A problem from RMO 2004

Denoting the faces of the cube $a,b,c,d,e,f$ , we get $abc + abf + dec + def + bdc + bdf + eaf + eac = 2004 ; (c + f)(ab + de + bd + ae)$ $2004$ factors to $6 × 334$ , so let $c + f$ equal $6$ . Then $ab + de + bd + ae = 334$ . Here SFFT comes in handy: $(a + b)(d + e) = 334$ . The prime factorisation of $334$ is $167 × 2$ , as $167$ and $2$ are obviously prime. Then letting $a + b$ equal $2$ and letting $d + e$ equal $167$ , we are now able to find the desired result : $a + b + c + d + e + f$ , which is going to equal $167 + 2 + 6 = 175$ .