There is a cuboidal dice with each of its faces numbered with a positive integer. The product of the numbers on the faces converging at each corner, is written at the respective corner. The sum of all such products is 1,995. If the sum of all the numbers (on the six faces) is s, how many values can s take?
Question
There is a cuboidal dice with each of its faces numbered with a positive integer. The product of the numbers on the faces converging at each corner, is written at the respective corner. The sum of all such products is 1,995. If the sum of all the numbers (on the six faces) is s, how many values can s take?
Solution
This problem can be solved by using the principles of algebra and geometry.
Let's denote the numbers on the faces of the cuboid as a, b, c, d, e, and f.
Since the product of the numbers on the faces converging at each corner is written at the respective corner, we can write the sum of all such products as:
4abc + 4def = 1995
This simplifies to:
abc + def = 498.75
Since a, b, c, d, e, and f are all positive integers, the sum of abc and def must also be an integer. However, 498.75 is not an integer, which contradicts our assumption.
Therefore, there are no possible values for s, the sum of all the numbers on the six faces of the cuboid.
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