Put numbers 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 at the vertices of a cube, such that the sum of any three numbers on any face is not less than 10 . The minimum sum of the four number on a face is k , then k

Question

Put numbers 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 at the vertices of a cube, such that the sum of any three numbers on any face is not less than 10 . The minimum sum of the four number on a face is k , then k /2 is equal to

TestHub · Accepted Answer

Suppose that the four numbers on face of the cube is $a_{1}, a_{2}, a_{3}, a_{4}$ such that their sum reaches the minimum and $a_{1} < a_{2} < a_{3} < a_{4}$ . Since the maximum sum of any three numbers less than 5 is 9 , we have $a_{4} \geq 6$ and $a_{1} + a_{2} + a_{3} + a_{4} \geq 16$ .

As seen in figure, we have

$2 + 3 + 5 + 6 = 16$

and that means minimum sum of four numbers on a face is 16 .

Mathematics - Permutation & Combination Question with Solution | TestHub

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