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[求助]请用矩阵法求下列五元一次不定方程通解
题:求下式通解
1036v+425w+210x+40y+30z=24567 (1)
(其中v,w,x,y,z均为整数)
解:
易知当v=5m+2 , w=2n+1时,(1)式可化简为
518m+85n+21x+4y+3z=2207 (2)
(2)式变形为
3z=2207-(518m+85n+4y)-21x (3)
故原题等价于使2207-(518m+85n+4y)为3的整倍数
因
2207≡518≡2 mod(3),85≡4≡1 mod(3)
当m=3a+1,n+y=3b
当m=3a+2,n+y=3b+1
当m=3a+3,n+y=3b+2
即当n+y-m+1=3(b-a)时2207-(518m+85n+4y)为3的整倍数
令p=b-a
即当y=3p+m-n-1时2207-(518m+85n+4y)为3的整倍数
将v=5m+2,w=2n+1,y=3p+m-n-1代入(3)式得
z=737-(174m+27n+7x+4p)
故
v=5m+2,w=2n+1,x=x,y=3p+m-n-1,z=737-(174m+27n+7x+4p)
为(1)式通解
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