The smallest four-digit number that has exactly 14 divisors (including 1 and itself), and one of its prime factors has a units digit of 3 is $\qquad$ .

Solution:
5.1458.

Let this four-digit number be $n$.
Since $14=14 \times 1=7 \times 2$, we have $n=p^{13}$ or $p^{6} q$ (where $p$ and $q$ are different prime numbers).
If $n=p^{13}$, by the given condition, the unit digit of $p$ is 3, so, $p \geqslant 3$.
Thus, $n \geqslant 3^{13}=1594323$, which is a contradiction.
Therefore, $n=p^{6} q$.
If $p \geqslant 5$, then the number of digits in $n$ will exceed 4, which is a contradiction.
If $p=3$, then $p^{6}=729$, and when $q=2$, $n=$ $p^{6} q=1458$, which satisfies the condition.

If $p=2$, then $p^{6}=64$, in this case, $q$ must be chosen from 3, 13, 23, 43, ... these prime numbers with a unit digit of 3.
Multiplying 64 by these numbers, the smallest four-digit number obtained is
$$
p^{6} q=64 \times 23=1472>1458 .
$$

Therefore, the smallest four-digit number that satisfies the condition is 1458.

Answer: 1458


If $x \in R-\{-7\}$, determine the smallest value of the expression
$$\frac{2x^2 + 98}{(x + 7)^2}$$

Solution:
To determine the smallest value of the expression
\[ \frac{2x^2 + 98}{(x + 7)^2}, \]
we will analyze the given expression and use calculus to find its minimum value.

1. **Rewrite the expression:**
   \[ f(x) = \frac{2x^2 + 98}{(x + 7)^2}. \]

2. **Simplify the expression:**
   \[ f(x) = \frac{2x^2 + 98}{x^2 + 14x + 49}. \]

3. **Find the critical points by taking the derivative:**
   Let \( f(x) = \frac{2x^2 + 98}{x^2 + 14x + 49} \).
   We use the quotient rule for differentiation:
   \[ f'(x) = \frac{(2x^2 + 98)'(x^2 + 14x + 49) - (2x^2 + 98)(x^2 + 14x + 49)'}{(x^2 + 14x + 49)^2}. \]

4. **Compute the derivatives:**
   \[ (2x^2 + 98)' = 4x, \]
   \[ (x^2 + 14x + 49)' = 2x + 14. \]

5. **Substitute the derivatives into the quotient rule:**
   \[ f'(x) = \frac{4x(x^2 + 14x + 49) - (2x^2 + 98)(2x + 14)}{(x^2 + 14x + 49)^2}. \]

6. **Simplify the numerator:**
   \[ 4x(x^2 + 14x + 49) = 4x^3 + 56x^2 + 196x, \]
   \[ (2x^2 + 98)(2x + 14) = 4x^3 + 28x^2 + 196x + 1372. \]

7. **Combine the terms:**
   \[ f'(x) = \frac{4x^3 + 56x^2 + 196x - 4x^3 - 28x^2 - 196x - 1372}{(x^2 + 14x + 49)^2}, \]
   \[ f'(x) = \frac{28x^2 - 1372}{(x^2 + 14x + 49)^2}. \]

8. **Set the derivative equal to zero to find critical points:**
   \[ 28x^2 - 1372 = 0, \]
   \[ x^2 = \frac{1372}{28}, \]
   \[ x^2 = 49, \]
   \[ x = \pm 7. \]

9. **Evaluate the function at the critical points:**
   Since \( x = -7 \) is not in the domain, we only consider \( x = 7 \):
   \[ f(7) = \frac{2(7)^2 + 98}{(7 + 7)^2} = \frac{2 \cdot 49 + 98}{14^2} = \frac{98 + 98}{196} = \frac{196}{196} = 1. \]

10. **Check the behavior as \( x \to \infty \) and \( x \to -\infty \):**
    \[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{2x^2 + 98}{x^2 + 14x + 49} = 2. \]
    \[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{2x^2 + 98}{x^2 + 14x + 49} = 2. \]

11. **Conclusion:**
    The smallest value of the expression is achieved at \( x = 7 \), and the value is \( 1 \).

The final answer is \( \boxed{1} \).

Answer: 1


The maximum value of the function $f(x)=7 \sin x+\sin 2x$ is $\qquad$.

Solution:
2. $\frac{15 \sqrt{15}}{8}$.

From the condition, we know
$$
\begin{array}{l}
f^{\prime}(x)=7 \cos x+2 \cos 2 x \\
=4 \cos ^{2} x+7 \cos x-2 \\
=(4 \cos x-1)(\cos x+2) .
\end{array}
$$

Let $\cos \varphi=\frac{1}{4}\left(0<\varphi<\frac{\pi}{2}\right)$. Then
$$
\sin \varphi=\frac{\sqrt{15}}{4} \text {. }
$$

It is easy to see that $f(x)$ is monotonically increasing in the interval $(-\varphi+2 k \pi, \varphi+2 k \pi)$ $(k \in \mathbf{Z})$, and monotonically decreasing in the interval $(\varphi+2 k \pi, 2 \pi-\varphi+2 k \pi)(k \in \mathbf{Z})$.
Therefore, the maximum value of $f(x)$ is
$$
f(\varphi+2 k \pi)=\frac{15 \sqrt{15}}{8} .
$$

Answer: \frac{15 \sqrt{15}}{8}


There are $100$ students who want to sign up for the class Introduction to Acting. There are three class sections for Introduction to Acting, each of which will fit exactly $20$ students. The $100$ students, including Alex and Zhu, are put in a lottery, and 60 of them are randomly selected to fill up the classes. What is the probability that Alex and Zhu end up getting into the same section for the class?

Solution:
1. **Determine the total number of ways to select 60 students out of 100:**
   \[
   \binom{100}{60}
   \]

2. **Determine the number of ways to select 58 students out of the remaining 98 students (excluding Alex and Zhu):**
   \[
   \binom{98}{58}
   \]

3. **Calculate the probability that Alex and Zhu are both selected among the 60 students:**
   \[
   \frac{\binom{98}{58}}{\binom{100}{60}}
   \]

4. **Given that Alex and Zhu are both selected, we need to find the probability that they end up in the same section. There are 3 sections, each with 20 students.**

5. **Calculate the number of ways to place Alex and Zhu in the same section:**
   - Choose 1 section out of 3 for Alex and Zhu:
     \[
     \binom{3}{1} = 3
     \]
   - Choose 18 more students out of the remaining 58 students to fill the chosen section:
     \[
     \binom{58}{18}
     \]

6. **Calculate the number of ways to place the remaining 40 students into the other 2 sections:**
   - Choose 20 students out of the remaining 40 students for one section:
     \[
     \binom{40}{20}
     \]
   - The remaining 20 students automatically go into the last section.

7. **Calculate the total number of ways to place Alex and Zhu in the same section:**
   \[
   3 \times \binom{58}{18} \times \binom{40}{20}
   \]

8. **Calculate the total number of ways to place 60 students into 3 sections of 20 students each:**
   \[
   \binom{60}{20} \times \binom{40}{20}
   \]

9. **Calculate the probability that Alex and Zhu end up in the same section given that they are both selected:**
   \[
   \frac{3 \times \binom{58}{18} \times \binom{40}{20}}{\binom{60}{20} \times \binom{40}{20}}
   \]

10. **Combine the probabilities:**
    \[
    \text{Total probability} = \left( \frac{\binom{98}{58}}{\binom{100}{60}} \right) \times \left( \frac{3 \times \binom{58}{18} \times \binom{40}{20}}{\binom{60}{20} \times \binom{40}{20}} \right)
    \]

11. **Simplify the expression:**
    \[
    \text{Total probability} = \frac{3 \times \binom{58}{18}}{\binom{60}{20}}
    \]

12. **Further simplification:**
    \[
    \text{Total probability} = \frac{3 \times \frac{58!}{18!40!}}{\frac{60!}{20!20!20!}} = \frac{3 \times 58! \times 20! \times 20! \times 20!}{18! \times 40! \times 60!}
    \]

13. **Final simplification:**
    \[
    \text{Total probability} = \frac{3 \times 58! \times 20! \times 20! \times 20!}{18! \times 40! \times 60!}
    \]

The final answer is \(\boxed{\frac{19}{165}}\).

Answer: \frac{19}{165}


The side lengths $a,b,c$ of a triangle $ABC$ are positive integers. Let:

\[T_{n}=(a+b+c)^{2n}-(a-b+c)^{2n}-(a+b-c)^{2n}+(a-b-c)^{2n}\]

for any positive integer $n$. If $\frac{T_{2}}{2T_{1}}=2023$ and $a>b>c$ , determine all possible perimeters of the triangle $ABC$.

Solution:
1. We start with the given expression for \( T_n \):

\[
T_n = (a+b+c)^{2n} - (a-b+c)^{2n} - (a+b-c)^{2n} + (a-b-c)^{2n}
\]

2. We need to find \( T_1 \) and \( T_2 \):

\[
T_1 = (a+b+c)^2 - (a-b+c)^2 - (a+b-c)^2 + (a-b-c)^2
\]

3. Expand each term:

\[
(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca
\]
\[
(a-b+c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc - 2ca
\]
\[
(a+b-c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca
\]
\[
(a-b-c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca
\]

4. Substitute these into the expression for \( T_1 \):

\[
T_1 = (a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) - (a^2 + b^2 + c^2 - 2ab + 2bc - 2ca) - (a^2 + b^2 + c^2 + 2ab - 2bc - 2ca) + (a^2 + b^2 + c^2 - 2ab - 2bc + 2ca)
\]

5. Simplify the expression:

\[
T_1 = (a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) - (a^2 + b^2 + c^2 - 2ab + 2bc - 2ca) - (a^2 + b^2 + c^2 + 2ab - 2bc - 2ca) + (a^2 + b^2 + c^2 - 2ab - 2bc + 2ca)
\]

\[
T_1 = (a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) - (a^2 + b^2 + c^2 - 2ab + 2bc - 2ca) - (a^2 + b^2 + c^2 + 2ab - 2bc - 2ca) + (a^2 + b^2 + c^2 - 2ab - 2bc + 2ca)
\]

\[
T_1 = 4bc
\]

6. Now, find \( T_2 \):

\[
T_2 = (a+b+c)^4 - (a-b+c)^4 - (a+b-c)^4 + (a-b-c)^4
\]

7. Use the binomial theorem to expand each term:

\[
(a+b+c)^4 = (a^2 + b^2 + c^2 + 2ab + 2bc + 2ca)^2
\]
\[
(a-b+c)^4 = (a^2 + b^2 + c^2 - 2ab + 2bc - 2ca)^2
\]
\[
(a+b-c)^4 = (a^2 + b^2 + c^2 + 2ab - 2bc - 2ca)^2
\]
\[
(a-b-c)^4 = (a^2 + b^2 + c^2 - 2ab - 2bc + 2ca)^2
\]

8. Substitute these into the expression for \( T_2 \):

\[
T_2 = (a^2 + b^2 + c^2 + 2ab + 2bc + 2ca)^2 - (a^2 + b^2 + c^2 - 2ab + 2bc - 2ca)^2 - (a^2 + b^2 + c^2 + 2ab - 2bc - 2ca)^2 + (a^2 + b^2 + c^2 - 2ab - 2bc + 2ca)^2
\]

9. Simplify the expression:

\[
T_2 = 4bc(6a^2 + 2b^2 + 2c^2)
\]

10. Given that \(\frac{T_2}{2T_1} = 2023\):

\[
\frac{4bc(6a^2 + 2b^2 + 2c^2)}{2 \cdot 4bc} = 2023
\]

\[
3a^2 + b^2 + c^2 = 2023
\]

11. Solve for \(a, b, c\) under the condition \(a > b > c\):

\[
3a^2 + b^2 + c^2 = 2023
\]

12. Check possible values for \(a\):

\[
a \leq \sqrt{\frac{2023}{3}} \approx 25
\]

13. Test \(a = 23\):

\[
3(23)^2 + b^2 + c^2 = 2023
\]

\[
3(529) + b^2 + c^2 = 2023
\]

\[
1587 + b^2 + c^2 = 2023
\]

\[
b^2 + c^2 = 436
\]

14. Check possible values for \(b\) and \(c\):

\[
b = 20, c = 6
\]

15. Verify:

\[
20^2 + 6^2 = 400 + 36 = 436
\]

16. The perimeter of the triangle is:

\[
a + b + c = 23 + 20 + 6 = 49
\]

The final answer is \(\boxed{49}\).

Answer: 49