更新时间2021-09-30 13:37:50
求整数解
除以4,35x + 57y + 71z = 13045
若 z = 0,则 y = 5n,7x + 57n = 2609,n 是整数;
7x = 2609 - 57n,x = 372 - 8n + ( 5 - n )/7
x 是整数,所以 ( 5 - n )/7 是整数,故应令 n = 7m - 2,m 是整数;
代入,x = 372 - 8( 7m - 2 ) + [ 5 - ( 7m - 2 ) ]/7 = 372 + 16 - 56m + 1 - m = 389 - 57m;
y = 5n = 5( 7m - 2 ) = 35m - 10;
即 z = 0 时,x、y 的通解为 x = 389 - 57m,y = 35m - 10;
71z 可以为 71( 35p + 57q ),p、q 是整数;
对应的,35x 应减去 71 * 35p,y 应减去 71 * 57q
即 35x = 35( 389 - 57m - 71p ),57y = 57( 35m - 10 - 71q )
譬如,m = 1,p = 1,q = 1,
则 x = 389 - 57 - 71 = 261,y = 35 - 10 - 71 = -46,z = 35 + 57 = 92;
35x + 57y + 71z = 9135 - 2622 + 6532 = 13045 。
则 35x = 35( 389 - 57m ) ≤ 13045,1995m ≥ 570,m ≥ 1;
57y = 57( 35m - 10 ) ≤ 13045,1995m ≤ 13615,m ≤ 6;
即 m 取值范围是 1 ≤ m ≤ 6;
将 m 值带入 x、y ,即可求得 p、q 的取值;
例如,m = 1,则 x = 389 - 57 * 1 - 71p ≥ 0,71p ≤ 332,p ≤ 4;
y = 35 * 1 - 10 - 71q ≥ 0,71q ≤ 25,q ≤ 0;
z = 35p + 57q ≥ 0,p ≥ -2q;
取 m = 1,q = -1,p = 2,则 x = 190,y = 96,z = 13 。
140x+228y+284z=52180
35x+57y+71z=13045
7(5x+2z)+57(y+z)=7*1831+57*4
7(5x+2z-1831)+57(y+z-4)=0
设y+z-4=7m①
则5x+2z-1831=-57m
5x=1831-57m-2z=1835-55m-(4+2m+2z)②
设4+2m+2z=10n,z=-2-m+5n③
③代入①,y-2-m+5n-4=7m,y=6+8m-5n
③代入②,5x=1835-55m-10n,x=367-11m-2n
(x,y,z)=(367-11m-2n,6+8m-5n,-2-m+5n)
当m与n是整数时,x、y、z都是整数。