Question

Please help

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

Answer 1

`x+y+z \Rightarrow 2+2+6 = 10`

Explanation:

First we have to factor the equation:

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

The commom factor of first three terms and
`105x` is
`3x`. Note that
`105x=3x(35)`

`3x(2yz+10y+7z+35)+2yz+10y+7z=812`

`3x(2yz+10y+7z+35)+(2yz+10y+7z)=812`

In order to have two equal factors add 35 both sides:

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

Let's factor
`(2yz+10y+7z+35)`

`(2yz+10y+7z+35) \Rightarrow 2y\left(z+5\right)+7\left(z+5\right) \Rightarrow \left(z+5\right)\left(2y+7\right)`

Now we have:

`3x(z+5)\left(2y+7\right)+\left(z+5\right)\left(2y+7\right)=847`

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

Now it is the interesting part: You have to figure out that 3 numbers multiplied by each other will result in 847.

We have
`847 = 7 \cdot 11 \cdot 11`

Let's try

`2y+7=7\\2y=0`

I will stop right here. This is not true.

Let's try then

`2y+7=11\\2y=4\\y=2`

Now let's try

`z+5 = 11\\z=6`

Now let's try

`3x+1=7\\3x=6\\x=2`

Note that in this case I didn't considered
`3x+1=11` because it would not be an integer.

Therefore,

`x+y+z \Rightarrow 2+2+6 = 10`