Question

find the HCF of the following pairs of integer by the prime factorization method : a) 963 & 657 ; b) 506& 1155 ; c) 1288 & 575​

Answer 1

(i) 963 and 6567 By applying Euclid’s division lemma 963 = 657 × 1 + 306 …(i) Since remainder ≠ 0, apply division lemma on divisor 657 and remainder 306 657 = 306 × 2 + 45 ….. (ii) Since remainder ≠ 0, apply division lemma on divisor 306 and remainder 4 306 = 45 × 6 + 36 …..(iii) Since remainder ≠ 0, apply division lemma on divisor 45 and remainder 36 45 = 36 × 1 + 9 …… (iv) Since remainder ≠ 0, apply division lemma on divisor 36 and remainder 9 36 = 9 × 4 + 0 ∴ HCF = 9 Now 9 = 45 – 36 × 1 [from (iv)] = 45 – [306 – 45 × 6] × 1 [from (iii)] = 45 – 306 × 1 + 45 × 6 = 45 × 7 – 306 × 1 = 657 × 7 – 306 × 14 – 306 × 1 [from (ii)] = 657 × 7 – 306 × 15 = 657 × 7 – [963 – 657 × 1] × 15 [from (i)] = 657 × 22 – 963 × 15

PLZ PLZ PLZ AS BRAINLIEST

Answer 2

Answer:

The HCF of 963&657 is 2×3

506&1155 is 11×2×3

1288&575 is 23×2×3

See your answer friend.