If x, y, and z are positive integers such that 6xyz+30xy+21xz+2yz+105x+10y+7z=812, find x+y+z.
\(\displaystyle 6xyz +30xy+21xz+2yz+105x+10y+7z=812.\)
Factorise.
\(\displaystyle (3x+1)(2y+7)(z+5)=812 = 2\times 2\times7\times29.\)
The only x, y, z that fit are x =1, y = 11 and z = 2, so x + y + z = 14.
