1. **Problem 57:** Given for each integer $n \geq 1$, define $I_n(x) = \int_1^x (\log t)^n \, dt$ for all $x > 1$. Determine which formula relating $I_n$ for $n \geq 3$ and all $x > 1$ holds. 2. To find a recursive relation, use integration by parts: Let $u = (\log t)^n$, then $du = \frac{n(\log t)^{n-1}}{t} \, dt$; Let $dv = dt$, then $v = t$. 3. By integration by parts: $$ I_n(x) = \int_1^x (\log t)^n dt = [t(\log t)^n]_1^x - \int_1^x t \cdot \frac{n(\log t)^{n-1}}{t} dt = x(\log x)^n - n \int_1^x (\log t)^{n-1} dt = x(\log x)^n - nI_{n-1}(x). $$ 4. So the correct relation is: $$ I_n(x) = x(\log x)^n - nI_{n-1}(x) $$ which corresponds to option (B). --- 5. **Problem 58:** The number of ways to distribute 25 identical trucks into 5 cities is the number of integer solutions to $x_1 + x_2 + x_3 + x_4 + x_5 = 25$ with each $x_i \geq 0$. 6. Total distributions are $\binom{25+5-1}{5-1} = \binom{29}{4}$. 7. The problem involves distributions for which at least one city gets 0 trucks, so the complement is when each city gets at least 1 truck. 8. Distributions with each city at least 1 truck equals the number of integer solutions to $x_1 + \cdots + x_5=25$ with each $x_i \geq 1$. By substitution $y_i=x_i-1 \geq 0$, sum $y_i=20$. 9. Number of such solutions is $\binom{20+5-1}{5-1} = \binom{24}{4}$. 10. Thus, number of distributions with at least 1 city receiving 0 trucks is total minus those with all $x_i \geq 1$: $$ \binom{29}{4} - \binom{24}{4}. $$ 11. Hence, the answer is option (A). --- 12. **Problem 59:** In $\mathbb{Z}_{30}$, define $U_{30}$ as the group of invertible elements mod 30. 13. Given a group homomorphism $\phi: U_{30} \to U_{30}$ with kernel $\ker(\phi) = \{1, 11\}$ and $\phi(7) = 7$, find which element maps to 7 as well. 14. Since $\ker(\phi) = \{1, 11\}$, it has order 2, so for $u \in U_{30}$, $\phi(u) = \phi(v)$ iff $u v^{-1} \in \ker(\phi)$. 15. Find all $u$ such that $\phi(u) = \phi(7) = 7$: $$ u = u, \quad \phi(u) = \phi(7) \Rightarrow u7^{-1} \in \ker(\phi).$$ 16. Since $\ker(\phi) = \{1, 11\}$, possibilities are: - $u7^{-1} = 1 \implies u=7$ - $u7^{-1} = 11 \implies u = 7 \cdot 11 = 77 \equiv 17 \pmod{30}.$ 17. Thus, $u=17$ also maps to 7. 18. Answer: (C). --- 19. **Problem 60:** Solve $Mx = b$ with $M = LU$ where $$ L = \begin{bmatrix}1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & -1 & 0 & 1 \end{bmatrix}, U = \begin{bmatrix}1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 2 \end{bmatrix}, \quad b = \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}. $$ 20. Note: $b$ is a 3-vector but $M$ is 4x4, likely $b$ should have 4 entries; assume $b = [0, 1, 0, 0]^T$ with last zero. 21. Solve $Ly = b$: - $y_1 = 0$ - $L_{2,:}: 1\cdot y_1 + 0 + 0 + 0 = y_1 = b_2 = 1$ implies $y_1=1$, contradicts above. 22. Checking $L$ matrix second row is $[1,0,0,0]$, so $1 \cdot y_1 = b_2$: - $y_1 = b_2 = 1$. 23. First row equation: $1 \cdot y_1 = b_1 = 0$ implies $y_1=0$, conflict. 24. The given $L$ is singular (second row same as first), which is unusual. But proceed with solving via substitution: Set $y=[y_1, y_2, y_3, y_4]^T$. From $Ly = b$: - Row 1: $y_1 = 0$ - Row 2: $y_1 = 1$ contradicts 25. Possible misprint: second row in $L$ likely should be different. 26. Assuming problem meant $b = [0,1,0,0]^T$ and treat rows as is, solve despite singularity: Row 1: $y_1=0$ Row 2: $y_1=1$ contradiction, so system inconsistent or problem typo. 27. If treat $y_1=0$ from Row 1, then compute $x$ by $Ux = y$: $$ U x = y,\quad x = U^{-1} y. $$ 28. Without consistent $y$, cannot proceed. Assuming $y$ from $Ly=b$ is $y=[0,1,0,0]^T$ compatible. Row 1: $y_1=0$ Row 2: $1 \cdot y_1 = 0$ but $b_2=1$ wrong 29. If we swap rows 1 and 2 of $L$ or similarly, or ignore contradiction and solve $U x = y = [0,1,0,0]^T$: From $Ux=y$: - Row 4: $2 x_4 = 0 \Rightarrow x_4 = 0$ - Row 3: $x_3 + 2x_4 = 0 \Rightarrow x_3=0$ - Row 2: $x_2 = 1$ - Row 1: $x_1 + x_3 = 0 \Rightarrow x_1 =0$ 30. So $x = [0, 1, 0, 0]^T$. The first coordinate is 0. 31. Final answer: (C).