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  • In addition to easily finding the gcd the Euclidean algorithm does something else.Let's

    In addition to easily finding the gcd the Euclidean algorithm does something else.Let's
    see how this works for our example.We now know that (3589,2522) = 97 which means
    that we know that there must be integers m and n such that 3589m+ 2522n = 97 .The
    proof of this is a purely theoretical result that gives no hint as to how to find these
    integers.But the Euclidean algorithm finds them for us!Let's see how this works.
    We will start with the last equation 388 = 291×1+ 97 .From this equation we were able to
    conclude that (388,291) = 97 but it is also easy to see that we can use the equation to
    write 97 as a linear combination of 388 and 291.We just solve for 97 to get:
    (*) 97 = 388×(1) + 291×(-1)
    We can now use the previous equation 1067 = 388× 2 + 291 to express 291 as a linear
    combination of 1067 and 388.
    (**) 291 =1067×(1) + 388×(-2)
    We now use (**) to substitute into (*) to express 97 as a linear combination of 388 and
    1067.
    (***) 97 = 388×(1) + [1067×(1) + 388×(-2)] ×(-1) = 388(3) +1067 ×(-1)
    We can now use the equation 2522 =1067 ×2 + 388 to express 388 as a linear
    combination of 2522 and 1067 and then substitute that into (***) to express 97 as a linear
    combination of 2522 and 1067.
    最后問題是:
    Exercise 9:Continue this process until you have found the integers m and n so that
    3589m+ 2522n = 97
    英語人氣:964 ℃時間:2020-06-09 16:30:54
    優(yōu)質(zhì)解答
    m=-7,n=10
    因為2522=1067 ×2 + 388...(1) ;同理3589=1067 ×3+ 388.(2)
    又因為97=388×3 +1067 ×(-1)
    3589m+2522n=97 .(3)將(1) (2)代入(3)得
    (2n+3m)×1067+(n+m)×388=97 所以
    2n+3m=-1
    n+m=3得到答案
    也可以直接根據(jù)這個程序 一直運算
    我來回答
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