Question

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

Answer 1

Answer:

10

Step-by-step explanation:

When we simplify we get $z(6xy+21x+2y+7)+30xy+105x+10y=812.$

Then we continue to factor to get: $(z+5)(6xy+21x+2y+7)&=847.$

We then see that we can factor $(6xy+21x+2y+7)$ into $(z+5)(3x+1)(2y+7)&=847.$

we then do the prime factorization of 847, which i think is,  $7*11*11$. we have to find the numbers that multiply to 847 and then plug them into z+5, 3x=1,2y+7.

It has to be a positive, non-negative integer, right?

We also see that 3x+1=11 so we see that x=10/3 (which wont work).

So 3x+1=7, so x=2.

So 11 has to be in another term. It has to be in 2y+7=11 so y=2

for the last term we get z+5=11 so z=6

2+2+6=10

Hope this helps and if you want please consider giving me brainliest. :)