instruction_seed stringlengths 2 10.7k | solution stringlengths 0 10.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
values | question_type stringclasses 5
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## Problem 5.
Given an integer $k \geq 1$, we define $a_{k}$ as the integer that in base ten is written
$$
\begin{gathered}
\square k^{k} \\
a_{k}=11 \ldots 1
\end{gathered}
$$
(that is, a 1 repeated $k$ times). Prove that $a_{k}$ divides $a_{l}$ if and only if $k$ divides $l$. | ## Solution:
If $l=d k+r$ where $r$ is smaller than $k$, then
$$
a_{l}=a_{k} 10^{r}\left(1+10^{k}+10^{2 k}+\ldots+10^{(d-1) k}\right)+a_{r}
$$
(We will set $a_{0}=0$.) If $k$ divides $l$, then $r=0$ and it is clear that $a_{r}=0$ so that $a_{k}$ divides $a_{l}$. Conversely, if $a_{k}$ divides $a_{l}$, then it must b... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
If the maximum value of the function $y=\log_{2}x$ on the interval $[1, a]$ (where $a > 1$) is 2, then $a$ equals ( )
A: $\frac {3}{2}$
B: 2
C: 4
D: 8 | Since the function $y=\log_{2}x$ is monotonically increasing on the interval $[1, a]$,
the maximum value is $\log_{2}a=2$,
thus, $a=4$.
Option C should be chosen.
Based on the monotonicity of the logarithmic function and the given information, we can solve $\log_{2}a=2$ to find the answer.
This question exami... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Let $g(x)$ be a function such that $g(0) = 2$ and
\[g(xy) = g(x^2 + y^2) + 2(x - y)^2\]
for all real numbers $x$ and $y$. Find $g(x)$. | 1. Setting $y = 0$, we obtain:
\[g(0) = g(x^2) + 2x^2.\]
Given $g(0) = 2$, it follows that:
\[2 = g(x^2) + 2x^2 \Rightarrow g(x^2) = 2 - 2x^2.\]
Because $x^2$ is always non-negative, $g(u) = 2 - 2u$ for all $u \geq 0$.
2. To verify this solution is consistent for other values of $y$, we can test $y = 1$:
... | 2 - 2x | Algebra | math-word-problem | Yes | Incomplete | synthetic_math | true |
The solution set of the inequality $\dfrac{x}{{x}^{2}-8x+15}\geqslant 2$ is . | **Analysis**
This problem mainly examines the solution methods for fractional inequalities and quadratic inequalities, reflecting the mathematical idea of equivalent transformation. Note the application where the denominator is not $0$.
**Solution**
Transform the inequality to: $\dfrac{2{x}^{2}-17x+30}{{x}^{2}-8x+15... | \left[ \dfrac{5}{2},3\right) \cup (5,6] | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false |
A sequence $\{a_n\}$ has a partial sum $S_n (n \in \mathbb{N}^*)$ and $a_1=\frac{1}{2}, S_n=n^2a_n (n \in \mathbb{N}^*)$. Use mathematical induction to conjecture the general term formula of $\{a_n\}$. The options are:
A: $a_n=\frac{2n-4}{3^n}$
B: $a_n=\frac{1}{n(n+1)}(n \in \mathbb{N}^*)$
C: $a_n=\frac{1}{2n}$
D: $a_n... | [Analysis]
This problem examines the recursive relationship and mathematical induction of a sequence. First, obtain the first few terms of the sequence based on its recursive relationship, analyze, and conjecture the general term formula. This is a basic problem.
[Answer]
According to the problem, we know that $a_1=\f... | a_n=\frac{1}{n(n+1)}(n \in \mathbb{N}^*) | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Let \(y\) be an angle such that \(\tan y = \frac{2c}{3d}\) and \(\tan 2y = \frac{3d}{2c + 3d}\). Find the least positive value of \(y\) in the form \(y = \tan^{-1} m\). Compute \(m\). | Starting with the given information:
\[
\tan 2y = \frac{3d}{2c + 3d} = \frac{1}{\frac{2c}{3d} + 1} = \frac{1}{\tan y + 1}
\]
Thus,
\[
(\tan y + 1) \tan 2y = 1
\]
Using the double angle identity for tangent:
\[
(\tan y + 1) \cdot \frac{2 \tan y}{1 - \tan^2 y} = 1
\]
This simplifies to:
\[
2 \tan y (\tan y + 1) = 1 - \ta... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
the probability that a number selected at random from the first 50 natural numbers is a composite number is - . | To find the probability that a number selected at random from the first 50 natural numbers is a composite number, we first need to determine how many of those numbers are composite.
The first 50 natural numbers are: 1, 2, 3, 4, 5, ..., 50.
We know that:
- 1 is neither prime nor composite.
- Prime numbers are numbers ... | 0.68 | Number Theory | math-word-problem | Yes | Yes | orca_math | true |
Example 9 Let $a, b, c$ be positive real numbers, and satisfy $abc=1$, prove: $\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)$. $\left(-c-1+\frac{1}{a}\right) \leqslant 1$. (41st IMO Problem) | Prove that for $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, where $x, y, z$ are positive real numbers, the original inequality is equivalent to
$$(x+y-z)(y+z-x)(x+z-y) \leqslant x y z$$
Let $u=x+z-y, v=x+y-z, w=y+z-x$. Since the sum of any two of these numbers is positive, at most one of them is not positive.
If ex... | proof | Inequalities | proof | Incomplete | Yes | inequalities | false |
Given that $(x)$ is a number defined on $R$, and the function $f'(x)$, when $x \in (-0]$, $f(x)$ has a unique zero point $(-3)$ with $xf'(x) < f(x)$, the range of real number values $x$ that satisfy the inequality $\frac{f(\quad\quad)}{x} \leqslant 0$ is ____. (Express the result using interval notation) | $f(x)$ is an even function defined on $R$, thus $f(x) = f(-x)$,
$f(x)$ is a monotonic function on $(0, +\infty)$,
Since $f(x)$ is an even function,
when $x \in (-\infty, 0]$, we always have $xf(-x) 0$: $\frac{f(x)}{x} - f(\quad\quad) 0$, $F(x) \leqslant F(\quad\quad)$, we get: $x \geqslant 3$
Since $(-x) = \frac{f(-x... | [-3, 0) \cup [3, +\infty) | Calculus | math-word-problem | Incomplete | Yes | cn_k12 | false |
## Task 3
On a school playground, a large circle is drawn. The center of the circle is point $O$, and points $A, B$, and $C$ are marked on the circumference. Angles $\measuredangle A O B$ and $\measuredangle B O C$ are acute, and the lengths $\overline{O A}$ and $\overline{O C}$ are perpendicular to each other. The pa... | ## Solution:
The orange and green circular segments together form a quarter of the circle bounded by an arc (or circular arc) of length $240: 4=60$ feet.
The green segment is bounded by two radii and the circular arc $\widehat{A B}$, and the orange segment by two radii and the circular arc $\widehat{B C}$.
The perim... | 6.3\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Given the equation $(x^{2}-2x+m)(x^{2}-2x+n)=0$, the four roots form an arithmetic sequence with the first term being $\frac{1}{4}$. Determine the value of $|m-n|$.
A: $1$
B: $\frac{3}{4}$
C: $\frac{1}{2}$
D: $\frac{3}{8}$ | Let the four roots be $x_1$, $x_2$, $x_3$, and $x_4$.
Then, $x_1 + x_2 = 2$ and $x_3 + x_4 = 2$.
By the properties of an arithmetic sequence, when $m+n=p+q$, $a_m + a_n = a_p + a_q$.
Let $x_1$ be the first term and $x_2$ be the fourth term. The sequence is $\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}$.
Thus, $m ... | \frac{1}{2} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
A function $f(x)$ is defined on $\mathbb{R}$ with its derivative $f''(x)$. The graph of $f(x)$ is symmetric about the line $x=1$, and $(x-1)f''(x) 2$, determine the relationship between $f(x\_1)$ and $f(x\_2)$.
A) $f(x\_1) > f(x\_2)$
B) $f(x\_1) < f(x\_2)$
C) $f(x\_1) = f(x\_2)$
D) Indeterminate | Given that the graph of $f(x)$ is symmetric about the line $x=1$ and $(x-1)f''(x) 1$, then $f(x\_1) > f(x\_2)$ due to $x\_1 2$, then $1 - x\_1 f(x\_2)$.
Therefore, in both cases, $f(x\_1) > f(x\_2)$.
The answer is: $\boxed{\text{A}}$.
To solve this problem, find the monotonic intervals of the function $f(x)$, and... | \text{A} | Calculus | MCQ | Yes | Yes | cn_k12 | false |
8. The 8th "上" character has $\qquad$ ・.
The translation preserves the original text's line breaks and format. | $41$ | 41 | Other | math-word-problem | Incomplete | Problem not solved | olympiads | false |
The solution set of the inequality $2^{x-2} > 1$ is. | The inequality $2^{x-2} > 1$ can be rewritten as $2^{x-2} > 2^0$,
Therefore, $x-2 > 0$, which means $x > 2$,
Hence, the solution set of the inequality is $(2, +\infty)$,
Thus, the answer is $\boxed{(2, +\infty)}$. | (2, +\infty) | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false |
Given $\overrightarrow{a}=(\cos α,1,\sin α)$ and $\overrightarrow{b}=(\sin α,1,\cos α)$, find the angle between the vectors $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$.
A: $90^{\circ}$
B: $60^{\circ}$
C: $30^{\circ}$
D: $0^{\circ}$ | We have $\overrightarrow{a}=(\cos α,1,\sin α)$ and $\overrightarrow{b}=(\sin α,1,\cos α)$.
First, let's find the vectors $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$:
$$
\begin{align}
\overrightarrow{a}+\overrightarrow{b} &= (\cos α,1,\sin α) + (\sin α,1,\cos α) \\
&= (\cos α + ... | \text{A: } 90^{\circ} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Given the universal set U={x|-2≤x≤6, x∈Z}, and set A={x|x=2n, n≤3, n∈N}, find the complement of A in U (represented by listing). | According to the problem, the universal set U={x|-2≤x≤6, x∈Z} = {-2, -1, 0, 1, 2, 3, 4, 5, 6},
and set A={x|x=2n, n≤3, n∈N} = {0, 2, 4, 6}.
The complement of A in U, denoted by $∁_U A$, is the set of elements in U that are not in A. Thus, we have:
$∁_U A$ = {-2, -1, 1, 3, 5}
Therefore, the answer is $\boxed{-2, -1, ... | -2, -1, 1, 3, 5 | Number Theory | math-word-problem | Yes | Yes | cn_k12 | false |
The length of the bridge, which a train 150 m long and traveling at 45 km/hr can cross in 30 sec, is? | First, we need to convert the speed of the train from km/hr to m/s because the length of the train and the time to cross the bridge are given in meters and seconds, respectively.
Speed in m/s = (Speed in km/hr) * (1000 m / 1 km) * (1 hr / 3600 s)
Speed in m/s = 45 * (1000 / 3600)
Speed in m/s = 45 * (5 / 18)
Speed in ... | 225 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
71. Given that $a, b, c$ are all positive numbers, prove: $\frac{a^{3}}{(a+b)^{3}}+\frac{b^{3}}{(b+c)^{3}}+\frac{c^{3}}{(c+a)^{3}} \geqslant \frac{3}{8}$. (2005 Vietnam Mathematical Olympiad) | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | proof | Inequalities | proof | Yes | Incomplete | inequalities | false |
A certain bacteria colony grows at a specific rate every day for 25 days, at which point it reaches the limit of its habitat and can no longer grow. If two bacteria colonies start growing simultaneously, they will take 24 days to reach the habitat's limit. What is the growth rate of the bacteria colony? | Let's denote the daily growth rate of the bacteria colony as \( r \), where \( r \) is a multiplier that represents the growth from one day to the next. For example, if \( r = 1.1 \), the colony grows by 10% each day.
If one colony takes 25 days to reach the habitat's limit, we can express the final size of the colony... | 2, | Algebra | math-word-problem | Yes | Yes | orca_math | true |
What quantity of water should be taken out to concentrate a certain amount of 40% acidic liquid to 60% acidic liquid, if after taking out the water, there are 5 liters less liquid? | Let's denote the initial volume of the 40% acidic liquid as V liters. This means that there are 0.40V liters of acid in the solution.
When water is removed, the volume of the liquid decreases by 5 liters, so the new volume of the liquid is V - 5 liters. We want the concentration of acid in this new volume to be 60%.
... | 5 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
9.022. $\frac{1}{2-x}+\frac{5}{2+x}<1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
9.022. $\frac{1}{2-x}+\frac{5}{2+x}<1$. | ## Solution.
We have
$$
\frac{1}{2-x}+\frac{5}{2+x}-10
$$
In the obtained inequality $x^{2}-4 x+8>0$ for $x \in R$, therefore it is equivalent to the inequality $(x-2)(x+2)>0$. Using the interval method, we find $x2$.
\cup(2;\infty) | Inequalities | math-word-problem | Yes | Incomplete | olympiads | false |
A train of a certain length is traveling at 45 km/hr and can cross a bridge in 30 seconds. The bridge is 285 meters long. What is the length of the train? | To find the length of the train, we first need to determine the distance the train travels while crossing the bridge. Since the train takes 30 seconds to cross the bridge and is traveling at a speed of 45 km/hr, we can convert the speed to meters per second and then multiply by the time to find the distance.
First, co... | 90 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Solve the inequality $ax^{2}-(a+1)x+1 \lt 0$ with respect to $x$ where $a\in \mathbb{R}$. | To solve the inequality $ax^{2}-(a+1)x+1 0\]
\[x > 1\]
Thus, the solution set is $\boxed{(1,+\infty)}$.
3. **Case 2: $a > 0$**
\[ax^{2}-(a+1)x+1 = (ax-1)(x-1) 1$.
Thus, the solution set is $\boxed{(1, \frac{1}{a})}$.
- **Subcase 2.2: $a=1$**
Here, $\frac{1}{a} = 1$.
The roots coincide, hence the solution set ... | (-\infty, \frac{1}{a})\cup (1,+\infty) | Inequalities | math-word-problem | Yes | Incomplete | cn_k12 | false |
74 (1008). A tourist who set out from $A$ to $B$ walked the first half of the distance at a speed of $v_{1}$, and the second half at a speed of $v_{2}\left(v_{1} \neq v_{2}\right)$. On the return trip, he walked half of the total time at a speed of $v_{1}$, and the other half at a speed of $v_{2}$. In which case did th... | Solution. I method. Let the time spent by the tourist on the journey from $B$ to $A$ be $2 t$. Then half of the time ($t$) he walked at a speed of $v_{1}$, and the second half of the time at a speed of $v_{2}$. The entire distance, therefore, is $v_{1} t + v_{2} t$, and half the distance is $\frac{v_{1} t + v_{2} t}{2}... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The sum of the squares of two positive integers is 100. The product of the two integers is 32. What is the sum of the two integers? | Let's denote the two integers by \(x\) and \(y\). Based on the information given, we have the equations:
1. \(x^2 + y^2 = 100\)
2. \(xy = 32\)
We need to find \(x + y\). We can use the identity \((x + y)^2 = x^2 + y^2 + 2xy\). Substituting the values from the problem, we find:
\[
(x + y)^2 = 100 + 2 \times 32 = 164
\]... | 2\sqrt{41} | Algebra | math-word-problem | Yes | Incomplete | synthetic_math | true |
Check if $x=4$ is a root of the polynomial $x^3 - 5x^2 + 7x - 12$. If it is not, find the correct numerical coefficient of $x$ in place of $7$ so that $x=4$ would be a root of the polynomial $x^3 - 5x^2 + ax - 12$. | 1. Test whether $x=4$ is a root of $x^3 - 5x^2 + 7x - 12$:
\[
f(4) = (4)^3 - 5(4)^2 + 7*4 - 12 = 64 - 80 + 28 - 12 = 0.
\]
Here, $f(4) = 0$, showing that $x=4$ is indeed a root.
2. The polynomial simplifies to $x^3 - 5x^2 + 7x - 12 = (x-4)(x^2 - ax + b)$, where we need to find $a$ and $b$. Since $4$ is a r... | 7 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
How many integers greater than $10$ and less than $100$, written in base-$10$ notation, are increased by $9$ when their digits are reversed?
$\textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 1 \qquad
\textbf{(C)}\ 8 \qquad
\textbf{(D)}\ 9 \qquad
\textbf{(E)}\ 10$ | 1. Let the two-digit number be represented as \(10x + y\), where \(x\) and \(y\) are the digits of the number, and \(x\) is the tens digit and \(y\) is the units digit. Given that the number is increased by 9 when its digits are reversed, we can write the equation:
\[
10x + y + 9 = 10y + x
\]
2. Simplify the ... | 8 | Number Theory | MCQ | Yes | Yes | aops_forum | false |
Find the greatest common divisor of $8!$ and $(6!)^2 \cdot 2^3.$ | First, calculate the prime factorizations:
$$ 8! = 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 2^7 \cdot 3^2 \cdot 5^1 \cdot 7^1 $$
$$ (6!)^2 = (6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)^2 = 2^8 \cdot 3^4 \cdot 5^2 $$
$$ (6!)^2 \cdot 2^3 = 2^{11} \cdot 3^4 \cdot 5^2 $$
Now find the gcd of $8!$ and $(6... | 5760 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
Given plane vectors $\overrightarrow{a}=(3,1)$, $\overrightarrow{b}=(-1,1)$, $\overrightarrow{c}=(x,-6)$. If $(\overrightarrow{a}+2\overrightarrow{b})∥\overrightarrow{c}$, then $x=\_\_\_\_\_\_$. | Given the vectors $\overrightarrow{a}=(3,1)$, $\overrightarrow{b}=(-1,1)$, and $\overrightarrow{c}=(x,-6)$, we are tasked with finding the value of $x$ given that $(\overrightarrow{a}+2\overrightarrow{b})∥\overrightarrow{c}$.
Step 1: Calculate $\overrightarrow{a}+2\overrightarrow{b}$.
We have $\overrightarrow{a} = (3,... | -2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
3. Find the smallest real number $M$, such that for all real numbers $a, b, c$, we have
$$
\begin{array}{l}
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \\
\leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2} .
\end{array}
$$ | 3. First consider
$$
P(t)=t b\left(t^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c t\left(c^{2}-t^{2}\right) .
$$
It is easy to see that $P(b)=P(c)=P(-c-b)=0$. Therefore, we have
$$
\begin{array}{l}
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \\
=|P(a)|=|(b-c)(a-b)... | \frac{9 \sqrt{2}}{32} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
A trader sells some meters of cloth for Rs. 9890 at the profit of Rs. 24 per meter of cloth. The cost price of one meter of cloth is Rs. 83.5. How many meters of cloth did the trader sell? | Let's denote the number of meters of cloth sold as \( x \).
The trader makes a profit of Rs. 24 per meter of cloth. Therefore, the selling price (SP) per meter of cloth is the cost price (CP) plus the profit per meter, which can be expressed as:
\[ SP = CP + \text{Profit per meter} \]
\[ SP = Rs. 83.5 + Rs. 24 \]
\[ ... | 92 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
3. In $\triangle A B C$, it is known that $B_{2}$ is the point symmetric to the excenter $B_{1}$ of side $A C$ with respect to the midpoint of $A C$, and $C_{2}$ is the point symmetric to the excenter $C_{1}$ of side $A B$ with respect to the midpoint of $A B$. The excircle of side $B C$ touches side $B C$ at point $D$... | 3. Auxiliary lines as shown in Figure 2. Let the excenter of side $B C$ be $A_{1}$.
By the properties of the excenter, $B_{1} 、 A 、 C_{1}, A_{1} 、 C 、 B_{1}, C_{1} 、 B$ 、 $A_{1}$ are collinear in threes, and $A_{1} A \perp B_{1} C_{1}$.
Take a point $P$ on the plane such that $\overrightarrow{C_{2} P}=\overrightarrow... | proof | Geometry | proof | Yes | Yes | cn_contest | false |
4. Let point $P$ be a point on the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$. A line passing through $P$ intersects the two asymptotes at $P_{1}$ and $P_{2}$, and $\overrightarrow{P_{1} P} \overrightarrow{P P_{2}}$ $=3$. Let $O$ be the origin. Then the area of $\triangle O P_{1} P_{2}$ is $\qquad$. | 4. 16 Detailed Explanation: The equations of the asymptotes of the hyperbola are $y= \pm \frac{4}{3} x$. Let $P_{1}\left(x_{1}, \frac{4}{3} x_{1}\right), P_{2}\left(x_{2},-\frac{4}{3} x_{2}\right), P\left(x_{0}, y_{0}\right)$. Given $\lambda=\frac{\overrightarrow{P_{1} P}}{\overrightarrow{P P_{2}}}=3$, we get $P\left(\... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Simplify the expression $\frac{5^5 + 5^3 + 5}{5^4 - 2\cdot5^2 + 5}$. | First, factor out a common factor of $5$ from both the numerator and the denominator:
\[
\frac{5^5 + 5^3 + 5}{5^4 - 2\cdot5^2 + 5} = \frac{5(5^4 + 5^2 + 1)}{5(5^3 - 2\cdot5 + 1)}
\]
This simplifies to:
\[
\frac{5^4 + 5^2 + 1}{5^3 - 2\cdot5 + 1}
\]
Calculate the numerator and the denominator separately:
- Numerator: $5^... | \frac{651}{116} | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
The average (arithmetic mean) of all scores on a certain algebra test was 90. If the average of the 8 male students' grades was 84, and the average of the female students' grades was 92, how many female students took the test? | Let's denote the number of female students as F.
The total score for the male students is 8 * 84, because there are 8 male students and each has an average score of 84.
The total score for the female students is F * 92, because there are F female students and each has an average score of 92.
The overall average scor... | 24 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Compute: $\frac{2\cos 10^{\circ} - \sin 20^{\circ}}{\cos 20^{\circ}} =$ \_\_\_\_\_\_. | Given $2\cos 10^{\circ} = 2\sin 80^{\circ} = 2\sin (60^{\circ}+20^{\circ})$, we can use the sine of the sum of two angles formula to expand and simplify the expression.
$2\sin (60^{\circ}+20^{\circ}) = 2(\frac{\sqrt{3}}{2}\cos 20^{\circ} + \frac{1}{2}\sin 20^{\circ}) = \sqrt{3}\cos 20^{\circ} + \sin 20^{\circ}$
Thus,... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Without using a calculator, determine the largest prime factor of $16^4 + 2 \times 16^2 + 1 - 13^4$. | First, rewrite the expression using the difference of squares factorization formula:
\[
16^4 + 2 \times 16^2 + 1 - 13^4 = (16^2 + 1)^2 - (13^2)^2.
\]
Expanding the terms within the parentheses:
\[
(16^2 + 1) = 256 + 1 = 257, \quad (13^2) = 169.
\]
Substituting these into the formula:
\[
(257)^2 - (169)^2 = (257 - 169)(... | 71 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
Lucy went to the grocery store. She bought some packs of cookies and 12 packs of cake. In the end, she had 56 amount of change remaining. She bought 14 packs of groceries in all. How many packs of cookies did she buy? | Lucy bought 14 packs of groceries in total, and we know that 12 of those were packs of cake. To find out how many packs of cookies she bought, we subtract the number of cake packs from the total number of grocery packs.
Number of cookie packs = Total number of grocery packs - Number of cake packs
Number of cookie pack... | 2 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Let $\forall n \in \mathbb{N}^{*} S_{n}=1^{2}+2^{2}+\cdots+n^{2}$, show by induction that $\forall n \in \mathbb{N}^{*} S_{n}=$ $\frac{n(n+1)(2 n+1)}{6}$
The formula for the sum of cubes is a bit special.
Indeed, we notice that it is exactly the square of the sum of the linear terms.
The goal of the following two ex... | - Initialization: for $n=1$ we have $S_{1}=1^{2}=1=\frac{1 \cdot 2 \cdot(2 \cdot 1+1)}{6}$
- Heredity: Let $n \in \mathbb{N}^{*}$ such that $S_{n}=\frac{n(n+1)(2 n+1)}{6}$, then we have:
$$
\begin{aligned}
& S_{n+1}=1^{2}+\cdots+n^{2}+(n+1)^{2}=S_{n}+(n+1)^{2} \underset{H . R .}{=} \frac{n(n+1)(2 n+1)}{6}+(n+1)^{2} \\... | proof | Algebra | proof | Yes | Yes | olympiads | false |
In a hostel, the number of students decreased by some percentage and the price of food increased by 20% over the previous year. If each student consumes the same amount of food then the consumption of food should be cut short by 7.4074074074074066% by every student, so that the total cost of the food remains the same a... | Let's denote the initial number of students as \( N \) and the initial price of food per student as \( P \). The total cost of food for the previous year would then be \( N \times P \).
Given that the price of food increased by 20%, the new price of food per student would be \( P + 0.20P = 1.20P \).
Let's denote the ... | 10\% | Algebra | math-word-problem | Yes | Yes | orca_math | true |
what is the unit ’ s digit of 7 ^ 6 | To find the unit's digit of \(7^6\), we can look at the pattern of the unit's digits of the powers of 7.
\(7^1 = 7\) (unit's digit is 7)
\(7^2 = 49\) (unit's digit is 9)
\(7^3 = 343\) (unit's digit is 3)
\(7^4 = 2401\) (unit's digit is 1)
\(7^5 = 16807\) (unit's digit is 7)
\(7^6 = 117649\) (unit's digit is 9)
So, th... | 9 | Number Theory | math-word-problem | Yes | Yes | orca_math | true |
# Task 8. Development of a Complex Sorting Algorithm
Task Description. The management of a mineral processing terminal has set a task for marking incoming samples and has entrusted you with the development of the first part of the marker. The following technical specification has been formulated: the value of the mass... | # Solution to the problem.
Example of program implementation in Python
```
def order_weight(string):
weights = string.split(' ')
order_weights = {}
for weight in weights:
sum = 0
for digit in weight:
sum += int(digit)
order_weights[weight] = sum
weights.sort()
w... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine the value of $b$ for which $\frac{1}{\log_3 b} + \frac{1}{\log_5 b} + \frac{1}{\log_6 b} = 1$. | Using the change-of-base formula, the given equation can be rewritten as:
\[
\log_b 3 + \log_b 5 + \log_b 6 = 1.
\]
We combine the logarithms:
\[
\log_b (3 \times 5 \times 6) = 1.
\]
Calculating the product inside the logarithm:
\[
\log_b 90 = 1.
\]
Therefore, we solve for $b$:
\[
b^{1} = 90 \quad \Rightarrow \quad b =... | 90 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
What is the value of $[\log_{10}(5\log_{10}100)]^2$?
$\textbf{(A)}\ \log_{10}50 \qquad \textbf{(B)}\ 25\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 1$ | $[\log_{10}(5\log_{10}100)]^2$
Since $10^2 = 100$, we have $\log_{10} 100 = 2$, so:
$[\log_{10}(5\cdot 2)]^2$
Multiply:
$[\log_{10}(10)]^2$
Since $10^1 = 10$, we have $\log_{10} 10 = 1$, so:
$[1]^2$
$1$
$\fbox{E}$. | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
9.5. The numbers $1,2,3, \ldots, 2017,2018,2019$ are written on a board. It is allowed to erase any two numbers and replace them with the absolute value of their difference. In the end, one number will remain on the board. Can it be zero? Explain. | 9.5. Solution. The sum of all numbers written on the board will be odd (the sum of 1010 odd numbers and 1009 even numbers is an odd number). When erasing 2 numbers, there can be the following 3 cases:
a) Two even numbers are erased, then the absolute difference will be an even number, and the new sum will be an odd nu... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\operatorname{arctg}\left(x \cos \left(\frac{1}{5 x}\right)\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | f^{\}(0) | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Let $ABC$ be a triangle with semiperimeter $s$ and inradius $r$. The semicircles with diameters $BC$, $CA$, $AB$ are drawn on the outside of the triangle $ABC$. The circle tangent to all of these three semicircles has radius $t$. Prove that
\[\frac{s}{2}<t\le\frac{s}{2}+\left(1-\frac{\sqrt{3}}{2}\right)r. \]
[i]Alterna... | 1. **Define the semicircles and their properties:**
Let $A', B', C'$ denote the midpoints of $BC, CA, AB$ respectively. The semicircles with diameters $BC, CA, AB$ have radii $\frac{a}{2}, \frac{b}{2}, \frac{c}{2}$ respectively, where $a, b, c$ are the lengths of the sides opposite vertices $A, B, C$ of triangle $AB... | \frac{s}{2} < t \leq \frac{s}{2} + \left(1 - \frac{\sqrt{3}}{2}\right)r | Geometry | proof | Yes | Yes | aops_forum | false |
The interval in which the function $f(x) = x^3 - 3x^2 + 1$ is decreasing is ( )
A: $(2, +\infty)$
B: $(-\infty, 2)$
C: $(-\infty, 0)$
D: $(0, 2)$ | By calculating $f'(x) = 3x^2 - 6x < 0$, we get $0 < x < 2$.
Therefore, the interval in which the function $f(x) = x^3 - 3x^2 + 1$ is decreasing is $(0, 2)$.
Hence, the answer is $\boxed{D}$.
**Analysis:** To find the interval in which the function is decreasing, we simply need to calculate $f'(x)$ and set it les... | D | Calculus | MCQ | Yes | Yes | cn_k12 | false |
9. When Xiaoyan goes to school, she rides a bike, and when she returns home, she walks, which takes a total of 50 minutes. If she walks both ways, the entire journey would take 70 minutes. Therefore, if Xiaoyan rides a bike both ways, the time needed would be $\qquad$ minutes. | answer: 30 | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
If α and β are two interior angles of a certain triangle, and it satisfies $\sin\alpha=\cos\beta$, then the shape of this triangle must be ( )
A: Right-angled triangle
B: Acute-angled triangle
C: Isosceles triangle
D: Right-angled triangle or acute-angled triangle | Since α and β are two angles of a triangle and it is given that $\sin\alpha=\cos\beta$, we can use the complementary angle identity stating that $\sin\theta=\cos(90^\circ-\theta)$ for any angle θ. Hence, we can infer that α and β are complementary, which implies:
$$\alpha + \beta = 90^\circ$$
Considering that the sum... | \text{A: Right-angled triangle} | Geometry | MCQ | Yes | Yes | cn_k12 | false |
The ratio of flour to water to sugar in a recipe is 8 : 4 : 3. The ratio in a new recipe calls for a doubling of the ratio of flour to water from the original recipe and a halving of the ratio of flour to sugar. The new recipe calls for a certain amount of water and it requires 6 cups of sugar. How much water does the ... | Let's first understand the original recipe's ratios:
Flour : Water : Sugar = 8 : 4 : 3
Now, according to the new recipe, we need to double the ratio of flour to water and halve the ratio of flour to sugar.
Original flour to water ratio is 8:4, which simplifies to 2:1. Doubling this ratio gives us 4:1.
Original flou... | 2 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Given the hyperbola $mx^{2}-y^{2}=m (m > 0)$, one of its asymptotes has an angle of inclination that is twice the angle of inclination of the line $x- \sqrt {3}y=0$. Find the value of $m$ ($\,\,$),
A: $3$
B: $\sqrt {3}$
C: $2$
D: $\sqrt {2}$ | From the given information, we know that the asymptote of the hyperbola $\frac{x^{2}}{1}-\frac{y^{2}}{m}=1 (m > 0)$ has an angle of inclination of $60^{\circ}$.
The slope of an asymptote of a hyperbola is given by $\sqrt{m}$, and the tangent of the angle of inclination is equal to the slope. Therefore, we can write th... | m=3 | Algebra | MCQ | Yes | Yes | cn_k12 | false |
In $\triangle ABC$, if the sides opposite to angles $A$, $B$, $C$ are $a$, $b$, $c$ respectively, and $\sqrt {2}a=2b\sin A$, then angle $B=$ \_\_\_\_\_\_. | Given that $\sqrt {2}a=2b\sin A$, by the sine law we have:
$$
\sqrt {2}\sin A=2\sin B\sin A
$$
Since $\sin A\neq 0$, we can simplify the equation to:
$$
\sin B= \frac { \sqrt {2}}{2}
$$
As $B\in (0,\pi)$, we have two possible solutions:
$$
B= \frac {π}{4} \quad \text{or} \quad B= \frac {3π}{4}
$$
Therefore, the possibl... | B=\frac{3\pi}{4} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
If the fifth term of the expansion of the binomial $$(x \sqrt {x}- \frac {1}{x})^{6}$$ is 5, then the value of $x$ is \_\_\_\_\_\_. | To solve this, we use the general term of the binomial expansion, which is $T_{r+1} = C_{6}^{r}(x \sqrt {x})^{6-r} \cdot (- \frac {1}{x})^{r} = (-1)^{r}C_{6}^{r}x^{\frac {18-5r}{2}}$.
Given that the fifth term of the expansion is 5, we have $T_{5} = (-1)^{4}C_{6}^{4}x^{-1} = 5$.
This simplifies to $15x^{-1} = 5$.
So... | 3 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Given $f(x)$ is an even function, and when $x < 0$, $f(x) = x^2 + x$, then $f(3) = \ ?$ | Since when $x < 0$, $f(x) = x^2 + x$, we have $f(-3) = 6$.
Because $f(x)$ is an even function,
we have $f(3) = f(-3) = 6$.
Therefore, the answer is $\boxed{6}$.
**Analysis:** From the given information, when $x < 0$, $f(x) = x^2 + x$, we can find $f(-3) = 6$. According to the properties of an even function, we... | 6 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
At Challenger Charter School, to pass a geometry test you must score at least $85\%$. The test consists of 50 problems. What is the maximum number of problems a student can miss and still pass the test? | To find the maximum number of problems missed while still passing:
1. Calculate the allowable percentage to miss which is \(100\% - 85\% = 15\%\).
2. Calculate \(15\%\) of 50 to find out how many problems can be missed:
\[
0.15 \times 50 = 7.5
\]
3. Since a student can't miss half a problem, round down to ensu... | 7 | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
Tameka is selling boxes of crackers for her scout troop. On Friday, she sold 40 boxes. On Saturday, she sold 10 fewer than twice that number. And on Sunday, she sold x% as many as Sunday. How many boxes did she sell over the three days?
If we know the answer to the above question is 145, what is the value of unknown va... | To determine the total number of boxes Tameka sold over the three days, we need to calculate the number of boxes sold each day and then sum them up.
First, let's find out how many boxes Tameka sold on Saturday:
- On Friday, she sold 40 boxes.
- On Saturday, she sold 10 fewer than twice the number of boxes sold on Frid... | 87.5 | Algebra | math-word-problem | Incomplete | Yes | metamath | true |
Point $A$ is on the number line. The number corresponding to point $A$ is represented by $2a+1$, and the distance from point $A$ to the origin is $3$. Find the value of $a$. | Given that the distance from point $A$ to the origin is represented by the absolute value $|2a+1|=3$, we proceed as follows:
1. We start with the equation given by the absolute value:
\[|2a+1|=3\]
2. This leads to two possible equations due to the nature of absolute values:
\[2a+1 = 3 \quad \text{or} \quad 2a+1 = -3\... | a = 1 \text{ or } a = -2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Given that the function $f(x)$ is an even function, and when $x 0$, $f(x) = $ ( )
A: $3x - 1$
B: $3x + 1$
C: $-3x - 1$
D: $-3x + 1$ | If $x > 0$, then $-x 0$ to $x < 0$ allows us to find the expression of the function. | null | Algebra | MCQ | Incomplete | Problem not solved | cn_k12 | false |
The function $f(n)$ is defined on the nonnegative integers $n$ by: $f(0) = 0, f(1) = 1$, and
\[f(n) = f\left(n -\frac{1}{2}m(m - 1)\right)-f\left(\frac{1}{2}m(m+ 1)-n\right)\]
for $\frac{1}{2}m(m - 1) < n \le \frac{1}{2}m(m+ 1), m \ge 2$. Find the smallest integer $n$ for which $f(n) = 5$. | null | null | Other | math-word-problem | Yes | Yes | aops_forum | false |
Find all $10$-digit whole numbers $N$, such that first $10$ digits of $N^2$ coincide with the digits of $N$ (in the same order). | 1. Let \( N = 10^9 K + X \) where \( 0 \le X \le 10^9 - 1 \). This representation helps us understand the structure of \( N \) in terms of its digits.
2. Given \( 10^9 K \le 10^9 K + X < 10^9 (K+1) \), we know that \( 10^{18} K^2 \le N^2 < 10^{18} (K+1)^2 \). This inequality helps us bound \( N^2 \).
3. Since \( K \le ... | 1000000000 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
How can you go from the number 11 to 25 by only multiplying with 2 or decreasing with 3 in a minimum number of steps? | null | null | Logic and Puzzles | math-word-problem | Yes | Yes | aops_forum | false |
Given the function \(f(x) = \ln x + \frac{a}{x}\) \((a\in \mathbb{R})\),
(1) If the function \(f(x)\) has two zeros, determine the range of the real number \(a\).
(2) If the function \(f(x)\) has extreme values, and the line \(y=m\) intersects the graph of the function \(y=f(x)\) at two points \((x_1, m)\) and \((x_2... | (1) Since \(f(x) = \ln x + \frac{a}{x}\) \((a\in \mathbb{R})\) and \(x > 0\),
The first derivative of \(f\) is given by
\[
f'(x) = \frac{1}{x} - \frac{a}{x^2} = \frac{x-a}{x^2}.
\]
Therefore, when \(x > a\), \(f'(x) > 0\), and \(f(x)\) is increasing.
When \(x a,
\]
Therefore,
\[
\boxed{x_1+x_2 > 2a}.
\] | x_1+x_2 > 2a | Calculus | math-word-problem | Yes | Yes | cn_k12 | false |
## Task 9 - 020829
The following statement is to be proven:
The centers of the squares constructed on the sides of any parallelogram, such that the squares lie outside the parallelogram, form a square when connected sequentially.
(It is not sufficient to just draw the diagram; that is not a proof! The properties of ... | Let the midpoints over the sides $a, b, c$ and $d$ be denoted by $M_{a}, M_{b}, M_{c}$ and $M_{d}$ in this order. Then one can show the congruence of, for example, triangle $A M_{a} M_{d}$ and $B M_{a} M_{b}$.
It holds that
$A M_{a}=B M_{a}$ (half diagonals in the same square),
$A M_{d}=B M_{b}$ (half diagonals in c... | proof | Geometry | proof | Yes | Yes | olympiads | false |
$4 \cdot 231$ Find all real numbers $a$ such that there exist non-negative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfying
$$
\sum_{k=1}^{5} k x_{k}=a, \sum_{k=1}^{5} k^{3} \cdot x_{k}=a^{2}, \sum_{k=1}^{5} k^{5} x_{k}=a^{3} .
$$ | [Solution] Let there be non-negative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ that satisfy the conditions. Then, by the Cauchy-Schwarz inequality,
$$
\begin{array}{l}
a^{4}=\left(\sum_{k=1}^{5} k^{3} x_{k}\right)^{2}=\left[\sum_{k=1}^{5}\left(k^{\frac{1}{2}} x_{k}^{\frac{1}{2}}\right)\left(k^{\frac{5}{2}} x_{k}... | 0,1,4,9,16,25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
If $x_{1}$ and $x_{2}$ are the two roots of the quadratic equation $x^{2}+x-2=0$, then the value of $x_{1}\cdot x_{2}$ is:
A: $-1$
B: $-2$
C: $1$
D: $2$ | Given the quadratic equation $x^{2} + x - 2 = 0$, we can identify $a = 1$, $b = 1$, and $c = -2$ based on the standard form of a quadratic equation $ax^{2} + bx + c = 0$.
The product of the roots of a quadratic equation, $x_{1} \cdot x_{2}$, can be found using the formula derived from Vieta's formulas, which is $x_{1}... | \text{B} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
A group of 8 friends were playing a video game. In the game, each player started with some lives. They had a total of 64 lives. How many lives did each player start with? | To find out how many lives each player started with, you would divide the total number of lives by the number of players.
So, 64 lives ÷ 8 players = 8 lives per player.
Each player started with $\boxed{8}$ lives. | 8 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
If the complex number $z$ satisfies $iz=2-i$, then $\overline{z}=\left(\ \ \right)$
A: $2+i$
B: $-1-2i$
C: $-1+2i$
D: $1+2i$ | To solve the given problem, we start by manipulating the given equation for $z$ as follows:
\[iz = 2 - i\]
To find $z$, we isolate it on one side:
\[z = \frac{2 - i}{i}\]
To simplify the fraction, we multiply both numerator and denominator by $i$:
\[z = \frac{(2 - i) \cdot i}{i \cdot i} = \frac{2i - i^2}{i^2}\]
W... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Given an integer $n \ge 3$, prove that the diameter of a convex $n$-gon (interior and boundary) containing a disc of radius $r$ is (strictly) greater than $r(1 + 1/ \cos( \pi /n))$.
The Editors | 1. Let \( K \) be a convex \( n \)-gon (interior and boundary) containing a disc \( D \) of radius \( r \), centered at some point \( O \). Let \( a_1, a_2, \ldots, a_n \) be the vertices of \( K \), labeled in a circular order around the boundary. Let \( 2\alpha_i \) be the angular span of \( D \) from \( a_i \), for ... | r \left(1 + \frac{1}{\cos \left(\frac{\pi}{n}\right)}\right) | Geometry | proof | Yes | Yes | aops_forum | false |
At Galbraith H.S., the lockers are arranged in banks of 20 lockers. Each bank of lockers consists of six columns of lockers; the first two columns in each bank consist of two larger lockers and the last four columns in each bank consist of four smaller lockers. The lockers are numbered consecutively starting at 1 , mov... | (a) In the second bank of lockers, each of the first two columns consists of two lockers.
The first column consists of lockers 21 and 22, and the second column consists of lockers 23 and 24 .
Thus, the sum of the locker numbers in the column containing locker 24 is $23+24=47$.
(b) Suppose that a column contains two ... | 47,6162,133,134,135,136,proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Which are the two consecutive odd numbers in the number sequence, the sum of whose squares is of the form $\frac{n(n+1)}{2}$, where $n$ is a natural number. | Let the two adjacent odd numbers be $2k-1$ and $2k+1$, then
$$
(2k-1)^{2}+(2k+1)^{2}=\frac{n(n+1)}{2}
$$
which means
$$
16k^{2}+4=n(n+1)
$$
Multiplying this by 4 and adding 1,
$$
64k^{2}+17=(2n+1)^{2}
$$
Denoting $(2n+1)$ by $m$, we get
$$
m^{2}-64k^{2}=(m+8k)(m-8k)=17
$$
Here, by assumption, $m$ is positive, s... | 1^{2}+3^{2}=\frac{4\cdot5}{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
If the temperature rises by $1^{\circ}\mathrm{C}$, it is denoted as $+1^{\circ}\mathrm{C}$. Then if the temperature decreases by $2^{\circ}\mathrm{C}$, it is denoted as
A: $-2^{\circ}\mathrm{C}$
B: $+2^{\circ}\mathrm{C}$
C: $-1^{\circ}\mathrm{C}$
D: $+1^{\circ}\mathrm{C}$ | To solve this problem, we follow the logic provided in the question and solution:
1. A temperature increase is denoted with a positive sign. For example, a rise of $1^{\circ}\mathrm{C}$ is represented as $+1^{\circ}\mathrm{C}$.
2. Following this logic, a temperature decrease should be denoted with a negative sign.
3. ... | \text{A} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Determine the smallest positive integer \( n \) such that
\[
\sum_{k = 0}^n \log_3 \left( 1 + \frac{1}{3^{3^k}} \right) \ge 1 + \log_3 \frac{4030}{4031}.
\] | First, transform the sum of logarithms to a single logarithm:
\[
\sum_{k = 0}^n \log_3 \left( 1 + \frac{1}{3^{3^k}} \right) = \log_3 \left[ \prod_{k = 0}^n \left( 1 + \frac{1}{3^{3^k}} \right) \right].
\]
The product inside the logarithm can be simplified using:
\[
(1 + x)(1 + x^3)(1 + x^9) \dotsm (1 + x^{3^n}) = \frac... | 1 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Given the following functions: (1) $f(x) = \frac{x^2-1}{x-1}$, $g(x)=x+1$; (2) $f(x)=|x|$, $g(x)=\sqrt{x^2}$; (3) $f(x)=x^2-2x-1$, $g(t)=t^2-2t-1$. Among them, those that represent the same function are ( )
A. ①②③
B. ①③
C. ②③
D. ② | For (1), the function $f(x)= \frac{x^2-1}{x-1}$ simplifies to $f(x)=x+1$ when $x\neq 1$, which is different from the function $g(x)=x+1$ for $x \in \mathbb{R}$ due to their domains. Hence, they do not represent the same function.
For (2), the function $f(x)=|x|$ for $x \in \mathbb{R}$ is identical to $g(x)=\sqrt{x^2}$... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false |
In $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. Given that $2b\cos B=a\cos C+c\cos A$ and $b^{2}=3ac$, find the measure of angle $A$. | In $\triangle ABC$, we have $2b\cos B=a\cos C+c\cos A$. By the sine law, we can write this as $2\sin B\cos B=\sin A\cos C+\sin C\cos A$. This simplifies to $\sin 2B=\sin (A+C)$.
Since $0^\circ < B < 180^\circ$, we have $2B=A+C$ (if $2B=180^{\circ}-(A+C)$, then $B=0^{\circ}$ which is not possible given the problem).
W... | A=\frac{7\pi}{12} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
A positive five-digit integer is in the form $AB,CBA$; where $A$, $B$, and $C$ are each distinct digits. What is the greatest possible value of $AB,CBA$ that is both divisible by eleven and by three? | 1. The integer $AB,CBA$ can be expressed as $10001A + 1010B + 100C$. For divisibility by $11$, we have $2A - 2B + C \equiv 0 \pmod{11}$.
2. For divisibility by $3$, the sum of the digits $A + B + C + B + A = 2A + 2B + C$ must be divisible by $3$.
3. Attempting with $A=9$ (maximizing the five-digit integer), the conditi... | 96,569 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
My three-digit code is 045. Reckha cannot choose a code that matches mine in two or more of the three digit-positions, nor can she choose a code that is just a permutation of the digits of my code (so 450 and 504 are forbidden, but 405 is fine). Reckha can otherwise choose any three-digit code where each digit is in th... | 1. Calculate the total number of three-digit codes:
$$ 10^3 = 1000 \text{ codes} $$
2. Subtract codes that differ by exactly one digit from "045":
- There are 3 positions to change, and 9 possible new digits for each position (0-9 except the original digit):
$$ 3 \times 9 = 27 \text{ codes} $$
3. Subtract co... | 970 | Combinatorics | math-word-problem | Yes | Yes | synthetic_math | true |
Given a function defined on $(-1,1)$, $f(x)=x-\sin x$, if $f(a-2)+f(4-a^2) < 0$, then the range of $a$ is ( )
A: $(-\infty ,-1)\cup (2,+\infty )$
B: $(\sqrt{3},\sqrt{5})$
C: $(2,\sqrt{5})$
D: $(0,2)$ | **Analysis**
Analyze the odd and even properties and the monotonicity of the function. Use the function's odd and even properties to transform the inequality into $f(a-2) < f(a^2-4)$, and then solve it by its monotonicity.
**Solution**
Since $f(x)=x-\sin x$, it is easy to see that $f(x)$ is an odd function. From $f(... | \text{C} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
10. Satisfy $0 \leqslant k_{i} \leqslant 20(i=1,2,3,4)$, and $k_{1}+k_{3}=k_{2}+k_{4}$ of the ordered integer tuples $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$ the number is $\qquad$ . | 10. 6181.
For $0 \leqslant m \leqslant 20$, the non-negative integer solutions satisfying $x+y=m$ and $0 \leqslant x, y \leqslant 20$ are
$$
(x, y)=(j, m-j)(0 \leqslant j \leqslant m),
$$
there are $m+1$ solutions;
When $20<m \leqslant 40$, the non-negative integer solutions satisfying $x+y=m$ and $0 \leqslant x, y \... | 6181 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given the function $f(x) = x^2 - 2x$, where $x \in [a, b]$, and its range is $[-1, 3]$, find the range of possible values for $b - a$. | The function can be rewritten as $f(x) = (x - 1)^2 - 1$. Since the range of $x \in [a, b]$ is $[-1, 3]$, when $a = -1$, we have $1 \leq b \leq 3$; when $b = 3$, we have $-1 \leq a \leq 1$. Therefore, the range of $b - a$ is $\boxed{[2, 4]}$. | [2, 4] | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
A right cylinder with a base radius of 4 units is inscribed in a sphere of radius 7 units. Calculate the total volume, in cubic units, of the space inside the sphere and outside the cylinder, expressed as \( V\pi \). Find \( V \) as a common fraction. | First, visualize the cylinder inscribed within the sphere. The height of the cylinder \( h \) and the cylinder's radius \( r \) create a right triangle with the sphere's radius \( R \) being the hypotenuse. The radius of the cylinder is given as 4 units and the radius of the sphere is 7 units.
We use the Pythagorean t... | V = \frac{1372}{3} - 32\sqrt{33} | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
Example 1 Given
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac... | Solve: Consider the given system of equations as a fractional equation in terms of $t$
$$
\begin{array}{c}
\frac{x^{2}}{t-1^{2}}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1 \\
\Rightarrow\left(t-1^{2}\right)\left(t-3^{2}\right)\left(t-5^{2}\right)\left(t-7^{2}\right)- \\
x^{2}\left(t-3^{2}\right... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. For the cyclic quadrilateral $ABCD$, the lengths of the four sides in sequence are $AB=2, BC=7, CD=6, DA=9$. Then the area of the quadrilateral is $\qquad$ . | 3. 30 .
Since $7^{2}+6^{2}=85=9^{2}+2^{2}$, that is,
$$
B C^{2}+C D^{2}=D A^{2}+A B^{2} \text {, }
$$
thus, $\triangle B C D$ and $\triangle D A B$ are both right triangles.
Therefore, the area of the quadrilateral is
$$
S_{\triangle B C D}+S_{\triangle D A B}=\frac{1}{2}(7 \times 6+9 \times 2)=30 \text {. }
$$ | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Terrell typically lifts two 30-pound weights 10 times during his workout. If he decides to use two 20-pound weights instead, how many times must he lift them to match the same total weight lifted previously? | First, calculate the total weight Terrell lifts with the original setup of two 30-pound weights lifted 10 times:
\[2 \cdot 30 \cdot 10 = 600 \text{ pounds}.\]
Now, set up an equation for the new scenario where Terrell uses two 20-pound weights, lifted $n$ times, to match the previous total weight:
\[2 \cdot 20 \cdot n... | 15 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Find all positive integers $d$ with the following property: there exists a polynomial $P$ of degree $d$ with integer coefficients such that $\left|P(m)\right|=1$ for at least $d+1$ different integers $m$. | 1. **Identify the problem constraints:**
We need to find all positive integers \( d \) such that there exists a polynomial \( P \) of degree \( d \) with integer coefficients, and \(|P(m)| = 1\) for at least \( d+1 \) different integers \( m \).
2. **Construct examples for small \( d \):**
- For \( d = 1 \), con... | d \leq 3 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
In a modified game of Deal or No Deal, participants choose a box at random from a set of $30$ boxes. Each box contains one of the following values, each appearing once: $0.01, 1, 5, 10, 25, 50, 100, 200, 300, 400, 500, 750, 1,000, 5,000, 10,000, 25,000, 50,000, 75,000, 100,000, 200,000, 300,000, 400,000, 500,000, 750,0... | This problem requires calculating how many of the $30$ boxes contain at least $200,000$. The values equal to or above $200,000$ are $200,000, 300,000, 400,000, 500,000, 750,000, 1,000,000$. This means 7 boxes meet the condition.
To have a 50% chance of having at least $200,000$, the participant must be left with a tot... | 16 | Combinatorics | math-word-problem | Yes | Yes | synthetic_math | true |
Let $S=\{1,2,\cdots ,n\}$ where $n$ is an odd integer. Let $f$ be a function defined on $\{(i,j): i\in S, j \in S\}$ taking values in $S$ such that
(i) $f(s,r)=f(r,s)$ for all $r,s \in S$
(ii) $\{f(r,s): s\in S\}=S$ for all $r\in S$
Show that $\{f(r,r): r\in S\}=S$ | 1. **Define the problem and given conditions:**
- Let \( S = \{1, 2, \ldots, n\} \) where \( n \) is an odd integer.
- Define a function \( f \) on \( \{(i, j) : i \in S, j \in S\} \) taking values in \( S \) such that:
- \( f(s, r) = f(r, s) \) for all \( r, s \in S \) (symmetry condition).
- \( \{f(r,... | \{f(r, r) : r \in S\} = S | Logic and Puzzles | proof | Yes | Yes | aops_forum | false |
Determine all acute angle values of $x$ such that an equilateral triangle with sides defined as $\cos x$, $\cos x$, and $\cos 5x$ is possible. Assume angle measures are in degrees. | For an equilateral triangle, all sides must be equal, so we must have $\cos x = \cos 5x$.
Using the properties of cosine:
\[
\cos x = \cos 5x \implies 5x = 360^\circ k \pm x \text{ for some integer } k.
\]
Solving for $x$:
1. Case $5x = 360^\circ k + x$:
\[
4x = 360^\circ k \implies x = 90^\circ k.
\]
Since $x$ must b... | 60^\circ | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
2.4. Find all values of $x$ for which the smallest of the numbers $8-x^{2}$ and $\operatorname{ctg} x$ is not less than -1. In the answer, record the total length of the found intervals on the number line, rounding it to hundredths if necessary. | Answer: 4.57 (exact value: $3+\frac{\pi}{2}$ ). | 4.57 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The residents of the Country of Fairy Tales have invited Baba Yaga, Koschei, and Zmey Gorynych as honored guests to their celebration. The residents have learned the following. If Baba Yaga comes to the celebration, then Koschei will not come. If Koschei comes, then Zmey Gorynych will not come. If Zmey Gorynych come... | # Solution
The condition "If Baba Yaga comes to the party, then Koschei will not come" is equivalent to the statement that both Baba Yaga and Koschei come to the party - an impossible event.
By similar conditions, it is also impossible for both Koschei and Zmey Gorynych, or both Zmey Gorynych and Baba Yaga, to come t... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false |
A certain country is divided into 5 provinces. Each province consists entirely of progressives and traditionalists. If each province contains the same number of traditionalists and the number of traditionalists in any given province is 1/15 the total number of progressives in the entire country, what fraction of the co... | Let's denote the number of traditionalists in each province as T. Since there are 5 provinces, the total number of traditionalists in the country is 5T.
According to the problem, T is 1/15 the total number of progressives in the entire country. Let's denote the total number of progressives in the country as P. So we h... | 1/4 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Compute the unique positive integer $m$ such that
\[1 \cdot 2^1 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + m \cdot 2^m = 2^{m + 8}.\] | Let
\[S = 1 \cdot 2^1 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + m \cdot 2^m.\]
Then
\[2S = 1 \cdot 2^2 + 2 \cdot 2^3 + 3 \cdot 2^4 + \dots + m \cdot 2^{m+1}.\]
Subtracting these equations, we get
\[
S = 1 \cdot 2^2 + 2 \cdot 2^3 + 3 \cdot 2^4 + \dots + m \cdot 2^{m+1} - (1 \cdot 2^1 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + m ... | 129 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
When the product \( (4x + 3y + 2)(2x + 5y + 6) \) is expanded, what is the sum of the coefficients of the terms which contain a nonzero power of \( y \)? | We multiply out using the distributive property:
\[
\begin{align*}
&(4x + 3y + 2)(2x + 5y + 6) \\
&= 4x(2x + 5y + 6) + 3y(2x + 5y + 6) + 2(2x + 5y + 6) \\
&= 8x^2 + 20xy + 24x + 6xy + 15y^2 + 18y + 4x + 10y + 12 \\
&= 8x^2 + 26xy + 28x + 15y^2 + 28y + 12.
\end{align*}
\]
The terms containing some power of \( y \) are... | 69 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Circles $c_1, c_2$ with centers $O_1, O_2$, respectively, intersect at points $P$ and $Q$ and touch circle c internally at points $A_1$ and $A_2$, respectively. Line $PQ$ intersects circle c at points $B$ and $D$. Lines $A_1B$ and $A_1D$ intersect circle $c_1$ the second time at points $E_1$ and $F_1$, respectively, an... | null | null | Geometry | proof | Yes | Yes | aops_forum | false |
Jennifer decides to share her sweets between herself and her 3 friends. She has 212 green sweets, 310 blue sweets and 502 yellow sweets. How many sweets will Jennifer and her friends get each? | To find out how many sweets Jennifer and her friends will get each, we need to add up all the sweets and then divide them by the number of people sharing them.
Jennifer has:
- 212 green sweets
- 310 blue sweets
- 502 yellow sweets
First, let's add up all the sweets:
212 + 310 + 502 = 1024 sweets in total
Now, Jennif... | 256 | Logic and Puzzles | math-word-problem | Yes | Yes | orca_math | true |
4. Let $p$ be a prime, and $f(x)$ be a $d$-degree polynomial with integer coefficients satisfying:
(1) $f(0)=0, f(1)=1$;
(2) For any positive integer $n, f(n)$ when divided by $p$ leaves a remainder of 0 or 1.
Prove: $d \geqslant p-1$. | Hint Refer to Lemma 2 of Example 5. | proof | Number Theory | proof | Yes | Incomplete | olympiads | false |
12. (15 points) Two squares $A B C D$ and $B E F G$ with side lengths of $8 \mathrm{~cm}$ and $6 \mathrm{~cm}$ respectively are placed side by side as shown in the figure. Connecting $D E$ intersects $B G$ at $P$. What is the area of the shaded region $A P E G$ in the figure? | 【Analysis】First, we need to find the relationship between the shaded part and the side length of the known square. We can transform $\triangle A P G$ into $\triangle D P G$ with the same base and height, and then connect it to the square with a side length of 6 using the principle of equal area transformation.
【Soluti... | 18\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11 |
| :---: | :---: | :---: |
| | [ Area of a triangle (through two sides and the angle between | |
| | The inscribed angle is half of the central angle | |
| | Cauchy's inequality | |
A circle is circumscribed around a unit square \(ABCD\), and a point \(M\) is chosen on this circle.
What is the maximum va... | Let $M$ not coincide with any of the vertices of the square (otherwise, the value of the product is zero and cannot be the greatest) and lie on the arc $BC$, $O$ - the center of the circumscribed circle (see figure). Note that $\angle A M B = 1/2 \angle A O B = 45^{\circ}$. Since $S_{A M B} = 1/2 M A \cdot M B \sin \an... | 0.5 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
A rectangle with width $30$ inches has the property that all points in the rectangle are within $12$ inches of at least one of the diagonals of the rectangle. Find the maximum possible length for the rectangle in inches. | null | null | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
$11 \cdot 17$ plane, is there 100 different lines that have exactly 1985 different intersection points? | ［Solution］There exists. Construct the lines
$$
\begin{array}{l}
l_{j}=\{(x, y) \mid x=j\}, j=0,1, \cdots, 72 ; \\
l_{j}=\{(x, y) \mid y=j-73\}, j=73,74, \cdots, 98,
\end{array}
$$
Then the number of different intersection points among these lines is $73 \times 26=1898$.
Next, select the 100th line such that it interse... | 1985 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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