Question

use Euclid's division algorithm to find the HCF of 273 and 1032​

Answer 1

Since 1032 > 273

we know that, b = aq + r

⇒1032 = 273 × 3 + 213

Here, the remainder 213 ≠ 0

⇒273 = 213 × 1 + 60

Here, again the remainder 60 ≠ 0

⇒213 = 60 × 3 + 33

Again, the remainder 33 ≠ 0

⇒60 = 33 × 27 + 27

Again, the remainder 27 ≠ 0

⇒33 = 27 × 1 + 6

Again, the remainder 6 ≠ 0

⇒27 = 6 × 4 + 3

Again, the remainder 3 ≠ 0.

⇒6 = 3 × 2 + 0

The remainder has now become zero with divisor 3, so procedure stops there.

Since the divisor at this stage is 3, the HCF of 274 and 1033 is 3.

Answer 2

$273 = 3 \times 7 \times 13$

And,

[tex]1032 = 2 × 2× 2×3 × 43

3 is common number in both 273,1032

so,

273, 1032 HCF = 3