Consider a cube such that the product of the three faces of the cube forms the vertex.. The sum of all vertex is 1001. Then what will be the sum of numbers in all the faces of the cube

• Let the value of faces be x
  Let the value of vertex be y
  Given: x*x*x=y
  Totally there are 8 vertices in a cube
  So 8y=1001
  =>y=1001/8
  x^3=1001/8
  x=cuberoot(1001/8)
  So the answer is 6*cuberoot(1001)/2
  =>3*cuberoot(1001)
  =>30(Appro)
• 7 years agoHelpfull: Yes(5) No(0)
• Let a, b and c, d and e, f be written on opposite faces, then
  
  ace + ade + acf + adf + bce + bde + bcf + bdf = 1001
  ae(c + d) + af(c + d) + be(c + d) + bf(c + d) = 1001
  (c + d)(ae + af + be + bf) = 1001
  (a + b)(c + d)(e + f) = 1001 = 7*11*13
  
  a + b + c + d + e + f = 31
  Ans:31
• 7 years agoHelpfull: Yes(2) No(0)
• Ans is 31
  6 faces making 8 vertex
  Group these faces into 3 each having 2 opposite faces. Take one face from each group form a vetex. Multiply the values of faces gives the value of vertex. Sum of all vertex is the multiplication of each group sums. Summing the group sum gives us result. Therefore factorize 1001 into 3 (3 only possible) and sum it will result 31. Try it with ur own.
• 7 years agoHelpfull: Yes(1) No(1)