if hcf of 384 and 26 is 2 find m,n such that 2=m*384+n*26 where m and n are integers
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HCF of 384 and 26 is 2
Because prime factors of 26 = 2 × 13
And prime factors of 384 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
now, take common factors ,
e.g., common factors = 2 = HCF
Now, 2 = 384m + 26n
1 = 192m + 13n
put m = 4 and n = -59
you get , RHS = 192 × 4 - 13 × 59 = 768 - 767 = 1 = LHS
Hence, one solution is m = 4 and n = -59
Because prime factors of 26 = 2 × 13
And prime factors of 384 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
now, take common factors ,
e.g., common factors = 2 = HCF
Now, 2 = 384m + 26n
1 = 192m + 13n
put m = 4 and n = -59
you get , RHS = 192 × 4 - 13 × 59 = 768 - 767 = 1 = LHS
Hence, one solution is m = 4 and n = -59
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