Introduction / Context:
When a divisor leaves different remainders with different numbers, subtracting the respective remainders from those numbers produces a set of integers all divisible by the same unknown divisor. The problem then reduces to computing the HCF of these adjusted numbers.
Given Data / Assumptions:
- N divides 130 with remainder 6 ⇒ N | (130 − 6) = 124.
- N divides 305 with remainder 9 ⇒ N | (305 − 9) = 296.
- N divides 245 with remainder 17 ⇒ N | (245 − 17) = 228.
Concept / Approach:
The required divisor N is the HCF of 124, 296, and 228. Apply the Euclidean algorithm sequentially to find the HCF efficiently.
Step-by-Step Solution:
HCF(124, 296): 296 − 2*124 = 48; HCF(124,48).HCF(124,48): 124 mod 48 = 28; 48 mod 28 = 20; 28 mod 20 = 8; 20 mod 8 = 4; 8 mod 4 = 0 ⇒ HCF = 4.Now HCF(4, 228): since 228 mod 4 = 0, overall HCF remains 4.
Verification / Alternative check:
Check directly: 130 ≡ 6 (mod 4); 305 ≡ 1 (mod 4) but given remainder 9 reduces to 1 mod 4; 245 ≡ 1 (mod 4) while remainder 17 reduces to 1 mod 4. Consistent.
Why Other Options Are Wrong:
- 5, 14, 24 are not common divisors of 124, 296, and 228; they fail at least one divisibility check.
Common Pitfalls:
- Mistaking the LCM approach; here we must use HCF after subtracting remainders.
Final Answer:
4
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