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