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 1011+1, then evaluate the respective minimum and maximum sum of the numbers on the faces.
Letter the faces a,b,c,d,e,f. Then the sum of the vertex products can be easily factored as (a+f)(b+d)(c+e).
10^11 + 1 = 11*11*23*4093*8779 so it's necessary to compress those factors into three.
(a+f)=11*11*23 gives the minimum sum of faces as 2783+4093+8779=15655.
(a+f)=23*4093*8779 gives the maximum sum of faces as 826446281+11+11=826446303.
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Posted by xdog
on 2023-05-17 07:55:15 |
solution