Question

Express the HCF of 48 and 18 as a linear combination

Answer 1

A=bq+r, where   o ≤  r < b
48=18x2+12
18=12x1+6
12=6x2+0 ∴ HCF  (18,48) = 6 now   6= 18-12x1
           6= 18-(48-18x2)
  6= 18-48x1+18x2
  6= 18x3-48x1   6= 18x3+48x(-1)
i.e.       6= 18x +48y
     
   ∴   

6= 18×3 +48×(-1)
=18×3 +48×(-1) + 18×48-18×48
=18(3+48)+48(-1-18)
=18×51+48×(-19)
6=18x+48
 
   ∴     


Hence, x and y are not unique.

Answer 2

Answer:

Step-by-step explanation:

A=bq+r, where   o ≤  r < b

48=18x2+12

18=12x1+6

12=6x2+0 ∴ HCF  (18,48) = 6 now   6= 18-12x1

          6= 18-(48-18x2)

 6= 18-48x1+18x2

 6= 18x3-48x1   6= 18x3+48x(-1)

i.e.       6= 18x +48y

    

  ∴    

6= 18×3 +48×(-1)

=18×3 +48×(-1) + 18×48-18×48

=18(3+48)+48(-1-18)

=18×51+48×(-19)

6=18x+48