writes down different numbers? | At least 2 people write down different numbers twice.
Let the $n$ people be denoted as $A_{1}, A_{2}, \ldots, A_{n}$.
Consider the following friendship relationship: $A_{i}$ is a friend of $A_{j}$ if and only if $|i-j|=1 (1 \leq i, j \leq n)$.
Under this friendship relationship, $A_{1}$ writes down the number $1$, $A_{i}$ writes down the number $i+1 (2 \leq i \leq n-1)$, and $A_{n}$ writes down the number $\frac{n+1}{2}$. Then after one pass,
the numbers written down by $A_{2}, A_{3}, \ldots, A_{n-1}$ remain unchanged. However, the number written down by $A_{1}$ changes from $1$ to $\frac{3}{2}$, and the number written down by $A_{n}$ changes from $\frac{n+1}{2}$ to $\frac{n}{2}$.
Hence, at this point, there are 2 people who write down different numbers twice.
Next, we prove that: at least 1 person writes down a smaller number the second time.
For each person, define their value as the ratio of the number they wrote down the first time to the number of their friends. Let the value of the person with the highest value among all be $M$.
Suppose person $B$ has a value of $M$, and $B$ has $k$ friends. Then the number written down by $B$ the first time is $kM$, and according to the rules of the game, the number written down by $B$ the second time is at most $kM$. If the number written down by $B$ the second time is smaller than $kM$, then the conclusion holds; if the number written down by $B$ the second time is $kM$, then the value of each of his friends is also $M$.
Construct a graph $G$, where the vertices are the $n$ people, and there is an edge between two vertices if and only if the two people are friends. Consider the connected component $G'$ where $B$ is located.
From the above discussion, if the conclusion does not hold, then every person represented by a vertex in $G'$ has a value of $M$. Since the degree of each vertex in $G'$ is at least $1$ and at most $|V|-1$ (where $|V|$ is the number of vertices in $G'$), by the pigeonhole principle, there must be two vertices with the same degree, which means the two people represented by these vertices wrote down the same number the first time, contradicting the condition. Thus, the conclusion holds.
Note that the sum of the numbers written down by all people the first time is equal to the sum of the numbers written down by all people the second time. From the conclusion, we know that at least one person writes down a smaller number the second time, thus, there must be another person who writes down a larger number the second time, so at least 2 people write down different numbers twice. | Combinary-18 | 2 |
A restaurant can offer 9 types of appetizers, 9 types of main courses, 9 types of desserts, and 9 types of wines. A company is having a dinner party at this restaurant, and each guest can choose one appetizer, one main course, one dessert, and one type of wine. It is known that any two people's choices of the four dishes are not completely identical, and it is also impossible to find four people on the spot who have three identical choices but differ pairwise in the fourth choice (for example, there are no 9 people who have the same appetizer, main course, and dessert, but differ pairwise in the wine). Then, at most how many guests can there be? | Consider the general case, replacing 9 with $n$.
Consider unordered triples $(x, y, z)$, where $x, y, z$ belong to three different categories of dishes. There are $\binom{4}{3} \times n \times n \times n = 4n^3$ such triples.
Because no $n$ guests can simultaneously choose three identical dishes, for any triple $(x, y, z)$, it can belong to at most $n-1$ guests.
Moreover, each guest chooses one dish from each of the four categories, so each guest has 4 triples.
Let there be $x$ guests.
Then $4x \leq 4n^3(n-1) \Rightarrow x \leq n^3(n-1)$.
Next, we prove that when there are $n^3(n-1)$ guests, the conditions of the problem can be satisfied.
We consider four categories of dishes as four sets $A, B, C, D$, and each set contains $n$ dishes labeled as $1, 2, \ldots, n$.
We will remove the following selections: the sum of the four numbers chosen is exactly divisible by $n$. There are $n^3$ such selections to remove. The remaining number of selections is $n^4 - n^3$, which corresponds to $n^4 - n^3$ guests, each with one selection.
According to the selection rules, we know that no two people choose the same four dishes. (1)
If a particular triple appears more than $n-1$ times, select $n$ guests containing this triple. According to conclusion (1), there exists a category of dishes such that the $n$ dishes chosen by these guests from this category are all different.
Because each category of dishes has only $n$ dishes, these $n$ guests cover all dishes in this category.
Since $1, 2, \ldots, n$ form a complete residue system modulo $n$, there must be an integer $i (1 \leq i \leq n)$ such that $n \equiv x + y + z + i \pmod{n}$.
However, such selections have been removed, which is a contradiction.
Therefore, the selections of these $n^4 - n^3$ dishes satisfy the conditions of the problem.
Thus, the maximum possible number of guests is $n^4 - n^3$.
In this case, when $n = 9$, the maximum number of guests is $9^4 - 9^3 = 5832$. | Combinary-19 | 5832 |
Alice and Bob are playing hide-and-seek. Initially, Bob selects a point $B$ inside a unit square (without informing Alice). Then, Alice sequentially selects points $P_0, P_1, \ldots, P_n$ on the plane. After each selection of a point $P_k (1 \leq k \leq n$, and at this point, Alice has not yet chosen the next point), Bob informs Alice which of the points $P_k$ and $P_{k-1}$ is closer to point $B$. After Alice selects $P_n$ and receives Bob's response, she chooses a final point $A$. If the distance between $A$ and $B$ does not exceed $\frac{1}{2020}$, Alice wins. Otherwise, Bob wins. When $n=18$, which of the following options is correct?
A. Alice cannot guarantee victory.
B. Alice can guarantee victory. | It suffices to prove that even under optimal conditions for each step, Alice still cannot guarantee victory.
Note that if we draw the perpendicular bisector of $P_k P_{k-1}$ for each $k (1 \leq k \leq n)$, Alice can determine on which side of the perpendicular bisector point $B$ lies. Consequently, starting from the selection of $P_1$, after each point is chosen, Alice can narrow down the range where point $B$ is guaranteed to be by at most $\frac{1}{2}$. Thus, after selecting $n$ points, Alice can ensure that point $B$ lies within an area of no less than $\frac{1}{2^n}$.
If Alice can guarantee victory, she should be able to cover an area of no less than $\frac{1}{2^n}$ with a disk of radius $\frac{1}{2020}$.
Therefore, we must have $\pi \left(\frac{1}{2020^2}\right) < 4 \times \frac{1}{2^{20}} = \frac{1}{2^{18}} = \frac{1}{2^n}$ when $n=18$.
Thus, inequality (1) does not hold.
Hence, Alice cannot guarantee victory. | Combinary-20 | A |
Anna, Carl take turns selecting numbers from the set $\{1,2, \cdots, p-1\}$ (where $p$ is a prime greater than 3). Anna goes first, and each number can only be selected once. Each number chosen by Anna is multiplied by the number Carl selects next. Carl wins if, after any round, the sum of all products computed so far is divisible by $p$. Anna wins if, after all numbers are chosen, Carl has not won. Which of the following options is correct?
A. Anna has a winning strategy.
B. Carl has a winning strategy.
C. Both players have no winning strategy. | Carl has a winning strategy.
Carl's winning strategy is to choose the number $p - a$ whenever Anna selects the number $a$ in a round.
Next, we prove that Carl's strategy guarantees his victory.
Conversely, if Carl's selection strategy does not lead to victory, specifically, if he still has not won after all numbers are chosen, then it implies that $p$ does not divide $1(p-1)+2(p-2)+\cdots+\frac{p-1}{2}\left(p-\frac{p-1}{2}\right)$.
Let $S$ denote:
\[S = 1(p-1)+2(p-2)+\cdots+\frac{p-1}{2}\left(p-\frac{p-1}{2}\right)\]
\[= p\left(1+2+\cdots+\frac{p-1}{2}\right)-\left(1^{2}+2^{2}+\cdots+\left(\frac{p-1}{2}\right)^{2}\right)\]
\[= p\left(1+2+\cdots+\frac{p-1}{2}\right)-\frac{1}{6} \cdot \frac{p-1}{2} \cdot \frac{p+1}{2} p\]
Thus, $p$ does not divide $S$ if and only if $24 \mid (p^{2}-1) \cdot (2)$.
Since $p$ is an odd prime number,
\[8 \mid (p^{2}-1) \Rightarrow 8 \mid (p-1)(p+1)\]
Combining equations (1) and (2), we have
\[3 \nmid (p^{2}-1) \Rightarrow 3 \nmid (p-1)(p+1)\]
Thus, $3 \mid p$, which means $p=3$, contradicting the conditions.
Therefore, Carl's strategy guarantees his victory. | Combinary-21 | B |
Xiao Ming is playing a coin game with three doors. Each time he opens a door, it costs him 2 coins. After opening the first door, he can see the second door. Upon opening the second door, two equally likely options appear: either return to the outside of the first door or proceed to the third door. Upon opening the third door, three equally likely options appear: either return to the outside of the first door, stay in place and need to reopen the third door, or pass the game. If Xiao Ming wants to pass the game, on average, he needs to spend how many coins? | List all possible paths:
(1) First door $\rightarrow$ Second door $\rightarrow$ First door, using 4 coins, with a probability of $\frac{1}{2}$;
(2) First door $\rightarrow$ Second door $\rightarrow$ Third door $\rightarrow$ First door, using 6 coins, with a probability of $\frac{1}{6}$;
(3) First door $\rightarrow$ Second door $\rightarrow$ Third door $\rightarrow$ Third door, using 6 coins, with a probability of $\frac{1}{6}$;
(4) First door $\rightarrow$ Second door $\rightarrow$ Third door $\rightarrow$ Pass, using 6 coins, with a probability of $\frac{1}{6}$.
Let $E_{1}$ be the average number of coins needed to pass from the first door, and $E_{2}$ be the average number of coins needed to pass from the third door. According to the problem:
$E_{1}=\frac{1}{2}\left(4+E_{1}\right)+\frac{1}{2}\left(4+E_{2}\right)$
$E_{2}=\frac{1}{3}\left(2+E_{1}\right)+\frac{1}{3}\left(2+E_{2}\right)+\frac{1}{3} \times 2$.
Solving yields $E_{1}=22, E_{2}=14$.
Thus, on average, it takes 22 coins to pass from the first door. | Combinary-23 | 22 |
The school offers 10 elective courses, and each student can enroll in any number of courses. The director selects $k$ students, where although each student's combination of courses is different, any two students have at least one course in common. At this point, it is found that any student outside these $k$ students cannot be classmates with these $k$ students regardless of how they enroll (having one course in common is enough to be classmates). Then $k=$ \_\_\_\_\_. | Let $S$ be a set with ten elements. According to the problem, $A_{1}, A_{2}, \cdots, A_{k}$ are subsets of $S$, each pairwise intersecting non-empty and mutually distinct. Any other subset of $S$ cannot intersect all of $A_{1}, A_{2}, \cdots, A_{k}$.
First, note that there are $2^{10}$ subsets of $S$, and they can be paired up to form $2^{9}$ pairs of complements. Thus, $k \leq 2^{9}$.
Secondly, if $k<2^{9}$, then besides $A_{1}, A_{2}, \cdots, A_{k}$, all the other subsets must contain a pair of complementary subsets, denoted as $C$ and $D$. Hence, there also exist $A_{i} \cap C=\varnothing$ and $A_{j} \cap D=\varnothing$.
Since $C$ and $D$ are complementary, it follows that $A_{i} \subset D$ and $A_{j} \subset C$. Thus, $A_{i} \cap A_{j}=\varnothing$, which is a contradiction.
Therefore, $k=2^{9}$ or 512. | Combinary-24 | 512 |
Let $n$ be a positive integer. Now, a frog starts jumping from the origin of the number line and makes $2^{n}-1$ jumps. The process satisfies the following conditions:
(1) The frog will jump to each point in the set $\left\{1,2,3, \cdots, 2^{n}-1\right\}$ exactly once, without missing any.
(2) Each time the frog jumps, it can choose a step length from the set $\left\{2^{0}, 2^{1}, 2^{2}, \cdots\right\}$, and it can jump either left or right.
Let $T$ be the reciprocal sum of the step lengths of the frog. When $n=2024$, the minimum value of $T$ is \_\_\_\_\_. | For a positive integer $n$, we prove that the minimum value of $T$ is $n$.
First, we provide an estimation.
Initially, we notice that the frog's jump length must be one of the terms in the set $\left\{2^{0}, 2^{1}, \cdots, 2^{n-1}\right\}$; otherwise, it would jump out of bounds.
Let $a_{i}$ ($0 \leq i \leq n-1$) denote the number of times the frog jumps $2^{i}$ steps. According to the conditions, we have $a_{0}+a_{1}+\cdots+a_{n-1}=2^{n}-1$.
Lemma: For any $k=1,2, \cdots, n$, we have $a_{n-1}+\cdots+a_{n-k} \leq 2^{n}-2^{n-k}$.
To prove this, let $m=n-k$ and consider the jump lengths modulo $2^{m}$. We categorize jumps less than $2^{m-1}$ as "small jumps" and those greater than or equal to $2^{m}$ as "big jumps". It is notable that small jumps change the residue class modulo $2^{m}$, while big jumps do not.
For each residue class modulo $2^{m}$, the frog can make at most $\frac{2^{n}}{2^{m}}-1$ big jumps. Therefore, the number of big jumps is at most $2^{m}\left(\frac{2^{n}}{2^{m}}-1\right)=2^{n}-2^{m}$.
The lemma is proved.
For example, when $n=3$: for $k=1$, $a_{2} \leq 2^{3}-2^{2}$ implies that there can be at most four jumps of length 4; for $k=2$, $a_{2}+a_{1} \leq 2^{3}-2^{1}$ implies that there can be at most six jumps of length 2 or 4; for $k=3$, $a_{2}+a_{1}+a_{0} \leq 2^{3}-2^{0}$ implies a total of at most 7 jumps. To achieve the maximum $S$, it is observed that when $a_{2}=4, a_{1}=2, a_{0}=1$, the maximum is attained. Therefore, it is speculated that the maximum value is attained when $a_{m}=2^{m}$.
Let $A=a_{0}+a_{1}+\cdots+a_{n-1}=2^{n}-1$, and $T=a_{0}+\frac{a_{1}}{2}+\cdots+\frac{a_{n-1}}{2^{n-1}}$. Then, according to the lemma, we have:
$A-T \leq 2^{n}-n-1$
$\Rightarrow T \geq n$
Equality holds when $a^{m}=2^{m}$.
Next, we use induction to construct two types of paths such that the frog jumps to $\left\{0,1, \cdots, 2^{n}-1\right\}$ exactly once, stopping at $x$ where $x \in\{1,2 n-1\}$.
When $n=2$, there are two paths: $\{0,2,1,3\}$ and $\{0,2,3,1\}$.
Assume the claim holds for $n$, we prove it for $n+1$.
(i) By the induction hypothesis, there is a path from 0 to 2: $\left\{0,2,4, \cdots, 2^{n+1}-2\right\}$.
(ii) Similarly, there is a path from 1 to $2^{n+1}-1$: $\left\{1,3,5, \cdots, 2^{n+1}-1\right\}$.
(iii) Connecting these two paths requires one step from 2 to 1.
Therefore, we can use the path $0 \rightarrow\left(2^{n+1}-2\right) \rightarrow\left(2^{n+1}-1\right) \rightarrow 1$.
Hence, the minimum value of $T$ is $n$.
In this question, when $n=2024$, the answer is 2024. | Combinary-25 | 2024 |
Given that there are 66 dwarves with a total of 111 hats, each hat belonging to a specific dwarf, and each hat is dyed in one of 66 colors. During the holiday, each dwarf wears his own hat. It is known that for any holiday, the colors of the hats worn by all dwarves are different. For any two holidays, there is at least one dwarf who wears hats of different colors on the two occasions. The question is: how many holidays can the dwarves celebrate at most? | The problem examines the maximum number of different perfect matches for a bipartite graph with two sets of the same number of vertices under the condition of a specified number of edges, and clarifies the basic structure of the graph when obtaining the maximum value.
The problem structure is described using a graph.
Let \(V=\{v_{1}, v_{2}, \cdots, v_{66}\}\) be the set of 66 dwarves, and \(U=\{u_{1}, u_{2}, \cdots, u_{66}\}\) be the set of 66 colors.
If dwarf \(v_{i}\) has a hat of color \(u_{j}\), then draw an edge between \(v_{i}\) and \(u_{j}\), and let \(E\) be the set of all such edges, obtaining the graph \(G(V, U ; E)\). Each way of wearing hats that satisfies the requirements for every holiday is a perfect match of \(G\). The problem is to find the maximum number \(f(G)\) of different perfect matches for graph \(G\).
The lemma states that for a bipartite graph \(G\left(V_{1}, V_{2} ; E\right)\) with \(\left|V_{1}\right|=\left|V_{2}\right|\), it holds that \(f(G) \leq 2^{\left[\frac{|E|-\left|E_{1}\right|}{2}\right]}\), where \([x]\) represents the largest integer not exceeding the real number \(x\).
Proof is by mathematical induction.
It is easy to prove that when \(\left|V_{1}\right|=\left|V_{2}\right|=1\), the conclusions hold true.
Suppose the conclusions hold when \(\left|V_{1}\right|=\left|V_{2}\right|=n-1(n \geq 2)\). Consider the case when \(\left|V_{1}\right|=\left|V_{2}\right|=n\).
Consider the bipartite graph \(G\left(V_{1}, V_{2} ; E\right)\).
If \(G\) has isolated vertices, then \(f(G)=0\), and the conclusions hold true. Otherwise, let \(a \in V_{1}\) be the vertex in \(G\) with the smallest degree \(N(N \geq 1)\), and let \(b_{1}, b_{2}, \cdots, b_{N}\) be the neighbors of \(a\). Let \(G_{i}=\left(U_{i}, U_{i}^{\prime} ; E_{i}\right)\) denote the bipartite graph obtained by deleting vertex \(a\) and vertices \(b_{i}\), as well as the edges incident to \(a\) or \(b_{i}\).
Then \(\left|U_{i}\right|=\left|U_{i}^{\prime}\right|=n-1\), and \(\left|E_{i}\right| \leq|E|-(2 N-1)\).
By the induction hypothesis, it follows that \(f\left(G_{i}\right) \leq 2^{\left[\frac{\left|E_{i}\right|-\left|U_{i}\right|}{2}\right]} \leq 2^{\left[\frac{|E|-(2 N-1)-(n-1)}{2}\right]}=2^{\left[\frac{|E|-n}{2}\right]} \frac{1}{2^{N-1}}\).
Therefore, \(f(G)=f\left(G_{1}\right)+f\left(G_{2}\right)+\cdots+f\left(G_{N}\right) \leq \frac{N}{2^{N-1}} 2^{\left[\frac{|E|-n}{2}\right]}\).
By Bernoulli's inequality, \(2^{N-1} \geq 1+(N-1)=N\). Thus, \(f(G) \leq \frac{N}{2^{N}} 2^{\left[\frac{|E|-n}{2}\right]} \leq 2^{\left[\frac{|E|-n}{2}\right]}\).
The lemma is proved.
In particular, when \(n=66,|E| \leq 111\),
\(f(G) \leq 2^{\left[\frac{111-66}{2}\right]}=2^{22}\).
On the other hand, let graph \(G\) be composed of 22 complete bipartite graphs \(K_{2,2}\) and one bipartite graph \(K_{1,1}\). In this case, \(|E|=110\), and each complete bipartite graph \(K_{2,2}\) has 2 perfect matches, so \(f(G)=2^{22}\).
Therefore, the maximum number of different perfect matches is \(2^{22}\). | Combinary-26 | 2^{22} |
There are $n$ cards, each labeled with the numbers $1 , 2, \cdots, n$. These $n$ cards are distributed among 17 people, with each person receiving at least 1 card. Then, there is always one person who receives cards with numbers $x$ and $y$, where $x>y$, and $118x \leqslant 119y$. The smallest positive integer $n$ that satisfies this condition is \_\_\_\_\_. | Consider the residue classes modulo 17, where the residue class with residue $i$ is represented as $17k+i$ ($k$ is a natural number $1 \leqslant i \leqslant 17 $).
Let $A_i$ denote the residue class with residue $i$. For any x and y in each $A_i$, we have $x-y \geqslant 17 $.
Then
\[
\begin{array}{l}
y \leqslant x-17, x \leqslant n \\
\Rightarrow 119y-118x \leqslant 119(x-17)-118x \\
\quad=x-2023 \leqslant n-2023 .
\end{array}
\]
If $n<2023$ , then
$11 y-118x \leqslant x-2023<0 \Rightarrow 119y<118x$,
and $118 x \leqslant 119 y$, contradiction.
Therefore, when $n<2023$, the condition is not satisfied.
When $n=2023$, since $2023=17 \times 119 $, it is known that $118 \times 17+i(i=0,1, \cdots, 17)$ are 18 numbers not exceeding $n$. By the Pigeonhole Principle, among these 18 numbers $\times 17+i(i=0,1, \cdots, 17)$ , there must be two numbers x and $(x>y)$ in the possession of the same person.
Let $x=118 \times 17+x_{1}$ and $y=118 \times 17+y_{1}$, where $ 17 \geqslant
x_{1} \geqslant y_{1} \geqslant 0$ .
then
\[
\begin{array}{l}
118x-119y\\
=118\left(118 \times 17+x_{1}\right)-119\left(118 \times 17+y_{1}\right) \\
=\left(118 x_{1}-119 y_{1}\right)-118 \times 17 \\
\leqslant 118 x_{1}-118 \times 17 \\
\leqslant 118 \times 17-118 \times 17=0 .
\end{array}
\]
Therefore, when n=2023, the condition is satisfied.
In conclusion, the smallest positive integer $n$ satisfying the condition is 2023. | Combinary-27 | 2023 |
let $a_{1}, a_{2}, \cdots, a_{9}$ be a permutation of $1,2, \cdots, 9$. If the permutation $C=\left(a_{1}, a_{2}, \cdots, a_{9}\right)$ can be obtained from $1,2, \cdots, 9$, by swapping two elements at most 4 times, the total number of permutations satisfying this condition is \_\_\_\_\_. | Since there are 9 single cycles in the permutation $(1,2, \cdots, 9)$, if the permutation
$C=\left(a_{1}, a_{2}, \cdots, a_{9}\right)$ can be obtained from the permutation $ (1,2, \cdots, 9)$ by swapping at most 4 times, it is known from analysis that the number of cycles in C must be no less than 5. Below is the classification counting:
(1) 9 cycles, there is 1.
(2) 8 cycles, then 1 2-cycle, there are $\mathrm{C}_{9}^{2} \cdot(2- 1)!=36$ permutations.
(3) 7 cycles, as $3+6 \times 1=2+2+5 \times 1$ , there are $C_{9}^{3} \cdot (3-1)!+\frac{C_{9}^{2}C_{7}^{2}}{2 !} -546$ permutations.
(4) 6 cycles, as 4+5×1=3+2+4×1=2+2+2+3×1 there are
$\mathrm{C}_{9}^{4} \cdot 3 !+\mathrm{C}_{9}^{3} \mathrm{C}_{7}^{2} \cdot 2 !+\frac{\mathrm{C}_{9}^{2} \mathrm{C}_{7}^{2} \mathrm{C}_{9}^{2}}{3 !}=4536$ permutations.
(5) 5 circles, as $5+4 \times 1=4+2+3 \times 1=3+3+3 \times 1 =3+2+2+2 \times 1=2+2+2+2+2+1, C_9^5 \cdot 4 !+C_{9}^{4} C_{5}^{2} \cdot 3 !+C_{9}^{3} C_{6}^{3} \cdot 2 !+\frac{C_{9}^{3} \cdot 2 ! C_{6}^{2} C_{4}^{2}}{2 !} + \frac{C_{9}^{2} C_{7}^{2} C_{5}^{2} C_{3}^{2}}{4 !}
=22449 $
permutations.
In conclusion, there are 27568 permutations satisfying the requirements. | Combinary-28 | 27568 |
Let sequence $a_{1}, a_{2}, \cdots, a_{10}$ satisfy $1 \leqslant a_{1} \leqslant{l} a_{2} \leqslant \cdots \leqslant a_{10} \leqslant 40$, $a_{4} \geqslant6$, and $\log _{2}\left(\left|a_{i}-i\right|+1\right) \in \mathbf{N}(i=1,2, \cdots, 10)$ The number of such sequences is \_\_\_\_\_. | Consider $b_{i}=a_{i}-i$. Then $b_{i} \in\{0, \pm 1, \pm 3 , \pm 7, \cdots\} $, and $ b_{i+1}-b_{i} \geqslant-1 $. Thus, the sequence $\left\{b_{i}\right\}$ has the following structure: there exists an integer $s \in[0,10]$, $b_{1}, \cdots, b_{s} \in\{-1,0,1\}$, $3 \leqslant b_{s+1} \leqslant \cdots \leqslant b_{10}$ and any adjacent pair $b_{1}, \cdots, b_{s}$ is not 1 or -1.
Duo to $a_{1} \geqslant 1, a_{10} \leqslant 4$ and $a_{4} \geqslant 6$, it is known that $b_{1} \neq-1 $ and $b_{s+1}, \cdots, b_{10} \in\{3,7,15\}(0 \leqslant s \leqslant 3)$ .
Conversely, for each sequence $\left\{b_{i}\right\}$, that meets the first two conditions, there exists a corresponding sequence $\left\{a_{i}\right\} $.
For a fixed $s \in\{0,1,2,3\} $, the number of ways to choose , $b_{1}, \cdots, b_{s}$ is $1,2,5$ (because $\left.\left(b_{1}, b_{2}\right)\neq(1,-1)\right)$ and $5 \times 3- 2=13\left(\right.$ Because $\left.\left(b_{1}, b_{2}, b_{3}\right) \neq(0,1,-1), (1,1,-1)\right).$
Now let's find the number of ways to choose $b_{s+1}, \cdots, b_{10}$. Suppose there are x occurrence of 3, y occurrence of 7, z occurrence of 15. Then the non-negative integer solutions to $x+y+z=10-s$ are $\mathrm{C}_{12-\mathrm{s}}^{2}$.
Therefore the number of sequences $\left\{a_{i}\right\}$ is
$1C_{12}^{2}+2 C_{11}^{2}+5 C_{10}^{2}+13 C_{9}^{2}=869$. | Combinary-29 | 869 |
A student walks through a hallway with a row of closed lockers numbered from 1 to 1024. He starts by opening locker number 1, then proceeds forward, alternately leaving untouched or opening one closed locker. When he reaches the end of the hallway, he turns around and walks back, opening the first closed locker he encounters. The student continues this back and forth journey until every locker is opened. The number of the last locker he opens is \_\_\_\_\_. | Assuming there are $2^{h}$ closed lockers in a row, the number of the last locker opened by the student is denoted as $a_{k}$. When the student first reaches the end of the hallway, $2^{k-1}$ lockers remain closed. These closed lockers are all even-numbered, arranged in decreasing order from the student's standing position towards the other end. Now, these lockers are renumbered from 1 to $2^{k-1}$. Note that the originally numbered locker $n$ (where $n$ is even) becomes locker number $2^{k-1}-\frac{n}{2}+1$. According to the new numbering, the number of the locker the student opens last should be $a_{k-1}$ , while the originally numbered $a_{k}$ is now numbered as $a_{k-1}$ . Therefore
\[
\begin{array}{l}
a_{k-1}=2^{k-1}-\frac{a_{k}}{2}+1 \\
\Leftrightarrow a_{k}=2^{k}+2-2 a_{k-1} \\
=2^{k}+2-2\left(2^{k-1}+2-2 a_{k-2}\right) \\
=4 a_{k-2}-2 \\
\Rightarrow a_{k}-\frac{2}{3}=4\left(a_{k-2}-\frac{2}{3}\right)
\end{array}
\]
Given that $a_{0}=1$, and $a_{1}=2 $, we have: \\
When k is even,
$a_{k}-\frac{2}{3}=\left(a_{0}-\frac{2}{3}\right) 4^{\frac{k}{2}} \Rightarrow a_{k}=\frac{1}{3}\left(4^{\frac{k}{2}}+2\right) ;$
When k is odd,,
$a_{k}-\frac{2}{3}=\left(a_{1}-\frac{2}{3}\right) 4^{\frac{k-1}{2}} \Rightarrow a_{k}=\frac{1}{3}\left(4^{\frac{k+1}{2}}+2\right).$
Here [x] represents the greatest integer not exceeding x.
Thus $a_{k}=\frac{1}{3}\left(4^{\left[\frac{k+1}{2}\right]}+2\right)(k \in \mathbf{N}) . $
For this problem, when $k=10, a_{10}=\frac{1}{3}\left(4^{5}+2\right)=342 $, indicating that the last locker opened is numbered 342. | Combinary-30 | 342 |
Let $A=\{-3,-2, \cdots, 4\}, a, b, c \in A$ be distinct elements of A. If the angle of inclination of the line: $a x+b y+c=0$ is acute, then the number of such distinct lines is \_\_\_\_\_. | Because the angle of inclination of line ll is acute, ab<0 . Counting is carried out in the following two cases:
(1) If $a$, $b$, and $c$ are all not equal to 4, then $a, b, c \in\{-3 , -2, \cdots, 3\} $. Without loss of generality, assume $a>0$ and $b<0$.
If $c=0$, there are 7 lines; if $c \neq 0$ , there $3 \times 3 \times 4=36$ lines. In this case, there are a total of 43 distinct lines.
(2) If one of $a$, $b$, or $c$ is 4. Let $a=4$, then $b<0$, there are 3 choices for b, and 6 choices for $c$, resulting in 18 lines (with duplicates). Among them, $4x-2y=0,4x-2y+ 2=0$ are counted twice, so there are only 16 distinct lines. Similarly, when $b=4$ or $c=4$, there are also 16 distinct lines for each case.
In conclusion, there are $43+16×3=9143+16×3=91$ distinct lines with acute angles of inclination. | Combinary-31 | 91 |
In a sequence of length 15 consisting of $a$ and $b$, if exactly five "aa"s occur and both "ab" and "ba" and "bb" occur exactly three times, there are \_\_\_\_\_ such sequences. | Because there are three occurrences each of "ab" and "ba", there are two possible cases for such sequences:
(1).(a)(b)(a)(b)(a)(b)(a) ;
(2).(b)(a)(b)(a)(b)(a)(b) (here, (a) denotes a segment composed entirely of aa, and (b) denotes a similar segment composed entirely of b).
For case (1), since there are five "aa"s and three "bb"s, the sequence contains a total of 9 aa's and 6 bb's. The 9 a's can be partitioned into 4 segments, which equals the number of positive integer solutions to the equation $x_{1}+x_{2}+x_{3}+x_{4}=9 $ which is $\mathrm{C}_{9-1}^{4-1}=\mathrm{C}_{8}^{3}$. Similarly, the 6 b's can be partitioned into 3 segments, which is $\mathrm{C}_{8}^{3} \mathrm{C}_{5}^{2}$.
For case (2), there are 8 aa's and 7 bb's, resulting in $\mathrm{C}_{7}^{2} \mathrm{C}_{6}^{3}$ possibilities.
Therefore, the total count is
$\mathrm{C}_{8}^{3} \mathrm{C}_{5}^{2}+\mathrm{C}_{7}^{2} \mathrm{C}_{6}^{3}=980 $. | Combinary-32 | 980 |
Among the five-digit numbers formed by the digits 1, 2, ..., 6, the number of five-digit numbers satisfying the condition of having at least three different digits and 1 and 6 not being adjacent is \_\_\_\_\_. | Using a recursive approach, let $S_{n}$, denote the number of n-digit numbers formed by the digits $1,2, \cdots, 6$ such that 1 and 6 are not adjacent.
Clearly, $S_{1}=6$ and $S_{2}=6^{2}-2=34$. Now we establish the recursive formula for $S_n$. We divide $S_n$ into three categories:
Let $a_{n}$ denote the number of nn-digit numbers with the first digit being 1, $b_{n}$ denote the number of nn-digit numbers with the first digit being 2, 3, 4, or 5, and $c_{n}$
It's evident that $S_{n}=a_{n}+b_{n}+c_{n}$, and
\[
\begin{array}{l}
a_{n}=a_{n-1}+b_{n-1}, \\
b_{n}=4\left(a_{n-1}+b_{n-1}+c_{n-1}\right), \\
c_{n}=b_{n-1}+c_{n-1} .
\end{array}
\]
Then $S_{n}=5\left(a_{n-1}+b_{n-1}+c_{n-1}\right)+b_{n-1}$
\[
\begin{array}{l}
=5 S_{n-1}+4 S_{n-2}(n \geqslant 3) . \\
\end{array}
\]
From $S_{1}=6$ and $S_{2}=34$ and we get
$S_{3}=194, S_{4}=1106, S_{5}=6306 $.
Subtracting the cases where there is exactly one digit or exactly two digits, and 1 and 6 are not adjacent, from $S_{5}$, there are $\left(\mathrm{C}_{6}^{2}-1\right)\left(2^{5}-2\right)=420$.
Therefore, the number of five-digit numbers satisfying the condition is 6306-6-420=5880. | Combinary-33 | 5880 |
How many numbers can be selected at most from 1 to 100 to ensure that the quotient of the least common multiple and greatest common divisor of any two numbers is not a perfect square? | Let \( a = mx \) and \( b = my \) (\( x \) and \( y \) are coprime). Then we have \( \frac{[a, b]}{(a, b)} = xy \). To ensure that \( xy \) is not a perfect square, \( x \) and \( y \) cannot both be square numbers. In other words, the two numbers cannot both be perfect squares after dividing them by their greatest common divisor. Based on this principle, construct the following sets: \( (1,4,9,16,25,36,49,64,81,100), (2,8,18,32,50,72,98), (3,12,27,48,75), (5,20,45,80), (6,24,54,80), (7,28,63), (10,40,90), (11,44,99), (13,52), (14,56), (15,60), (17,68), (19,76), (21,84), (22,88), (23,92) \), and there are 45 individual numbers. Selecting one number from each set, a maximum of \( 16 + 45 = 61 \) numbers can be chosen to meet the requirement. | Combinary-34 | 61 |
From the natural numbers 1 to 100, choose any \( m \) numbers such that among these \( m \) numbers, there exists one number that can divide the product of the remaining \( m-1 \) numbers. The minimum value of \( m \) is \_\_\_\_\_. | First, when selecting 25 prime numbers, none of them can divide the product of the remaining 24 numbers. When \( m = 25 \), the condition cannot be satisfied, so \( m \) must be greater than 25.
Next, we prove that by selecting any 26 numbers from 1 to 100, there will always be one number that can divide the product of the remaining 25 numbers.
Let these 26 numbers be \( a, b, c, \ldots, z \). Since there are only 25 prime numbers from 1 to 100, we can express these 26 numbers in their prime factorization form:
\[ a = 2^{a_1} \times 3^{a_2} \times 5^{a_3} \times \cdots \times 97^{a_{25}} \]
\[ b = 2^{b_1} \times 3^{b_2} \times 5^{b_3} \times \cdots \times 97^{b_{25}} \]
\[ z = 2^{z_1} \times 3^{z_2} \times 5^{z_3} \times \cdots \times 97^{z_{25}} \]
From these 26 numbers, first, remove the one with the highest exponent of 2 (if there are ties, remove any one of them). Then, from the remaining 25 numbers, remove the one with the highest exponent of 3, then remove the one with the highest exponent of 5, and so on.
After removing 25 numbers, the remaining number will have exponents for each prime factor that do not exceed the maximum exponent among the 26 numbers. Therefore, this remaining number will definitely divide the product of the remaining 25 numbers. | Combinary-35 | 26 |
Several teams are participating in a friendly football match, where any two teams play at most one match against each other. It is known that each team has played 4 matches, and there are no draws. A team is considered a "weak team" if it loses at least 2 out of the 4 matches it plays. If there are only 3 "weak teams" in this friendly match, then at most how many teams could have participated in the matches? | Assume in a tournament with n teams, there are 3 weak teams, meaning the other n-3 teams are strong teams. There will be a total of $4n/2=2n$ games, resulting in 2n losses. Each weak team can lose up to 4 games, and each strong team can lose up to 1 game. Therefore, we have: $2n \leq 3×4 + n-3$, from which we derive: $n \leq 9$. Hence, there can be at most 9 teams.
The construction is as follows: Teams 1, 2, 3, and 4 win against Team 7; Teams 3, 4, 5, and 6 win against Team 8; Teams 5, 6, 1, and 2 win against Team 9; Team 1 beats Team 2, Team 2 beats Team 3, Team 3 beats Team 4, Team 4 beats Team 5, Team 5 beats Team 6, and Team 6 beats Team 1; Teams 1 through 6, each with 3 wins and 1 loss, are the strong teams, while Teams 7, 8, and 9, each with 4 losses, are the weak teams. | Combinary-36 | 9 |
In the equation, the same letter represents the same digit, and different letters represent different digits. Then, the four-digit number \(\overline{a b c d}\) is given by:
$$
(\overline{a b})^{c} \times \overline{a c d}=\overline{a b c a c d}
$$ | The original expression can be transformed using the principle of place value into:
$$
(\overline{a b})^{c} \times \overline{a c d}=\overline{a b c a c d}
$$
According to this equation, where the same letter represents the same digit, and different letters represent different digits, let's analyze the possibilities:
First, we consider the case where \(c=3\):
Since \(20^{3} \times 200=1600000\), exceeding six digits, we conclude that \(a=1\). Thus, the equation becomes:
$$
\left[(\overline{1 b})^{3}-1\right] \times \overline{13 d}=\overline{1 b 3} \times 1000
$$
However, since there are 3 factors of 5 on the right side of the equation, and \(\overline{13 d}\) can only provide at most 1 factor of 5, we infer that \((\overline{1 b})^{3}-1\) must be a multiple of 25. This implies that the last two digits of \((\overline{1})^{3}\) can be \(01, 26, 51,\) or \(76\). By considering the units place, we find that \(b=1\) or \(6\), but neither of these options satisfies the conditions.
Next, let's explore the case where \(c=2\). This yields:
$$
\left[(\overline{a b})^{2}-1\right] \times \overline{a 2 d}=\overline{a b 2000}
$$
From this equation, we deduce that:
$$(\overline{a b})^{2}-1=1000 \times \frac{\overline{a b 2}}{\overline{a 2 d}}$$
Because \(0.5<\frac{\overline{a b 2}}{\overline{a 2 d}}<2\), we conclude that \(500<(\overline{a b})^{2}-1<2000\). Hence, \(a\) can be \(2, 3,\) or \(4\). However, since \(\overline{a 2 d}\) cannot be 125 or 625, it can provide at most 2 factors of 5. Therefore, \((\overline{a b})^{2}-1\) must be a multiple of 5, leading to the possible values \(b=1, 9, 4,\) or \(6\).
Based on the equation \(\left[(\overline{a b})^{2}-1\right] \times \overline{a 2 d}=\overline{a b 2000}\), we have:
(1) \((\overline{a b}+1) \times \overline{a 2 d}=\overline{a b 2000} \div(\overline{a b}-1)>1000\)
(2) \((\overline{a b}-1) \times \overline{a 2 d}=\overline{a b 2000} \div(\overline{a b}+1)<1000\)
Consequently, \(a\) can only be 3. Then, \((\overline{a b}-1) \times(\overline{a b}+1)\) cannot be a multiple of 25, implying that \(\overline{a 2 d}\) must be a multiple of 25. Thus, \(d=5\), and \(\overline{a 2 d}=325\).
Now, \((\overline{3 b}-1) \times(\overline{3 b}+1) \times 325=\overline{3 b 2000}\), where \(\overline{3 b 2000}\) is a multiple of 16. This means that either \((\overline{3 b}-1)\) or \((\overline{3 b}+1)\) must be a multiple of 8. Therefore, \(b\) can be 1 or 3. Since \(a=3\), \(b\) can only be 1.
After verifying: \(31^{2} \times 325=312325\), so \(\overline{a b c d}=3125\). | Combinary-37 | 3125 |
Let $n$ be represented as the difference of squares of two nonzero natural numbers, then there are $F(n)$ ways to do so.
For example, $15=8^{2}-7^{2}=4^{2}-1^{2}$, so $F(15)=2$; whereas 2 cannot be represented, hence $F(2)=0$. Then, the calculation result of $F(1)+F(2)+F(3)+\cdots+F(100)$ is \_\_\_\_\_. | For a nonzero natural number $a$, if it can be expressed as the difference of squares of two nonzero natural numbers, let $a=m^{2}-n^{2}$ (here $m$ and $n$ are both nonzero natural numbers, and $m>n$). Since $m^{2}-n^{2}=(m-n)(m+n)$, both $m-n$ and $m+n$ are factors of the natural number $a$. Considering that the parity of $m-n$ and $m+n$ is the same, thus for any pair of positive integers $p$ and $q$ with the same parity $(p>q)$, their product $pq$ can always be expressed as the difference of squares of two nonzero natural numbers, $\left(\frac{p+q}{?}\right)^{2}-\left(\frac{p-q}{?}\right)^{2}$ is the corresponding representation. This indicates that as long as we can find out how many pairs of positive integers $(p, q)$ with the same parity exist such that $p>q$ and $1 \leq pq \leq 100$, the number of such pairs exactly equals the value of $F(1)+F(2)+F(3)+\cdots+F(100)$. Classification calculation:
When $q=1$, $p$ can take 3, 5, 7, $3, 99$, there are 49 pairs $(p, q)$ satisfying the condition;
When $q=2$, $p$ can take $4, 6, 8, \ldots, 50$, there are 24 pairs $(p, q)$ satisfying the condition;
When $q=3$, $p$ can take 5, 7, 9, ...33, there are 15 pairs $(p, q)$ satisfying the condition;
When $q=4$, $p$ can take $6, 8, 10, \ldots, 24$, there are 10 pairs $(p, q)$ satisfying the condition;
When $q=5$, $p$ can take $7, 9, 11, \ldots, 19$, there are 7 pairs $(p, q)$ satisfying the condition;
When $q=6$, $p$ can take $8, 10, 12, 14, 16$, there are 5 pairs $(p, q)$ satisfying the condition;
When $q=7$, $p$ can take $9, 11, 13$, there are 3 pairs $(p, q)$ satisfying the condition;
When $q=8$, $p$ can take $10, 12$, there are 2 pairs $(p, q)$ satisfying the condition;
When $q=9$, $p$ can only take 11, there is only 1 pair $(p, q)$ satisfying the condition;
When $q>9$, there are no pairs $(p, q)$ satisfying the condition.
In conclusion, there are a total of 116 pairs satisfying the condition, which is the desired result. | Combinary-39 | 116 |
What is the simplified value of the expression, $8x^3-3xy+\sqrt{p}$, if $p=121$, $x=-2$, and $y=\frac{3}{2}$?
\begin{align*}
\text{A)}\ & 84 &
\text{B)}\ & 73+\sqrt{11}\\
\text{C)}\ & -28 &
\text{D)}\ & -44\\
\end{align*} | By plugging in the variables for x, y, and z, we arrive at the equation $8\times (-2)3-3\times (-2)\times 32+121$ which evaluates to $-64+9+11$, leading to the answer -44. | Alg1 | D |
Which expression best represents “the product of twice a quantity x and the difference of that quantity and 7”?
\begin{align*}
\text{A)}\ & 2x(7-x) &
\text{B)}\ & 2x(x-7) \\
\text{C)}\ & 2x-(x-7) &
\text{D)}\ & 2(x-7)\\
\end{align*} | The phrase “the product of” refers to multiplying the next two groupings mentioned. In this case, our next two groupings are “twice a quantity x” and “the difference of that quantity and 7”. We can evaluate this first group as 2x and the second group as $(x-7)$. Multiplying the two results in $2x(x-7)$. | Alg2 | B |
The formula for the area of a triangle is $A = \frac{1}{2}bh$. The area of a triangle is 62 square meters, and its height is 4 meters. What is the length of the base?
\noindent Options:\\
A) 15.5 m\\
B) 27 m\\
C) 31 m\\
D) 62 m | Given that the area is 62 square meters and the height is 4 meters, we can arrive at the equation $62 = \frac{1}{2}b \cdot 4$. This can be simplified to $62 = 2b$ and further to $b = 31$. | AI-Geometry1 | C |
Simplify: $(-m^2n^{-3})^3 \cdot (4m^{-1}n^2p^3)^2$
\begin{align*}
\text{A)}\ & \frac{-16m^4p^6}{n^5} &
\text{B)}\ & \frac{16m^3p^5}{n^2} \\
\text{C)}\ & \frac{8m^3p^5}{n^2} &
\text{D)}\ & \frac{-16m^6p^9}{n^{23}}\\
\end{align*} | First simplify the expression by expanding the exponents through the equation: $(-m^6n^{-9})\cdot(16m^{-2}n^4p^6)$. Combining like terms results in $-16\times m^4\times n^{-5}\times p^6$. | Alg4 | A |
If $M=24a^{-2}b^{-3}c^5$ and $N=18a^{-7}b^6c^{-4}$, then $\frac{N}{M}=$
\begin{align*}
\text{A)}\ & \frac{4a^5c^9}{3b^9} &
\text{B)}\ & \frac{4a^5c}{3b^3} \\
\text{C)}\ & \frac{3b^9}{4a^5c^9} &
\text{D)}\ & \frac{3b^3}{4a^5c}\\
\end{align*} | Using law of exponents and combining like terms arrives at answer C. | Alg5 | C |
The volume of a rectangular prism is $2x^5 + 16x^4 + 24x^3$. If the height of the rectangular prism is $2x^3$, which of the following could represent one of the other dimensions of the rectangular prism?
\begin{align*}
\text{A)}\ & (x+2) &
\text{B)}\ & (x+3) \\
\text{C)}\ & (x+4) &
\text{D)}\ & (x+12)\\
\end{align*} | we can first divide the volume of the rectangular prism by $2x^3$ to get $x^2+8x+12$. Factoring this expression results in $(x+6)(x+2)$. Therefore, the answer is A | Alg6 | A |
Which expression is equivalent to $(3x - 1)(2x^2 + 1)$?
\noindent Options:\\
A) $6x^3 - 2x^2 - 3x + 1$\\
B) $6x^3 - 2x^2 + 3x - 1$\\
C) $5x^3 - 2x^2 + 3x - 1$\\
D) $5x^3 - x^2 + 4x$ | Multiplying out the expression with FOIL (first outer inner last) nets $6x^3 + 3x - 2x^2 - 1$. | AI-Algebra2 | B |
Factor \( 64b^2 - 16b + 1 \) completely.
\noindent Options:\\
A) \( (32b - 1)(32b + 1) \)\\
B) \( (b - 8)^2 \)\\
C) \( (8b - 1)^2 \)\\
D) \( (8b + 1)^2 \) | We first observe that this follows the pattern for a perfect square binomial. This leads us to the answer \( (8b - 1)^2 \). | AI-Algebra3 | C |
Simplify $6\sqrt[3]{64}-\sqrt{12}\cdot 2\sqrt{27}$
\begin{align*}
\text{A)}\ & 12 &
\text{B)}\ & -12 \\
\text{C)}\ & 24-12\sqrt{3} &
\text{D)}\ & 48-72\sqrt{3}\\
\end{align*} | $6\sqrt[3]{64}$ can be simplified into 24 by basic arithmetic rules. $12\cdot2\sqrt{27}$ can be reduced to $2\sqrt{324}$ which is equivalent to 36. Finally $24-36= -12$. | Alg9 | B |
What is the simplified form of $\sqrt{80x^5y^2z^3}$? Assume all variables are positive.
\begin{align*}
\text{A)}\ & 16x^2yz\sqrt{5xz} &
\text{B)}\ & 16xyz\sqrt{5x^3z} \\
\text{C)}\ & 4x^2yz\sqrt{5xz} &
\text{D)}\ & 4\sqrt{5x^5y^2z^3}\\
\end{align*} | Take out any variable or constant that has a squared factor: $4x^2yz\sqrt{5xz}$ | Alg10 | C |
Which of the following is NOT equivalent to $-4$?
\begin{align*}
\text{A)}\ & 2\sqrt{9}-5\sqrt[3]{8} &
\text{B)}\ & 3\sqrt[3]{64}-2\sqrt{64} \\
\text{C)}\ & 2\sqrt{121}-3\sqrt[3]{216} &
\text{D)}\ & 4\sqrt{25}-8\sqrt[3]{27}\\
\end{align*} | A,B,D all equivalent to -4 except C leads to 3. | Alg11 | C |
What are the solutions of the quadratic equation $15x^2=2x+8$
\begin{align*}
\text{A)}\ & \{-\frac{4}{3},-\frac{3}{2}\} &
\text{B)}\ & \{-\frac{4}{5},\frac{2}{3}\} \\
\text{C)}\ & \{-\frac{3}{2},\frac{4}{5}\} &
\text{D)}\ & \{-\frac{2}{3},\frac{4}{5}\}\
\end{align*} | First move all terms to one side: $15x^2-2x-8=0$. Then factor into $(5x-4)(3x+2)=0$. Setting $5x-4$ to zero results in a solution of $x = \frac{4}{5}$ and setting $3x+2$ to zero results in a solution of $x = -\frac{2}{3}$. | Alg12 | D |
Find the solution of $4(3y-5)=2(7y+3)$
\begin{align*}
\text{A)}\ & -13 &
\text{B)}\ & -4 \\
\text{C)}\ & \frac{11}{2} &
\text{D)}\ & 13\
\end{align*} | First distribute and expand to $12y-20 = 14y+6$. Combine like terms to arrive at $2y = -26$. Thus, $y = -13$. | Alg13 | A |
Solve the equation $x^2-7+2x=0$
\begin{align*}
\text{A)}\ & \{-1-2\sqrt{2},-1+2\sqrt{2}\} &
\text{B)}\ & \{1-2\sqrt{2},1+2\sqrt{2}\}\\
\text{C)}\ & \{\frac{-7-\sqrt{41}}{2},\frac{-7+\sqrt{41}}{2}\} &
\text{D)}\ & \{\frac{7-\sqrt{41}}{2},\frac{7+\sqrt{41}}{2}\}\
\end{align*} | Utilize the quadratic formula with $a = 1$, $b=2$, and $c = -7$. $\frac{-2\pm\sqrt{2^2+28}}{2} = -1\pm \sqrt{8} = -1\pm 2\sqrt{2}$. | Alg14 | A |
At a movie theater, the adult ticket price is \$8 and the child ticket price is \$6. For a certain movie, 210 tickets were sold and \$1500 was collected. How many adult tickets were sold?
\begin{align*}
\text{A)}\ & 30 &
\text{B)}\ & 90\\
\text{C)}\ & 120 &
\text{D)}\ & 150\\
\end{align*} | We can create a system of equations $8x+6y = 1500$ and $x+y=210$. By rearranging the second equation to $y = 210-x$ and plugging $210-x$ in for y in the first equation, we get 8x+1260-6x = 1500 which evaluates to $2x = 240$. Therefore $x = 120$ and $y = 90$. | Alg15 | C |
Solve for $y:x-yz+5=8$
\begin{align*}
\text{A)}\ & y=\frac{x-3}{z} &
\text{B)}\ & y=\frac{x-13}{z} \\
\text{C)}\ & y=\frac{3-x}{z} &
\text{D)}\ & y=\frac{13-x}{z}\\
\end{align*} | Attempt to isolate y by first isolating the yz term: $yz=x-3$. Then divide by $z$ for $y=\frac{x-3}{z}$. | Alg16 | A |
Describe the type of solution for the linear system of equations defined by
\[
\left\{
\begin{array}{rcl}
2y - 3x &=& 20\\
-\frac{3}{2}x + y &=& 10
\end{array}
\right.
\]
\noindent Options:\\
A) no solution\\
B) infinite solutions\\
C) one solution\\
D) two solutions | First, rearrange the second equation to \( y = \frac{3}{2}x + 10 \). Then plug this into the first equation for \( 3x + 20 - 3x = 20 \). This is obviously true so there must be infinite solutions. | AI-Algebra4 | B |
Identify all of the following equations that have a solution of -2?
\begin{align*}
\text{I)}\ & 3(x+7)=5(x+5) &
\text{II)}\ & x^2+x-6=0\\
\text{III)}\ & 2(x-4)=x-10 &
\text{IV)}\ & x^2=4 \\
\end{align*}
\begin{align*}
\text{A)}\ & I and III &
\text{B)}\ & II and IV\\
\text{C)}\ & I, II and IV &
\text{D)}\ & I, III and IV\\
\end{align*} | Plug in -2 for $x$ in each equation and see if it works
I: $3\times 5=5\times 3$ works.\\
II: $4-2-6=0$ does not work.\\
III:$2\times-6=-2-10$ works. \\
IV: $4=4$ works. | Alg18 | D |
What is the solution of the system of linear equations?
\begin{align*}
\left\{ \begin{array}{l}
7x-2y=5\\
-3x-4y=7
\end{array}\right.
\end{align*}
\begin{align*}
\text{A)}\ & (\frac{3}{17},-\frac{32}{17}) &
\text{B)}\ & (-\frac{3}{17}, -\frac{32}{17})\\
\text{C)}\ & (\frac{3}{17},\frac{32}{17}) &
\text{D)}\ & (-\frac{3}{17},\frac{32}{17})\\
\end{align*} | We will use the elimination method to get rid of $y$ by multiplying the first equation by 2 and subtracting the two equations from each other. This results in $17x=3$ so $x = \frac{3}{17}$ and $y = -\frac{32}{17}$. | Alg19 | A |
What values of x make the inequality true? $4(x-2)-10x\geq -3x+13$
\begin{align*}
\text{A)}\ & \{x: x\geq 1\} &
\text{B)}\ & \{x: x\geq -7\}\\
\text{C)}\ & \{x: x\leq 1\} &
\text{D)}\ & \{x: x\leq -7\}\\
\end{align*} | Simplify the inequality into $-6x-8\geq -3x+13$. This is equivalent to $-3x\geq 21$ so $x\leq -7$. | Alg20 | D |
What is the equation of the line that passes through the points $(4,-4)$ and $(-5, 14)$?
\begin{align*}
\text{A)}\ & x+2y=2 &
\text{B)}\ & 2x+3y=12\\
\text{C)}\ & 2x+y=4 &
\text{D)}\ & 3x-2y=-6\\
\end{align*} | First find slope $\frac{14+4}{-5-4}=-2$, then use point-slope form on one of the points. $y+4 = -2(x-4)$. This simplifies to $y = -2x+4$. Convert to standard form. | Alg22 | C |
The cost $c$ per person to participate in a guided mountain biking tour depends on the number of people $n$ participating in the tour. This relationship can be described by the function $c = -3n + 60$, where $0 < n < 12$. What is the rate of change described by this function?
\begin{align*}
\text{A)}\ & \text{20 people per tour} &
\text{B)}\ & \text{-3 people per tour}\\
\text{C)}\ & \text{20\$ per person} &
\text{D)}\ & \text{-3\$ per person}\\
\end{align*} | $c= -3n +60$ can be interpreted as the $\text{cost per person}=-3\times \text{number of people} +60$. So as more people join the cost per person decreases by 3 dollars per person joining. | Alg23 | D |
Find $f(-3)$ if $f(x)=6x^2-x-2$.
\begin{align*}
\text{A)}\ & 55 &
\text{B)}\ & 53\\
\text{C)}\ & -53 &
\text{D)}\ & -59\\
\end{align*} | Plug in $-3$ for $x$ to arrive at $6\times (-3)^2+3-2 = 55$ | Alg25 | A |
When Darcy’s school bus travels at 30 miles per hour, it gets from her home to school in 12 minutes. What is the speed of Darcy’s bus if it makes the same trip in
18 minutes?
\begin{align*}
\text{A)}\ & 20 \ \text{mph} &
\text{B)}\ & 28 \ \text{mph}\\
\text{C)}\ & 36 \ mph &
\text{D)}\ & 45\ mph\\
\end{align*} | If the bus travels for 12 minutes at 30 mph, the bus travels $30
\times 12/60$ miles or 6 miles. If the bus does the same trip in 18 minutes, it would be at a speed of $6/(18/60)$ or 20 miles per hour. | Alg26 | A |
The price of a package varies directly with the number of stickers in the package. If a package contains 650 stickers and sells for \$26.00, what is the constant of variation? How much will 800 stickers cost?
\begin{align*}
\text{A)}\ & k = 0.04; \$32.00 &
\text{B)}\ & k = 0.40; \$320.00
\\
\text{C)}\ & k = 6.24; \$806.24
&
\text{D)}\ & k = 25; \$20,000.00
\\
\end{align*} | Find the constant rate by taking $26/650 = 0.04$. Then $800\times 0.04$ results in \$32. | Alg27 | A |
Which function does NOT have an x-intercept?
\begin{align*}
\text{A)}\ &y=\frac{1}{2}x-7&
\text{B)}\ & y=-\frac{1}{3}x-5\\
\text{C)}\ & y=-x^2+2x+5 &
\text{D)}\ & y=x^2-2x+5\\
\end{align*} | A and B are both linear equations with a non zero slope so they must have an x intercept. Plug in $y = 0 $for the other two equations and attempt to solve a solution. D results in no solutions. | Alg28 | D |
What is the x-intercept and Y-intercept of the graph of $5x – 3y = -30$?
\begin{align*}
\text{A)}\ &\text{The x-intercept is 6, and the y-intercept is -10}\\
\text{B)}\ &\text{The x-intercept is -6, and the y-intercept is 10}.
\\
\text{C)}\ &\text{The x-intercept is 10, and the y-intercept is -6}.
\\
\text{D)}\ & \text{The x-intercept is -10, and the y-intercept is 6}.
\\
\end{align*} | Plug in $y = 0$ to solve for $x$ intercept: $5x-3\times 0 = -30, \ x = -6$. Plug in $x = 0$ to solve for $y$ intercept: $5\times 0-3y = -30,\ y = 10$. | Alg31 | B |
Is the point $(1,-3)$ a solution to the equation $f(x) = x^2 +4x – 8$?
\begin{align*}
\text{A)}\ & \text{Yes} &
\text{B)}\ & \text{No}\\
\end{align*} | Plug in the point: $-3 = 12+4\times 1-8$. True. | Alg32 | A |
If y varies inversely with x, and $x = 18$ when $y = 4$, find y when $x = 12$.
\begin{align*}
\text{A)}\ & y=3 &
\text{B)}\ & y=6\\
\text{C)}\ & y=9 &
\text{D)}\ & y=54 \\
\end{align*} | Inverse variation means that $x\times y = c$. here $18\times 4 =72$ so we know our $c = 72$. Now we can plug in $x=12$ for $12\times y=72$. We get $y=6$. | Alg33 | B |
If 10 workers can build a house in 12 weeks, how long will it take 15 workers to build the same house?
\begin{align*}
\text{A)}\ &6 &
\text{B)}\ & 16\\
\text{C)}\ & 8 &
\text{D)}\ & 18\\
\end{align*} | First calculate total weeks to build a house. With 10 workers and 12 weeks, we have a total of 120 weeks. Taking 120 and dividing by 15 workers results in 8 weeks needed. | Alg34 | C |
Write the equation of the line that has a y–intercept of –3 and is parallel to the line $y = -5x +1$. | Parallel lines have the same slope but different y-intercepts. | Alg35 | y=-5x-3 |
A model house was built that states that 3 inches represents 10 ft. If the width of the door on the model is 1.2 inches, what is the width of the actual door? | Use proportional representation to set up $3in/10ft = 1.2in/Xft$, cross multiplication $3x=12, x =4ft$. | Alg36 | 4ft |
The product of 4 more than a number and 6 is 30 more than 8 times the number. What is the number? | $6(x+4) = 8x + 30\Rightarrow
6x+24 = 8x +30\Rightarrow
2x = - 6\Rightarrow
x = -3$ | Alg37 | -3 |
Solve $4(x+4)=24+3(2x-2)$. | Use distributive property to both sides of the equations, and then simplify to $2x = -2$, then $x = -1$. | Alg39 | x=-1 |
Three times as many robins as cardinals visited a bird feeder. If a total of 20 robins and cardinals
visited the feeder, write a system of equations to represent the situation and solve how many were robins? | Define x as the number of robins, and define y as the number of cardinals, then set up system of equations.
$x + y = 20$
, $y = 3x$
Use substitution method to solve, $x + 3x = 20$, $x = 5$.
\section*{Functions and Applications} | Alg40 | 5 Robins. |
Describe all the transformations of the function: $f(x)=-|x-3|+1$
\begin{align*}
\text{A)}\ & \text{Translated 3 units left, 1 unit up and reflected over the x-axis} \\
\text{B)}\ & \text{Translated 1 unit right, 3 units down and reflected over the x-axis}\\
\text{C)}\ & \text{Translated 3 units right and 1 unit up and reflected over the y-axis}\\
\text{D)}\ & \text{Translated 3 units right and 1 unit up and reflected over the x-axis}\\
\end{align*} | With the parent function as the absolute value function, we can see that the +1 causes a one unit translation up and the x-3 causes a 3 unit translation right. Because there is a negative sign in front of the absolute value there is also a reflection over the x axis | Alg2-2 | D |
If the value of the discriminant for the function $f(x)=2x^2-5x+6$ is equal to -23, which of the following correctly describes the graph of $f(x)$?
\begin{align*}
\text{I)}\ & f(x)\text{ has real roots.}&
\text{II)}\ & f(x) \text{ has imaginary roots.} \\
\text{III)}\ & f(x) \text{ has two solutions.} &
\text{IV)}\ & f(x) \text{ has one solution.}\\
\end{align*}
\begin{align*}
\text{A)}\ & \text{I and III only} &
\text{B)}\ & \text{I and IV only}\\
\text{C)}\ & \text{II and III only} &
\text{D)}\ & \text{IV only}\\
\end{align*} | Since our discriminant is negative there must be imaginary roots and two solutions. | Alg2-3 | C |
What is the axis of symmetry of the function $y=-3(x+1)^2+4$?
\begin{align*}
\text{A)}\ & x=-3 &
\text{B)}\ & x=-1 \\
\text{C)}\ & x=1 &
\text{D)}\ & x=4\\
\end{align*} | This function is a parabola and there is a shift of 1 to the left, this makes the axis of symmetry at $x = -1$. | Alg2-7 | B |
Which equation shows the function $f(x)=12x^2+36x+27$ in intercept form?
\begin{align*}
\text{A)}\ & f(x)=3(2x-3)^2 &
\text{B)}\ & f(x)=3(2x+3)^2 \\
\text{C)}\ & f(x)=(6x-9)(2x-3) &
\text{D)}\ & f(x)=3(2x+3)(2x-3)\\
\end{align*} | Factoring 3 out of the function results in $f(x) = 3(4x^2+12x+9)$. This is a perfect square of $2x+3$. | Alg2-8 | B |
Which of the following represents the function in intercept form $y=64x^2-49$?
\begin{align*}
\text{A)}\ & y=(64x+1)(x-49) &
\text{B)}\ & y=(8x+7)(8x-7) \\
\text{C)}\ & y=(8x+7)(8x+7) &
\text{D)}\ & y=(8x-7)(8x-7)\\
\end{align*} | Since the function is in the form $S^2-s^2$, we can simplify to $(S+s)(S-s)$. | Alg2-9 | B |
Use factoring to find the solutions to the equation $x^2+24x=-144$.
\begin{align*}
\text{A)}\ & -12 &
\text{B)}\ & 12 \\
\text{C)}\ & -12 \ and \ 12 &
\text{D)}\ & 9 \ and \ 16\\
\end{align*} | $x^2+24x+144=0$, then $(x+12)^2 =0$. | Alg2-11 | A |
Solve the equation $-x^2-11=-2x^2+5$ for the variable $x$.
\begin{align*}
\text{A)}\ & \pm2 &
\text{B)}\ & \pm4 \\
\text{C)}\ & \pm\sqrt{2} &
\text{D)}\ & \pm\sqrt{6}\\
\end{align*} | Rearrange equation to $x^2-16=0$. solving results in $x=\pm 4$. | Alg2-12 | B |
What must be added to the equation \( x^2 + 20x = 0 \) to complete the square?
\noindent Options:\\
A) 10\\
B) 25\\
C) 40\\
D) 100 | For \( x^2 + 20x + c \) to be a complete square, \( c \) must be \( \left(\frac{20}{2}\right)^2 = 100 \). | AI-Algebra5 | D |
If $f(x)=x^2+4$ and $g(x)=\sqrt{10-x}$, what is the value of $f(g(1))$?
\begin{align*}
\text{A)}\ & 1 &
\text{B)}\ & 0 \\
\text{C)}\ & \sqrt{5} &
\text{D)}\ & 13\\
\end{align*} | first plug 1 in as $x$ for the $g$ equation, $g(1) = 3$. Then plug 3 in for $x$ in the $f$ function, $f(3) = 13$. | Alg2-14 | D |
Find all the solutions to the function: $0=(-4x+9)(x-1)(3x-5)$
\begin{align*}
\text{A)}\ & x=-\frac{7}{4}, x=1, x=\frac{7}{3} &
\text{B)}\ & x=\frac{9}{4}, x=1, x=\frac{5}{3} \\
\text{C)}\ & x=9, x=-1, x = -\frac{5}{3} &
\text{D)}\ & x=\frac{9}{8}, x =\frac{1}{2}, x=\frac{5}{6}\\
\end{align*} | By setting each part in parenthesis to zero and solving, we can find the three solutions. | Alg2-15 | B |
What are the solutions to the equation $0.5x^2-0.45x-0.3=0$?
\begin{align*}
\text{A)}\ & x=-1.35 \ and \ x=0.45 &
\text{B)}\ & x=1.35 \ and \ x=-0.45 \\
\text{C)}\ & x=-1.35 \ and \ x=-0.45 &
\text{D)}\ & x=1.35 \ and \ x=0.45\\
\end{align*} | Plug into the quadratic formula. | Alg2-16 | B |
Find the remainder when $f(x)=5x^4+2x^2-3x+1$ is divided by $x-2$.
\begin{align*}
\text{A)}\ & 95 &
\text{B)}\ & 43 \\
\text{C)}\ & 83 &
\text{D)}\ & -25\\
\end{align*} | \begin{align*}
\frac{5x^4+2x^2-3x+1}{x-2} & = 5x^3+\frac{10x^3+2x^2-3x+1}{x-2}\\
& = 5x^3+10x^2+\frac{22x^2-3x+1}{x-2}\\
& = 5x^3+10x^2+22x+\frac{41x+1}{x-2}\\
& = 5x^3+10x^2+22x+41+\frac{83}{x-2}
\end{align*} | Alg2-17 | C |
The path of an object falling to Earth is represented by the equation $h(t)=-16t^2+vt+s$. What is the equation of an object that is shot up into the air from 150 feet above the ground and has an initial velocity of 62 feet per second?
\begin{align*}
\text{A)}\ & h(t)=-16t^2+62t+150 &
\text{B)}\ & h(t)=-16t^2+150t+62 \\
\text{C)}\ & h(t)=-16t^2+62t &
\text{D)}\ & h(t)=-16t^2+150t\\
\end{align*} | Plug in 150 for $s$ and 62 for $v$. | Alg2-18 | A |
The graph above shows a portion of a system of equations where $f(x)$ has $a>0$ and $g(x)$ has $a<0$. Which of the following satisfies the equation $f(x)=g(x)$?
\begin{align*}
\text{A)}\ & \{(0,-1);(-3,2)\} &
\text{B)}\ & \{(-1,2);(-2,3)\} \\
\text{C)}\ & \{(1,2);(2,7)\} &
\text{D)}\ & \{(-3.7,0);(0.4,0)\}\\
\end{align*} | Only the points of intersection satisfy the equation. | Alg2-19 | A |
Which of the following is the conjugate of the expression $\frac{2}{3-\sqrt{2}}$?
\begin{align*}
\text{A)}\ & 3-\sqrt{2} &
\text{B)}\ & 3+\sqrt{2} \\
\text{C)}\ & \sqrt{2} &
\text{D)}\ & -\sqrt{2}\\
\end{align*} | The conjugate is changing the sign in the denominator. | Alg2-20 | B |
Solve the equation $\frac{2}{x+5}+\frac{3}{x-5}=\frac{7 x-9}{x^{2}-25}$ for the variable $x$.
\begin{align*}
\text{A)}\ & x=5 &
\text{B)}\ & x=\frac{9}{2} \\
\text{C)}\ & x=2 &
\text{D)}\ & x=7\\
\end{align*} | $\frac{2(x-5)+3(x+5)}{x^{2}-25}=\frac{7 x-9}{x^{2}-25} .5 x+5=7 x-9 . x=7$ | Alg2-21 | D |
Choose all the following that have an end behavior as $x \rightarrow \infty, f(x) \rightarrow \infty$. NOTE: You may choose more than one.
\begin{align*}
\text{A)}\ & f(x)=-x^{4}+3 x^{2}-x-7 &
\text{B)}\ & f(x)=x^{3}+5 x+1 \\
\text{C)}\ & f(x)=-3 x^{5}+2 x^{3}+9 x-4 &
\text{D)}\ & f(x)=-2(x-7)^{2}+4\\
\text{E)}\ & f(x)=-3 x+2 &
\text{F)}\ & f(x)=(x-8)^{2}+2\\
\end{align*} | Functions with the desired end behavior must either be positive even functions or positive odd functions. | Alg2-22 | B, F |