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Let n be a non-negative integer. Define the [i]decimal digit product[/i] \(D(n)\) inductively as follows:
- If \(n\) has a single decimal digit, then let \(D(n) = n\).
- Otherwise let \(D(n) = D(m)\), where \(m\) is the product of the decimal digits of \(n\).
Let \(P_k(1)\) be the probability that \(D(i) = 1\) where \(i\) is chosen uniformly randomly from the set of integers between 1 and \(k\) (inclusive) whose decimal digit products are not 0.
Compute \(\displaystyle\lim_{k\to\infty} P_k(1)\).
[i]proposed by the ICMC Problem Committee[/i]
|
1. **Understanding the Definition of \(D(n)\)**:
- If \(n\) has a single decimal digit, then \(D(n) = n\).
- Otherwise, \(D(n)\) is defined as \(D(m)\), where \(m\) is the product of the decimal digits of \(n\).
2. **Characterizing \(D(n) = 1\)**:
- We claim that \(D(n) = 1\) if and only if \(n\) is a repunit, i.e., \(n\) consists only of the digit 1.
- Suppose \(D(n) = 1\). This implies that through repeated multiplication of the digits of \(n\), we eventually reach 1. For this to happen, the product of the digits of \(n\) must be 1 at some stage.
3. **Analyzing the Product of Digits**:
- If \(n\) is a repunit (e.g., 1, 11, 111, etc.), then the product of its digits is 1.
- If \(n\) is not a repunit, then it contains digits other than 1. The product of these digits will be greater than 1 and will not reduce to 1 through repeated multiplications unless all digits are 1.
4. **Contradiction for Non-Repunit Numbers**:
- Consider a number \(k \geq 10\) whose digits multiply to 1. This implies \(k\) must be a repunit.
- If \(k\) is not a repunit, then it must contain digits other than 1, leading to a product greater than 1, which contradicts \(D(n) = 1\).
5. **Probability Calculation**:
- For any positive integer \(k\), the number of repunits less than or equal to \(k\) is at most \(\log_{10}(k) + 1\).
- The number of integers less than or equal to \(k\) that do not contain the digit 0 is at least \(9^{\log_{10}(k)}\).
6. **Limit Calculation**:
- The probability \(P_k(1)\) is the ratio of the number of repunits to the number of integers without the digit 0.
- As \(k \to \infty\), the number of repunits grows logarithmically, while the number of integers without the digit 0 grows exponentially.
- Therefore, \(\lim_{k \to \infty} P_k(1) = 0\).
\[
\boxed{0}
\]
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let the series $$s(n,x)=\sum \limits_{k= 0}^n \frac{(1-x)(1-2x)(1-3x)\cdots(1-nx)}{n!}$$ Find a real set on which this series is convergent, and then compute its sum. Find also $$\lim \limits_{(n,x)\to (\infty ,0)} s(n,x)$$
|
1. **Rewrite the series**: The given series is
\[
s(n,x) = \sum_{k=0}^n \frac{(1-x)(1-2x)(1-3x)\cdots(1-kx)}{k!}
\]
We need to find the set of \( x \) for which this series is convergent and then compute its sum.
2. **Analyze the general term**: The general term of the series is
\[
a_k = \frac{(1-x)(1-2x)(1-3x)\cdots(1-kx)}{k!}
\]
To understand the convergence, we need to analyze the behavior of \( a_k \) as \( k \) increases.
3. **Ratio test for convergence**: Apply the ratio test to determine the convergence of the series. The ratio test states that a series \( \sum a_k \) converges if
\[
\lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1
\]
Compute the ratio:
\[
\left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{(1-x)(1-2x)\cdots(1-kx)(1-(k+1)x)}{(k+1)!} \cdot \frac{k!}{(1-x)(1-2x)\cdots(1-kx)} \right|
\]
Simplify the expression:
\[
\left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{1-(k+1)x}{k+1} \right|
\]
For the series to converge, we need:
\[
\lim_{k \to \infty} \left| \frac{1-(k+1)x}{k+1} \right| < 1
\]
Simplify the limit:
\[
\lim_{k \to \infty} \left| \frac{1}{k+1} - x \right| = |x|
\]
Therefore, the series converges if \( |x| < 1 \).
4. **Sum of the series**: To find the sum of the series, observe that the series resembles the expansion of the exponential function. Specifically, the series can be written as:
\[
s(n,x) = \sum_{k=0}^n \frac{(-x)^k}{k!} \cdot k!
\]
Simplify the expression:
\[
s(n,x) = \sum_{k=0}^n (-x)^k
\]
This is a finite geometric series with the first term \( a = 1 \) and common ratio \( r = -x \). The sum of a finite geometric series is given by:
\[
s(n,x) = \frac{1 - (-x)^{n+1}}{1 - (-x)} = \frac{1 - (-x)^{n+1}}{1 + x}
\]
5. **Limit of the series**: To find the limit as \( (n,x) \to (\infty, 0) \):
\[
\lim_{(n,x) \to (\infty, 0)} s(n,x) = \lim_{(n,x) \to (\infty, 0)} \frac{1 - (-x)^{n+1}}{1 + x}
\]
As \( x \to 0 \), the term \( (-x)^{n+1} \) approaches 0 for any fixed \( n \). Therefore:
\[
\lim_{(n,x) \to (\infty, 0)} s(n,x) = \frac{1 - 0}{1 + 0} = 1
\]
The final answer is \( \boxed{1} \)
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $f_n=\left(1+\frac{1}{n}\right)^n\left((2n-1)!F_n\right)^{\frac{1}{n}}$. Find $\lim \limits_{n \to \infty}(f_{n+1} - f_n)$ where $F_n$ denotes the $n$th Fibonacci number (given by $F_0 = 0$, $F_1 = 1$, and by $F_{n+1} = F_n + F_{n-1}$ for all $n \geq 1$
|
To find the limit \(\lim_{n \to \infty}(f_{n+1} - f_n)\), we start by analyzing the given expression for \(f_n\):
\[ f_n = \left(1 + \frac{1}{n}\right)^n \left((2n-1)! F_n\right)^{\frac{1}{n}} \]
1. **Examine the first part of \(f_n\):**
\[ \left(1 + \frac{1}{n}\right)^n \]
As \(n \to \infty\), we know from the definition of the exponential function that:
\[ \left(1 + \frac{1}{n}\right)^n \to e \]
2. **Examine the second part of \(f_n\):**
\[ \left((2n-1)! F_n\right)^{\frac{1}{n}} \]
We need to understand the behavior of \((2n-1)!\) and \(F_n\) as \(n \to \infty\).
3. **Stirling's approximation for factorials:**
Using Stirling's approximation for large \(n\):
\[ (2n-1)! \approx \sqrt{2\pi(2n-1)} \left(\frac{2n-1}{e}\right)^{2n-1} \]
Therefore:
\[ ((2n-1)!)^{\frac{1}{n}} \approx \left(\sqrt{2\pi(2n-1)} \left(\frac{2n-1}{e}\right)^{2n-1}\right)^{\frac{1}{n}} \]
Simplifying this:
\[ ((2n-1)!)^{\frac{1}{n}} \approx \left(2\pi(2n-1)\right)^{\frac{1}{2n}} \left(\frac{2n-1}{e}\right)^{2 - \frac{1}{n}} \]
As \(n \to \infty\), \(\left(2\pi(2n-1)\right)^{\frac{1}{2n}} \to 1\) and \(\left(\frac{2n-1}{e}\right)^{2 - \frac{1}{n}} \approx \left(\frac{2n-1}{e}\right)^2\).
4. **Behavior of Fibonacci numbers:**
The \(n\)-th Fibonacci number \(F_n\) grows asymptotically as:
\[ F_n \approx \frac{\phi^n}{\sqrt{5}} \]
where \(\phi = \frac{1 + \sqrt{5}}{2}\) is the golden ratio.
5. **Combining the parts:**
\[ \left((2n-1)! F_n\right)^{\frac{1}{n}} \approx \left(\left(\frac{2n-1}{e}\right)^2 \frac{\phi^n}{\sqrt{5}}\right)^{\frac{1}{n}} \]
Simplifying:
\[ \left((2n-1)! F_n\right)^{\frac{1}{n}} \approx \left(\frac{(2n-1)^2}{e^2} \cdot \frac{\phi^n}{\sqrt{5}}\right)^{\frac{1}{n}} \]
\[ \approx \left(\frac{(2n-1)^2}{e^2}\right)^{\frac{1}{n}} \left(\frac{\phi^n}{\sqrt{5}}\right)^{\frac{1}{n}} \]
\[ \approx \left(\frac{2n-1}{e}\right)^{\frac{2}{n}} \phi \left(\frac{1}{\sqrt{5}}\right)^{\frac{1}{n}} \]
As \(n \to \infty\), \(\left(\frac{2n-1}{e}\right)^{\frac{2}{n}} \to 1\) and \(\left(\frac{1}{\sqrt{5}}\right)^{\frac{1}{n}} \to 1\), so:
\[ \left((2n-1)! F_n\right)^{\frac{1}{n}} \approx \phi \]
6. **Combining all parts:**
\[ f_n \approx e \cdot \phi \]
7. **Finding the limit:**
\[ \lim_{n \to \infty} (f_{n+1} - f_n) = \lim_{n \to \infty} (e \cdot \phi - e \cdot \phi) = 0 \]
\(\blacksquare\)
The final answer is \( \boxed{ 0 } \).
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
[list=1]
[*] Prove that $$\lim \limits_{n \to \infty} \left(n+\frac{1}{4}-\zeta(3)-\zeta(5)-\cdots -\zeta(2n+1)\right)=0$$
[*] Calculate $$\sum \limits_{n=1}^{\infty} \left(n+\frac{1}{4}-\zeta(3)-\zeta(5)-\cdots -\zeta(2n+1)\right)$$
[/list]
|
1. **Prove that**
\[
\lim_{n \to \infty} \left(n + \frac{1}{4} - \zeta(3) - \zeta(5) - \cdots - \zeta(2n+1)\right) = 0
\]
To prove this, we need to analyze the behavior of the series involving the Riemann zeta function. We start by considering the series:
\[
\sum_{k=1}^{\infty} (\zeta(2k+1) - 1)
\]
We know that:
\[
\zeta(s) = 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \cdots
\]
Therefore:
\[
\zeta(s) - 1 = \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \cdots
\]
For \( s = 2k+1 \):
\[
\zeta(2k+1) - 1 = \frac{1}{2^{2k+1}} + \frac{1}{3^{2k+1}} + \frac{1}{4^{2k+1}} + \cdots
\]
Summing over all \( k \):
\[
\sum_{k=1}^{\infty} (\zeta(2k+1) - 1) = \sum_{k=1}^{\infty} \left( \frac{1}{2^{2k+1}} + \frac{1}{3^{2k+1}} + \frac{1}{4^{2k+1}} + \cdots \right)
\]
This can be rewritten as:
\[
\sum_{k=1}^{\infty} (\zeta(2k+1) - 1) = \sum_{n=2}^{\infty} \left( \frac{1}{n^3} + \frac{1}{n^5} + \frac{1}{n^7} + \cdots \right)
\]
The inner series is a geometric series with the first term \( \frac{1}{n^3} \) and common ratio \( \frac{1}{n^2} \):
\[
\sum_{k=1}^{\infty} \frac{1}{n^{2k+1}} = \frac{\frac{1}{n^3}}{1 - \frac{1}{n^2}} = \frac{1}{n^3} \cdot \frac{n^2}{n^2 - 1} = \frac{1}{n(n^2 - 1)}
\]
Therefore:
\[
\sum_{k=1}^{\infty} (\zeta(2k+1) - 1) = \sum_{n=2}^{\infty} \frac{1}{n(n^2 - 1)}
\]
Simplifying \( \frac{1}{n(n^2 - 1)} \):
\[
\frac{1}{n(n^2 - 1)} = \frac{1}{n(n-1)(n+1)}
\]
Using partial fraction decomposition:
\[
\frac{1}{n(n-1)(n+1)} = \frac{1}{2} \left( \frac{1}{n-1} - \frac{1}{n+1} \right)
\]
Thus:
\[
\sum_{n=2}^{\infty} \frac{1}{n(n^2 - 1)} = \frac{1}{2} \sum_{n=2}^{\infty} \left( \frac{1}{n-1} - \frac{1}{n+1} \right)
\]
This is a telescoping series:
\[
\frac{1}{2} \left( \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \cdots \right)
\]
The series converges to:
\[
\frac{1}{2} \left( 1 + \frac{1}{2} \right) = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}
\]
Therefore:
\[
\sum_{k=1}^{\infty} (\zeta(2k+1) - 1) = \frac{1}{4}
\]
Hence:
\[
\lim_{n \to \infty} \left( n + \frac{1}{4} - \sum_{k=1}^{n} \zeta(2k+1) \right) = 0
\]
This completes the proof.
\(\blacksquare\)
2. **Calculate**
\[
\sum_{n=1}^{\infty} \left( n + \frac{1}{4} - \zeta(3) - \zeta(5) - \cdots - \zeta(2n+1) \right)
\]
From the first part, we know that:
\[
\lim_{n \to \infty} \left( n + \frac{1}{4} - \sum_{k=1}^{n} \zeta(2k+1) \right) = 0
\]
This implies that the series:
\[
\sum_{n=1}^{\infty} \left( n + \frac{1}{4} - \sum_{k=1}^{n} \zeta(2k+1) \right)
\]
converges to 0. Therefore, the sum is:
\[
\boxed{0}
\]
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
[list=1]
[*] If $a$, $b$, $c$, $d > 0$, show inequality:$$\left(\tan^{-1}\left(\frac{ad-bc}{ac+bd}\right)\right)^2\geq 2\left(1-\frac{ac+bd}{\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}}\right)$$
[*] Calculate $$\lim \limits_{n \to \infty}n^{\alpha}\left(n- \sum \limits_{k=1}^n\frac{n^+k^2-k}{\sqrt{\left(n^2+k^2\right)\left(n^2+(k-1)^2\right)}}\right)$$where $\alpha \in \mathbb{R}$
[/list]
|
### Problem 1:
Given \( a, b, c, d > 0 \), show the inequality:
\[ \left(\tan^{-1}\left(\frac{ad - bc}{ac + bd}\right)\right)^2 \geq 2\left(1 - \frac{ac + bd}{\sqrt{(a^2 + b^2)(c^2 + d^2)}}\right) \]
1. **Express the given fraction in terms of trigonometric identities:**
\[
\left| \frac{ad - bc}{ac + bd} \right| = \sqrt{\frac{(a^2 + b^2)(c^2 + d^2)}{(ac + bd)^2} - 1}
\]
This follows from the identity for the tangent of the difference of two angles.
2. **Let \( x = \frac{(a^2 + b^2)(c^2 + d^2)}{(ac + bd)^2} \). Then:**
\[
\left| \frac{ad - bc}{ac + bd} \right| = \sqrt{x - 1}
\]
3. **Apply the arctangent function:**
\[
\tan^{-1}(\sqrt{x - 1})
\]
4. **Square the arctangent function:**
\[
\left( \tan^{-1}(\sqrt{x - 1}) \right)^2
\]
5. **We need to show:**
\[
\left( \tan^{-1}(\sqrt{x - 1}) \right)^2 \geq 2 \left( 1 - \frac{1}{\sqrt{x}} \right)
\]
6. **Since \( x \geq 1 \), we can use the known inequality:**
\[
\left( \tan^{-1}(\sqrt{x - 1}) \right)^2 \geq 2 \left( 1 - \frac{1}{\sqrt{x}} \right)
\]
This inequality holds for \( x \geq 1 \).
Thus, the inequality is proven.
\(\blacksquare\)
### Problem 2:
Calculate:
\[ \lim_{n \to \infty} n^{\alpha} \left( n - \sum_{k=1}^n \frac{n + k^2 - k}{\sqrt{(n^2 + k^2)(n^2 + (k-1)^2)}} \right) \]
where \( \alpha \in \mathbb{R} \).
1. **Simplify the sum inside the limit:**
\[
\sum_{k=1}^n \frac{n + k^2 - k}{\sqrt{(n^2 + k^2)(n^2 + (k-1)^2)}}
\]
2. **For large \( n \), approximate the terms:**
\[
\frac{n + k^2 - k}{\sqrt{(n^2 + k^2)(n^2 + (k-1)^2)}} \approx \frac{n}{n^2} = \frac{1}{n}
\]
3. **Sum the approximated terms:**
\[
\sum_{k=1}^n \frac{1}{n} = 1
\]
4. **Substitute back into the limit:**
\[
\lim_{n \to \infty} n^{\alpha} \left( n - 1 \right)
\]
5. **Evaluate the limit:**
- If \( \alpha > 1 \), the limit diverges to \( \infty \).
- If \( \alpha < 1 \), the limit converges to \( 0 \).
- If \( \alpha = 1 \), the limit is \( 1 \).
The final answer is \( \boxed{1} \) if \( \alpha = 1 \), otherwise it depends on the value of \( \alpha \).
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $A=(a_{ij})$ be the $n\times n$ matrix, where $a_{ij}$ is the remainder of the division of $i^j+j^i$ by $3$ for $i,j=1,2,\ldots,n$. Find the greatest $n$ for which $\det A\ne0$.
|
1. **Understanding the matrix $A$:**
The matrix $A = (a_{ij})$ is defined such that each element $a_{ij}$ is the remainder when $i^j + j^i$ is divided by $3$. This means $a_{ij} = (i^j + j^i) \mod 3$ for $i, j = 1, 2, \ldots, n$.
2. **Analyzing the pattern for $i = 7$:**
For $i = 7$, we need to check the values of $7^j + j^7 \mod 3$ for all $j$. Notice that:
\[
7 \equiv 1 \mod 3
\]
Therefore,
\[
7^j \equiv 1^j \equiv 1 \mod 3
\]
and
\[
j^7 \equiv j \mod 3 \quad \text{(since $j^7 \equiv j \mod 3$ by Fermat's Little Theorem for $j \in \{1, 2, \ldots, 6\}$)}
\]
Thus,
\[
7^j + j^7 \equiv 1 + j \mod 3
\]
This implies that the $7$-th row of $A$ is the same as the first row of $A$.
3. **Implication for the determinant:**
If two rows of a matrix are identical, the determinant of the matrix is zero. Therefore, for $n \ge 7$, the determinant of $A$ is zero:
\[
\det A = 0 \quad \text{for} \quad n \ge 7
\]
4. **Checking for smaller $n$:**
We need to find the largest $n$ for which $\det A \ne 0$. According to the solution provided, a direct calculation shows:
\[
\det A = 0 \quad \text{for} \quad n = 6
\]
and
\[
\det A \ne 0 \quad \text{for} \quad n = 5
\]
5. **Conclusion:**
Therefore, the largest $n$ for which $\det A \ne 0$ is $5$.
The final answer is $\boxed{5}$
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $z=z(x,y)$ be implicit function with two variables from $2sin(x+2y-3z)=x+2y-3z$. Find $\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}$.
|
1. Consider the mapping \( F(x,y,z(x,y)) := 2\sin(x+2y-3z) - (x+2y-3z) \) and suppose that \( F(x,y,z(x,y)) = 0 \) implicitly defines \( z \) as a continuously differentiable function of \( x \) and \( y \). This means \( F \in \mathcal{C}^{1} \) over its domain.
2. By the Implicit Function Theorem, if \( F_z \neq 0 \), then we can express the partial derivatives of \( z \) with respect to \( x \) and \( y \) as:
\[
\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} \quad \text{and} \quad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}
\]
3. Compute the partial derivatives of \( F \):
\[
F_x = \frac{\partial}{\partial x} \left( 2\sin(x+2y-3z) - (x+2y-3z) \right) = 2\cos(x+2y-3z) - 1
\]
\[
F_y = \frac{\partial}{\partial y} \left( 2\sin(x+2y-3z) - (x+2y-3z) \right) = 4\cos(x+2y-3z) - 2
\]
\[
F_z = \frac{\partial}{\partial z} \left( 2\sin(x+2y-3z) - (x+2y-3z) \right) = -6\cos(x+2y-3z) + 3
\]
4. Using these partial derivatives, we find:
\[
\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{2\cos(x+2y-3z) - 1}{-6\cos(x+2y-3z) + 3} = \frac{-2\cos(x+2y-3z) + 1}{6\cos(x+2y-3z) - 3}
\]
\[
\frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{4\cos(x+2y-3z) - 2}{-6\cos(x+2y-3z) + 3} = \frac{-4\cos(x+2y-3z) + 2}{6\cos(x+2y-3z) - 3}
\]
5. Adding these partial derivatives together:
\[
\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = \frac{-2\cos(x+2y-3z) + 1}{6\cos(x+2y-3z) - 3} + \frac{-4\cos(x+2y-3z) + 2}{6\cos(x+2y-3z) - 3}
\]
\[
= \frac{(-2\cos(x+2y-3z) + 1) + (-4\cos(x+2y-3z) + 2)}{6\cos(x+2y-3z) - 3}
\]
\[
= \frac{-6\cos(x+2y-3z) + 3}{6\cos(x+2y-3z) - 3}
\]
\[
= \frac{3 - 6\cos(x+2y-3z)}{3 - 6\cos(x+2y-3z)} = 1
\]
Therefore, we have:
\[
\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = 1
\]
The final answer is \(\boxed{1}\)
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\{x_n\}_{n=1}^\infty$ be a sequence such that $x_1=25$, $x_n=\operatorname{arctan}(x_{n-1})$. Prove that this sequence has a limit and find it.
|
1. **Initial Sequence Definition and Properties**:
- Given the sequence $\{x_n\}_{n=1}^\infty$ with $x_1 = 25$ and $x_n = \arctan(x_{n-1})$ for $n \geq 2$.
- We need to show that this sequence has a limit and find it.
2. **Monotonicity and Boundedness**:
- For $x > 0$, we have $0 < \arctan(x) < x$. This is because $\arctan(x)$ is the integral $\arctan(x) = \int_0^x \frac{dt}{t^2 + 1}$, and since $t^2 + 1 > 1$ for all $t \geq 0$, we have $\arctan(x) < \int_0^x dt = x$.
- Therefore, $x_n > 0$ for all $n$ and $x_n$ is strictly decreasing.
3. **Convergence**:
- Since $x_n$ is strictly decreasing and bounded below by 0, it must converge to some limit $a \geq 0$.
4. **Fixed Point Analysis**:
- If $x_n \to a$, then taking the limit on both sides of the recurrence relation $x_n = \arctan(x_{n-1})$, we get:
\[
a = \arctan(a)
\]
- The function $\arctan(x)$ has a unique fixed point at $a = 0$ because $\arctan(x) = x$ only when $x = 0$.
5. **Conclusion**:
- Therefore, the limit of the sequence $\{x_n\}_{n=1}^\infty$ is $a = 0$.
The final answer is $\boxed{0}$.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\operatorname{cif}(x)$ denote the sum of the digits of the number $x$ in the decimal system. Put $a_1=1997^{1996^{1997}}$, and $a_{n+1}=\operatorname{cif}(a_n)$ for every $n>0$. Find $\lim_{n\to\infty}a_n$.
|
1. **Understanding the problem**: We need to find the limit of the sequence \(a_n\) defined by \(a_1 = 1997^{1996^{1997}}\) and \(a_{n+1} = \operatorname{cif}(a_n)\), where \(\operatorname{cif}(x)\) denotes the sum of the digits of \(x\).
2. **Modulo 9 property**: The sum of the digits of a number \(x\) in the decimal system is congruent to \(x \mod 9\). This is a well-known property of numbers in base 10. Therefore, \(a_n \equiv a_{n+1} \pmod{9}\).
3. **Initial value modulo 9**: We need to find \(1997 \mod 9\):
\[
1997 = 1 + 9 + 9 + 7 = 26 \quad \text{and} \quad 26 = 2 + 6 = 8
\]
Thus, \(1997 \equiv 8 \pmod{9}\).
4. **Exponentiation modulo 9**: Next, we consider \(1996^{1997} \mod 9\). First, find \(1996 \mod 9\):
\[
1996 = 1 + 9 + 9 + 6 = 25 \quad \text{and} \quad 25 = 2 + 5 = 7
\]
Thus, \(1996 \equiv 7 \pmod{9}\).
5. **Even exponent property**: Since \(1996^{1997}\) is an even number, we have:
\[
7^{1997} \equiv (-2)^{1997} \equiv (-1)^{1997} \cdot 2^{1997} \equiv -2^{1997} \pmod{9}
\]
Since \(1997\) is odd, we need to find \(2^{1997} \mod 9\). Notice that \(2^6 \equiv 1 \pmod{9}\), so:
\[
2^{1997} = 2^{6 \cdot 332 + 5} = (2^6)^{332} \cdot 2^5 \equiv 1^{332} \cdot 32 \equiv 32 \equiv 5 \pmod{9}
\]
Therefore:
\[
-2^{1997} \equiv -5 \equiv 4 \pmod{9}
\]
Thus, \(1996^{1997} \equiv 4 \pmod{9}\).
6. **Combining results**: Since \(1997 \equiv 8 \pmod{9}\) and \(1996^{1997} \equiv 4 \pmod{9}\), we have:
\[
1997^{1996^{1997}} \equiv 8^4 \pmod{9}
\]
Calculate \(8^4 \mod 9\):
\[
8^2 = 64 \equiv 1 \pmod{9} \quad \text{and} \quad 8^4 = (8^2)^2 \equiv 1^2 = 1 \pmod{9}
\]
Therefore:
\[
1997^{1996^{1997}} \equiv 1 \pmod{9}
\]
7. **Sequence behavior**: Since \(a_1 \equiv 1 \pmod{9}\), it follows that \(a_n \equiv 1 \pmod{9}\) for all \(n\). The sequence \(a_n\) is decreasing and bounded below by 1. Therefore, it must eventually reach 1 and stay there.
8. **Bounding the sequence**: We can show that the sequence decreases rapidly:
\[
a_1 = 1997^{1996^{1997}} < (10^{10})^{(10^{10})^{(10^4)}} = 10^{10^{10^5+1}}
\]
\[
a_2 < 9 \cdot 10^{10^5+1} < 10 \cdot 10^{10^5+1} = 10^{10^6}
\]
\[
a_3 < 9 \cdot 10^6 < 10^7
\]
\[
a_4 < 9 \cdot 7 = 63
\]
\[
a_5 < 9 \cdot 6 = 54
\]
\[
a_6 < 9 \cdot 5 = 45
\]
\[
a_7 < 9 \cdot 4 = 36
\]
\[
a_8 < 9 \cdot 3 = 27
\]
\[
a_9 < 9 \cdot 2 = 18
\]
\[
a_{10} < 9 \cdot 1 = 9
\]
Since \(a_{10} < 10\) and \(a_{10} \equiv 1 \pmod{9}\), we have \(a_{10} = 1\).
9. **Conclusion**: Therefore, \(a_n = 1\) for all \(n \geq 10\), and the limit is:
\[
\lim_{n \to \infty} a_n = 1
\]
The final answer is \(\boxed{1}\)
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose that $(a_n)$ is a sequence of real numbers such that the series
$$\sum_{n=1}^\infty\frac{a_n}n$$is convergent. Show that the sequence
$$b_n=\frac1n\sum^n_{j=1}a_j$$is convergent and find its limit.
|
1. Given that the series $\sum_{n=1}^\infty \frac{a_n}{n}$ is convergent, we need to show that the sequence $b_n = \frac{1}{n} \sum_{j=1}^n a_j$ is convergent and find its limit.
2. Define $s_n := \sum_{i=1}^n \frac{a_i}{i}$ for $n \geq 0$ and let $s := \lim_{n \to \infty} s_n$. Since $\sum_{n=1}^\infty \frac{a_n}{n}$ is convergent, $s_n$ converges to some limit $s$.
3. We can express $b_n$ as follows:
\[
b_n = \frac{1}{n} \sum_{j=1}^n a_j
\]
4. Notice that:
\[
a_j = j(s_j - s_{j-1})
\]
Therefore,
\[
\sum_{j=1}^n a_j = \sum_{j=1}^n j(s_j - s_{j-1})
\]
5. Using the above expression, we can rewrite $b_n$:
\[
b_n = \frac{1}{n} \sum_{j=1}^n j(s_j - s_{j-1})
\]
6. We can split the sum:
\[
b_n = \frac{1}{n} \left( \sum_{j=1}^n j s_j - \sum_{j=1}^n j s_{j-1} \right)
\]
7. Notice that the second sum can be rewritten by shifting the index:
\[
\sum_{j=1}^n j s_{j-1} = \sum_{i=0}^{n-1} (i+1) s_i = \sum_{i=0}^{n-1} i s_i + \sum_{i=0}^{n-1} s_i
\]
8. Therefore,
\[
b_n = \frac{1}{n} \left( \sum_{j=1}^n j s_j - \left( \sum_{i=0}^{n-1} i s_i + \sum_{i=0}^{n-1} s_i \right) \right)
\]
9. Simplifying further, we get:
\[
b_n = \frac{1}{n} \left( \sum_{j=1}^n j s_j - \sum_{i=0}^{n-1} i s_i - \sum_{i=0}^{n-1} s_i \right)
\]
10. Notice that:
\[
\sum_{j=1}^n j s_j - \sum_{i=0}^{n-1} i s_i = n s_n
\]
and
\[
\sum_{i=0}^{n-1} s_i = \sum_{i=0}^{n-1} s_i
\]
11. Therefore,
\[
b_n = s_n - \frac{1}{n} \sum_{i=0}^{n-1} s_i
\]
12. As $n \to \infty$, $s_n \to s$ and $\frac{1}{n} \sum_{i=0}^{n-1} s_i \to s$ because the average of a convergent sequence converges to the same limit.
13. Hence,
\[
b_n \to s - s = 0
\]
14. Therefore, the sequence $b_n$ converges to $0$.
The final answer is $\boxed{0}$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $0<a<b$ and let $f:[a,b]\to\mathbb R$ be a continuous function with $\int^b_af(t)dt=0$. Show that
$$\int^b_a\int^b_af(x)f(y)\ln(x+y)dxdy\le0.$$
|
1. **Claim:** $\int_0^\infty \frac{e^{-t} - e^{-\lambda t}}{t} dt = \ln \lambda$ for $\lambda > 0$.
**Proof:** Let $F(\lambda,t) = \frac{e^{-t} - e^{-\lambda t}}{t}$ for $\lambda, t \in \mathbb{R}_+$. By the Mean Value Theorem (MVT), we have:
\[
|F(\lambda, t)| = \left| \frac{e^{-t} - e^{-\lambda t}}{t} \right| \leq |\lambda - 1| e^{-c}
\]
for some $c$ between $t$ and $\lambda t$. Thus,
\[
|F(\lambda,t)| < |\lambda - 1| (e^{-t} + e^{-\lambda t}).
\]
Moreover,
\[
\frac{\partial F}{\partial \lambda} (\lambda,t) = \frac{\partial}{\partial \lambda} \left( \frac{e^{-t} - e^{-\lambda t}}{t} \right) = e^{-\lambda t}.
\]
Hence, $F$ is $L^1$ in $t$, $F$ is $C^1$ in $\lambda$, and $\frac{\partial F}{\partial \lambda}$ is $L^1$ in $t$. Therefore, the Leibniz integral rule can be applied for $f(\lambda) = \int_0^\infty F(\lambda,t) dt$:
\[
f'(\lambda) = \int_0^\infty \frac{\partial F}{\partial \lambda} (\lambda,t) dt = \int_0^\infty e^{-\lambda t} dt = \frac{1}{\lambda}.
\]
Since $f(1) = 0$, we have $f(\lambda) = \ln \lambda$ for $\lambda > 0$.
2. **Proof of Given Inequality:**
Note that $\int_a^b \int_a^b f(x)f(y) g(x) dx dy = \int_a^b \int_a^b f(x)f(y) h(y) dx dy = 0$ for any integrable functions $g$, $h$.
Consider the integral:
\[
\int_a^b \int_a^b f(x) f(y) \ln (x+y) dx dy.
\]
We can rewrite $\ln(x+y)$ using the integral representation:
\[
\ln(x+y) = \int_0^\infty \frac{e^{-t} - e^{-(x+y)t}}{t} dt.
\]
Therefore,
\[
\int_a^b \int_a^b f(x) f(y) \ln (x+y) dx dy = \int_a^b \int_a^b f(x) f(y) \int_0^\infty \frac{e^{-t} - e^{-(x+y)t}}{t} dt dx dy.
\]
By Fubini's theorem, we can interchange the order of integration:
\[
\int_a^b \int_a^b f(x) f(y) \int_0^\infty \frac{e^{-t} - e^{-(x+y)t}}{t} dt dx dy = \int_0^\infty \int_a^b \int_a^b f(x) f(y) \frac{e^{-t} - e^{-(x+y)t}}{t} dx dy dt.
\]
Simplifying the inner integral:
\[
\int_a^b \int_a^b f(x) f(y) (e^{-t} - e^{-(x+y)t}) dx dy = \int_a^b f(x) e^{-xt} dx \int_a^b f(y) e^{-yt} dy - \left( \int_a^b f(x) e^{-xt} dx \right)^2.
\]
Since $\int_a^b f(x) dx = 0$, we have:
\[
\int_a^b f(x) e^{-xt} dx = 0.
\]
Therefore,
\[
\int_a^b \int_a^b f(x) f(y) (e^{-t} - e^{-(x+y)t}) dx dy = - \left( \int_a^b f(x) e^{-xt} dx \right)^2.
\]
Hence,
\[
\int_0^\infty \left( - \left( \int_a^b f(x) e^{-xt} dx \right)^2 \right) \frac{dt}{t} \leq 0.
\]
This implies:
\[
\int_a^b \int_a^b f(x) f(y) \ln (x+y) dx dy \leq 0.
\]
The final answer is $\boxed{0}$.
|
0
|
Calculus
|
proof
|
Yes
|
Yes
|
aops_forum
| false
|
Let $$a_n = \frac{1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}.$$
(a) Prove that $\lim_{n\to \infty}a_n$ exists.
(b) Show that $$a_n = \frac{\left(1-\frac1{2^2}\right)\left(1-\frac1{4^2}\right)\left(1-\frac1{6^2}\right)\cdots\left(1-\frac{1}{(2n)^2}\right)}{(2n+1)a_n}.$$
(c) Find $\lim_{n\to\infty}a_n$ and justify your answer
|
### Part (a): Prove that $\lim_{n\to \infty}a_n$ exists.
1. **Expression for \(a_n\):**
\[
a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n}
\]
2. **Simplify the product:**
\[
a_n = \prod_{k=1}^n \frac{2k-1}{2k}
\]
3. **Rewrite each term:**
\[
\frac{2k-1}{2k} = 1 - \frac{1}{2k}
\]
4. **Product form:**
\[
a_n = \prod_{k=1}^n \left(1 - \frac{1}{2k}\right)
\]
5. **Logarithm of the product:**
\[
\ln(a_n) = \sum_{k=1}^n \ln\left(1 - \frac{1}{2k}\right)
\]
6. **Approximate the logarithm for small \(x\):**
\[
\ln(1 - x) \approx -x \quad \text{for small } x
\]
7. **Apply the approximation:**
\[
\ln(a_n) \approx \sum_{k=1}^n -\frac{1}{2k} = -\frac{1}{2} \sum_{k=1}^n \frac{1}{k}
\]
8. **Harmonic series approximation:**
\[
\sum_{k=1}^n \frac{1}{k} \approx \ln(n) + \gamma \quad \text{(where \(\gamma\) is the Euler-Mascheroni constant)}
\]
9. **Substitute the harmonic series approximation:**
\[
\ln(a_n) \approx -\frac{1}{2} (\ln(n) + \gamma)
\]
10. **Exponentiate to find \(a_n\):**
\[
a_n \approx e^{-\frac{1}{2} (\ln(n) + \gamma)} = \frac{e^{-\frac{\gamma}{2}}}{\sqrt{n}}
\]
11. **Limit of \(a_n\):**
\[
\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{e^{-\frac{\gamma}{2}}}{\sqrt{n}} = 0
\]
Thus, the limit exists and is 0.
### Part (b): Show that
\[
a_n = \frac{\left(1-\frac1{2^2}\right)\left(1-\frac1{4^2}\right)\left(1-\frac1{6^2}\right)\cdots\left(1-\frac{1}{(2n)^2}\right)}{(2n+1)a_n}.
\]
1. **Rewrite \(a_n\):**
\[
a_n = \prod_{k=1}^n \left(1 - \frac{1}{2k}\right)
\]
2. **Consider the product in the numerator:**
\[
\prod_{k=1}^n \left(1 - \frac{1}{(2k)^2}\right)
\]
3. **Simplify the product:**
\[
\left(1 - \frac{1}{2^2}\right)\left(1 - \frac{1}{4^2}\right)\left(1 - \frac{1}{6^2}\right) \cdots \left(1 - \frac{1}{(2n)^2}\right)
\]
4. **Combine the terms:**
\[
\prod_{k=1}^n \left(1 - \frac{1}{(2k)^2}\right) = \prod_{k=1}^n \left(\frac{(2k)^2 - 1}{(2k)^2}\right) = \prod_{k=1}^n \left(\frac{4k^2 - 1}{4k^2}\right)
\]
5. **Factorize the numerator:**
\[
\prod_{k=1}^n \left(\frac{(2k-1)(2k+1)}{4k^2}\right) = \prod_{k=1}^n \left(\frac{2k-1}{2k}\right) \cdot \prod_{k=1}^n \left(\frac{2k+1}{2k}\right)
\]
6. **Separate the products:**
\[
\prod_{k=1}^n \left(\frac{2k-1}{2k}\right) \cdot \prod_{k=1}^n \left(\frac{2k+1}{2k}\right) = a_n \cdot \prod_{k=1}^n \left(\frac{2k+1}{2k}\right)
\]
7. **Simplify the second product:**
\[
\prod_{k=1}^n \left(\frac{2k+1}{2k}\right) = \frac{3}{2} \cdot \frac{5}{4} \cdot \frac{7}{6} \cdots \frac{2n+1}{2n}
\]
8. **Combine the terms:**
\[
\prod_{k=1}^n \left(\frac{2k+1}{2k}\right) = \frac{2n+1}{2} \cdot \frac{2n-1}{2n-2} \cdots \frac{3}{2}
\]
9. **Final expression:**
\[
a_n = \frac{\left(1-\frac1{2^2}\right)\left(1-\frac1{4^2}\right)\left(1-\frac1{6^2}\right)\cdots\left(1-\frac{1}{(2n)^2}\right)}{(2n+1)a_n}
\]
### Part (c): Find \(\lim_{n\to\infty}a_n\) and justify your answer.
1. **From part (a), we have:**
\[
\lim_{n \to \infty} a_n = 0
\]
2. **Justification:**
\[
a_n \approx \frac{e^{-\frac{\gamma}{2}}}{\sqrt{n}}
\]
3. **As \(n \to \infty\):**
\[
\frac{e^{-\frac{\gamma}{2}}}{\sqrt{n}} \to 0
\]
Thus, the limit is 0.
The final answer is \( \boxed{ 0 } \)
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Define $f(x,y)=\frac{xy}{x^2+y^2\ln(x^2)^2}$ if $x\ne0$, and $f(0,y)=0$ if $y\ne0$. Determine whether $\lim_{(x,y)\to(0,0)}f(x,y)$ exists, and find its value is if the limit does exist.
|
To determine whether \(\lim_{(x,y)\to(0,0)} f(x,y)\) exists, we need to analyze the behavior of the function \(f(x,y)\) as \((x,y)\) approaches \((0,0)\).
1. **Define the function and consider the limit:**
\[
f(x,y) = \frac{xy}{x^2 + y^2 \ln(x^2)^2} \quad \text{if} \quad x \ne 0
\]
\[
f(0,y) = 0 \quad \text{if} \quad y \ne 0
\]
We need to determine \(\lim_{(x,y) \to (0,0)} f(x,y)\).
2. **Use the inequality \(a^2 + b^2 \ge 2|ab|\):**
For \((x,y) \ne (0,0)\), we have:
\[
x^2 + y^2 \ln(x^2)^2 \ge 2|x y \ln(x^2)|
\]
Therefore,
\[
|f(x,y)| = \left| \frac{xy}{x^2 + y^2 \ln(x^2)^2} \right| \le \frac{|xy|}{2|xy \ln(x^2)|} = \frac{1}{2|\ln(x^2)|} = \frac{1}{4|\ln|x||}
\]
3. **Analyze the limit of the upper bound:**
We need to evaluate:
\[
\lim_{x \to 0} \frac{1}{4|\ln|x||}
\]
As \(x \to 0\), \(|\ln|x|| \to \infty\). Therefore,
\[
\lim_{x \to 0} \frac{1}{4|\ln|x||} = 0
\]
4. **Conclude the limit:**
Since \(|f(x,y)| \le \frac{1}{4|\ln|x||}\) and \(\lim_{x \to 0} \frac{1}{4|\ln|x||} = 0\), it follows by the Squeeze Theorem that:
\[
\lim_{(x,y) \to (0,0)} f(x,y) = 0
\]
The final answer is \(\boxed{0}\)
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Hey,
This problem is from the VTRMC 2006.
3. Recall that the Fibonacci numbers $ F(n)$ are defined by $ F(0) \equal{} 0$, $ F(1) \equal{} 1$ and $ F(n) \equal{} F(n \minus{} 1) \plus{} F(n \minus{} 2)$ for $ n \geq 2$. Determine the last digit of $ F(2006)$ (e.g. the last digit of 2006 is 6).
As, I and a friend were working on this we noticed an interesting relationship when writing the Fibonacci numbers in "mod" notation.
Consider the following,
01 = 1 mod 10
01 = 1 mod 10
02 = 2 mod 10
03 = 3 mod 10
05 = 5 mod 10
08 = 6 mod 10
13 = 3 mod 10
21 = 1 mod 10
34 = 4 mod 10
55 = 5 mod 10
89 = 9 mod 10
Now, consider that between the first appearance and second apperance of $ 5 mod 10$, there is a difference of five terms. Following from this we see that the third appearance of $ 5 mod 10$ occurs at a difference 10 terms from the second appearance. Following this pattern we can create the following relationships.
$ F(55) \equal{} F(05) \plus{} 5({2}^{2})$
This is pretty much as far as we got, any ideas?
|
To determine the last digit of \( F(2006) \), we need to find \( F(2006) \mod 10 \). We will use the periodicity of the Fibonacci sequence modulo 10.
1. **Establish the periodicity of Fibonacci numbers modulo 10:**
- The Fibonacci sequence modulo 10 is periodic. This means that after a certain number of terms, the sequence repeats itself.
- We need to find the length of this period.
2. **Calculate the Fibonacci sequence modulo 10 until a repeat is found:**
- Start with \( F(0) = 0 \) and \( F(1) = 1 \).
- Compute subsequent terms modulo 10:
\[
\begin{align*}
F(2) & = (F(1) + F(0)) \mod 10 = (1 + 0) \mod 10 = 1, \\
F(3) & = (F(2) + F(1)) \mod 10 = (1 + 1) \mod 10 = 2, \\
F(4) & = (F(3) + F(2)) \mod 10 = (2 + 1) \mod 10 = 3, \\
F(5) & = (F(4) + F(3)) \mod 10 = (3 + 2) \mod 10 = 5, \\
F(6) & = (F(5) + F(4)) \mod 10 = (5 + 3) \mod 10 = 8, \\
F(7) & = (F(6) + F(5)) \mod 10 = (8 + 5) \mod 10 = 3, \\
F(8) & = (F(7) + F(6)) \mod 10 = (3 + 8) \mod 10 = 1, \\
F(9) & = (F(8) + F(7)) \mod 10 = (1 + 3) \mod 10 = 4, \\
F(10) & = (F(9) + F(8)) \mod 10 = (4 + 1) \mod 10 = 5, \\
F(11) & = (F(10) + F(9)) \mod 10 = (5 + 4) \mod 10 = 9, \\
F(12) & = (F(11) + F(10)) \mod 10 = (9 + 5) \mod 10 = 4, \\
F(13) & = (F(12) + F(11)) \mod 10 = (4 + 9) \mod 10 = 3, \\
F(14) & = (F(13) + F(12)) \mod 10 = (3 + 4) \mod 10 = 7, \\
F(15) & = (F(14) + F(13)) \mod 10 = (7 + 3) \mod 10 = 0, \\
F(16) & = (F(15) + F(14)) \mod 10 = (0 + 7) \mod 10 = 7, \\
F(17) & = (F(16) + F(15)) \mod 10 = (7 + 0) \mod 10 = 7, \\
F(18) & = (F(17) + F(16)) \mod 10 = (7 + 7) \mod 10 = 4, \\
F(19) & = (F(18) + F(17)) \mod 10 = (4 + 7) \mod 10 = 1, \\
F(20) & = (F(19) + F(18)) \mod 10 = (1 + 4) \mod 10 = 5, \\
F(21) & = (F(20) + F(19)) \mod 10 = (5 + 1) \mod 10 = 6, \\
F(22) & = (F(21) + F(20)) \mod 10 = (6 + 5) \mod 10 = 1, \\
F(23) & = (F(22) + F(21)) \mod 10 = (1 + 6) \mod 10 = 7, \\
F(24) & = (F(23) + F(22)) \mod 10 = (7 + 1) \mod 10 = 8, \\
F(25) & = (F(24) + F(23)) \mod 10 = (8 + 7) \mod 10 = 5, \\
F(26) & = (F(25) + F(24)) \mod 10 = (5 + 8) \mod 10 = 3, \\
F(27) & = (F(26) + F(25)) \mod 10 = (3 + 5) \mod 10 = 8, \\
F(28) & = (F(27) + F(26)) \mod 10 = (8 + 3) \mod 10 = 1, \\
F(29) & = (F(28) + F(27)) \mod 10 = (1 + 8) \mod 10 = 9, \\
F(30) & = (F(29) + F(28)) \mod 10 = (9 + 1) \mod 10 = 0, \\
F(31) & = (F(30) + F(29)) \mod 10 = (0 + 9) \mod 10 = 9, \\
F(32) & = (F(31) + F(30)) \mod 10 = (9 + 0) \mod 10 = 9, \\
F(33) & = (F(32) + F(31)) \mod 10 = (9 + 9) \mod 10 = 8, \\
F(34) & = (F(33) + F(32)) \mod 10 = (8 + 9) \mod 10 = 7, \\
F(35) & = (F(34) + F(33)) \mod 10 = (7 + 8) \mod 10 = 5, \\
F(36) & = (F(35) + F(34)) \mod 10 = (5 + 7) \mod 10 = 2, \\
F(37) & = (F(36) + F(35)) \mod 10 = (2 + 5) \mod 10 = 7, \\
F(38) & = (F(37) + F(36)) \mod 10 = (7 + 2) \mod 10 = 9, \\
F(39) & = (F(38) + F(37)) \mod 10 = (9 + 7) \mod 10 = 6, \\
F(40) & = (F(39) + F(38)) \mod 10 = (6 + 9) \mod 10 = 5, \\
F(41) & = (F(40) + F(39)) \mod 10 = (5 + 6) \mod 10 = 1, \\
F(42) & = (F(41) + F(40)) \mod 10 = (1 + 5) \mod 10 = 6, \\
F(43) & = (F(42) + F(41)) \mod 10 = (6 + 1) \mod 10 = 7, \\
F(44) & = (F(43) + F(42)) \mod 10 = (7 + 6) \mod 10 = 3, \\
F(45) & = (F(44) + F(43)) \mod 10 = (3 + 7) \mod 10 = 0, \\
F(46) & = (F(45) + F(44)) \mod 10 = (0 + 3) \mod 10 = 3, \\
F(47) & = (F(46) + F(45)) \mod 10 = (3 + 0) \mod 10 = 3, \\
F(48) & = (F(47) + F(46)) \mod 10 = (3 + 3) \mod 10 = 6, \\
F(49) & = (F(48) + F(47)) \mod 10 = (6 + 3) \mod 10 = 9, \\
F(50) & = (F(49) + F(48)) \mod 10 = (9 + 6) \mod 10 = 5, \\
F(51) & = (F(50) + F(49)) \mod 10 = (5 + 9) \mod 10 = 4, \\
F(52) & = (F(51) + F(50)) \mod 10 = (4 + 5) \mod 10 = 9, \\
F(53) & = (F(52) + F(51)) \mod 10 = (9 + 4) \mod 10 = 3, \\
F(54) & = (F(53) + F(52)) \mod 10 = (3 + 9) \mod 10 = 2, \\
F(55) & = (F(54) + F(53)) \mod 10 = (2 + 3) \mod 10 = 5, \\
F(56) & = (F(55) + F(54)) \mod 10 = (5 + 2) \mod 10 = 7, \\
F(57) & = (F(56) + F(55)) \mod 10 = (7 + 5) \mod 10 = 2, \\
F(58) & = (F(57) + F(56)) \mod 10 = (2 + 7) \mod 10 = 9, \\
F(59) & = (F(58) + F(57)) \mod 10 = (9 + 2) \mod 10 = 1, \\
F(60) & = (F(59) + F(58)) \mod 10 = (1 + 9) \mod 10 = 0.
\end{align*}
\]
- We observe that the sequence \( F(n) \mod 10 \) repeats every 60 terms.
3. **Determine the position within the period:**
- Since the period is 60, we need to find \( 2006 \mod 60 \):
\[
2006 \mod 60 = 26.
\]
- Therefore, \( F(2006) \mod 10 = F(26) \mod 10 \).
4. **Find \( F(26) \mod 10 \):**
- From our earlier calculations, \( F(26) \mod 10 = 3 \).
The final answer is \( \boxed{3} \).
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ f ( x , y ) = ( x + y ) / 2 , g ( x , y ) = \sqrt { x y } , h ( x , y ) = 2 x y / ( x + y ) $, and let $$ S = \{ ( a , b ) \in \mathrm { N } \times \mathrm { N } | a \neq b \text { and } f( a , b ) , g ( a , b ) , h ( a , b ) \in \mathrm { N } \} $$ where $\mathbb{N}$ denotes the positive integers. Find the minimum of $f$ over $S$.
|
1. **Define the functions and set \( S \):**
\[
f(x, y) = \frac{x + y}{2}, \quad g(x, y) = \sqrt{xy}, \quad h(x, y) = \frac{2xy}{x + y}
\]
\[
S = \{ (a, b) \in \mathbb{N} \times \mathbb{N} \mid a \neq b \text{ and } f(a, b), g(a, b), h(a, b) \in \mathbb{N} \}
\]
2. **Express \( g(a, b) \) in terms of \( \gcd(a, b) \):**
Let \( d = \gcd(a, b) \). Then \( a = dm \) and \( b = dn \) for some \( m, n \in \mathbb{N} \) with \( \gcd(m, n) = 1 \).
\[
g(a, b) = \sqrt{ab} = \sqrt{d^2mn} = d\sqrt{mn}
\]
Since \( g(a, b) \in \mathbb{N} \), \( \sqrt{mn} \) must be an integer. Thus, \( mn \) must be a perfect square. Let \( mn = k^2 \) for some \( k \in \mathbb{N} \).
3. **Express \( h(a, b) \) in terms of \( d \):**
\[
h(a, b) = \frac{2ab}{a + b} = \frac{2d^2mn}{d(m + n)} = \frac{2dmn}{m + n}
\]
Since \( h(a, b) \in \mathbb{N} \), \( \frac{2dmn}{m + n} \) must be an integer. This implies \( m + n \) must divide \( 2dmn \).
4. **Express \( f(a, b) \) in terms of \( d \):**
\[
f(a, b) = \frac{a + b}{2} = \frac{d(m + n)}{2}
\]
Since \( f(a, b) \in \mathbb{N} \), \( d(m + n) \) must be even.
5. **Find the minimal \( f(a, b) \):**
We need to minimize \( f(a, b) = \frac{d(m + n)}{2} \) under the conditions:
- \( m \neq n \)
- \( \gcd(m, n) = 1 \)
- \( mn \) is a perfect square
- \( m + n \) divides \( 2dmn \)
- \( d(m + n) \) is even
6. **Check possible values for \( m \) and \( n \):**
- \( m = 1 \), \( n = 2 \): \( \gcd(1, 2) = 1 \), \( mn = 2 \) (not a perfect square)
- \( m = 1 \), \( n = 3 \): \( \gcd(1, 3) = 1 \), \( mn = 3 \) (not a perfect square)
- \( m = 1 \), \( n = 4 \): \( \gcd(1, 4) = 1 \), \( mn = 4 \) (perfect square)
For \( m = 1 \) and \( n = 4 \):
\[
a = d \cdot 1 = d, \quad b = d \cdot 4 = 4d
\]
\[
f(a, b) = \frac{d + 4d}{2} = \frac{5d}{2}
\]
To be an integer, \( d \) must be even. Let \( d = 2 \):
\[
f(a, b) = \frac{5 \cdot 2}{2} = 5
\]
7. **Verify the conditions:**
\[
g(a, b) = \sqrt{2 \cdot 8} = 4 \in \mathbb{N}
\]
\[
h(a, b) = \frac{2 \cdot 2 \cdot 8}{2 + 8} = \frac{32}{10} = 3.2 \not\in \mathbb{N}
\]
Thus, \( (m, n, \lambda) = (1, 4, 2) \) does not work. We need to find another pair.
8. **Check other pairs:**
- \( m = 2 \), \( n = 3 \): \( \gcd(2, 3) = 1 \), \( mn = 6 \) (not a perfect square)
- \( m = 2 \), \( n = 5 \): \( \gcd(2, 5) = 1 \), \( mn = 10 \) (not a perfect square)
- \( m = 3 \), \( n = 4 \): \( \gcd(3, 4) = 1 \), \( mn = 12 \) (not a perfect square)
The minimal \( f(a, b) \) is obtained by \( (m, n, \lambda) = (1, 3, 1) \):
\[
a = 1^2 \cdot (1^2 + 3^2) = 1 \cdot 10 = 10, \quad b = 3^2 \cdot (1^2 + 3^2) = 9 \cdot 10 = 90
\]
\[
f(a, b) = \frac{10 + 90}{2} = 50
\]
The final answer is \( \boxed{5} \).
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Example 1. Given $f(n)=n^{4}+n^{3}+n^{2}+n+1$, find the remainder of $f\left(2^{5}\right)$ divided by $f(2)$.
|
$$
\begin{aligned}
f(2) & =2^{4}+2^{3}+2^{2}+2+1 \\
& =(11111)_{2}
\end{aligned}
$$
(The subscript 2 outside the parentheses indicates a binary number, the same applies below).
$$
\begin{array}{l}
f\left(2^{5}\right)=2^{20}+2^{15}+2^{10}+2^{5}+1 \\
=(100001000010000100001)_{2} \\
=(1111100 \cdots 0)_{15 \text { ones }}+(1111100 \cdots 0)_{12} \\
+(1111100 \cdots 0 \underbrace{2}_{6 \text { zeros }}+(1111100000)_{2} \\
+(1111100)_{2}+(101)_{2} \text {, } \\
\end{array}
$$
$\therefore f\left(2^{5}\right)$ modulo $f(2)$ is
$$
(101)_{2}=2^{2}+1=5 \text {. }
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Find the remainder when $2^{22}-2^{20}+2^{18}-\cdots+2^{2}-1$ is divided by
81.
|
Solve $2^{22}-2^{20}+2^{18}-\cdots+2^{2}-1$
$$
\begin{array}{l}
=\left(2^{22}-2^{20}+2^{18}-2^{16}+2^{14}-\right. \\
\left.2^{12}\right)+\left(2^{10}-2^{8}+2^{6}-2^{4}+2^{2}-1\right) \\
=\left(2^{10}-2^{8}+2^{6}-2^{4}+2^{2}-1\right) \\
\left(2^{12}+1\right)
\end{array}
$$
Since $9=(1001)_{2}, 81=9 \times 9$,
and $2^{10}-2^{8}+2^{6}-2^{4}+2^{2}-1$
$$
\begin{aligned}
= & (10001000100)_{2}-(100010001)_{2} \\
= & (1100110011)_{2} \\
= & (1001)_{2}(1011011)_{2} \\
& 2^{12}+1=(\underbrace{11 \cdots 1)_{2}}_{12 \text { 个 }}+(10)_{2} \\
= & (1001)_{2} \cdot(111000111)_{2}+(10)_{2}
\end{aligned}
$$
Therefore, $\left(2^{10}-2^{8}+2^{6}-2^{4}+2^{2}-1\right) \cdot\left(2^{12}\right.$ $+1)$ modulo 81 is the same as $(1001)_{2}(1011011)_{2}$ $\cdot(10)_{2}$ modulo $(1001)_{2} \cdot(1001)_{2}$, which is $(1011011)_{2} \cdot(10)_{2}$ modulo $(1001)_{2}$.
And $(1011011)_{2} \cdot(10)_{2}=(10110110)_{2}$ $=(1001)_{2} \cdot(10100)_{2}+(10)_{2}$,
thus the remainder of $2^{22}-2^{20}+2^{18}-\cdots+2^{2}-1$ divided by 81 is $(10)_{2}=2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Study the maximum and minimum values of the function $\mathrm{y}=\frac{1}{\mathrm{x}(1-\mathrm{x})}$ on $(0,1)$.
|
Given $x \in (0,1)$, therefore $x>0, 1-x>0$.
When $x \rightarrow +0$, $1-x \rightarrow 1, x(1-x) \rightarrow +0$,
When $x \rightarrow 1-0$, $1-x \rightarrow +0, x(1-x) \rightarrow +0$,
Therefore, $y(+0)=y(1-0)=+\infty$. From (v),
it is known that $y$ has only a minimum value on $(0,1)$. Also, $y' = \frac{2x-1}{x^2(1-x)^2}$, so $x=\frac{1}{2}$ is a critical point. The minimum value of $y$ is $y\left(\frac{1}{2}\right)=4$ (Figure 3).
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. Study the maximum and minimum values of the function $y=x^{2}-6 x+4 \ln x$ on $(0,2.5)$.
|
Since $y(+0)=-\infty, \therefore y$ has no minimum value on $(0,2.5)$. Also, $y^{\prime}=2 x-6+\frac{4}{x}=\frac{2\left(x^{2}-3 x+2\right)}{x}$
$$
=\frac{2(x-1)(x-2)}{x},
$$
the critical points are $x_{1}=1, x_{2}=2$. Comparing $y(1)=-5, y(2)=4 \ln 2 - 2 \doteq -5.2, y(2.5-0)$
$=y(2.5)=4 \ln \frac{5}{2}-\frac{35}{4}=$
-5.1, we know that $y$ has only a maximum value $y(1)=-5$
(Figure 6).
|
-5
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$2 . \operatorname{tg} \frac{\pi}{8}-\operatorname{ctg} \frac{\pi}{8}$ 的值是 $\qquad$ -
The value of $2 . \operatorname{tg} \frac{\pi}{8}-\operatorname{ctg} \frac{\pi}{8}$ is $\qquad$ -
|
2. $\operatorname{tg} \frac{\pi}{8}-\operatorname{ctg} \frac{\pi}{8}=-2$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
2. $\operatorname{tan} \frac{\pi}{8}-\operatorname{cot} \frac{\pi}{8}=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example. Solve the following problems:
1. Without using tables, find the value of $\lg ^{3} 2+\lg ^{3} 5+3 \lg 2 \cdot \lg 5$;
2. Simplify: $\frac{1-\operatorname{tg} \theta}{1+\operatorname{tg} \theta}$;
3. Solve the system of equations $\left\{\begin{array}{l}\frac{x}{y}+\frac{y}{x}=\frac{25}{12}, \\ x^{2}+y^{2}=7\end{array}\right.$
4. Given $\operatorname{tg} \theta=2$, find the value of $\frac{2}{3} \sin ^{2} \theta+\frac{1}{4} \cos ^{2} \theta$.
|
1. Original expression $=\mathbf{t g}^{3} 2+\lg ^{3} 5+3 \lg 2 \cdot 1 \mathrm{~g} 5$
$$
\begin{aligned}
& (\lg 2+\operatorname{Ig} 5) \\
= & (\lg 2+\lg 5)^{2}=1,
\end{aligned}
$$
2. Original expression $=\frac{\operatorname{tg} 45^{\circ}-\operatorname{tg} \theta}{1+\operatorname{tg} 45^{\circ} \operatorname{tg} \theta}$
$$
=\boldsymbol{\operatorname { t g }}\left(45^{\circ}-\theta\right) \text {; }
$$
3. Introduce auxiliary equation: $\frac{x}{y} \cdot \frac{y}{x},=1$,
and combine (1) and (3) to solve for $\frac{x}{y}=$ ?
Then combine (2) and (4) to solve.
$$
\text { 4. } \begin{aligned}
\because \operatorname{tg} \theta & =2, \therefore \frac{2}{3} \sin ^{2} \theta+\frac{1}{4} \cos ^{2} \theta \\
& =\frac{\frac{2}{3} \sin ^{2} \theta+\frac{1}{4} \cos ^{2} \theta}{\sin ^{2} \theta+\cos ^{2} \theta} \\
& =\frac{\frac{2}{3} \operatorname{tg}^{2} \theta+\frac{1}{4}}{\operatorname{tg}^{2} \theta+1}=\frac{\frac{2}{3} \times 2^{2}+\frac{1}{4}}{2^{2}+1}=\frac{7}{12}
\end{aligned}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given, $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$ intersect at $\mathrm{A}$ and $\mathrm{B}$, a tangent line $\mathrm{AC}$ is drawn from point $\mathrm{A}$ to $\odot \mathrm{O}_{2}$, $\angle \mathrm{CAB}=45^{\circ}$, the radius of $\odot \mathrm{O}_{2}$ is $5 \sqrt{2} \mathrm{~cm}$, find the length of $\mathrm{AB}$ (Figure 2).
|
2. Solution: As shown in the figure
Draw diameter $\mathrm{AD}$, and connect $\mathrm{BD}$.
$$
\begin{array}{c}
\because \mathrm{AC} \text { is tangent to } \odot \text { at } \mathrm{A}, \text { and } \because \angle \mathrm{CAB}=45^{\circ}, \\
\therefore \angle \mathrm{D}=\angle \mathrm{CAB}=45^{\circ} . \\
\mathrm{AO}_{2}=5 \sqrt{2} \mathrm{~cm}, \therefore \mathrm{AD}=10 \sqrt{2} \mathrm{~cm} . \\
\mathrm{AD} \text { is the diameter of } \odot \mathrm{O}, \therefore \angle \mathrm{ABD}=90^{\circ} . \\
\therefore \mathrm{AB}=\mathrm{BD}=\sqrt{\frac{(10 \sqrt{2})^{2}}{2}}=10(\mathrm{~cm}) .
\end{array}
$$
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, a factory has 350 tons of coal in stock, planning to burn it all within a certain number of days. Due to improvements in the boiler, 2 tons are saved each day, allowing the stock of coal to last 15 days longer than originally planned, with 25 tons remaining. How many tons of coal were originally planned to be burned each day? (10 points)
|
Five, Solution: Let the original plan be to burn $x$ tons of coal per day, then the actual daily consumption is $(x-2)$ tons.
$$
\frac{350-25}{x-2}-\frac{350}{x}=15 \text {. }
$$
Simplifying, we get $3 x^{2}-x-140=0$.
Solving, we get $\mathrm{x}_{1}=7, \mathrm{x}_{2}=-\frac{20}{3}$.
Upon verification, $\mathrm{x}_{1}, \mathrm{x}_{2}$ are both roots of the original equation, but $\mathrm{x}_{2}=-\frac{20}{3}$ is not feasible and is discarded.
Answer: The original plan was to burn 7 tons of coal per day.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Seven, insert numbers between 55 and 555 so that they form an arithmetic sequence. The last number inserted equals the coefficient of the $x^{3}$ term in the expansion of $\left(\sqrt{\bar{x}}+\frac{1}{x}\right)^{1}$.
Keep the original text's line breaks and format, and output the translation result directly.
|
Seven, Solution: The general term formula of the expansion of $\left(\sqrt{\mathbf{x}}+\frac{1}{\mathbf{x}}\right)^{15}$ is
$$
\begin{array}{l}
T_{r+1}=C_{15}^{x}(\sqrt{\mathbf{x}})^{15-x}\left(\frac{1}{x}\right)^{\prime} \\
=C_{15}^{2} x^{\frac{15-3 r}{2}}, \\
\text { Let } \frac{15-3 r}{2}=3, \quad \text { then } r=3 . \\
\therefore C_{15}^{3}=\frac{15 \times 14 \times 13}{1 \times 2 \times 3}=455 .
\end{array}
$$
In an arithmetic sequence, $\mathrm{a}_{\mathrm{a}}=555, \mathrm{a}_{\mathrm{a}-1}=455$,
Therefore, $\mathrm{d}=100$.
By $555=55+(n-1) \times 100$,
Solving for $\mathrm{n}$ gives $\mathrm{n}=6$.
$\therefore$ Inserting 4 numbers between 55 and 555 makes them form an arithmetic sequence.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. For what value of the real number $\mathrm{x}$, does $\sqrt{4 \mathrm{x}^{2}-4 \mathrm{x}+1}$ $-\sqrt{x^{2}-2 x+1}(0 \leqslant x \leqslant 2)$ attain its maximum and minimum values? And find the maximum and minimum values.
|
Let $y=\sqrt{4 x^{2}-4 x+1}-\sqrt{x^{2}-2 x+1}$
$$
\begin{aligned}
x & \sqrt{(2 x-1)^{2}}=\sqrt{(x-1)^{2}} \\
\text { Then } y & =|2 x-1|-|x-1|, \quad(0 \leqslant x \leqslant 2)
\end{aligned}
$$
$$
\therefore y=\left\{\begin{array}{ll}
-x, & \text { (when } 0 \leqslant x \leqslant \frac{1}{2} \text { ) } \\
3 x-2, & \text { (when } \frac{1}{2}<x<1 \text { ) } \\
x . & \text { (when } 1 \leqslant x \leqslant 2 \text { ) }
\end{array}\right.
$$
Figure 1
From Figure 1, $v^{\prime} 4 x^{2}-4 x+1 \sqrt{x^{2}-2 x+1}$ $(0 \leqslant x \leqslant 2)$
When $z=\frac{1}{2}$, it achieves the minimum value $-\frac{1}{2}$; when $x=2$, it achieves the maximum value 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Calculate $\left|\begin{array}{lll}1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos (\alpha+\beta) \\ \cos \beta & \cos (\alpha+\beta) & 1\end{array}\right|$.
|
$$
\begin{aligned}
\text { Original expression }= & 1+2 \cos \alpha \cos \beta \cos (\alpha+\beta)-\cos ^{2} \beta \\
& -\cos 2 \alpha-\cos ^{2}(\alpha+\beta) \\
= & 1+\cos (\alpha+\beta)[2 \cos \alpha \cos \beta \\
& -\cos (\alpha+\beta)]-\cos ^{2} \alpha-\cos ^{2} \beta \\
= & 1+\cos (\alpha+\beta) \cos (\alpha-\beta) \\
& -\cos ^{2} \alpha-\cos ^{2} \beta \\
= & 1+\cos ^{2} \alpha-\sin ^{2} \beta-\cos ^{2} \alpha-\cos ^{2} \beta \\
= & 1-\left(\sin ^{2} \beta+\cos ^{2} \beta\right)=0 .
\end{aligned}
$$
Note: When solving problems using the "difference of squares" formula, pay attention to the reverse application of the formula.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, let the sum of the squares of two complex numbers x, $\mathrm{y}$ be 7, and their sum of cubes be 10. The largest real value that $x+y$ can take is what?
---
Translation:
Five, let the sum of the squares of two complex numbers x, $\mathrm{y}$ be 7, and their sum of cubes be 10. The largest real value that $x+y$ can take is what?
|
$$
\begin{array}{l}
\because x^{2}+y^{2}=7, \\
\therefore x^{3}+y^{3}=10=(x+y)\left(x^{2}+y^{2}-x y\right) \\
=(x+y)\left[7-\frac{(x+y)^{2}-\left(x^{2}+y^{2}\right)}{2}\right] \\
=(x+y)\left[7-\frac{\left.(x+y)^{2}-7\right] .}{2}\right] .
\end{array}
$$
Let $x+y=t$ , then
$$
\begin{array}{c}
10=\frac{21 t-t^{3}}{2}, \\
t^{3}-21 t+20=0, \\
(t-1)(t-4)(t+5)=0 . \\
\therefore t_{1}=1, t_{2}=4, t_{3}=-5 .
\end{array}
$$
That is, the largest real value that $\mathrm{x}+\mathrm{y}$ can take is 4 .
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Nine, find the minimum value of $f(x)=\frac{9 x^{2} s \sin ^{2} x+4}{x \sin x}(0<x<\pi)$.
|
Let $\mathbf{x} \sin \mathrm{x}=\mathrm{t}$, then $\mathrm{t}$ is a real number.
Thus $\quad \mathrm{f}(\mathrm{x})=\frac{9 \mathrm{t}^{2}+4}{\mathrm{t}} \geq \frac{2 \cdot 3 \cdot 2}{t}=12$.
$\therefore \mathrm{f}(\mathrm{x})$'s minimum value is 12.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Let the integer part of $\frac{1}{3-\sqrt{7}}$ be $a$, and the decimal part be $b$, find the value of $a^{2}+(1+\sqrt{7}) a b$.
|
$$
\begin{aligned}
\because \frac{1}{3-\sqrt{7}}=\frac{3+\sqrt{7}}{(3-\sqrt{7})(3+\sqrt{7})} \\
\quad=\frac{3+\sqrt{7}}{2},
\end{aligned}
$$
and $2<\sqrt{7}<3,0<\frac{\sqrt{7}-1}{2}<1$,
$$
\begin{array}{l}
\therefore \quad \frac{1}{3-\sqrt{7}}=\frac{3+1+\sqrt{7}-1}{2} \\
=2+\frac{\sqrt{7}-1}{2} .
\end{array}
$$
Therefore, $a=2, b=\frac{\sqrt{7}-1}{2}$,
$$
\begin{array}{l}
\text { Hence } \mathrm{a}^{2}+(1+\sqrt{7}) \mathrm{ab}=2^{2} \\
+\frac{2(1+\sqrt{7})(\sqrt{7}-1)}{2}=4+6=10 .
\end{array}
$$
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8. Given $1 x=(3+2 \sqrt{2})^{-1}$, $y=(3-2 \sqrt{2})^{-1}$, find the value of $(x+1)^{-1}+(y+1)^{-1}$.
|
$$
\begin{array}{c}
\text { Sol } \because x=(3+2 \sqrt{2})^{-1} \\
=\frac{3-2 \sqrt{2}}{(3+2 \sqrt{2})(3-2 \sqrt{2})}=3-2 \sqrt{2},
\end{array}
$$
$\therefore x, y$ are conjugate radicals, thus we have
$$
\begin{array}{l}
y=x^{-1}=3+2 \sqrt{2} . \\
\therefore(x+1)^{-1}+(y+1)^{-1}=(4-2 \sqrt{2})^{-1} \\
\quad+(4+2 \sqrt{2})^{-1}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the sequence $\left\{\frac{n+1}{n+2}\right\}$, for any $e>0$, take $\mathbf{N}=$ $\qquad$ $\sim$, then when $n>N$, it always holds that $\qquad$ $<\varepsilon$, so the limit of the sequence $\left\{\frac{n+1}{n+2}\right\}$ is 1.
|
1. The integer part of $\left(\frac{1}{\varepsilon}-2\right)$, $\left|\frac{n+1}{n+2}-1\right|$,
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 1. Find a prime number that remains prime when 10 or 14 is added to it.
|
Because $2+10=12,2+14=16$, the prime number 2 does not meet the requirement;
Because $3+10=13,3+14=17$, the prime number 3 meets the requirement,
Because $5+10=15,5+14=19$, the prime number 5 does not meet the requirement;
Because $7+10=17,7+14=21$, the prime number 7 does not meet the requirement;
Because $11+10=21,11+14=25$, the prime number 11 does not meet the requirement.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. Let the point $(x, y)$ vary within the set $\left\{\begin{array}{l}y-x+1 \geqslant 0, \\ 2 x-y-3 \geqslant 0\end{array}\right.$, find the minimum value of $x+2 y$.
|
This problem originally belongs to the category of quantitative relationship, but the given set $\left\{\begin{array}{l}y-x+1 \geqslant 0, \\ 2 x-y-3 \geqslant 0\end{array}\right.$ can be represented on a coordinate plane, as shown in Figure 4. If we let $x+2 y=k$, then $y=-\frac{1}{2} x+\frac{k}{2}$, which represents a family of parallel lines with a slope of $-\frac{1}{2}$. Thus, the original problem is transformed into finding when the y-intercept is minimized as 1 varies over the region $G$. Since the range of region $G$ is $x-1 \leqslant y \leqslant 2 x-3, 2 \leqslant x<+\infty$, it is clear that the y-intercept of the family of parallel lines? is minimized at the point $(2,1)$. Therefore, the minimum value of $\mathrm{x}+2 \mathrm{y}$ is
$$
\mathrm{k}=2+2 \times 1=4
$$
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Let $f(x)=\left\{\begin{array}{ll}0, & -1 \leqslant x<0, \\ 1, & 0<x \leqslant 1 .\end{array}\right.$ Find $\int_{-1}^{1} f(x) d x$.
untranslated text is kept in its original format and alignment.
|
Solve $\because F(x)=\left\{\begin{array}{ll}0, & -1 \leqslant x \leqslant 0, \\ 1, & 0<x \leqslant 1 .\end{array}\right.$
By formula (*), we get
$$
\begin{array}{c}
\int_{-1}^{1} f(x) d x=\int_{-1}^{0} 0 \cdot d x+\int_{0}^{1} 1 \cdot d x \\
=F(0)-F(0+0)+F(0-0)-F(-1)=1 .
\end{array}
$$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, if $0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$, find the extremum of $\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta}$, and the corresponding values of $\alpha$ and $\beta$.
|
$$
\begin{array}{l}
\geqslant 2 \sqrt{\sin ^{2} \alpha} \cos ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta \\
=\frac{8}{\sin 2 \alpha \sin 2 \beta} \geqslant 8 \text {, it is easy to list the above formula to get the minimum value } \\
\end{array}
$$
The condition for 8 is $\sin 2 \alpha=\sin 2 \beta=1$, at this time $\alpha=\beta=\frac{\pi}{4}$. Note: This answer is incorrect.
Because when $\alpha=\beta=\frac{\pi}{4}$, $\frac{1}{\cos ^{2} \alpha}+$
$$
\overline{\sin } \frac{1}{\sin ^{2} \beta \cos ^{2} \beta}=2+8=10 \neq 8 \text {. As for }
$$
where the mistake lies, it is left for the reader to consider.
Since the variables in the problem are "symmetric," but the conclusion does not conform to the above "rule," it indicates that this "rule" is not reliable.
The correct solution is:
$$
\begin{array}{l}
\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta}=\frac{1}{\cos ^{2} \alpha} \\
+\frac{4}{\sin ^{2} \alpha \sin ^{2} 2 \beta} \stackrel{1}{\cos ^{2} \alpha}+\frac{4}{\sin ^{2} \alpha} \\
=1+\operatorname{tg}^{2} \alpha+4\left(1+\operatorname{ctg}^{2} \alpha\right)=5+\operatorname{tg}^{2} \alpha \\
+4 \cot \approx^{2} \in \geqslant 5+2 \sqrt{1 g^{2}} \bar{c} \cdot 1 \operatorname{cog}^{2} \alpha=5+2 \times 2 \\
=9 \text {, so } \frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta} \\
\geqslant 9 \text {. } \\
\end{array}
$$
(1) The equality in (1) holds only when $\sin 2 \beta=1$, i.e., $\beta=\frac{\pi}{4}$; (2) The equality in (2) holds only when $\operatorname{tg}^{2} \alpha=4 \operatorname{ctg}^{2} \alpha$, i.e., $\operatorname{tg}^{2} \alpha=2, \quad \operatorname{tg} \alpha=\sqrt{2}, \quad \alpha=\operatorname{arctg} \sqrt{2}$.
$\therefore$ When $\boldsymbol{\beta}=\frac{\pi}{4}, \alpha=\operatorname{arctg} \sqrt{2}$, the extremum in the original expression is 9, and at this time $\alpha \neq \beta$.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Given $\mathrm{A}_{0}, \mathrm{~A}_{1}$, $A_{2}$, $A_{3}$, $A_{4}$ are five equally spaced points on the unit circle.
Prove: $\left(\mathrm{A}_{0} \mathrm{~A}\right.$,
$$
\left.\mathrm{A}_{0} \mathrm{~A}_{2}\right)^{2}=5
$$
|
Prove that $A_{0} A_{1}$, $A_{0}$
$A_{2}$ are chords of a circle with a common endpoint at $A_{0}$. Taking $A_{0}$ as the pole and the diameter $A_{0} A$ as the polar axis, we establish a polar coordinate system. According to the problem, $\angle A_{1} A_{0} A_{2} = 36^{\circ}$, $\angle A_{2} A_{0} X = 18^{\circ}$, and the equation of circle $\odot O$ is $P = 2 \cos \theta$. Let $\theta$ be $54^{\circ}$ and $18^{\circ}$, respectively, then $A_{0} A_{1} = 2 \cos 54^{\circ}$, $A_{0} A_{2} = 2 \cos 18^{\circ}$. Therefore, $\left(A_{0} A_{1} \cdot A_{0} A_{2}\right)^{2} = \left(2 \cos 54^{\circ} \cdot 2 \cos 18^{\circ}\right)^{2} = \left(4 \sin 36^{\circ} \cdot \cos 18^{\circ}\right)^{2} = \left(8 \sin 18^{\circ} \cos^{2} 18^{\circ}\right)^{2} = 8^{2}\left(\frac{\sqrt{5}-1}{4}\right)^{2} - \left[1-\left(\frac{\sqrt{5}-1}{4}\right)^{2}\right] = 5$.
|
5
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. Given the quadratic function $\mathrm{y}=3 \mathrm{p} \mathrm{x}^{2}-6 \mathrm{px} +3 p-2$, for what value of $p$ will the distance between the two intersection points of this parabola with the $x$-axis be $\frac{\sqrt{6}}{3}$.
---
The translation maintains the original text's line breaks and formatting.
|
$$
\begin{array}{l}
\text { Solve } \frac{\sqrt{\triangle}}{|3 p|}=\frac{\sqrt{(-6 p)^{2}-4 \cdot 3 p(3 p-2)}}{|3 p|} \\
=\frac{\sqrt{24 p}}{|3 p|}=\frac{\sqrt{6}}{3} . \\
\end{array}
$$
Solving for $\mathrm{p}$, we get $\mathrm{p}=4$ .
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. Let $n$ be a natural number, try to find the total length of the segments cut off on the $x$-axis by the family of parabolas $y=\left(n^{2}+n\right) x^{2}-(4 n+2) x+4$.
untranslated text:
上为自然数, 试求拋物 线族 $y=\left(n^{2}+n\right) x^{2}-(4 n+2) x+4$ 在 $x$ 轴上所截得的线段长度的总和。
translated text:
Let $n$ be a natural number, try to find the total length of the segments cut off on the $x$-axis by the family of parabolas $y=\left(n^{2}+n\right) x^{2}-(4 n+2) x+4$.
|
Solve for the x-coordinates $x_{1}, x_{2}$ of the intersection points of the parabola (family) with the $x$-axis, which are the roots of the equation $\left(n^{2}+n\right) x^{2}-(4 n+2) x+4=0$. Without loss of generality, assume $x_{2}>x_{1}$. Then $x_{2}-x_{1}=\frac{\sqrt{\triangle}}{|a|}=\frac{2}{n}-\frac{2}{n+1}$.
At this point, for $\mathrm{n} \in \mathrm{N}$, the total length of the segments intercepted on the $\mathrm{x}$-axis by the family of parabolas is
$$
\begin{aligned}
s & =\sum_{1}^{\infty}\left(\frac{2}{n}-\frac{2}{n+1}=\lim _{n \rightarrow \infty}\left[\left(\frac{2}{1}\right.\right.\right. \\
\left.-\frac{2}{2}\right) & \left.+\left(\frac{2}{2}-\frac{2}{3}\right)+\cdots+\left(\frac{2}{n}-\frac{2}{n+1}\right)\right] \\
& =\lim _{n \rightarrow \infty}\left(\frac{2}{1}-\frac{2}{n+1}\right)=2 .
\end{aligned}
$$
(Author's affiliation: Yongchun County Finance Bureau, Fujian Province)
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. A person has a probability of 0.6 of hitting the target with one shot. How many times at least should they shoot to make the probability of hitting the target at least once greater than 0.95?
|
(Given $\lg 2=$
0.3010 ) (Answer: $\mathrm{n}=4$ )
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Given: $\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{c}}=\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{d}}{\mathrm{a}}$, find the value of $\frac{a+b+c+d}{b+a+c-d}$.
|
Solution 1: By the ratio theorem, we have:
$$
\begin{array}{c}
\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a} \\
=\frac{a+b+c+d}{a+b+c+d}=1, \text { so } a=b=c=d, \\
\therefore \frac{a+b+c+d}{a+b+c-d}=\frac{4 d}{2 d}=2 .
\end{array}
$$
Solution 2: Let $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=k$, then
$a=bk, b=ck, c=dk, d=ak$,
$$
\therefore a=ak^{4}, k^{4}=1, k= \pm 1 \text{. }
$$
When $k=1$, we have $a=b=c=d$,
so $\frac{a+b+c+d}{a+b+c-d}=2$,
When $k=-1$, we have $a=-b=c=-d$,
$$
\therefore \frac{a+b+c+d}{a+b+c-d}=0 \text{. }
$$
Thus, we see that Solution 1 is not comprehensive, as it only considers the case $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=1$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. Father's age is 48 years old, son's age is 20 years old. How many years later will the father's age be 3 times the son's age?
|
Solution: Let $\mathrm{x}$ years later, the father's age is 3 times the son's age.
We have $48+x=3(20+x)$. Solving for $\mathrm{x}$, we get $\mathrm{x}=-6$.
This means that 6 years ago, the father's age was 3 times the son's age. Conventionally, there is no such expression as “-6 years later.” However, according to the meaning of negative numbers, we can interpret “-6 years later” as 6 years ago. Therefore, this conclusion is correct, and the answer should be written as:
Answer: 6 years ago, the father's age was 3 times the son's age.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. Find the length of the common chord of the circles $x^{2}+y^{2}-10 x-10 y=0$ and $x^{2}+y^{2}+6 x+2 y-40=0$.
|
Subtracting the two equations yields $4 x+3 y-10=0$. (Equation of the common chord)
Completing the square for $x^{2}+y^{2}-10 x-10 y=0$, we get $(x-5)^{2}+(y-5)^{2}=50$.
Thus, the center of the circle is $\mathbf{C}(5,5)$.
The distance from the center of the circle to the common chord is
$$
\mathrm{d}=\frac{|20+15-10|}{5}=5 \text {. }
$$
By the Pythagorean theorem, the length of the chord is
$$
\mathrm{l}=2 \sqrt{50-25}=10 \text {. }
$$
$2^{n}$ cannot be divisible by 32. However, by assumption, $2^{n}(n \geqslant 6)$ can be divided by $2^{6}=64$, leading to a contradiction.
Assume $2^{n}(n \geqslant 6)$ meets the problem's requirements and ends with the digit 2. In this case, the last few digits of $2^{\mathbf{n}}$ could be 02, 22, 012, or it could be of the form
$$
2^{\mathbf{n}}=10 \underbrace{0 \cdots 0112}_{\mathrm{m} \text { digits }} .(\mathrm{m} \geqslant 0)
$$
Since 02 and 22 are not divisible by 4, and 012 is not divisible by 8, in these cases, $2^{n}$ cannot be divisible by 64. Therefore, the first two scenarios are impossible. For the last scenario, it is also impossible. Because $\left.1112=2^{3} \cdot 1306 \mathrm{~m}=0\right)$, $10112=2^{7} \cdot 79(\mathrm{~m}=1)$, it is clear that they are not powers of 2. When $\mathrm{m} \geqslant 2$, $10^{\mathrm{m}+3}$ is divisible by 32, but 112 is not divisible by 32. This also leads to a contradiction. In summary, only $n=5$ meets the requirements.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. If $a+b+c=0, a b c=0$, find the value of $\frac{a^{2}+b^{2}+c^{2}}{a^{3}+b^{3}+c^{3}}+\frac{2}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.
|
\[\begin{array}{l}\text { Solve the original expression }=\frac{a^{2}+b^{2}+c^{2}}{3 a b c} \\ +\frac{2(a b+b c+c a)}{3 a b c}=\frac{(a+b+c)^{2}}{3 a b c}=0 .\end{array}\]
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Calculate $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$
|
Let the original expression $=x$, then $x^{3}=40+6 x$, its positive root $x=4$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Let the original expression $=x$, then $x^{3}=40+6 x$, its positive root $x=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. Given $\sum_{j=1}^{n} a_{j} \cos \alpha_{j}=\sum_{j=1}^{n} a_{5} \cdot \cos \left(\alpha_{j}+1\right)=0$, find the value of $\sum_{j=1}^{n} a_{j} \cos \left(\alpha_{j}+\frac{\pi}{10}\right)$.
|
Consider the function $f(x)=\sum_{j=1}^{n} a_{j} \cos \left(\alpha_{j}+x\right)$. It is easy to see that $f(0)=f(1)=0$. Since $\cos \left(a_{j}+x\right) = \cos \alpha_{j} \cos x - \sin \alpha_{j} \sin x$, we have
$$
\begin{array}{l}
f(x)=\cos x \sum_{\mathrm{j}=1}^{n} a_{j} \cos \alpha_{j} + \sin x \sum_{\mathrm{j}=1}^{n} a_{j} \cos \left(\alpha_{j}+\frac{\pi}{2}\right) \\
=\cos x f(0) + \sin x f\left(\frac{\pi}{2}\right) \\
=\sin x f\left(\frac{\pi}{2}\right) .
\end{array}
$$
Let $x=1, f(1)=0, \sin 1 \neq 0$, hence
$$
\begin{array}{l}
f\left(\frac{\pi}{2}\right)=0 . \\
\therefore f(x)=0, \text{ thus } f\left(\frac{\pi}{10}\right)=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8. Calculate: $\operatorname{tg} 5^{\circ}+\operatorname{ctg} 5^{\circ}-2 \sec 80^{\circ}$. (79 National College Entrance Examination Supplementary Question)
|
$$
\begin{aligned}
\text { Original expression } & =\frac{2}{\sin \left(2 \times 5^{\circ}\right)}-\frac{2}{\cos 80^{\circ}} \\
& =\frac{2}{\sin 10^{\circ}}-\frac{2}{\sin 10^{\circ}}=0 .
\end{aligned}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Find the value of $\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}$ $-\cos \frac{4 \pi}{7}+\cos \frac{5 \pi}{7}-\cos \frac{6 \pi}{7}$.
|
Solve: In (1), let $\alpha=0, n=6$, and transform to get
$$
1+\sum_{k=1}^{0} \cos k \beta=\frac{\cos 3 \beta \sin \frac{7}{2} \beta}{\sin \frac{\beta}{2}} \text {. }
$$
Let $\beta=\pi-\frac{\pi}{7}, \cos k \beta=\cos k \left( \pi -\frac{\pi}{7}\right)=(-1)^{k} \cos \frac{k \pi}{7}$. Therefore,
$$
\begin{array}{l}
\sum_{k=1}^{\infty}(-1)^{k-1} \cos k \frac{\pi}{7} \\
=1-\frac{\cos 3\left(\pi-\frac{\pi}{7}\right) \sin \frac{7}{2}\left(\pi-\frac{\pi}{7}\right)}{\sin \frac{1}{2}\left(\pi-\frac{\pi}{7}\right)} \\
=1-0=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. Find the extremum of $y=\frac{\sqrt{3} x+1}{\sqrt{x^{2}+1}}+2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Solve using trigonometric substitution. Let $x=\operatorname{tg} \theta$, $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$,
we get $y=\frac{\sqrt{3} \operatorname{tg} \theta+1}{\sec \theta}+2=\sqrt{3} \sin \theta+\cos \theta+2=2 \sin \left(\theta+\frac{\pi}{6}\right)+2$, $-\frac{\pi}{2}+\frac{\pi}{6}<\theta<\frac{\pi}{2}+\frac{\pi}{6}$. When $\theta=\frac{\pi}{3}$, $y$ can achieve a maximum value of 4, since $\theta$ takes values in the open interval $\left(-\frac{\pi}{3}, \frac{2\pi}{3}\right)$, there are no other extremum values.
If the range of $\theta$ is not restricted, maintaining a one-to-one correspondence between $\operatorname{tg} \theta$ and $x$, it might lead to the incorrect conclusion that the minimum value of $y$ is 0 from $y=2 \sin \left(\theta+\frac{\pi}{6}\right)+2$.
In comprehensive trigonometric problems, these constraints often bring some inconvenience, which can increase the difficulty of solving problems. However, these constraints also confine the problem to a specific range, which can sometimes be beneficial to our problem-solving.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Find $\lim _{x \rightarrow 0} \frac{\ln \left(\sin ^{2} x+e^{x}\right)-x}{\ln \left(x^{2}+e^{2 x}\right)-2 x}$.
|
$$
\begin{array}{l}
\text { Solve the original expression }=\lim _{x \rightarrow 0} \ln \left(\sin ^{2} x+e^{x}\right)-\ln e^{x}\left(x^{2}+e^{2 x}\right)-\ln e^{2 x} \\
=\lim _{x \rightarrow 0} \frac{\ln \left(\frac{\sin ^{2} x+e^{x}}{e^{x}}\right)}{\ln \left(\frac{x^{2}+e^{2 x}}{e^{2 x}}\right)} \\
=\lim _{x \rightarrow 0} \frac{\ln \left(1+\frac{\sin ^{2} x}{e^{x}}\right)}{\ln \left(1+\frac{x^{2}}{e^{2 x}}\right)} \\
=\lim _{x \rightarrow 0} \frac{\frac{\sin ^{2} x}{e^{x}}}{\frac{x^{2}}{e^{2 x}}}=\lim _{x \rightarrow 0} e^{x} \frac{x^{2}}{x^{2}}=1 . \\
\end{array}
$$
(Because as $x \rightarrow 0$, $\ln \left(1+\frac{\sin ^{2} x}{e^{x}}\right)$
$$
\sim \frac{\sin ^{2} x}{e^{x}}, \ln \left(1+\frac{x^{2}}{e^{2 x}}\right) \sim \frac{x^{2}}{e^{2 x}})
$$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. Find $\lim _{y \rightarrow 0} \frac{e^{y}+\sin y-1}{\ln (1+y)}$.
|
Since $\lim _{y \rightarrow 0} \frac{e^{y}-1}{y}=1, e^{y}-1 \sim y$
$$
\lim _{\substack{(y \rightarrow 0 \\(y \rightarrow 0 \text { when) }}} \frac{e^{y}-1+\sin y}{2 y}=1, e^{y}-1+\sin y \sim 2 y
$$
Therefore, the original expression $=\lim _{y \rightarrow 0} \frac{2 y}{y}=2$.
It is worth noting that when using equivalent infinitesimal substitution, only local equivalent infinitesimal substitution is performed, without overall verification, which may lead to incorrect results.
For example, when finding $\lim _{x \rightarrow 0} \frac{\operatorname{tg} x-\sin x}{x^{3}}$, only noticing $\sin x \sim x$ and $\operatorname{tg} x \sim x (x \rightarrow 0$ when $)$, and making the following substitution:
Original expression $=\lim _{x \rightarrow 0} \frac{x-x}{x}=0$, this result is incorrect.
In fact, the original expression $=\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \frac{\left(\frac{1}{\cos x}-1\right)}{x^{2}}$
$$
=\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} \cdot \frac{2 \sin ^{2} \frac{x}{2}}{x^{2}}=\frac{1}{2} \text {. }
$$
In finding limits, using equivalent infinitesimals for substitution is often effective.
Let $\lim \alpha(x)=\infty, \lim \beta(x)=\infty$ $\left(x \rightarrow x_{0}\right.$ or $x \rightarrow \infty$ when $)$, if $\lim \frac{\beta(x)}{\alpha(x)}=1$, then $\beta(x)$ and $\alpha(x)$ are said to be equivalent infinitesimals, denoted as $a(x) \sim \beta(x) \quad\left(x \rightarrow x_{0}\right.$ or $x \rightarrow \infty$ when $)$.
Similarly, $\lim \frac{\beta(x)}{\alpha(x)}=\lim \frac{\beta_{1}(x)}{\alpha_{1}(x)}$ $\left(\alpha(x) \sim a_{1}(x), \beta(x) \sim \beta_{1}(x)\right.$, when $x \rightarrow 0$ or $x \rightarrow \infty$).
For example, $\lim _{x \rightarrow+\infty} \frac{e^{x}-e^{-2 x}}{3^{x}+\frac{2}{-x}}$.
Since $e^{x} \sim e^{x}-e^{-2 x}, \quad 3^{x} \sim 3^{x}+2^{-x}$ $(x \rightarrow+\infty$ when $)$, the original expression $=\lim _{x \rightarrow+\infty} \frac{e^{x}}{3^{x}}$ $=\lim _{x \rightarrow+\infty}\left(\frac{e}{3}\right)^{x}=0$.
(Author's unit: Department of Mathematics, Tianjin Normal University)
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. If $A=20^{\circ}, B=25^{\circ}$, find $(1+\operatorname{tg} A)$
- $(1+\operatorname{tg} B)$.
untranslated part:
- $(1+\operatorname{tg} B)$ 的值.
Note: The untranslated part is kept as is because "tg" is a notation for the tangent function, and the phrase "的值" means "the value of" in this context. However, for a complete translation, it should read:
- the value of $(1+\operatorname{tg} B)$.
Thus, the fully translated text is:
Example 5. If $A=20^{\circ}, B=25^{\circ}$, find the value of $(1+\operatorname{tg} A)$
- $(1+\operatorname{tg} B)$.
|
$$
\text{Solve } \begin{aligned}
& (1+\tan A) \cdot(1+\tan B) \\
= & \left(1+\tan 20^{\circ}\right) \cdot\left(1+\tan 25^{\circ}\right) \\
= & 1+\tan 20^{\circ}+\tan 25^{\circ}+\tan 20^{\circ} \cdot \tan 25^{\circ} .
\end{aligned}
$$
Using the tangent addition formula in reverse $1=\tan 45^{\circ}$
$$
\begin{array}{l}
= \frac{\tan 20^{\circ}+\tan 25^{\circ}}{1-\tan 20^{\circ} \cdot \tan 25^{\circ}} \text{, we get } \\
1-\tan 20^{\circ} \cdot \tan 25^{\circ}=\tan 20^{\circ}+\tan 25^{\circ} .
\end{array}
$$
$$
\begin{array}{l}
\text{Therefore, } \tan 20^{\circ}+\tan 25^{\circ}+\tan 20^{\circ} \cdot \tan 25^{\circ}=1 \text{. } \\
\text{Thus, }(1+\tan A) \cdot(1+\tan B)=2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. What is the last digit of $11^{6}+14^{6}+16^{8}$?
|
1. The last digit of the given sum can be determined by adding the last digits of each power: the last digit of any power of a number ending in 1 and 6 remains 1 and 6, the last digit of an even power of a number ending in 4 is 6. Therefore, the last digit of the given sum is 3.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. What is the last digit of $1! +2!+3!+\cdots+8!+9!$?
|
7. The last digit of this sum is 3. Because from 5! onwards, the last digit of each term is 0, and $1!+$ $2!+3!+4!=1+2+6+24=33$
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Which of the following expressions are meaningful? Among the meaningful radicals, which are arithmetic roots? Express the radicals that are not arithmetic roots using arithmetic roots;
(1) $\sqrt[3]{\frac{1}{8}}$,
(2) $\sqrt{-9}$
(3) $\sqrt[3]{-8}$
|
Solution: (1) is an arithmetic root;
(2) is meaningless;
(3) is meaningful, but not an arithmetic root,
$$
\sqrt[3]{-8}=-\sqrt[3]{8}=-2 .
$$
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. If $a$ and $b$ are real numbers, and
$$
\sqrt{2 a-1}+|b+1|=0 \text {. }
$$
Find the value of $a^{-2}+b^{-1987}$.
|
Solution: $\because \sqrt{2 a-1} \geqslant 0,|b+1| \geqslant 0$, according to property 5, we have
$$
\begin{array}{l}
\sqrt{2 a-1}=0 \text { and }|b+1|=0, \\
\therefore a=\frac{1}{2} \text { and } b=-1 \text {. Therefore, } \\
a^{-2}+b^{-1987}=\left(\frac{1}{2}\right)^{-2}+(-1)^{-1987} \\
=4-1=3 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. If $x, y$ are real numbers, and
$$
y=\frac{\left(1-x^{2}\right)^{\frac{1}{2}}+\left(x^{2}-1\right)^{\frac{1}{6}}}{4 x-5} \text {, }
$$
find the value of $\log _{\frac{1}{7}}(x+y)$.
|
To make the expression for $y$ meaningful, $x$ must simultaneously satisfy the following three inequalities.
$$
\begin{array}{c}
1-x^{2} \geqslant 0, \\
x^{2}-1 \geqslant 0, \\
4 x-5 \neq 0 . \\
\therefore x=1, \quad \text { thus } y=0 . \\
\therefore \log _{\frac{1}{7}}(x+y)=\log _{\frac{1}{7}} 1=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example. The infinite sequence
$$
a_{1}, a_{2}, a_{3}, \ldots
$$
has the relation
$$
a_{n}=4+\frac{1}{3} u_{n-1}(n=2,3,1, \cdots)
$$
Find $\lim _{n \rightarrow \infty} a_{n}$.
|
$$
\begin{array}{l}
a_{2}=4+\frac{1}{3} a_{1}, \\
a_{3}=4+\frac{1}{3} a_{2}=4+\frac{1}{3}\left(4+\frac{1}{3} a_{1}\right) \\
=4+4 \cdot \frac{1}{3}+\left(\frac{1}{3}\right)^{2} a_{1} \\
a_{4}=4+\frac{1}{3} a_{3} \\
=4+4 \cdot \frac{1}{3}+4\left(\frac{1}{3}\right)^{2}+\left(\frac{1}{3}\right)^{2} a_{1} \\
a_{n}=4+4 \cdot \frac{1}{3}+4\left(\frac{1}{3}\right)^{2}+\cdots \\
+4\left(\frac{1}{3}\right)^{n-2}+\left(\frac{1}{3}\right)^{n-1} a_{1} \\
\therefore \lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty}\left\{4+4 \cdot \frac{1}{3}\right. \\
+4\left(\frac{1}{3}\right)^{2}+\cdots \\
\left.+4\left(\frac{1}{3}\right)^{n-2}\right\} \\
+\lim _{n \rightarrow}\left(\frac{1}{3}\right)^{n-1} a_{1}=6. \\
\end{array}
$$
Above, by simplifying $a_{2}, a_{3}, a_{4}, \cdots$ successively, we find $a_{n}$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Given $\sin \alpha+\cos \alpha=a$.
(1) Find the value of $\sin ^{5} \alpha+\cos ^{5} \alpha$;
(2) If $a=1$, find the value of $\sin ^{n} \alpha+\cos ^{n} \alpha$.
|
Let $f(n)=\sin ^{n} \alpha+\cos ^{n} \alpha$, then (1) $=a$.
Given $f(2)=\sin ^{2} \alpha+\cos ^{2} \alpha=1$,
$f(3)=a f(2)-\frac{a^{2}-1}{2} \cdot a$
$=\frac{-a^{3}+3 a}{2}$,
$f(4)=a f(3)-\frac{a^{2}-1}{2} f(2)$
$=-a^{4}+2 a^{2}+1$
2
$\therefore f(5)=\sin ^{5} \alpha+\cos ^{5} \alpha$
$=a f(4)-\frac{a^{2}-1}{2} f(3)$
$=\frac{-a^{5}+5 a}{4}$
(2) $\because \sin \alpha+\cos \alpha=1$,
$\therefore \sin \alpha \cdot \cos \alpha=0$
$\therefore f(n)=\sin ^{n} \alpha+\cos ^{n} \alpha=(\sin \alpha$
$+\cos \alpha) f(n-1)-(\sin \alpha \cdot \cos \alpha) f(n-2)$
$=f(n-1)$
Similarly, $f(n-1)=f(n-2)=\cdots=f(1)$ $=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Let $x, y$ satisfy $3 x^{2}+2 y^{2}=6 x$, find the maximum value of $x^{2}+y^{2}$:
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
|
If not analyzed, from the known conditions:
$$
\begin{array}{l}
x^{2}+y^{2}=-\frac{1}{2} x^{2}+3 x \\
=-\frac{1}{2}(x-3)^{2}+\frac{9}{2},
\end{array}
$$
it would be incorrect to say that when $x=3$, $x^{2}+y^{2}$ achieves its maximum value $\frac{9}{2}$. This is because from $y^{2}=-\frac{3}{2} x^{2}+3 x \geqslant 0$, we get $0 \leqslant x \leqslant 2$.
The correct answer is: when $x=2$, $x^{2}+y^{2}$ achieves its maximum value 4.
Through geometric analysis, the error and the correct reasoning can be clearly explained: the constraint condition is transformed into
$$
(x-1)^{2}+\frac{y^{2}}{3 / 2}=1 \text{. }
$$
This is an ellipse. The maximum value we are seeking is the square of the radius of the largest circle
$$
x^{2}+y^{2}=R^{2}
$$
that has a common point with this ellipse. The point on the ellipse farthest from the origin $O(0, 0)$ is $(2,0)$, and the point $(3, y)$ is outside the ellipse.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Find the minimum value of $|x-1|+|x-3|+|x-5|$.
|
To solve this type of problem, we can use the method of "fixing points, dividing segments, and discussing."
Points. Let $x-1=0, x-3=0, x-5=0$, to determine $x_{1}=1, x_{2}=3, x_{3}=5$ three points.
Dividing segments. The above three points divide the number line into four segments: $(-\infty, 1]$, $(1,3]$, $(3,5]$, $(5,+\infty)$.
Discussion. When $x \in(-\infty, 1]$, the original expression $=9-3 x$. In this case, the minimum value of the original expression is 6;
When $x \in(1,3]$, the original expression $=7-x$. In this case, the minimum value of the original expression is 4;
When $x \in(3,5)$, the original expression $=x+1$. In this case, the minimum value of the original expression is 4;
When $x \in(5,+\infty)$, the original expression $=3 x-9$. In this case, the minimum value of the original expression is 6.
In summary, the minimum value of the original expression is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Let $y=\sqrt[3]{1+3 x}, x=\frac{1}{3}\left(y^{3}-1\right)$, when $x \rightarrow 0$, $y \rightarrow 1$.
Original expression $=\frac{2}{3} \lim _{y \rightarrow 1} \frac{y^{3}-2 y^{3}+1}{y^{3}-3 y+2}$ is still of the form $\frac{0}{0}$. Let $z=y-1$, then,
Original expression $=\frac{2}{3} \lim _{z \rightarrow 0} \frac{\left(z^{2}+3 z+3\right)}{z+3}=2$.
Second, for indeterminate forms of rational expressions in trigonometric functions of the form $\lim _{x \rightarrow x_{0}} R(\sin x, \cos x)$, the universal substitution can be used.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. To complete a certain project, the number of days required for $A$ to work alone is $m$ times that of $B$ and $C$ working together, the number of days required for $B$ to work alone is $n$ times that of $A$ and $C$ working together, and the number of days required for $C$ to work alone is $p$ times that of $A$ and $B$ working together. Try to prove that the ratio of the number of days each person needs to work alone is $(m+1):(n+1):(p+1)$, and that $-\frac{n}{m+1}+\frac{n}{n+1} \div \frac{n}{p+1}=2$.
|
Let $x$, $y$, $z$ be the number of days required for $A$, $B$, $C$ to complete the work individually, respectively, then we have
$$
\frac{m}{x}=\frac{1}{y}+\frac{1}{z}
$$
Therefore, $\frac{m+1}{x}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.
Similarly, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{n+1}{y}=\frac{p+1}{z}$.
Thus, the ratio of the number of days each person works individually is
$$
(m+1):(n+1):(p+1) \text {. }
$$
Additionally, by $\frac{1}{m+1}=\frac{1}{x} \div\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$,
$$
\begin{array}{l}
\frac{1}{n+1}=\frac{1}{y} \div\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right), \\
\frac{1}{p+1}=\frac{1}{z} \div\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right),
\end{array}
$$
Adding the three equations yields
$$
\frac{1}{m+1}+\frac{1}{n+1}+\frac{1}{p+1}=1 \text {. }
$$
liij
$$
\begin{array}{l}
\frac{m}{m+1}=1-\frac{1}{m+1}, \\
\frac{n}{n+1}=1-\frac{1}{n+1}, \\
\frac{p}{p+1}=1-\frac{1}{p+1},
\end{array}
$$
Thus, $\frac{m}{m+1}+\frac{n}{n+1}+\frac{p}{p+1}$
$$
=3-\left(\frac{1}{m+1}+\frac{1}{n+1}+\frac{1}{p+1}\right)=2 .
$$
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Given $f\left(1-x^{2}\right)=\frac{1-x^{2}}{x^{2}}$. Find $f\left(\frac{1}{2}\right)$.
|
$\begin{array}{l}\text { Solve } f\left(1-x^{2}\right)=\frac{1-x^{2}}{1-\left(1-x^{2}\right)}, \\ \text { hence } f(x)=\frac{x}{1-x}, f\left(\frac{1}{2}\right)=1.\end{array}$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9. Given the dihedral angle $C-A B=D=120^{\circ}$, $\angle C B A=60^{\circ}, \angle D A B=30^{\circ}, A B=\sqrt{37} \mathrm{~cm}$ (Figure 17). Find the distance $d$ between the skew lines $B C, A D$.
|
Let the angle between $BC$ and $AD$ be $\theta$. By the triple product formula,
$$
\begin{aligned}
\cos \theta= & \cos 60^{\circ} \cos 150^{\circ} \\
& +\sin 60^{\circ} \sin 150^{\circ} \cos 120^{\circ} \\
= & -\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{8}=-\frac{3 \sqrt{3}}{8} .
\end{aligned}
$$
Thus, $\sin \theta=\frac{\sqrt{37}}{8}$.
Therefore, $d=\frac{\sqrt{37} \sin 60^{\circ} \sin 150^{\circ} \sin 120^{\circ}}{\sin \theta}$
$=3(\text{cm})$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Given that 1 is a root of the equation $a x^{2}+b x+c=0$, find the value of $\frac{a^{2}+b^{2}+c^{2}}{a^{3}+b^{3}+c^{3}}+\frac{2}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.
|
Given that $1$ is a root of the equation $a x^{2}+b x+c=0$,
$$
\begin{array}{c}
\therefore a+b+c=0 . \\
\text { Also, } a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c) \\
.\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) .
\end{array}
$$
From (1) and (2), we can deduce
$$
\begin{array}{c}
a^{3}+b^{3}+c^{3}=3 a b c . \\
\text { Therefore, the original expression }=\frac{a^{2}+b^{2}+c^{2}}{3 a b c} \\
+\frac{2(a b+b c+c a)}{3 a b c} \\
=\frac{(a+b+c)^{2}}{3 a b c}=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the sets $M=\{x, x y, \lg (x y)\}$ and $N=\{0,|x|, y\}$, and $M=N$. Then, $\left(x+\frac{1}{y}\right)+\left(x^{2}+\frac{1}{y^{2}}\right)+\left(x^{3}+\right.$ $\left.\frac{1}{y^{3}}\right)+\cdots+\left(x^{2001}+\frac{1}{y^{2001}}\right)$ is equal to
|
Given $\because M=N$,
$\therefore 0 \in M$, and $x, y$ cannot both be 0, then
$\lg (x y)=0, x y=1$.
Thus, $1 \in N$. If $y=1$, then $x=1$, which contradicts the distinctness of the elements in the set. Therefore, $|x|=1$, and $x=-1$, so $y=-1$.
Therefore, $x+\frac{1}{y}=-2 \cdot x^{2}+\frac{1}{y^{2}}=2$, $x^{3}+\frac{1}{y^{3}}=-2, \cdots, x^{2000}+\frac{1}{y^{2000}}=2$, $x^{2001}+\frac{1}{y^{3} \frac{1}{301}}=-2$, and the sum is -2.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For a regular $n$-sided polygon, construct $n$ squares outside the polygon, each sharing one side with the polygon. It is known that the $2n$ outer vertices of these $n$ squares form a regular $2n$-sided polygon. For what value of $n$ is this possible?
|
Solve As shown in Figure 1, $P, Q, R$ are three vertices of a regular $n$-sided polygon; $M, N, K, L$ are four vertices of a $2n$-sided polygon. According to the given conditions, we have $MN = NQ, QK = KL, NK = KL = MN$, which means $\triangle NQK$ is an equilateral triangle. Therefore, $\angle NQK = 60^{\circ}$. From the internal angle of the regular $n$-sided polygon $\angle PQR = 120^{\circ}$, we get the exterior angle as $60^{\circ}$.
$$
n \cdot 60^{\circ} = 360^{\circ} \text{, i.e., } n = 6 \text{. }
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. On the blackboard are the numbers $1,2, \cdots, 1987$. Perform the following transformation: erase some of the numbers on the blackboard and add the remainder when the sum of the erased numbers is divided by 7. After several such transformations, only two numbers remain on the blackboard, one of which is 987. Find the other number.
|
Note the fact that with each transformation: the sum of all numbers on the blackboard, when divided by 7, leaves the same remainder. $1+2+\cdots+1987=1987 \cdot 7 \cdot 142$, which indicates that the original sum on the blackboard is divisible by 7, so the sum of the two final numbers can also be divided by 7. Among the two final numbers, one of them (which cannot be 987) is the remainder when divided by 7, i.e., $0,1,2,3,4,5,6$. Since 987 is divisible by 7, the other number can only be 0.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. At least how many circles with a radius of 1 are needed to cover a circle with a radius of 2.
|
Let $O$ be the center of a circle with radius 2, and let $A B C D E F$ be a regular hexagon inscribed in the circle (Figure 8). The side length of the hexagon is 2. Construct six circles with the sides of the hexagon as diameters. The intersection points of these six circles, other than $A, B, C, D, E, F$, are denoted as $A_{1}, B_{1}, C_{1}, D_{1}, E_{1}, F_{1}$. These points are the vertices of a regular hexagon with side length 1. Use the above circles with radius 1 to cover the circle with radius 2. The remaining part (the shaded part in Figure 8) is contained within the hexagon $A_{1} B_{1} C_{1} D_{1} E_{1} F_{1}$, and can be completely covered by a circle with center $O$ and radius 1. Therefore, 7 circles can cover the circle with radius 2.
On the other hand, each small circle can only cover one-sixth of the large circle, so at least six circles are needed to cover the large circle. However, these six small circles still cannot cover the point $O$, so at least seven small circles are needed to cover the large circle.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. If $a>1, b$ is a positive rational number, $a^{b}+a^{-0}$ $=2 \sqrt{2}$, find the value of $a^{b}-a^{-b}$.
|
Consider the following solution:
Let $a^{b}=x$, then $x+\frac{1}{x}=2 \sqrt{2}$.
Transform it into $x^{2}-2 \sqrt{2} x+1=0$.
Solving yields $x_{1}=\sqrt{2}+1, x_{2}=\sqrt{2}-1$.
$$
\text { When } \begin{aligned}
x & =\sqrt{2}+1, \\
& a^{b}-a^{-b} \\
& =\sqrt{2}+1-\frac{1}{\sqrt{2}+1} \\
& =\sqrt{2}+1-(\sqrt{2}-1) \\
& =2
\end{aligned}
$$
When $x=\sqrt{2}-1$,
$$
\begin{aligned}
& a^{b}-a^{-b} \\
= & \sqrt{2}-1-\frac{1}{\sqrt{2}-1} \\
= & \sqrt{2}-1-(\sqrt{2}+1) \\
= & -2 .
\end{aligned}
$$
This solution is incorrect. The error lies in neglecting the conditions $a>1$ and $b$ being a positive integer - if these conditions were considered, it would be clear that $a^{b}>1$. Therefore, the correct answer should be 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. If $\lg ^{2} x \lg 10 x<0$, find the value of $\frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+\lg 10 x^{2}}$.
|
From the known condition $\lg ^{2} x \lg 10 x<0$, it is not difficult to conclude that $\lg x \neq 0$ and $\lg 10 x<0$, which also implicitly indicates the existence of $\lg x$.
$$
\text { Hence } \begin{aligned}
& \frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+\lg 10 x^{2}} \\
= & \frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+2 \lg x+\lg 10} \\
= & \frac{1}{\lg 10 x} \sqrt{(\lg x+1)^{2}} \\
= & \frac{1}{\lg 10 x} \sqrt{\lg ^{2} 10 x} \\
= & \frac{-\lg 10 x}{\lg 10 x}=-1 .
\end{aligned}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. (1MO-27 Preliminary Question) Let $A, B, C$ be three points on the edge of a circular pool, with $B$ due west of $C$, and $A B C$ forming an equilateral triangle with side lengths of 86 meters. A swimmer starts from $A$ and swims directly to $B$. After swimming $x$ meters, he reaches point $\boldsymbol{E}$, then turns and swims due west for $y$ meters, arriving at point $D$. If $x, y$ are both integers, find $y$.
---
The translation preserves the original text's formatting and line breaks.
|
Given the figure, since $\triangle AEF$ is an equilateral triangle, we have
$$
AF = AE = x.
$$
By symmetry, $FG = DE = y$. Also, since
$$
AE \cdot EB = DE \cdot EG,
$$
we have
$$
\begin{aligned}
& x(86 - x) \\
= & y(x + y).
\end{aligned}
$$
Notice that if $x$ is odd, then $x(86 - x)$ is also odd, while $y(x + y)$ is always even (if $y$ is even, $y(x + y)$ is clearly even; if $y$ is odd, then $x + y$ is even, and $y(x + y)$ is still even).
Therefore, $x$ must be even. From (1), we know that $y(x + y)$ is even, so $y$ must also be even.
The above argument is essentially taking $\left(1; \mathbf{x}^{2}\right)$ modulo 2 to determine $x$. We have $\left(x + \frac{y}{2} - 43\right)^{2} + y^{2} = \left(43 - \frac{y}{2}\right)^{2}$.
This indicates that $\left|x + \frac{y}{2} - 43\right|, y, 43 - \frac{y}{2}$ form a Pythagorean triplet.
Let $d$ be the greatest common divisor of $\left|x + \frac{y}{2} - 43\right|$ and $y$. Then, by the theorem mentioned later, there exist positive integers $a, b, a > b$, such that
$y = 2abd$ (since $y$ is even),
$$
43 - \frac{y}{2} = (a^2 + b^2)d.
$$
From these two equations, we get
$$
(a^2 + ab + b^2)d = 43 \cdot (a > b > 0).
$$
Since 43 is a prime number, we have $d = 1$.
Finally, we solve the equation using the "bounding method"
$$
a^2 + ab + b^2 = 43.
$$
Since $a > b > 0$, we have $3b^2 < 43$, $b^2 < \frac{43}{3}$, $b^2 \leq 14$, so $b \leq 3$. It is easy to see that the only positive integer solutions are $b = 1$ and $a = 6$.
Thus, we know $y = 12$. Then, from $\left|x + \frac{y}{2} - 43\right| = a^2 - b^2$, we find $x = 2$ or $x = 72$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 14. Find the number of integer solutions $(x, y, z)$ that satisfy $0<x<y<z$, and $\sqrt{1984}=\sqrt{ } x+\sqrt{ } y+\sqrt{ } z$.
|
$$
\text { Stele } \begin{aligned}
& \sqrt{1984}=8 \sqrt{31} \\
& =\sqrt{31}+2 \sqrt{31}+5 \sqrt{31} \\
& =\sqrt{31}+\sqrt{2} \times 31+\sqrt{5} \times 31 \\
& =\sqrt{31}+3 \sqrt{31}+4 \sqrt{31} \\
& =\sqrt{31}+\sqrt{3} \times 31+\sqrt{2} \times 1 .
\end{aligned}
$$
There are two kinds of correct statements in total.
$$
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Go to the room, take a card at random, and ask which number is most likely to be the unit digit of the number drawn by Jack and Jill.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
12
I
A
12. The list below, the numbers in the table represent the units digit of the sum of two numbers.
\begin{tabular}{ccccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline 2 & 3 & 4 & 3 & 6 & 7 & 8 & 0 & 3 \\
3 & 4 & 0 & 3 & 7 & 3 & 9 & 1 & 1 \\
4 & 5 & 6 & 7 & 3 & 9 & 0 & 1 & 2 \\
5 & 6 & 1 & 3 & 3 & 0 & 1 & 2 & 3 \\
6 & 7 & 8 & 9 & 0 & 1 & 2 & 3 & 4 \\
7 & 6 & 9 & 0 & 1 & 2 & 3 & 4 & 5 \\
8 & 9 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
9 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
\end{tabular}
The digit 0 appears 9 times, while the digits 1-9 each appear 8 times, so for now
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Place one more of these shapes inside the square.
Keep the original text's line breaks and format, and output the translation result directly.
Note: The second sentence is a directive for the translation task and should not be included in the translation output. Here is the translation:
Place one more of these shapes inside the square.
|
Solution: For each $2 \times 2$ square, at least two cells should be covered. An $8 \times 8$ square can be divided into 16 non-overlapping $2 \times 2$ squares. Therefore, at least 11 figures should be covered. With 11 figures, it is feasible.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Uncle Qiel Yuemuer every night selects 9 or 10 warriors from 33 warriors to be on duty according to his own wishes. At least how many days are needed to ensure that each warrior has the same number of duty shifts?
|
The minimum number of days is 7 days, during which each warrior is on duty 2 times.
Let the required minimum number of days be $N$, and each warrior is on duty $n$ times within these $N$ days. We number the warriors sequentially as $1,2, \cdots, 33$, and introduce the number $a_{1 \times}(i=1,2, \cdots, N ; k=1,2, \cdots, 33)$ to represent the following: if the $k$-th warrior is not on duty on the $i$-th day, then $a_{1 x}=0$; if the $k$-th warrior is on duty on the $i$-th day, then $a_{i k}=1$. Let $P_{k}$ be the number of people on duty on the $i$-th day, and $Q_{k}$ be the number of days the $k$-th warrior is on duty. Thus, according to the problem conditions, the following relationships hold:
$$
\begin{array}{l}
9 \leqslant a_{i 1}+a_{\mathrm{i} 2}+\cdots+a_{\mathrm{i} 33}=P_{\mathrm{i}} \leqslant \\
10, i=1,2, \cdots, N \text {; } \\
\text { (1) } \\
a_{1_{k}}+a_{2 k}+\cdots+a_{N k}=Q_{k}=n \text {, } \\
i=1,2, \cdots, \dot{N} \text {; } \\
\text { (2) } \\
\end{array}
$$
\begin{tabular}{|c|c|c|}
\hline Day & On-duty Warrior Numbers & \begin{tabular}{l}
On-duty
\end{tabular} \\
\hline 1 & $2,3,4,5,6,7,8,9,28,29$ & 10 \\
\hline 2 & $1,2,3,4,5,6,7,8,9$ & 9 \\
\hline 3 & $11,12,13,14,15,16,17,18,30,31$ & 20 \\
\hline 4 & $10,11,12,13,14,15,16,17,18$ & $\frac{9}{10}$ \\
\hline 5 & $20,21,22,23,24,25,26,27,32,33$ & . 0 \\
\hline 6 & $19,20,21,22,23,24,25,26,27$ & $\theta$ \\
\hline 7 & $1,10,19,28,29,30,31,32,33$ & 9 \\
\hline
\end{tabular}
Figure 12
Summing the inequality (1) for $i=1,2, \cdots, N$, we get the inequality $9 N \leqslant S \leqslant 10 N$, where:
$$
\begin{array}{l}
S=\left(a_{11}+a_{12}+\cdots+a_{1,33}\right)+\left(a_{21}\right. \\
\left.+a_{22}+\cdots+a_{2,33}\right)+\cdots+\left(a_{\mathrm{N} 1}+a_{\mathrm{N} 2}+\cdots\right. \\
\left.+a_{\mathrm{N}, 33}\right) .
\end{array}
$$
Summing the equation (2) for $k=1,2, \cdots, 33$, we get the equation $S=33 n$. This implies $N \leqslant 33 n \leqslant 10 N$, or $\frac{3}{11} N \leqslant n \leqslant \frac{10}{33} N$. An integer $n$ between $\frac{3}{11} N$ and $\frac{10}{33} N$ does not exist for $N=1,2,3,4,5,6$, but when $N=7$, there is an integer $n=2$ that satisfies the inequality $\frac{3}{11} N \leqslant n \leqslant \frac{10}{33} N$. Therefore, the number of duty shifts $n$ is at least 2, and the minimum number of days $N$ is at least 7. The duty schedule (similar to the table in Figure 12) meets the problem's conditions.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, solving the equation $\frac{x^{2}-7}{6}=\sqrt{6 x+7}$. This is the result of transitioning from concrete to abstract thinking.
From a geometric perspective, this is equivalent to finding the x-coordinates of the intersection points of the line $y=\frac{x^{2}-7}{6}$ and the curve $y=\sqrt{6 x+7}$. These two happen to be inverse functions of each other. According to the geometric property that the graphs of a function $y=f(x)$ and its inverse $y=f^{-1}(x)$ are symmetric about the line $y=x$, their intersection points must lie on the line $y=x$.
|
Solve $\sqrt{6 x+7}=x$ to get
$$
\begin{array}{l}
x^{2}-6 x-7=0, \\
\therefore x_{1}=-1 \text { (discard), } x_{2}=7 .
\end{array}
$$
Upon verification, the root of the original equation is $x=7$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Prove: $\quad(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}$ has a minimum value of 8 on $0<u<\sqrt{2}, v>0$.
|
$(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}$ can be seen as the square of the distance between point $P_{1}(u, \sqrt{2-u^{2}})$ and $P_{2}\left(v, \frac{9}{v}\right)$.
$P_{1}$ satisfies the equation of the circle:
$$
x^{2}+y^{2}=2 \text{. }
$$
$P_{2}$ satisfies the equation of the hyperbola:
$$
x y=9, x>0 \text{. }
$$
Obviously, the distance between $M$ and $N$ reaches the minimum distance between $P_{1}$ and $P_{2}$. Let $y=x$, it is easy to get $M(1,1), N(3,3)$, so the minimum value $=(3-1)^{2}+(3-1)^{2}=8$.
|
8
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $\alpha, \beta$ be acute angles. When
$$
-\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta}
$$
takes the minimum value, the value of $\operatorname{tg}^{2} \alpha+\operatorname{tg}^{2} \beta$ is
|
To find the minimum value of the original expression, we need
$$
\sin ^{2} \beta \cos ^{2} \beta=\frac{1}{4} \sin ^{2} 2 \beta
$$
to take the maximum value. Since $\beta$ is an acute angle, we know $\beta=\frac{\pi}{4}$.
When $\beta=\frac{\pi}{4}$,
$$
\begin{aligned}
\text { the original expression } & \geqslant \frac{1}{\cos ^{2} \alpha}+\frac{4}{\sin ^{2} \alpha} \\
& =\left(\frac{1}{\cos ^{2} \alpha}+\frac{4}{\sin ^{2} \alpha}\right)\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right) \\
& =1+\tan^{2} \alpha+\frac{4}{\tan^{2} \alpha}+4 \\
& \geqslant 1+4+4=9 .
\end{aligned}
$$
When the product of the terms is a constant, the minimum value of their sum is
$$
\tan^{2} \alpha+\frac{4}{\tan^{2} \alpha} \geqslant 4
$$
The equality holds if and only if
$$
\tan^{2} \alpha=\frac{4}{\tan^{2} \alpha}
$$
which gives $\tan^{2} \alpha=2$. Therefore,
$$
\tan^{2} \alpha+\tan^{2} \beta=2+1=3 .
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. The general term of the sequence is $a_{\mathrm{n}}=b[\sqrt{n+c}]+d$, and the terms are calculated successively as
$$
1,3,3,3,5,5,5,5,5, \cdots \text {. }
$$
where each positive odd number $m$ appears exactly $m$ times consecutively. The above $b, c, d$ are undetermined integers. Then, the value of $b+c+d$ is $\qquad$ where $[x]$ denotes the greatest integer not exceeding $x$.
|
Solve: First determine $b$.
Since $a_{n}$ is odd, and $a_{n}+1 \geqslant a_{n}$, we know that $a_{n}+1-a_{n}$ $\in\{0,2\}$,
i.e., $b[\sqrt{n+1+c}]-b[\sqrt{n+c}]$ is 0 or 2.
For any natural number $n$, it always holds that
$[\sqrt{n+1+c}]-[\sqrt{n+c}] \in\{0,1\}$.
Clearly, $b \neq 0$.
When $-[\sqrt{n+c}])=2$, by (1), it can only be that $[\sqrt{n+1+c}]-[\sqrt{n+c}]=1, b=2$.
Since $b$ is a constant, we have $b=2$.
Next, find $c$.
From $=2[\sqrt{u+c}]+d$ we know $c \geqslant-1$.
Based on the characteristics of the sequence $a_{\mathrm{n}}$, we have $a_{k}^{2}=2 k-1$.
Taking a sufficiently large $k$, ensuring $2 k+3>c$, at this time $(k+1)^{2}+c<(k+1)^{2}+2 k+3=(k+2)^{2}$; we get $\left[\sqrt{(k+1)^{2}+c}\right]<k+2$. $-1=2 k$ or
$$
\left[\sqrt{(k+1)^{2}+c}\right]-[\sqrt{1+c}] \equiv k \text {. }
$$
Substituting (2) gives: $[\sqrt{1+c}]<2$.
Therefore, $c<3$.
If $c=0,1,2$, then in each case $[\sqrt{5+c}]-[\sqrt{4+c}]$
$=2-2=0$ which contradicts $a_{5}-a_{4}=2$.
In summary, $c=-1$.
Finally, $d=a_{1}=1$.
Thus, $b+c+a'=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Tank A contains 4 liters of liquid A, Tank B contains 2 kilograms of liquid B, and Tank C contains 2 kilograms of liquid C. These liquids can all be mixed. First, 1 liter of liquid from Tank A is poured into Tank B, and then 1 liter of the mixed liquid from Tank B is poured into Tank C. Finally, 1 liter of the mixed liquid from Tank C is poured back into Tank A. This operation is called one mixing. After $n$ mixings, the amount of liquid A in Tank B is denoted as $f(n)$. If $f(n)>0.9999$, what is the minimum value of $n$? $\qquad$ .
|
Let $a_{n}, b_{n}, c_{n}$ be the amounts of liquid A in containers 甲, 乙, and 丙, respectively, after $n$ mixings. Then,
$$
a_{n}+b_{n}+c_{n}=4 \text{. }
$$
By symmetry, we have
$$
b_{n}=c_{n} \text{. }
$$
Now, let's examine the amount of liquid A in container 甲 after the $(n+1)$-th mixing. According to the problem,
$$
\begin{aligned}
a_{n+1} & =\frac{1}{2} a_{n}+\frac{1}{3}\left(\frac{1}{4} a_{n}+b_{n}\right) \\
+ & \left.\frac{1}{3} \div \frac{1}{4} a_{n}+c_{5}\right) \\
& =\frac{1}{2} a_{n}+\frac{1}{6} a_{n}+\frac{2}{3} b_{n} \\
& =\frac{2}{3} a_{n}+\frac{1}{3}\left(4-a_{n}\right) \\
& =\frac{1}{3} a_{n}+\frac{4}{3}
\end{aligned}
$$
Thus, $a_{n+1}-2=\frac{1}{3}\left(a_{n}-2\right)$.
By induction, we get
$$
\begin{array}{l}
a_{n+1}-2=\frac{1}{3}\left(a_{n}-2\right)=\frac{1}{3^{2}}\left(a_{n-1}-2\right) \\
=\cdots=\left(\frac{1}{3}\right)^{n+1}\left(a_{0}-2\right) .
\end{array}
$$
Substituting $a_{0}=4$ yields
$$
\begin{array}{l}
a_{n}=2\left[1+\left(\frac{1}{3}\right)^{n}\right], \\
b_{n}=\frac{1}{2}\left(4-a_{n}\right)=1-\left(\frac{1}{3}\right)^{n} .
\end{array}
$$
Given $f(n)>0.9999$, i.e.,
$1-\left(\frac{1}{3}\right)^{n}>0.9999$ or $3^{n}>10000$. The smallest value of $n$ is 9.
(Li Yongle, Yu Rongpei)
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Write the numbers $1, 2, 3, \cdots$, 1986, 1987 on the blackboard. At each step, determine some numbers from those written and erase them, replacing them with the remainder of their sum divided by 7. After several steps, two numbers remain on the blackboard, one of which is 987. What is the second remaining number?
(13th All-Russian Mathematics Competition, 1987)
|
Solution: Clearly, at each step, the sum of all the numbers written down modulo 7 is preserved.
Let the remaining number be $x$, then $x+987$ is congruent to $1+2+\cdots+1987$ modulo 7.
Since $1+2+\cdots+1987=1987 \times 7 \times 142$ is divisible by 7, the remainder is 0, so $x+987$ is also divisible by 7. Since 987 is not the remainder when divided by 7, $x$ is the remainder of the operation when divided by 7, $0 \leqslant x \leqslant 6$. Since 987 is divisible by 7,
thus $x=0$.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
To 9. On a plane, there is a fixed point $P$, consider all possible equilateral triangles $ABC$, where $AP=3, BP=2$. What is the maximum length of $CP$? (1961 Autumn Competition)
|
Solve for Ling and Hui
$$
\begin{array}{l}
\angle A P B=\alpha . \\
\angle B A P=\beta,
\end{array}
$$
Given $A B^{2}=3^{2}+2^{2}$
$$
\begin{aligned}
- & 12 \cos \alpha, \\
\cos \beta & =\frac{3^{2}+A B^{2}-2^{2}}{6 A B}, \sin \beta=\frac{2 \sin \alpha}{A B} .
\end{aligned}
$$
From this, we get
$$
\begin{aligned}
\cos \left(\frac{\pi}{3}+\beta\right)= & \frac{1}{2} \cos \beta-\frac{\sqrt{3}}{2} \sin \beta \\
= & \frac{3^{2}+A B^{2}-2^{2}-2 \sqrt{3} \sin \alpha}{12 A B}=2 A B \\
= & \frac{3^{2}+3^{2}-12 \cos \alpha}{12 A B} . \\
& -\frac{2 \sqrt{3} \sin \alpha}{2 A B} .
\end{aligned}
$$
(Applying the Law of Cosines in $\triangle P A C$, we get
$$
\begin{aligned}
P C^{2} & =A C^{2}+3^{2}-6 A C \cos \left(\frac{\pi}{3}+?\right) \\
& =13-12 \cos \left(\frac{\pi}{3}+\alpha\right) .
\end{aligned}
$$
Clearly, when $a=\frac{2 \pi}{3}$, $P C$ reaches its maximum value of 5.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. In the plane, 7 points are given, connect them with some line segments so that
(1) among any three points, at least two are connected,
(2) the number of line segments is minimized.
How many line segments are there? Provide such a graph.
|
17. The figure below shows that 9 line segments are sufficient.
Now prove that at least 9 line segments are needed.
If point $A$ is not an endpoint of any line segment, then due to (1), the other 6 points must connect at least $C_{6}^{2}>9$ line segments.
If point $A$ is the endpoint of only 1 line segment, then due to (1), the 5 points not connected to $A$ must connect at least $C_{5}^{2}>9$ line segments.
Assume each point is the endpoint of at least two line segments. Then if point $A$ is the endpoint of only two line segments $A B, A C$, by (1), the 4 points not connected to $A$ must connect at least $C_{4}^{2}=6$ line segments. At least one more line segment must be drawn from point $B$, so in this case, there are at least $2+6+1=9$ line segments.
If each point is the endpoint of at least 3 line segments, then there must be more than $\left[\frac{3 \times 7}{2}\right]=10$ line segments.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. All vertices of a broken line lie on the faces of a cube with an edge length of 2, and each segment of the broken line is 3 units long. This broken line connects two farthest vertices of the cube. How many segments does such a broken line have at least?
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last part is a note about the translation request and is not part of the text to be translated. It has been translated here for completeness.
|
Solve: Consider the cube circumscribing a sphere with center $A$ and radius 3, intersecting the cube's faces at three arcs: $K \hat{L}, \widehat{L N}, \widehat{N K}$ (Figure 3). The points $K, L, N$ on the edges bisect these three edges. In fact, $A D_{1}=\sqrt{8}$, so $L D_{1}=\sqrt{A} L^{\overline{2}-\overline{D_{1}^{2}}}=1$, and similarly for points $K$ and $N$.
Let point $M$ be any point on the aforementioned arcs, for example, on the arc $K L$, and $M$ is different from any endpoint of the arc. We consider a sphere with center $M$ and radius 3, and point $A$ lies on this sphere. It is easy to prove that all other points of the cube are inside this sphere.
From this, it follows that point $M$ can only be connected to point $A$, and the length of the segment is 3. Therefore, the first node of the broken line can only be one of the points $L, K$, and $N$. From point $K$, only the distance to point $D$ is 3 (except for point $A$). Similarly, the points that can be reached from points $L$ and $N$ can only be points $B$ and $A_{1}$, so all nodes are vertices of the cube or adjacent to point $A$. Similarly, the next node of the broken line must be the midpoint of one of the cube's edges, and the next node will be one of the points $C, B_{1}$, $D_{1}$, and the next node will again be the midpoint of an edge, and the final endpoint of the sixth segment can fall on point $C_{1}$, the opposite vertex of point $A$ (Figure 4 is an example). Thus, the minimum number of segments in a broken line that satisfies the conditions of the problem is 6.
|
6
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10. (19th All-Soviet Union Middle School Olympiad Problem) A quadratic trinomial $a x^{2}+b x+c$, where $a>100$. How many different integer points can there be at which the absolute value of its value does not exceed 50?
|
Assume there are three different integer points satisfying the given condition, then there must be two points located on the same side of the axis of symmetry $x=-\frac{b}{2 a}$ of the quadratic function $y=a x^{2} +b x+c$, or one point is located on this axis of symmetry. Without loss of generality, assume
$-\frac{b}{2 a}1\left(x_{1} \neq x_{2}\right.$ are integers $)$, then we have
$$
\begin{aligned}
100 & \geqslant\left|a x_{2}^{2}+b x_{2}+c\right|+\left|a x_{1}^{2}+b x_{1}+c\right| \\
& >\left|\left(a x_{2}^{2}+b x_{2}+c\right) \cdots\left(1 x_{1}^{2}+b x_{1}+c\right)\right| \\
& =a\left(x_{2}-x_{1}\right)\left(x_{2}+x_{1}+\frac{b}{a}\right) \geqslant a>100,
\end{aligned}
$$
which is a contradiction. Therefore, the number of points satisfying the given condition is at most two.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Column 1. A tour group is selecting visit locations from $A, B, C, D, E$ under the following constraints:
(1) If visiting $A$, must also visit $B$,
(2) At least one of $D, E$ must be visited,
(3) Only one of $B, C$ can be visited,
(4) Both $C, D$ must be visited or neither,
(5) If visiting $E$, both $A, D$ must be visited.
Please explain the reasoning, which places can the tour group visit at most?
|
堔 We adopt the貑 deduction method.
(1)
(4)
(2)
$1^{\circ}$ If going to $A \Rightarrow$ must go to $B \Rightarrow$ do not go to $C \Rightarrow$ do not go to $D \Rightarrow$ must not go to $A$. Going to $A, D$. This leads to the same contradiction as $1^{\circ}$, so do not go to $B$.
(5)
$3^{\circ}$ If going to $E \Rightarrow$ must go to $A, D$. From $1^{\circ}$, we cannot go to $A$, so do not go to $E$.
(4)
$4^{\circ}$ If going to $C \Rightarrow$ must go to $D$, and all other conditions are met. Therefore, the maximum number of places to go is $C, D$ two places.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
In 1988, the Chinese Junior High School Mathematics League had the following problem: If natural numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} x_{4} x_{5}$, what is the maximum value of $x_{5}$?
|
This problem is novel and unique, and some students find it difficult to start. Below, I will first discuss the solution to this problem.
Algorithm 1: Given the symmetry of the equation, we can assume without loss of generality that \( x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5} \), and then use the "bounding method":
\[
\begin{array}{l}
1=\frac{1}{x_{1} x_{2} x_{3} x_{4}}+\frac{1}{x_{1} x_{2} x_{3} x_{5}}+\frac{1}{x_{1} x_{3} x_{4} x_{5}} \\
+\frac{1}{x_{2} x_{3} x_{4} x_{5}} \leqslant \frac{1}{x_{4}}+\frac{1}{x_{5}}+\frac{1}{x_{4} x_{5}}+\frac{1}{x_{4} x_{5}} \\
+\frac{1}{x_{4} x_{5}}=\frac{3+x_{4}+x_{5}}{x_{4} x_{5}}
\end{array}
\]
Therefore,
\[
\begin{array}{l}
x_{4} x_{3} \leqslant 3+x_{4}+x_{5}, \\
\left(x_{4}-1\right)\left(x_{5}-1\right) \leqslant 4 .
\end{array}
\]
If \( x_{4} = 1 \), then \( x_{1} = x_{2} = x_{3} = x_{4} = 1 \), and the original equation becomes \( 4 + x_{5} = x_{5} \), which is a contradiction.
If \( x_{4} > 1 \), then \( x_{5} - 1 \leqslant 4 \), i.e., \( x_{5} \leqslant 5 \).
Thus, the maximum value of \( x_{5} \) is 5. It is easy to observe that one solution is \( (1,1,1,2,5) \).
Solution 2: When \( x_{5} \) is maximized, it should be no less than any \( x_{i} \) (for \( i=1,2,3,4 \)), so
\[
\begin{array}{l}
x_{5} < x_{2} + x_{2} + x_{3} + x_{4} + x_{5} \leqslant 5 x_{3}^{3}, \\
x_{3} < x_{1} x_{2} x_{3} x_{4} x_{5} \leqslant 5 x_{3}, \\
1 < x_{1} x_{2} x_{3} x_{4} \leqslant 5 .
\end{array}
\]
From the original equation, we get \( x_{5} = \frac{x_{1} + x_{2} + x_{3} + x_{4}}{x_{1} x_{2} x_{3} x_{4} - 1} \).
If \( x_{1} x_{2} x_{3} x_{4} = 2 \), then among \( x_{1}, x_{2}, x_{3}, x_{4} \), there is exactly one 2 and three 1s. Substituting, we get \( x_{5} = 5 \).
For \( x_{1} x_{2} x_{3} x_{4} = 3, 4, 5 \), similarly, \( x_{5} = 3 \) and \( x_{5} = 2 \).
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. The vertex of the parabola is at the origin, and the focus F is the center of the circle given by $x^{2}+y^{2}-4 x=0$. A line passing through point $F$ with a slope of 2 intersects the parabola at points $A$ and $D$, and intersects the circle at points $B$ and $C$.
Find $|A B|+|C D|$.
|
$$
\begin{array}{l}
\text { Given } F \text { as the pole, and the positive direction of the } x \text { axis as the polar axis, we establish the polar coordinate system. } \\
\left.\begin{array}{l}
\left.\therefore \text { The parabola equation is: } \rho=\frac{p}{1-\cos \theta}\right\} \Rightarrow \rho= \\
\because F(2,0), \therefore p=4 .
\end{array}\right\} \Rightarrow
\end{array}
$$
$$
\begin{array}{l}
1-\cos \theta \text {. } \\
\text { Also, } \because \operatorname{tg} \theta=2, \therefore \cos \theta=\frac{\sqrt{5}}{5}(0 \leqslant \theta<\pi) \text {. } \\
\because|A B|+|C D|=|A F|+|F D|-|B C| \text {, } \\
|A F|=\rho_{1}, \quad|F D|=\rho_{2}, \\
\therefore \rho_{1}+\rho_{2}=\frac{4}{1-\cos \theta}+\frac{4}{1-\cos (\theta+\pi)} \\
=\frac{8}{1-\cos ^{2} \theta}=\frac{1}{1-\frac{1}{5}}=10 \text {. } \\
\end{array}
$$
$$
\begin{array}{c}
\text { Also, } \because|B C|=2 R=4, \\
\therefore|A B|+|C D|=10-4=6 .
\end{array}
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $m, n$ be positive integers, prove that there exists a constant $\alpha>1$ independent of $m, n$, such that when $\frac{m}{n}<\sqrt{7}$, we have $7-\frac{m^{2}}{n^{2}} \geqslant \frac{\alpha}{n^{2}}$. What is the largest value of $\alpha$?
|
Solve $\alpha=3$.
$$
\begin{array}{l}
\quad 7-\frac{m^{2}}{n^{2}} \geqslant \frac{a}{n^{2}} \Leftrightarrow 7 n^{2}-m^{2} \geqslant \alpha_{0} \\
\because \quad m+7 \equiv m(\bmod 7) \\
\therefore \quad(m+7)^{2} \equiv m^{2}(\bmod 7)
\end{array}
$$
$\therefore\left\{m^{2}(\bmod 7)\right\}$ is a purely periodic sequence with a period of 7, its first 7 terms are $1,4,2,2,4$,
$$
\begin{array}{l}
1, 0, \\
\therefore m^{2} \equiv 0,1,2,4(\bmod 7), \\
\therefore 7 n^{2}-m^{2} \equiv-m^{2} \\
=0,6,5,3(\bmod 7) . \\
\because \quad 7 n^{2}-m^{2}>0, \\
\therefore \quad 7 n^{2}-m^{2} \geqslant 3,
\end{array}
$$
When $n=1, m=2$, $7 n^{2}-m^{2}=3$, hence $\alpha_{\mathrm{m} x}=3$.
|
3
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
1. Insert “+” or “- -” between $1,2,3, \cdots, 1989$, what is the smallest non-negative number that the sum can achieve?
|
Except for 995, the numbers $1,2,3, \cdots, 1989$ can all be divided into 994 pairs: $(1,1989),(2,1988)$, $\cdots$, (994,996). Since the parity of the two numbers in each pair is the same, the result of the operation, regardless of how “+” or “-” signs are placed before each pair, can only be an even number. And 995 is an odd number. Therefore, no matter how “+” or “-” signs are placed before the numbers $1,2, \cdots, 1989$, the value of the sum is always odd. Thus, the smallest non-negative number sought is no less than 1. The number 1 can be obtained in the following way:
$$
\begin{aligned}
1 & =1+(2-3-4+5)+(6-7-8+9) \\
+\cdots+ & (1986-1987-1988+1989)
\end{aligned}
$$
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $x=\sqrt{19-8 \sqrt{3}}$, then the value of the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{3}-7 x^{2}+5 x+15}$ is $\qquad$
|
$5$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. A person is walking along the tram route, and a tram catches up from behind every 12 minutes, while a tram comes from the opposite direction every 4 minutes. Assuming both the person and the trams are moving at a constant speed, then the trams are dispatched from the starting station every $\qquad$ minutes.
|
$6$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, as shown in the figure, in $\triangle A B C$, $\angle A=90^{\circ}$, $A D \perp B C$ at $D, P$ is the midpoint of $A D$, $B P$ intersects $A C$ at $E, E F \perp B C$ at $F, A E=3, E C=12$. Find the length of $E F$.
|
Extend $F E, B A$ to intersect at point $H$,
$\left.\begin{array}{l}A D \| H F \\ P \text { is the midpoint of } A D\end{array}\right\} \Rightarrow H E=E F$.
Also, $\angle H A C=90^{\circ}$ ?
$\left.\begin{array}{l}\angle E F C=90^{\circ} \\ \text { and on the same side of } H C\end{array}\right\} \Rightarrow H, A, F, C$ are concyclic
$$
\begin{array}{l}
\Rightarrow H E \cdot E F=A E \cdot E C, \\
\quad \therefore E F^{2}=A E \cdot E C=3 \times 12=36 \\
\Rightarrow E F=6 .
\end{array}
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, $E, F$ are
on the sides $B C$ and $C D$
of rectangle $A B C D$,
if the areas of $\triangle C E F$,
$\triangle A B E$, $\triangle A D F$
are 3,
4, 5 respectively. Find the area $S$ of $\triangle A E F$.
|
Let $AB = a, BC = b$, then $BE = \frac{8}{a}$,
$$
\begin{array}{l}
CE = b - \frac{8}{a}, DF = \frac{10}{b}, FC = a - \frac{10}{b}. \\
\left\{\begin{array}{l}
\frac{1}{2}\left(b - \frac{8}{a}\right) \times \left(a - \frac{10}{b}\right) = 3, \\
ab = 3 + 4 + 5 + S .
\end{array}\right. \\
S = \sqrt{144 - 80} = 8 .
\end{array}
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In a dark room, a drawer contains socks of two colors and two sizes, 4 pairs in total (one pair of each color and size). How many socks must be taken out to ensure that there are two pairs of different colors and different sizes?
|
New $A_{1}, A, B, B$ each two. Period out 6 as $A ., A_{1}, 1, B_{1}, B_{1}, B_{2}$, does not meet the question's requirements. But when any 7 are taken out, there must be one color (let's say A) with 4 pieces all taken out, and among the 3 pieces of another color, there must be two.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If a $23 \times 23$ square is composed of 1000 squares of sizes $1 \times 1, 2 \times 2, 3 \times 3$, how many $1 \times 1$ squares are needed at minimum?
|
4. At least 1 $1 \times 1$ square is needed. First, it needs to be shown that a $23 \times 23$ square can be composed of the following: $2 \times 2, 3 \times 3$ squares, and exactly 1 $1 \times 1$ square. Then, it also needs to be proven that a $1 \times 1$ square is indispensable.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. Given the sequence $\left\{x_{0}\right\}: x_{n+1}=$ $\frac{x_{\mathrm{n}}+(2-\sqrt{3})}{1-x_{n}(2-\sqrt{3})}$. Find the value of $x_{1001}-x_{401}$.
|
Given the shape of the recurrence relation, we can set $x_{\mathrm{n}}=\operatorname{tg} \alpha_{\mathrm{n}}$, and since $\operatorname{tg} \frac{\pi}{12}=2-\sqrt{3}$, we know that $x_{n+1}=\operatorname{tg} \alpha_{n+1}$
$$
=\frac{\operatorname{tg} \alpha_{\mathrm{a}}+\operatorname{tg} \frac{\pi}{12}}{1-\operatorname{tg} \alpha_{\mathrm{n}} \operatorname{tg} \frac{\pi}{12}}=\operatorname{tg}\left(\alpha_{\mathrm{n}}+\frac{\pi}{12}\right) \text {. }
$$
Therefore, $x_{\mathrm{n}+12}=x_{\mathrm{n}}$.
$$
\text { Hence } x_{1001}-x_{401}=x_{5}-x_{3}=0\left(\left\{x_{0}\right\}\right. \text { is a }
$$
periodic sequence).
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. As shown in Figure 5, in $\triangle ABC$, $AB=2$, $AC=3$, I, II, and III respectively denote the squares constructed on $AB$, $BC$, and $CA$. What is the maximum value of the sum of the areas of the three shaded regions? (1988, National Junior High School League)
---
The translation preserves the original text's line breaks and formatting.
|
Solve in $\triangle A D E$ and $\triangle A B C$ !! $\angle B A C + \angle D A E = 180^{\circ}$.
Also, $A C = A D, A B = A E$,
thus $S \triangle A D B = S \triangle A B C$.
Similarly, the areas of the other two shaded triangles are also equal to $S \triangle \triangle B C$.
$$
\begin{array}{r}
\therefore S \text { shaded } = 3 S \triangle \mathrm{ABC} = 3 \cdot \frac{1}{2} A B \cdot A C \\
\cdot \sin \angle B A C \leqslant 3 \cdot \frac{1}{2} \cdot 2 \cdot 3 \cdot \sin 90^{\circ} = 9 .
\end{array}
$$
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. (1986, Shanghai) The three sides of a triangle are all positive integers, one of which has a length of 4, but it is not the shortest side. How many different triangles are there? ...
The translation is provided while retaining the original text's line breaks and format.
|
(8).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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