Question 1 With which of the following triangle congruence shortcuts could be used to prove △ 𝑃 𝑅 𝑄 = △ 𝑇 𝑅 𝑆 △PRQ=△TRS? Options: a) Side-Side-Side Postulate (SSS) b) Side-Angle-Side Postulate (SAS) c) Angle-Side-Angle Postulate (ASA) d) Angle-Angle-Side Theorem (AAS) Answer 1 Step 1 Given that: ∠ 𝑄 ≅ ∠ 𝑆 ∠Q≅∠S 𝑄 𝑅 = 𝑆 𝑅 QR=SR Identify the Triangles In △ 𝑃 𝑄 𝑅 △PQR and △ 𝑅 𝑆 𝑇 △RST: ∠ 𝑄 ≅ ∠ 𝑆 ∠Q≅∠S (given) 𝑄 𝑅 = 𝑆 𝑅 QR=SR (equal sides given) ∠ 𝑃 𝑅 𝑄 = ∠ 𝐿 𝑆 𝑇 ∠PRQ=∠LST (vertically opposite angles) Step 2 Here, two angles and one side of both triangles are equal. Therefore, by Angle-Side-Angle (ASA) congruence, we conclude: △ 𝑃 𝑄 𝑅 ≅ △ 𝑅 𝑆 𝑇 △PQR≅△RST Answer: Hence, option (c) is correct. Question 2 Prove directly from the definition of congruence modulo 𝑛 n that if 𝑎 , 𝑐 a,c, and 𝑛 n are integers, 𝑛 > 1 n>1, and 𝑎 ≡ 𝑐 ( mod 𝑛 ) a≡c(mod n), then 𝑎 3 ≡ 𝑐 3 ( mod 𝑛 ) a 3 ≡c 3 (mod n). Answer 1 Step 1 Let 𝑎 , 𝑐 a,c, and 𝑛 n be integers such that 𝑛 > 1 n>1. We have to prove that if 𝑎 ≡ 𝑐 ( mod 𝑛 ) a≡c(mod n), then 𝑎 3 ≡ 𝑐 3 ( mod 𝑛 ) a 3 ≡c 3 (mod n) using the definition of congruence modulo 𝑛 n. Note that 𝑝 ≡ 𝑞 ( mod 𝑛 ) p≡q(mod n) means 𝑛 n divides 𝑝 − 𝑞 p−q where 𝑝 p and 𝑞 q are integers. Step 2 Since 𝑎 ≡ 𝑐 ( mod 𝑛 ) a≡c(mod n), it implies 𝑛 n divides ( 𝑎 − 𝑐 ) (a−c). Now, 𝑛 n divides any integer multiple of ( 𝑎 − 𝑐 ) (a−c), which implies that 𝑛 n divides ( 𝑎 − 𝑐 ) ( 𝑎 2 + 𝑎 𝑐 + 𝑐 2 ) (a−c)(a 2 +ac+c 2 ). By algebraic identity: ( 𝑎 3 − 𝑐 3 ) = ( 𝑎 − 𝑐 ) ( 𝑎 2 + 𝑎 𝑐 + 𝑐 2 ) (a 3 −c 3 )=(a−c)(a 2 +ac+c 2 ) So, 𝑛 n divides ( 𝑎 3 − 𝑐 3 ) (a 3 −c 3 ). Conclusion Therefore, by the definition of congruence modulo 𝑛 n, 𝑎 3 ≡ 𝑐 3 ( mod 𝑛 ) a 3 ≡c 3 (mod n) This completes the proof. Question 3 Find the smallest positive integer 𝑥 x that solves the congruence: 7 𝑥 ≡ 5 ( mod 52 ) 7x≡5(mod 52) Answer 1 Step 1 Given: 7 𝑥 ≡ 5 ( mod 52 ) 7x≡5(mod 52) To solve for 𝑥 x, we need to isolate 𝑥 x using modular arithmetic. Step 2 Since 7 𝑥 = 5 ( mod 52 ) 7x=5(mod 52), Using modular inverses or methods of solving linear congruences, we find that: 𝑥 = 5 7 ( mod 52 ) x= 7 5 (mod 52) Solution: The smallest positive integer satisfying the condition. Question 4 Solve the linear congruence: 𝑥 + 2 𝑦 ≡ 1 ( mod 5 ) x+2y≡1(mod 5) 2 𝑥 + 𝑦 ≡ 1 ( mod 5 ) 2x+y≡1(mod 5) Answer 1 Step 1 Given a system of linear congruences: 𝑥 + 2 𝑦 ≡ 1 ( mod 5 ) x+2y≡1(mod 5) 2 𝑥 + 𝑦 ≡ 1 ( mod 5 ) 2x+y≡1(mod 5) We will solve it. Step 2 Multiply the first congruence by 2: 2 𝑥 + 4 𝑦 ≡ 2 ( mod 5 ) 2x+4y≡2(mod 5) Add this with the second congruence: 4 𝑥 + 5 𝑦 ≡ 3 ( mod 5 ) 4x+5y≡3(mod 5) 4 𝑥 ≡ 3 ( mod 5 ) 4x≡3(mod 5) Step 3 Solve for values of 𝑥 x and 𝑦 y as specified. Question 5 Using the diagram below, complete the following congruence statement: △ 𝐴 𝐵 𝐶 ≅ ? △ABC≅? Answer 1 Step 1 From the given figure, it can be observed that: △ 𝐴 𝐵 𝐶 ≅ △ 𝐷 𝐶 𝐵 △ABC≅△DCB. This is due to corresponding congruent angles and sides. Conclusion Answer: △ 𝐴 𝐵 𝐶 ≅ △ 𝐷 𝐶 𝐵 △ABC≅△DCB, which is option (d)