求所有的整数数解(a,b,c,x,y,z)使得下列混合组成立:a≥b≥c≥1,x≥y≥z≥1,a+b+c= xyz, x+y+z= abc

a+b+c=xyz
abc=x+y+z
1/a+1/b+1/c=xyz/(x+y+z)
a,b,c是正整数，xyz/(x+y+z)<=1/3
由x>=y>=z,xyz/3x<=xyz/(x+y+z)<=1/3
所以yz<=9
同理,bc<=9
这就可以枚举b,c,y,z,求得a,x
程序如下(C++):
#include <stdio.h>
#include <stdlib.h>

int main()
{
int a,b,c,x,y,z;
for(b = 1;b <= 9;b++)
for(c = 1;c <= b && b*c <= 9;c++)
for(y = 1;y <= 9;y++)
for(z = 1;z <= y && y*z <= 9 ;z++)
{
if(b*c*y*z == 1)
continue;
if((b*c*c + b*b*c + y + z) % (b*c*y*z - 1))
continue;
x = (b*c*c + b*b*c + y + z) / (b*c*y*z - 1);
if(x < y || x < z)
continue;
a = x*y*z - b - c;
if(a < b || a < c)
continue;
printf("a=%d b=%d c=%d x=%d y=%d z=%d\n",a,b,c,x,y,z);
}
system("PAUSE");
return 0;
}
运行结果
a=8 b=1 c=1 x=5 y