1. Statement of the problem. We are asked to compute $\gcd(2101,1009)$ and to solve the congruences $55x \equiv 34 \pmod{89}$ and $105x \equiv 143 \pmod{100}$. 2. Compute $\gcd(2101,1009)$ using Euclid's algorithm. Apply Euclid's algorithm step by step. $2101 = 2\cdot1009 + 83$. $1009 = 12\cdot83 + 13$. $83 = 6\cdot13 + 5$. $13 = 2\cdot5 + 3$. $5 = 1\cdot3 + 2$. $3 = 1\cdot2 + 1$. $2 = 2\cdot1 + 0$. Therefore $\gcd(2101,1009) = 1$. 3. Solve (b)i $55x \equiv 34 \pmod{89}$. Since $\gcd(55,89)=1$ an inverse of 55 modulo 89 exists and the solution is unique modulo 89. Compute the inverse using the extended Euclidean algorithm. $89 = 1\cdot55 + 34$. $55 = 1\cdot34 + 21$. $34 = 1\cdot21 + 13$. $21 = 1\cdot13 + 8$. $13 = 1\cdot8 + 5$. $8 = 1\cdot5 + 3$. $5 = 1\cdot3 + 2$. $3 = 1\cdot2 + 1$. Back-substituting gives $-21\cdot89 + 34\cdot55 = 1$ so $34$ is the inverse of $55$ modulo $89$. Multiply both sides of the congruence by $34$ to get $x \equiv 34\cdot34 \equiv 1156 \equiv 88 \pmod{89}$. Thus the solution is $x \equiv 88 \pmod{89}$. 4. Solve (b)ii $105x \equiv 143 \pmod{100}$. Compute $\gcd(105,100)$ to determine solvability. $105 = 1\cdot100 + 5$. $100 = 20\cdot5 + 0$. Therefore $\gcd(105,100) = 5$. Since $5$ does not divide $143$ there is no solution. Therefore the congruence is unsolvable. Final answers. (a) $\gcd(2101,1009)=1$. (b)i $x \equiv 88 \pmod{89}$. (b)ii unsolvable.