Question

Use Euclid's algorithm to find the HCF of 4052 and 12576.

Answer 1

Question:

Use Euclid's algorithm to find the HCF of 4052 and 12576.

Answer:

4

Step-by-step explanation:

$\tt \: 12576=4052 × 3 + 420 \\ \tt \: 4052 = 420 \times 9 + 272\\ \tt \: 420 = 272 \times 1 + 148\\ \tt \: 272 = 148 \times 1 + 124\\ \tt \: 148 = 124 \times 1 + 24\\ \tt \: 124 = 24 \times 5 + 4\\ \tt \: 24 = 4 \times 6 + 0$

Hence,HCF of 4052 and 12576 is 4.

$\huge \bold {{@QianNiu}}$

Answer 2

Step-by-step explanation:

For solutions please refer to the attachment

12576 = 4052 x 3 + 420

4052 = 420 x 9 + 272

420 = 272 x 1 + 148

272 = 148 x 1 + 124

148 = 124 x 1 + 24

124 = 24 x 5 + 4

24 = 4 x 3 + 0

Hence the HCF of 12576 and 4052 is 4