1. Determine whether 21, 25, and 33 are pairwise relatively prime. Step 1: List prime factors. - 21 = 3 × 7 - 25 = 5 × 5 - 33 = 3 × 11 Step 2: Check gcd of each pair. - gcd(21, 25): No common factors, gcd = 1 - gcd(21, 33): Common factor 3, gcd = 3 - gcd(25, 33): No common factors, gcd = 1 Step 3: Since gcd(21,33) ≠ 1, the numbers are not pairwise relatively prime. 2. Evaluate (101112) × (1012). Assume numbers are in base 10: Step 1: Multiply 101112 × 1012. Step 2: Use standard multiplication: $$101112 \times 1012 = 101112 \times (1000 + 10 + 2) = 101112 \times 1000 + 101112 \times 10 + 101112 \times 2$$ Step 3: Calculate: $$101112 \times 1000 = 101112000$$ $$101112 \times 10 = 1011120$$ $$101112 \times 2 = 202224$$ Step 4: Sum: $$101112000 + 1011120 + 202224 = 104125344$$ 3. Find gcd(a,b) and lcm(a,b) given: $$a = 3^{75} \times 37^{3}$$ $$b = 2^{113} \times 55^{9}$$ First, express the prime factors of 55: $$55 = 5 \times 11$$ Therefore, $$b = 2^{113} \times 5^{9} \times 11^{9}$$ Step 1: Find gcd by taking minimum powers of common primes. Prime factors of a: 3, 37 Prime factors of b: 2, 5, 11 No common prime factors, so $$\gcd(a,b) = 1$$ Step 2: Find lcm by taking the maximum powers of all primes. $$\mathrm{lcm}(a,b) = 2^{113} \times 3^{75} \times 5^{9} \times 11^{9} \times 37^{3}$$ 4. Find decimal expansion of binary number (11010)_2. Step 1: Write place values: $$11010_2 = 1 \times 2^{4} + 1 \times 2^{3} + 0 \times 2^{2} + 1 \times 2^{1} + 0 \times 2^{0}$$ Step 2: Calculate: $$= 16 + 8 + 0 + 2 + 0 = 26$$ 5. Find equivalent octal form of hexadecimal number (C2)_{16}. Step 1: Convert hexadecimal to decimal. $$C = 12$$ so $$C2_{16} = 12 \times 16^{1} + 2 \times 16^{0} = 12 \times 16 + 2 = 192 + 2 = 194$$ Step 2: Convert decimal 194 to octal. Divide 194 by 8: $$194 \div 8 = 24 \text{ remainder } 2$$ $$24 \div 8 = 3 \text{ remainder } 0$$ $$3 \div 8 = 0 \text{ remainder } 3$$ Step 3: Read remainders bottom-up: $$302_8$$ 6. Use General Term equation to find the constant term in expansion of $(3x - \frac{1}{x})^{4}$. Step 1: General term in expansion of $(a + b)^n$ is: $$T_{k+1} = \binom{n}{k} a^{n-k} b^{k}$$ Here, $$a = 3x, \quad b = -\frac{1}{x}, \quad n=4$$ Step 2: Write general term: $$T_{k+1} = \binom{4}{k} (3x)^{4-k} \left(-\frac{1}{x}\right)^k = \binom{4}{k} 3^{4-k} x^{4-k} (-1)^k x^{-k} = \binom{4}{k} 3^{4-k} (-1)^k x^{4-k-k} = \binom{4}{k} 3^{4-k} (-1)^k x^{4-2k}$$ Step 3: Constant term means power of $x$ is zero: $$4 - 2k = 0 \implies k=2$$ Step 4: Calculate constant term: $$T_3 = \binom{4}{2} 3^{2} (-1)^2 = 6 \times 9 \times 1 = 54$$ 7. From a class of 20 boys and 30 girls, select a team of 3 boys and 2 girls. Step 1: Number of ways to select 3 boys from 20: $$\binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$$ Step 2: Number of ways to select 2 girls from 30: $$\binom{30}{2} = \frac{30 \times 29}{2 \times 1} = 435$$ Step 3: Total ways: $$1140 \times 435 = 495900$$ 8. Knights and Knaves logic puzzle: - Knights always tell the truth. - Knaves always lie. Given: - A says: "The two of us are both knights." - B says: "A is a knave." Step 1: Assume A is a knight (telling truth), then both are knights. But B says "A is a knave," which is false, so B would be lying. Contradiction since if both are knights, B must tell truth. Step 2: Assume A is a knave (lying). Then A's statement "The two of us are both knights" is false, which fits. B says "A is a knave," which is true, so B is a knight. Conclusion: A is a knave and B is a knight.