Question

Find the HCF of 120 and 448 and express it in the form of 120m+488n

Answer 1

Answer:

Hope this helps you!!!!

Answer 2

Answer:

8 =  120(15) + 448(-4), where m = 15, n = -4

Explanation:

448 = 120*3 + 88

120 = 88*1 + 32

88 = 32*2 + 24

32 = 24*1 + 8

24 = 8*3 = 0

Therefore, HCF(120,448) = 8

8 = 32 - 24

8 = 32 - 88 + 32*2

8 = 32(3) - 88

8 = {120 --88)3 - 88

8 = 120(3) - 88(4)

8 = 120(3) - {448 - 120(3)}(4)

8 = 120(3) - 448(4) + 120 (12)

8 = 120(15) - 448(4)

8 = 120m + 448n

m = 15, n= -4