A unique dice

Number Theory Level 4

A positive integer is written on each face of a cube. Each vertex is then assigned the product of the three numbers written on the three faces intersecting at that vertex. The sum of all the numbers assigned to the vertices is equal to 2431.

Find the smallest of the sums of opposite faces.

Clarification: There are 3 pairs of opposite faces on a cube.

The answer is 11.

1 solution

Brian Charlesworth
Mar 31, 2017

Let the pairs of opposite sides be $(a,c), (b,d)$ and $(e,f)$ , where $a,b,c,d,e,f$ are positive integers. Then the sum of the numbers assigned to the vertices is

$abe + abf + ade + adf + cbe + cbf + cde + cdf = (a + c)(b + d)(e + f)$ .

Now $2431 = 11 \times 13 \times 17$ , which is the only way of factoring $2431$ as the product of 3 integers each greater than $1$ . We can then assign these factors to $(a + c), (b + d)$ and $(e + f)$ in any order, each of which is the sum of opposite sides. The desired least sum is thus $\boxed{11}$ .

A unique dice

The answer is 11.

1 solution

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