Question

use Euclid's algorithm to find the HCF of 4052 and 12576​

Answer 1

Answer:

HCF is the largest number which exactly divides two or more positive integers.

Since 12576 > 4052

12576 = (4052 × 3) + 420

420 is a reminder which is not equal to zero (420 ≠ 0).

4052 = (420 × 9) + 272

271 is a reminder which is not equal to zero (272 ≠ 0).

Now consider the new divisor 272 and the new remainder 148.

272 = (148 × 1) + 124

Now consider the new divisor 148 and the new remainder 124.

148 = (124 × 1) + 24

Now consider the new divisor 124 and the new remainder 24.

124 = (24 × 5) + 4

Now consider the new divisor 24 and the new remainder 4.

24 = (4 × 6) + 0

Reminder = 0

Divisor = 4

HCF of 12576 and 4052 = 4.

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Answer 2

Answer:

by Eucild's division algorithm we have,

a=bq+r, where 0 ≤ r < b

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.

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