Here's the answer: 10
6xyz+30xy+21xz+2yz+105x+10y+7z=812
We can group them and factor:
6xy(z+5)+21x(z+5)+2y(z+5)+7z=812
(z+5)(6xy+21x+2y)+7z=812
Now if we add 35 to both sides, we can put 7z into parenthesis too,
(z+5)(6xy+21x+2y)+7z+35 --> 7(z+5)=812+35
(z+5)(6xy+21x+2y+7)=847
This simplifies to
(z+5)(3x+1)(2y+7)=847.
Now we can find the factors of 847, which are:
1, 7, 11, 77, 121, 847.
Now, we must find three positive integer values that multiply to 847. There are three of them, which are
and 7x11x11(they don't have to be distinct)
(z+5)(3x+1)(2y+7) take the place of the three integers.
Now we must test all three possibilities.
z+5=1
z=-4
Since it is negative, 1 does not work
3x+1=1
3x=0
x=0
But zero is neither positive or negative, so 1 does not work for this one either
2y+7=1
2y=-6
y=-3
Since this is also negative, it does not work
This means that
1x7x121
1x11x77
do not work.
That means that (z+5)(3x+1)(2y+7)=7x11x11
BUT...
Which is which?
We must test again...
z+5=7
z=2 This works...
3x+1=11
3x=10
x=10/3
This does NOT work because z must be an integer.
So
z+5 must equal 11
z=6
Now we need to find the other 11 and the 7
This is easy, because we already know that 3x+1 cannot equal 11, so it equals 7.
3x+1=7
3x=6
x=2
This means that
2y+7=11
2y=4
y=2
NOW WE JUST ADD THEM UP
2+2+6=10
The answer is 10
XD