Question

Use Euclids division algorithm find HCF of 4052 and 12576

Answer 1

Using Euclid's division algorithm on 4052, 12576:
here, 12576>4052
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
The remainder has now become 0.
∴, the H.C.F of 4052,12576 is 4.

Answer 2

Answer:

• The divisor at this stage, ie, 4 is the HCF of 12576 and 4052.

Step-by-step explanation:

Clearly, 12576 > 4052

Applying the Euclid's division lemma to 12576 and 4052, we get

12576 = 4052 x 3 + 420

Since the remainder 420 0, we apply the Euclid's division lemma to divisor 4052 and remainder 420 to get

4052 = 420 x 9 + 272

We consider the new divisor 420 and remainder 272 and apply the division lemma to get

420 = 272 x 1 + 148

We consider the new divisor 272 and remainder 148 and apply the division lemma to get

272 = 148 x 1 + 124

We consider the new divisor 148 and remainder 124 and apply the division lemma to get

148 = 124 x 1 + 24

We consider the new divisor 124 and remainder 24 and apply the division lemma to get

124 = 24 x 5 + 4

We consider the new divisor 24 and remainder 4 and apply the division lemma to get

24 = 4 x 6 + 0

Now, the remainder at this stage is 0.

So, the divisor at this stage, ie, 4 is the HCF of 12576 and 4052.