If x, y, and z are positive integers such that 6xyz+30xy+21xz+2yz+105x+10y+7z=812, find x+y+z

x=0;p=0; y=0;z=0;n=6*x*y*z+30*x*y+21*x*z+2*y*z+105*x+10*y+7*z;if(n==812, goto loop, goto next);loop:p=p+1;printp," =",x,y,z;next:x++;if(x<200, goto4,0);x=0;y++;if(y<200, goto4, 0);x=0;y=0;z++;if(z<200, goto4,0)

 

OUTPUT:  All the following 5 values will balance the equation, but I think only the 2nd one meets your condition of "positive integers".

 

       x      y        z

1  = 0     57      2
2  = 2     2       6
3  = 0     35     6
4  = 0     2     72
5  = 0     0     116
 

This can be done algebraicly, and without what seems to look like a not understandable copy and pasted answer.

 

Using some factoring:

6xyz + 30xy + 21xz + 2yz + 105x + 10y + 7z = 812

3x(2yz + 10y + 7z + 35) + (2yz + 10y + 7z) = 812      <------ factor out 3x, and "join like terms"

 

We see that the difference between the first and the second is "35", so we add 35 to both sides:

3x(2yz + 10y + 7z + 35) + (2yz + 10y + 7z + 35) = 847 

We can factor out a 2y and a 7 from the left:

3x(2y + 7)(z + 5) + 1(2y + 7)(z +5) = 847

Now, we can factor out (2y + 7)(z + 5).

(3x + 1)(2y + 7)(z + 5) = 847

 

The prime factorization of 847 is 7 * 11 * 11, which can be applied to our factored form (they are both factors....get it?):

So: 

3x + 1 = 7: x = 2

 

2y + 7 = 11: y = 2

 

z + 5 = 11: z = 6

 

Therefore, x + y + z = 10

 

:)