Question

1. Using EDL find the HCF of
482 and 9684
O 4
O2
O 6​

Answer 1

Answer:

4 Is the answer.

Since 12576 > 4052

12576 = 4052 × 3 + 420

Since the remainder 420 ≠ 0

4052 = 420 × 9 + 272

Consider the new divisor 420 and the new remainder 272

420 = 272 × 1 + 148

Consider the new divisor 272 and the new remainder 148

272 = 148 × 1 + 124

Consider the new divisor 148 and the new remainder 124

148 = 124 × 1 + 24

Consider the new divisor 124 and the new remainder 24

124 = 24 × 5 + 4

Consider the new divisor 24 and the new remainder 4

24 = 4 × 6 + 0

The remainder has now become zero, so procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.

Answer 2

Answer:

2 is the answer

Step-by-step explanation:

9684=482*20+44

482=44*10+42

44=42*1+2

42=2*21+;r=0, HCF=2