2011 AIME I Problems/Problem 5

First, we determine which possible combinations of digits 1 through 9 will yield sums that are multiples of 3. It is simplest to do this by looking at each of the digits mod 3. We see that the numbers 1, 4, and 7 are congruent to 1 (mod 3), that the numbers 2, 5, and 8 are congruent to 2 (mod 3), and that the numbers 3, 6, and 9 are congruent to 0 (mod 3). In order for a sum of three of these numbers to be a multiple of three, the mod 3 sum must be congruent to 0. Quick inspection reveals that the only possible combinations are 0+0+0, 1+1+1, 2+2+2, and 0+1+2. However, every set of three consecutive vertices must sum to a multiple of three, so using any of 0+0+0, 1+1+1, or 2+2+2 would cause an adjacent sum to include exactly 2 digits with the same mod 3 value, and this is an unacceptable arrangement. Thus the only possible groupings are composed of three digits congruent to three different mod 3 values. We see also that there are two possible arrangements for these trios on the nonagon: a digit congruent to 1 (mod 3) can be located counterclockwise of a digit congruent to 0 and clockwise of a digit congruent to 2, or the reverse can be true. The nonagon can be rotated, so if we find all possible strings beginning with one particular digit, we have found all indistinguishable arrangements. For each of the two trio arrangements, we find 72 possible strings: The first digit is predetermined as 3 because we want to avoid strings that rotate to become indistinguishable, so we have one option as a choice for the first digit. The other two 0 (mod 3) numbers can be arranged in 2!=2 ways. The three 1 (mod 3) and three 2 (mod 3) can both be arranged in 3!=6 ways. Therefore, the desired result is 2(2*6*6)=144.