Let aₙ and bₙ be two sequences such that aₙ=13+6(n−1) and bₙ=15+7(n−1) for all natural numbers n. Then, the largest three digit integer that is common to both these sequences, is:
Question:
Let aₙ and bₙ be two sequences such that aₙ=13+6(n−1) and bₙ=15+7(n−1) for all natural numbers n. Then, the largest three digit integer that is common to both these sequences, is:
Let aₙ and bₙ be two sequences such that aₙ=13+6(n−1) and bₙ=15+7(n−1) for all natural numbers n. Then, the largest three digit integer that is common to both these sequences, is:
Options
Answer: 967
Explanation:
Step 1: Write the general terms aₙ = 13 + 6(n − 1) = 6n + 7 bₙ = 15 + 7(n − 1) = 7n + 8 We want numbers common to both sequences: 6m+7=7n+86m + 7 = 7n + 86m+7=7n+8 for some integers m, n ≥ 1. Step 2: Solve the Diophantine equation 6m+7=7n+8 ⟹ 6m−7n=16m + 7 = 7n + 8 \implies 6m - 7n = 16m+7=7n+8⟹6m−7n=1 This is a linear Diophantine equation: 6m − 7n = 1 → 6m − 7n = 1 Step 3: Find integer solutions Solve modulo 7: 6m ≡ 1 (mod 7) 6 ≡ 6 (mod 7), inverse of 6 mod 7 is 6 (since 6×6 ≡ 36 ≡ 1 mod 7) So m ≡ 6 × 1 ≡ 6 mod 7 → m = 6 + 7k, k ≥ 0 Corresponding n: 6m − 7n = 1 → n = (6m − 1)/7 = (6(6 + 7k) − 1)/7 = (36 + 42k − 1)/7 = (35 + 42k)/7 = 5 + 6k Step 4: General formula for common numbers Common numbers = aₘ = 6m + 7 = 6(6 + 7k) + 7 = 36 + 42k + 7 = 43 + 42k So common numbers = 43, 85, 127, 169, … +42 each step Step 5: Find largest three-digit common number Largest three-digit number ≤ 999: 43+42k≤999 ⟹ 42k≤956 ⟹ k≤22.7643 + 42k ≤ 999 \implies 42k ≤ 956 \implies k ≤ 22.7643+42k≤999⟹42k≤956⟹k≤22.76 So largest integer k = 22 → number = 43 + 42×22 = 43 + 924 = 967 ✅ Answer 967
Explanation:
Step 1: Write the general terms aₙ = 13 + 6(n − 1) = 6n + 7 bₙ = 15 + 7(n − 1) = 7n + 8 We want numbers common to both sequences: 6m+7=7n+86m + 7 = 7n + 86m+7=7n+8 for some integers m, n ≥ 1. Step 2: Solve the Diophantine equation 6m+7=7n+8 ⟹ 6m−7n=16m + 7 = 7n + 8 \implies 6m - 7n = 16m+7=7n+8⟹6m−7n=1 This is a linear Diophantine equation: 6m − 7n = 1 → 6m − 7n = 1 Step 3: Find integer solutions Solve modulo 7: 6m ≡ 1 (mod 7) 6 ≡ 6 (mod 7), inverse of 6 mod 7 is 6 (since 6×6 ≡ 36 ≡ 1 mod 7) So m ≡ 6 × 1 ≡ 6 mod 7 → m = 6 + 7k, k ≥ 0 Corresponding n: 6m − 7n = 1 → n = (6m − 1)/7 = (6(6 + 7k) − 1)/7 = (36 + 42k − 1)/7 = (35 + 42k)/7 = 5 + 6k Step 4: General formula for common numbers Common numbers = aₘ = 6m + 7 = 6(6 + 7k) + 7 = 36 + 42k + 7 = 43 + 42k So common numbers = 43, 85, 127, 169, … +42 each step Step 5: Find largest three-digit common number Largest three-digit number ≤ 999: 43+42k≤999 ⟹ 42k≤956 ⟹ k≤22.7643 + 42k ≤ 999 \implies 42k ≤ 956 \implies k ≤ 22.7643+42k≤999⟹42k≤956⟹k≤22.76 So largest integer k = 22 → number = 43 + 42×22 = 43 + 924 = 967 ✅ Answer 967
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