https://file.aiquickdraw.com/mathbot/file/1726892497655zsznf1t7.jpg

Let's go through the problems step by step and solve them.

### Question 1: 
Find \( d = \gcd(24,175) \) and write \( d = s \cdot 24 + t \cdot 175 \), where \( s, t \in \mathbb{Z} \). Hence find \( \text{LCM}(24,175) \).

#### Step 1: Find the GCD using the Euclidean algorithm
We apply the Euclidean algorithm to find \( \gcd(24,175) \):

\[
175 = 7 \times 24 + 7 \quad \Rightarrow \quad 175 - 7 \times 24 = 7
\]
\[
24 = 3 \times 7 + 3 \quad \Rightarrow \quad 24 - 3 \times 7 = 3
\]
\[
7 = 2 \times 3 + 1 \quad \Rightarrow \quad 7 - 2 \times 3 = 1
\]
\[
3 = 3 \times 1 + 0
\]

Thus, \( \gcd(24, 175) = 1 \).

#### Step 2: Express \( d = 1 \) as a linear combination of 24 and 175
By working backward:

\[
1 = 7 - 2 \times 3 = 7 - 2 \times (24 - 3 \times 7) = 7 - 2 \times 24 + 6 \times 7 = 7 \times 7 - 2 \times 24
\]
\[
1 = (175 - 7 \times 24) \times 7 - 2 \times 24 = 7 \times 175 - 49 \times 24 - 2 \times 24 = 7 \times 175 - 51 \times 24
\]

Thus, \( 1 = 7 \times 175 - 51 \times 24 \). Therefore, \( s = -51 \) and \( t = 7 \).

#### Step 3: Find LCM
The formula to calculate the least common multiple is:
\[
\text{LCM}(a, b) = \frac{a \cdot b}{\gcd(a, b)}
\]
Since \( \gcd(24, 175) = 1 \):
\[
\text{LCM}(24, 175) = \frac{24 \times 175}{1} = 4200
\]

Thus, \( \text{LCM}(24, 175) = 4200 \).

---

### Question 2:
Solve the equation \( 2 \otimes x = 5 \) in \( \mathbb{Z}_{10} \).

We are solving \( 2x \equiv 5 \mod 10 \). To solve this, find the multiplicative inverse of 2 mod 10 (if it exists).

However, \( \gcd(2, 10) = 2 \), which means that 2 does not have a multiplicative inverse in \( \mathbb{Z}_{10} \), and thus this equation has no solution.

---

### Question 3:
Solve the following system of congruences:
\[
x \equiv 7 \pmod{9}, \quad x \equiv 3 \pmod{10}, \quad x \equiv 5 \pmod{11}
\]

This is a system of linear congruences that can be solved using the Chinese Remainder Theorem (CRT).

#### Step 1: Solve two congruences
Start by solving:
\[
x \equiv 7 \pmod{9}, \quad x \equiv 3 \pmod{10}
\]
Let \( x = 10k + 3 \). Substituting into the first equation:
\[
10k + 3 \equiv 7 \pmod{9} \quad \Rightarrow \quad 10k \equiv 4 \pmod{9}
\]
Since \( 10 \equiv 1 \pmod{9} \), this simplifies to:
\[
k \equiv 4 \pmod{9}
\]
Thus, \( k = 9m + 4 \). Substituting back into \( x = 10k + 3 \):
\[
x = 10(9m + 4) + 3 = 90m + 43
\]
So, \( x \equiv 43 \pmod{90} \).

#### Step 2: Solve with the third congruence
Now solve:
\[
x \equiv 43 \pmod{90}, \quad x \equiv 5 \pmod{11}
\]
Let \( x = 90n + 43 \). Substituting into the second equation:
\[
90n + 43 \equiv 5 \pmod{11} \quad \Rightarrow \quad 90n \equiv -38 \pmod{11}
\]
Since \( 90 \equiv 2 \pmod{11} \) and \( -38 \equiv 5 \pmod{11} \), this simplifies to:
\[
2n \equiv 5 \pmod{11}
\]
Multiplying both sides by the inverse of 2 mod 11, which is 6:
\[
n \equiv 5 \times 6 \pmod{11} \quad \Rightarrow \quad n \equiv 30 \pmod{11} \quad \Rightarrow \quad n \equiv 8 \pmod{11}
\]
Thus, \( n = 11p + 8 \). Substituting into \( x = 90n + 43 \):
\[
x = 90(11p + 8) + 43 = 990p + 763
\]
So, \( x \equiv 763 \pmod{990} \).

Thus, the solution is \( x \equiv 763 \pmod{990} \).

---

I'll solve the remaining problems next if needed

Solve problems related to GCD, LCM, modular arithmetic, differentiation, extrema, and volume changes.

This assignment covers a variety of mathematical topics including solving GCD and LCM, modular arithmetic equations, differentiating complex functions, finding critical points, and calculating changes in spherical volume. Suitable for undergraduate-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation