+0  
 
0
46
2
avatar

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

 Mar 3, 2020
 #2
avatar+111 
+1

\(\displaystyle 6xyz +30xy+21xz+2yz+105x+10y+7z=812.\)

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.

 Mar 4, 2020

29 Online Users

avatar