Question

4. Using Euclid's division algorithm, find the HCF of
(i) 612 and 1314
(ii) 1260 and 7344
(iii) 4052 and 12576​

Answer 1

Answer:

(iii) 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.

Step-by-step explanation:

HOPE IT'S HELP YOU

IF YOU SATISFIED MY ANSWER PLZ MARK BRAINLIST

Answer 2

Step-by-step explanation:

iii) 12576=4052×3+420

4052=420×9+272

420=272×1+148

272=148×1+124

148=124×1+24

124=24×5+4

24=4×6+0

HCF(4052,12567)=4

hope it's helpful for you