+212 If x, y, and z are positive integers such that 6xyz+30xy+21xz+2yz+105x+10y+7z=812, find x+y+z
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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
+1084 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
:)
