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Numbered Cube (Posted on 2007-01-26) Difficulty: 2 of 5
Six positive integers are placed on the faces of a cube. For each vertex of the cube we create a number by multiplying the numbers on the sides forming that vertex. If the sum of the numbers on the vertices is 1001 what is the sum of the numbers on the faces?

No Solution Yet Submitted by atheron    
Rating: 4.0000 (4 votes)

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Solution Quick solution | Comment 2 of 3 |
If we call the sides A, B, C, D, E, and F (A is opposite B, C is opposite D, and E is opposite F) the sum of the products can be reduced to (A+B)(C+D)(E+F)= 1001.

Since 1001=7x11x13 (only possibility), then A+B+C+D+E+F= 7+11+13= 31.

  Posted by Federico Kereki on 2007-01-26 20:17:01
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