Vumos wants to write the integers 1 to 9 in the nine boxes shown so that the sum of the integers in any three adjacent boxes is a multiple of 3. In how many ways can he do this?
A. 6 × 6 × 6 × 6
B. 6 × 6 × 6
C. 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
D. 6 × 5 × 4 × 3 × 2 × 1
E. 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
A.
Let a, b, c, d be the numbers in four adjacent boxes. Then both a + b + c and b + c + d must be multiples of 3. Therefore a − d is a multiple of 3. This applies to any entries three apart. So the numbers in the set {1, 4, 7} must be listed three apart; and the same applies to {2, 5, 8} and to {3, 6, 9}. This will automatically ensure that the sum of three adjacent numbers is a multiple of 3. There are 3 choices about which of these sets go in the first, fourth and seventh boxes, 2 choices for the next set and 1 for the third. Also there are 3 × 2 × 1 = 6 choices for which order the numbers in each set are given. That gives the total number of choices as 6 × 6 × 6 × 6.
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