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Q.

Find the HCF of 592 and 252 and express it as a linear combination of them.


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a

3;4 = 252 × 47 - 592 × 20

b

4;4 = 252 × 47 - 592 × 20

c

5;4 = 252 × 47 - 592 × 20

d

6;4 = 252 × 47 - 592 × 20 

answer is B.

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Detailed Solution

Given to find the HCF of two integers 592 and 252 and to express it as linear combination of 592 and 252.
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 592, 252
592=252×2+88   . . .  . . .(1)
∵  remainder ≠ 0,  apply Euclid’s division lemma on divisor 252 and remainder 88
252=88×2+76   . . . . . . (2)
∵  remainder ≠ 0,  apply Euclid’s division lemma on divisor 88 and remainder 76
88=76×1+12   . . . . . . (3)
  ∵  remainder ≠ 0,  apply Euclid’s division lemma on divisor 76 and remainder 12
76=12×6+4   . . . . . . .(4)
  ∵  remainder ≠ 0,  apply Euclid’s division lemma on divisor 12 and remainder 4
12=4×3+0   . . . . . .(5)
∴  remainder = 0  
Hence, HCF of 592 and 252 is 12
Now, from (4), we get,
4=76−12×6   . . . . . .(6)
Substitute (3) in (6),
4=76−(88−76×1)×6 ⇒76−88×6+76×6 ⇒76×7−88×6          . . . . . .…(7)  
Substitute (2) in (7)
4=(252−88×2)×7−88×6 ⇒252×7−88×14−88×6 ⇒252×7−88×20            ……(8)  
 Substitute (1) in (8)
4=252×7−(592−252×2)×20 ⇒252×7−592×20+252×40 ⇒252×47−592×20  
∴  HCF of 592 and 252 is 4 and can be expressed as a linear combination of 592 and 252 by,
4 = 252 × 47 - 592 × 20.
Hence the correct option is 2.
 
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