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April Challenge: How many permutations ( x 1 , x 2 , x 3 , x 4 ) of the set of integers {1, 2, 3, 4} have the property that the sum x 1 x 2 + x 2 x 3 + x 3 x 4 + x 4 x 1 is divisible by 3?

SOLUTION: There are 24 permutations for the set of integers {1, 2, 3, 4}. For each permutation, x 1 x 2 + x 2 x 3 + x 3 x 4 + x 4 x 1 is calculated in the table below, and the sum is tested for divisibility by 3.

x 1

x 2

x 3

x 4

x 1 x 2

x 2 x 3

x 3 x 4

x 4 x 1

Sum

1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4

2 2 3 3 4 4 1 1 3 3 4 4 1 1 2 2 4 4 1 1 2 2 3 3

3 4 2 4 2 3 3 4 1 4 1 3 2 4 1 4 1 2 2 3 1 3 1 2

4 3 4 2 3 2 4 3 4 1 3 1 4 2 4 1 2 1 3 2 3 1 2 1

2 2 3 3 4 4 2 2 6 6 8 8 3 3 6 6

6 8 6

12 12

4 3 4 2 3 2 8 6 8 2 6 2

24 25 21 25 21 24 25 24 21 24 21 25 25 21 24 21 24 25 24 21 25 21 25 24

8 8 6 6

12

8

12

3 4 3

12 12

4 4 3 3 8 8 4 4 2 2 6 6 3 3 2 2

12

4

12

2 4 2 8 4 8 2 3 2 6 3 6

12

6

12

3 6 3

12 12

4 4 8 8

12

8

12

4 8 4

12 12

The sums divisible by 3 are marked in blue. There are 16 permutations that are divisible by 3.

Virginia Mathematics Teacher vol. 46, no. 2

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