几道关于整除的题目(很急啊,好的追分)

1个回答

  • (1)

    因为3整除4X-Y

    也就是4X-Y=3k(k为整数)

    X+2Y+3(X-Y)=3k

    X+2Y=3(k-X+Y)

    4X²+7XY-2Y²

    =(4X-Y)(X+2Y)

    =9k(k-X+Y)

    能被9整除

    (2)①当A=3m(m为整数)时

    A(A+1)(2A+1) =3m(3m+1)(6m+1)

    若m=2n,(n为整数)

    则A(A+1)(2A+1)=6n(3m+1)(6m+1)得证

    若m=2n+1,则A(A+1)(2A+1)=3n(6n+4)(6m+1)=6n(3n+2)(6m+1) 得证

    ②当A=3m+1时,A(A+1)(2A+1)=(3m+1)(3m+2)(6m+3)

    =3(3m+1)(3m+2)(2m+1)

    同理,当m为奇数时,3(2m+1)能被6整除,

    当m为偶数时,3(3m+2)能被6整除,

    ③ 当A=3m+2时,A(A+1)(2A+1)=3(3m+2)(m+1)(6m+5)

    当m为奇数时,3(m+1)能被6整除,

    当m为偶数时,3(3m+2)能被6整除,

    于是此问得证

    (3)由题可设MA-B=Nx

    MC-D=Ny (x,y为整数)

    M=(Nx+B)/A=(Ny+D)/C

    于是 ANy+AD=NxC+BC

    AD-BC =N(xC-Ay)

    于是 N整除AD-BC