AIME 2011 I · 第 5 题

Solution 1

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 $\bmod{3}$.

We see that the numbers $1, 4,$ and $7$ are congruent to $1 \pmod{3}$, that the numbers $2, 5,$ and $8$ are congruent to $2 \pmod{3}$, and that the numbers $3, 6,$ and $9$ are congruent to $0 \pmod{3}$. 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 $\bmod{3}$ value, and this is an unacceptable arrangement. Thus the only possible groupings are composed of three digits congruent to three different $\bmod{3}$ values.

We see also that there are two possible arrangements for these trios on the nonagon: a digit congruent to $1 \pmod{3}$ can be located counterclockwise of a digit congruent to $0$ and clockwise of a digit congruent to $2 \pmod{3}$, or the reverse can be true.

We set the first digit as $3$ avoid overcounting rotations, so we have one option as a choice for the first digit. The other two $0 \pmod{3}$ numbers can be arranged in $2!=2$ ways. The three $1 \pmod{3}$ and three $2 \pmod{3}$ can both be arranged in $3!=6$ ways. Therefore, the desired result is $2(2 \times 6 \times 6)=\boxed{144}$.

Solution 2

Notice that there are three triplets of congruent integers $\mod 3$: $(1,4,7),$ $(2,5,8),$ and $(3,6,9).$ There are $3!$ ways to order each of the triplets individually and $3!$ ways to order the triplets as a group (see solution 1). Rotations are indistinguishable, so there are $(3!)^4/9=\boxed{144}$ total arrangements.

Video solution

https://www.youtube.com/watch?v=vkniYGN45F4