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Q.

Find the HCF of  506 and 1155 and express it as a linear combination of them.


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a

11;11=  506 × 16 - 1155 × 7

b

12;12=  506 × 16 - 1155 × 6

c

12;12=  506 × 16 - 1155 × 7

d

13;13=  506 × 16 - 1155 × 7 

answer is A.

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Detailed Solution

Given to find the HCF of two integers 506 and 1155 and to express it as linear combination of 506 and 1155.
Euclid's Division Lemma states that if two positive integers a and b exist, then there must be unique values of q and r that satisfy the formula a= bq + r, where  0 ≤ r < b.
Apply Euclid’s division lemma on 506, 1155   . . .  . . .(1)
1155=506×2+143  
∵  remainder ≠ 0,  apply Euclid’s division lemma on divisor 506 and remainder 143
506=143×3+77   . . . . . . (2)
∵  remainder ≠ 0,  apply Euclid’s division lemma on divisor 143 and remainder 77
143=77×1+66   . . . . . . (3)
∵  remainder ≠ 0,  apply Euclid’s division lemma on divisor 77 and remainder 66
77=66×1+11   . . . . . . .(4)
∵  remainder ≠ 0,  apply Euclid’s division lemma on divisor 66 and remainder 11
66=11×6+0   . . . . . .(5)
∴   HCF of 506, 1155 is 11
Now,from (4),
11=77−66×1   . . . . . .(6)
Substitute (3) in (6)
11=77−(143−77×1)×1 ⇒77−143×1+77×1 ⇒77×2−143×1             . . . . . . . . .(7)  
Substitute (2) in (7),
11=(506−143×3)×2−143×1 ⇒506×2−143×6−143×1 ⇒506×2−143×7                  … . . . . ..(8)  
Substitute (1) in (8),
11=506×2−(1155−506×2)×7 ⇒506×16−1155×7 ⇒506×2−1155×7+506×14  
∴  HCF of 506 and 1155 is 11 and can be expressed as a linear combination of 506 and 1155 by,
11=  506 × 16 - 1155 × 7.
Hence the correct option is 1.
 
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