find he hcf of 120 and 448 and express it in the form of 120m 448n
Answers
Answered by
55
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
trisharoy0208:
**Correction: In the HCF calculation, 32 = 24(1) + 8. And 24 = 8(3) + 0.
Answered by
10
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
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