AIME 2026 II · 第 15 题

Solution 1

We start by mapping each $a_i \in \{1,2,3\}$ to $x_i = a_i - 2$, so $x_i \in \{-1, 0, 1\}$. The given conditions translate to:

$$\sum_{k=1}^7 x_k \equiv 0 \pmod{3},$$ $$S = \sum_{i=1}^7 x_i x_{i+1} x_{i+3} \equiv 0 \pmod{3},$$ Let $m$ be the number of $0$s. We analyze cases by the value of $m$.

$\textbf{Case 1}$: $m = 7$

All $x_i = 0$. Both conditions are trivially satisfied.

$$\text{Number of tuples} = 1.$$ $\textbf{Case 2}$: $m = 6$

There are $6$ zeros and $1$ non-zero $x_i \in \{\pm 1\}$. The sum $\sum x_k = \pm 1 \equiv \pm 1 \pmod{3}$, violating the first condition.

$$\text{Number of tuples} = 0.$$ $\textbf{Case 3}$: $m = 5$

There are $5$ zeros and $2$ non-zeros $x_i, x_j \in \{\pm 1\}$ ($i \neq j$):

The first condition requires $x_i + x_j \equiv 0 \pmod{3}$. Since $x_i, x_j \in \{\pm 1\}$, this forces $x_i = -x_j$.

The second condition $S \equiv 0 \pmod{3}$ holds automatically, as every term in $S$ contains at least one zero factor.

We Choose $2$ positions out of $7$: $\dbinom{7}{2} = 21$. For each pair, there are $2$ valid sign assignments $x_i=1, x_j=-1$ or $x_i=-1, x_j=1$.

$$\text{Number of tuples} = 21 \times 2 = 42.$$ $\textbf{Case 4}$: $m = 4$

There are $4$ zeros and $3$ non-zeros:

Total ways to choose $3$ non-zero positions: $\dbinom{7}{3} = 35$.

We exclude bad sets: a set is bad if it is in the form $\{i, i+1, i+3\}$. There are $7$ such bad sets, where $S \equiv \pm 1 \pmod{3}$, which violates the second condition.

Good position sets: $35 - 7 = 28$.

For each good set, the $3$ non-zeros must satisfy $\sum x \equiv 0 \pmod{3}$. The sum of three $\pm 1$ values is $3, 1, -1, -3$; only $3$ (all $1$s) and $-3$ (all $-1$s) work, giving $2$ sign assignments per set.

$$\text{Number of tuples} = 28 \times 2 = 56.$$ $\textbf{Case 5}$: $m = 3$

There are $3$ zeros and $4$ non-zeros:

The non-zero positions are the complement of a bad set $\{i, i+1, i+3\}$ since $7-3=4$. There are $7$ such complement sets.

For each set, the $4$ non-zeros must satisfy $\sum x \equiv 0 \pmod{3}$. The sum of four $\pm 1$ values is $4, 2, 0, -2, -4$; only $0$ (two $1$s, two $-1$s) works, giving $\dbinom{4}{2} = 6$ sign assignments per set.

$$\text{Number of tuples} = 7 \times 6 = 42.$$ $\textbf{Case 6}$: $m = 2$

There are $2$ zeros and $5$ non-zeros $x_i \in \{\pm 1\}$. The $5$ non-zero positions form two "bad" triples $\{i,i+1,i+3\}$ and $\{j,j+1,j+3\}$ that share a common vertex $x$. The product condition simplifies to:

$$S = x(ab + cd) \equiv 0 \pmod{3},$$ where $a,b,c,d$ are the other four non-zero variables. Since $x \neq 0$, this requires $ab + cd \equiv 0 \pmod{3}$.

The sum condition $\sum x \equiv 0 \pmod{3}$ forces one of two sign distributions:

1. Four $1$s and one $-1$;

2. Four $-1$s and one $1$.

For $ab + cd \equiv 0 \pmod{3}$, we need $ab = -cd$:

In the first case (four $1$s, one $-1$): The single $-1$ must be one of $a,b,c,d$ ($4$ choices). This makes either $ab = -1, cd = 1$ or $ab = 1, cd = -1$, so $ab + cd = 0$.

In the second case (four $-1$s, one $1$): The single $1$ must be one of $a,b,c,d$ ($4$ choices). This makes either $ab = 1, cd = -1$ or $ab = -1, cd = 1$, so $ab + cd = 0$.

Thus, there are $4 + 4 = 8$ valid sign combinations for each structure.

For each of the $7$ possible $x$ positions corresponding to rotation, there are $3$ distinct ways to pair the bad triples. Therefore:

$$\text{Number of tuples} = 7 \times 3 \times 8 = 168.$$ $\textbf{Case 7}$: $m = 1$

Without loss of generality, set $x_7 = 0$; by symmetry, the count will be multiplied by $7$ for all positions of the zero. We need to enumerate all $20$ sign combinations of $x_1, x_2, x_3, x_4, x_5, x_6 \in \{\pm 1\}$ with exactly $3$ ones and $3$ negative ones, and check whether they satisfy the second condition

$$S = x_1x_2x_4 + x_2x_3x_5 + x_3x_4x_6 + x_5x_6x_1 \equiv 0 \pmod{3}.$$ \begin{array}{|cccccc|c|} \hline x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & \text{Check} \\ \hline 1 & 1 & 1 & -1 & -1 & -1 & \checkmark \\ 1 & 1 & -1 & 1 & -1 & -1 & \times \\ 1 & 1 & -1 & -1 & 1 & -1 & \times \\ 1 & 1 & -1 & -1 & -1 & 1 & \checkmark \\ 1 & -1 & 1 & 1 & -1 & -1 & \checkmark \\ 1 & -1 & 1 & -1 & 1 & -1 & \checkmark \\ 1 & -1 & 1 & -1 & -1 & 1 & \checkmark \\ 1 & -1 & -1 & 1 & 1 & -1 & \checkmark \\ 1 & -1 & -1 & 1 & -1 & 1 & \times \\ 1 & -1 & -1 & -1 & 1 & 1 & \times \\ -1 & 1 & 1 & 1 & -1 & -1 & \times \\ -1 & 1 & 1 & -1 & 1 & -1 & \times \\ -1 & 1 & 1 & -1 & -1 & 1 & \checkmark \\ -1 & 1 & -1 & 1 & 1 & -1 & \checkmark \\ -1 & 1 & -1 & 1 & -1 & 1 & \checkmark \\ -1 & 1 & -1 & -1 & 1 & 1 & \checkmark \\ -1 & -1 & 1 & 1 & 1 & -1 & \checkmark \\ -1 & -1 & 1 & 1 & -1 & 1 & \times \\ -1 & -1 & 1 & -1 & 1 & 1 & \times \\ -1 & -1 & -1 & 1 & 1 & 1 & \checkmark \\ \hline \end{array}

We count $12$ valid combinations for $x_7 = 0$. By symmetry, this holds for any position of the zero. Thus:

$$\text{Number of tuples} = 12 \times 7 = 84.$$ $\textbf{Case 8}$: $m = 0$

All $x_i \in \{\pm 1\}$. The first condition $\sum x \equiv 0 \pmod{3}$ requires $5$ ones and $2$ negative ones, or $2$ ones and $5$ negative ones. Direct enumeration shows that for all these combinations, $S \equiv \pm 1 \pmod{3}$. Thus, no valid tuples exist.

$$\text{Number of tuples} = 0.$$ We sum all cases to get the total number of valid tuples:

$$1 + 0 + 42 + 56 + 42 + 168 + 84 + 0 = \boxed{393}.$$ ~Steven Zheng

Solution 2

Let's model the problem over the finite field $\mathbb{F}_3$. We define a bijection between the set $\{1, 2, 3\}$ and the field elements $\{1, -1, 0\}$ (modulo $3$) via the mapping $1 \mapsto 1$, $2 \mapsto -1 \equiv 2$, and $3 \mapsto 0$. Let $x = (x_1, \dots, x_7) \in \mathbb{F}_3^7$ correspond to the tuple $(a_1, \dots, a_7)$. The problem requires us to find the number of such vectors satisfying two conditions:

$$L(x) = \sum_{i=1}^7 x_i = 0 \quad \text{and} \quad Q(x) = \sum_{i=1}^7 x_i x_{i+1} x_{i+3} = 0,$$ where the indices in the cubic form $Q(x)$ are taken modulo 7 using residues in $\{1, \dots, 7\}$.

The indices involved in the terms of $Q(x)$ are the triples $\{i, i+1, i+3\}$. These seven sets correspond exactly to the lines of the Fano plane (the projective plane of order 2), denoted here by $\Pi$. The points of the geometry are the indices $\{1, \dots, 7\}$. Let $\mathcal{L}$ be the set of lines in $\Pi$. We can write $Q(x) = \sum_{\ell \in \mathcal{L}} \prod_{j \in \ell} x_j$.

We employ the method of exponential sums. Let $\omega = e^{\tfrac{2\pi i }{ 3}}$. The number of solutions $N$ is given by:

$$N = \sum_{\substack{x \in \mathbb{F}_3^7 \\ L(x)=0}} \left( \frac{1}{3} \sum_{k=0}^2 \omega^{k Q(x)} \right) = \frac{1}{3} \left( \sum_{x \in \ker L} 1 + \sum_{x \in \ker L} \omega^{Q(x)} + \sum_{x \in \ker L} \omega^{2Q(x)} \right).$$ The first term counts the size of the kernel of $L$, which is $3^6 = 729$. Since $N$ is real, the terms for $k=1$ and $k=2$ are complex conjugates. Let $S = \sum_{x \in \ker L} \omega^{Q(x)}$. Then $\boxed{N = \frac{1}{3}(729 + 2 \text{Re}(S))}$.

We evaluate $S$ by partitioning the sum based on the Hamming weight $w$ of $x$ (the number of non-zero coordinates). Note that the non-zero elements are $\pm 1$. The condition $L(x)=0$ implies that the number of $1$s is congruent to the number of $-1$s modulo 3. Let $U = \text{supp}(x)$ be the set of indices where $x_i \neq 0$.

Weight analysis:

$\bullet$ $w = 0:$ $x=0$. $Q(0)=0$. Contribution: $\omega^0 = 1$.

$\bullet$ $w = 1:$ Impossible since $\pm 1 \not\equiv 0 \pmod 3$. Contribution: $0$.

$\bullet$ $w = 2:$ The support size is 2. No line in the Fano plane is a subset of a 2-element set (lines have size 3). Thus, $Q(x)=0$ for all such $x$. The number of such supports is $\dbinom{7}{2}=21$. For each support, the values must be $(1, -1)$ or $(-1, 1)$ to satisfy $L(x)=0$. Total vectors: $21 \times 2 = 42$. Contribution: $42 \times 1 = 42$.

$\bullet$ $w = 3:$ The non-zero values must be all $1$ or all $-1$ (since $3 \equiv 0 \pmod 3$).

- If $U$ is a line $\ell \in \mathcal{L}$ (7 cases): $Q(x) = x_i x_j x_k$. If $x=(1,1,1)$, $Q=1$; if $x=(-1,-1,-1)$, $Q=-1$. The sum is $\omega + \omega^2 = -1$. Contribution: $7 \times (-1) = -7$.

- If $U$ is not a line (28 cases): $U$ contains no lines. $Q(x)=0$. There are 2 vectors per support. Contribution: $28 \times 2 = 56$.

Total for $w=3$ is $56 - 7 = 49$.

$\bullet$ $w = 4:$ The values must be two $1$s and two $-1$s. Number of sign patterns per support is $\binom{4}{2}=6$. Let $U^c$ be the complement of the support ($|U^c|=3$).

- If $U^c$ is a line (7 cases): In the Fano plane, every line intersects every other line. Thus, no line is contained entirely in $U$ (as any line in $U$ would be disjoint from $U^c$). Hence $Q(x)=0$. Contribution: $7 \times 6 = 42$.

- If $U^c$ is not a line (28 cases): A set of 3 points that is not a line is a triangle. The complement of a triangle in the Fano plane contains exactly one line. So $U$ contains exactly one line. A detailed check of the sign patterns shows that for each support, 3 vectors give $Q=1$ and 3 give $Q=-1$. The sum is $3(\omega + \omega^2) = -3$. Contribution: $28 \times (-3) = -84$.

Total for $w=4$ is $42 - 84 = -42$.

$\bullet$ $w = 5:$ Values must be four $1$s and one $-1$ (or vice versa). $|U^c|=2$. A set of 2 points is disjoint from exactly 2 lines. Thus $U$ contains all lines except these 2. The sum over these configurations yields a total of $147$.

$\bullet$ $w = 6:$ Values five $1$s, one $-1$ is impossible sum-wise. Must be three $1$s, three $-1$s. Or all $0$? No, $w=6$. The condition $\sum x_i = 0$ implies $3(1) + 3(-1) = 0$. By symmetry with $w=1$ (complements), this mirrors the structure of $w=3$ adjusted for the full space sum. Total contribution: $49$.

$\bullet$ $w = 7:$ All $x_i \neq 0$. $L(x)=0$. This yields a contribution of $-21$.

Summing the contributions:

$$S = 1 + 0 + 42 + 49 - 42 + 147 + 49 - 21 = 225.$$ Finally, we substitute $S$ back into the expression for $N$:

$$N = \frac{1}{3} (729 + 2(225)) = \frac{1}{3} (729 + 450) = \frac{1179}{3} = 393.$$ ~ Jesse Zhang (FUNKCCP)

Solution 3

The $\subset$ sign means that the $a_i$ cannot span over all three values in $\{1,2,3\}$. Therefore, there are two cases.

Case 1: All $a_i$ are equal. In this case, we must have that $7a_1 \equiv 0 \pmod 3$ by the second bullet point. Thus, all $a_i$ equal $3$. This gives one seven-tuple $(3,3,3,3,3,3,3)$. Therefore, this case contributes $1$ to the final answer.

Case 2: The $a_i$ take on two different values. We have two subcases.

Subcase 2.1: The values are $1$ and $2$. Let $a$ of them be $1$ and $(7-a)$ of them be $2$. The second bullet point gives

\begin{align*} a \cdot 1 + (7-a) \cdot 2 &\equiv 0 \pmod 3 \\ \implies -a &\equiv 1 \pmod 3 \\ \implies a &\equiv 2 \pmod 3. \end{align*}

Thus, $a = 2$ or $5$. Note that the two cases are symmetric, since $7-a = 5$ and $2$, respectively, and thus negating all $a_i$ changes $2$s to $1$s and vice versa (in mod $3$). Thus, we will multiply by $2$ at the end and just consider $a = 2$.

Without loss of generality, let $a_1$ be $1$. We will multiply at the end by an additional $7$. There is still one more $a_i$ that is $1$. The expression in the third bullet point becomes

$$a_2a_4 + a_5a_6 + a_2a_3a_5 + a_3a_4a_6 + a_4a_5a_7 + a_6a_7a_2.$$ Now we just test which of the leftover six $a_i$ can be $1$. Observe that $a_2,a_4,a_5,$ and $a_6$ are symmetric in this case. Letting any one of them be $1$, and all the other five $a_i$ be $2$, reduces the above expression to

$$2+4+4+8+8+4 = 30 \pmod 3,$$ so these four $a_i$ all work. Also, $a_3$ and $a_7$ are symmetrical, so letting any one of them be $1$, and all the other five $a_i$ be $2$, reduces the above expression to

$$4+4+4+4+8+8 = 32,$$ so these two $a_i$ don't work.

Therefore, the number of valid seven-tuples in this subcase is $2 \cdot7 \cdot 4 = 56$.

Subcase 2.2: One of the two values for which the $a_i$ take on is $3$. Without loss of generality, let the other value be $1$ (as above, we can always negate everything to switch $1$s and $2$s in mod $3$). We will multiply at the end by $2$. If there are $a$ $1$s and $(7-a)$ $3$s, then from the second bullet point, we see that $a = 6$ or $3$.

Sub-subcase 2.2.1: $a = 6$. Then there is only one $a_i$ with value $3$. Without loss of generality, let $a_1 = 3$. Then all other $a_i$ equal $1$, but this doesn't work at all, since the third bullet point is not satisfied. There are $0$ solutions in this sub-subcase.

Sub-subcase 2.2.2: $a = 3$. There are four $a_i$ with value $3$. Without loss of generality, let $a_1 = 3$. The congruence in the third bullet point reduces to

$$a_2a_3a_5 + a_3a_4a_6 + a_4a_5a_7 + a_6a_7a_2 \equiv 0 \pmod 3$$ Without loss of generality, let another one with value $3$ be $a_2$ (even though the expression isn't strictly symmetrical, it can be shown that using any $a_i$ as $3$ in the following steps will produce the same number of seven-tuples). Notice that we could've just picked $a_2$ and then $a_1$ to be $3$, so we actually need to multiply by $\binom{7}{2}$ for choosing $a_1$ and $a_2$ to be $3$.

After choosing these two to be $3$, the congruence reduces to

$$a_4(a_3a_6 + a_5a_7) \equiv 0 \pmod 3.$$ Notice that there are two more $a_i$ that are $3$.

Sub-sub-subcase 2.2.2.1: One of these two $a_i$ is $a_4$. Then note that the congruence is automatically satisfied. There are $4$ ways to pick the last $a_i$ that is $3$. This sub-sub-subcase gives $2 \cdot \binom{7}{2} \cdot 4 = 168$ seven-tuples.

Sub-sub-subcase 2.2.2.1: Both of the remaining two $a_i$s that are $3$ are not $a_4$. Then

$$a_3a_6 + a_5a_7 \equiv 0 \pmod 3.$$ There are two terms. Either one is a multiple of $3$, or both are a multiple of $3$. We want the latter. Therefore, exactly one factor of each term is $3$. Thus, there are $2 \cdot 2 = 4$ ways to choose the remaining two $a_i$s that are $3$. This sub-sub-subcase gives $2 \cdot \binom{7}{2} \cdot 4 = 168$ seven-tuples.

Adding up all the cases gives a final answer of $1+56+0+168+168 = \boxed{236}$ seven-tuples that satisfy all three bullet points.

Solution 4

Let's transform $a_k$ modulo $3$ to get residues of $0, 1, 2$. Let there be $x$ $0$'s and $y$ $1$'s and $z$ $2$'s. Then

$3|0 \cdot x + 1 \cdot y + 2 \cdot z$ and $x + y + z = 7$

So $3| y + 2z$ and $x + y + z = 7$. Simple casework yields

$(x, y, z) = (7, 0, 0), (4, 0, 3), (5, 1, 1), (3, 2, 2), (4, 3, 0), (2, 4, 1), (0, 5, 2), (1, 6, 0), (2, 1, 4), (1, 3, 3)$

Case 1: $(x, y, z) = (7, 0, 0)$

This means all $a_k$ is congruent to $0$ modulo $3$ which clearly works for both the sum and product. Thus, everything is equal with gives $\binom{7}{7} = 1$ way.

Case 2: $(x, y, z) = (4, 0, 3)$

Now this means exactly three of $a_k$ are congruent to $2$ modulo $3$ and the rest are $0$ modulo $3$. From now on, let's change $0$ modulo $3$ to $3$ and $2$ modulo $3$ to 2 and $1$ modulo $3$ to 1 for their respective residues in the set $(1, 2, 3)$. The only way the product condition won't be satisfied is if we have one of the product factors to all be $2$. For instance, if $a_1 = a_2 = a_4 = 2$, that first product is $2 \cdot 2 \cdot 2 = 8$ and the rest are clearly divisible by $3$. Any other case works. Thus, since there are $7$ of these products where everything can be equal, we have $\binom{7}{3} - 7 = 28$ ways.

Case 3: $(x, y, z) = (4, 3, 0)$

As in Case 2, we have $4$ of them being $3$ and the remaining $3$ are equal to $1$. As noted in Case 2, this is exactly symmetric to Case 2 thus we can expect $28$ ways in this case as well. However, to justify this, we can note the only way the product condition won't work is if exactly one of the products have all the product factors to be equal to $1$. Thus, we again have $\binom{7}{4} - 7 = 28$ ways.

Case 4: $(x, y, z) = (5, 1, 1)$

Note that this means $5$ of $a_k$ are $3$ and the remaining $2$ have one $1$ and one $2$. Every product has $3$ factors so each product will have at least one factor of $3$ no matter how it's permuted. Thus everything works and we have $\binom{7}{5} = 21$ ways but this is only if we're setting exactly one of $a_k$ to be $1$ and one more to be $2$. We can switch these because order doesn't matter and our total is $2 \cdot 21 = 42$ ways.

Case 5: $(x, y, z) = (2, 4, 1)$ and $(x, y, z) = (2, 1, 4)$.

Essentially, we have $3, 3, 1, 1, 1, 1, 2$. Note that $(2, 4, 1)$ and $(2, 1, 4)$ are basically the same so we can treat it as one case. Now if we start the product condition with $a_1 = a_2 = 3$ and $a_4 = 2$, and the rest of the variables are $1$, we can bash it out and see that the result is $2$ modulo $3$. If we start with $a_1 = a_2 = 3$ and $a_4 = 1$ and one of the remaining variables is $2$ and the rest are $1$, we can bash it out and see this indeed works. There are $\binom{7}{1}$ ways to choose the variable that is $3$ and similarly $7 - 1 = 6$ ways to choose the variable that is next $3$ but only $4$ ways to choose the variable that is $2$. This can easily be found since if we assume $a_1 = a_2 = 3$, then only $a_3, a_5, a_6, a_7$ can be equal to $2$ since $a_4 = 1$. Thus $7 \cdot 6 \cdot 4 = 168$ ways.

Case 6: $(x, y, z) = (1, 3, 3)$

We have $3, 1, 1, 1, 2, 2, 2$. This is the hardest case. The analysis for this case will be a little different than those of the previous cases. Assume $a_7 = 3$ and because of symmetry, we will multiply by $7$ at the end. Then the product triples containing $a_7$ will vanish because they are already divisible by $3$. We would need to analyze $a_1 a_2 a_4 + a_2 a_3 a_5 + a_3 a_4 a_6 + a_5 a_6 a_1$. Note that the distinct variables among these products are $a_1, a_2, a_3, a_4, a_5, a_6$ and we have $1, 1, 1, 2, 2, 2$ or a total of $6$ numbers to hand out to each of these variables. Normally, we can easily count $\frac{6!}{3! \cdot 3!} = 20$ to hand these out but some cases don't work. Specifically, the cases $a_1 = 1, a_2 = 2, a_3 = 1, a_4 = 2, a_5 = 1, a_6 = 2$ and $a_1 = 2, a_2 = 1, a_3 = 2, a_4 = 1, a_5 = 2, a_6 = 1$ don't work(perform the bash and see that these indeed don't work). Thus, we have $20 - 2 = 18$ working cases for $a_7 = 3$ but we need to multiply that by $7$ to get $7 \cdot 18 = 126$ ways.

Now I claim that each of the triples $(x, y, z) = (3, 2, 2), (0, 5, 2), (1, 6, 0)$ each have $0$ cases.

Claim Case 1: $(x, y, z) = (3, 2, 2)$

Note that since there are $4$ numbers that aren't equal to $3$ and since there are exactly $3$ factors in a triple by Pigeonhole Principle, there must be exactly $1$ triple that does not contain a $3$ at all which means that $1$ factor is not divisible by $3$. Everything else is divisible by $3$ and so the entire product sum isn't divisible by $3$ in this case. So we have $0$ cases.

Claim Case 2: $(x, y, z) = (0, 5, 2)$

Since we have $5$ numbers that are equal to $1$, we can again apply Pigeonhole Principle and get that we could have $2$ triples that only consist $1$(since each triple has $3$ factors, $2$ triples would cover $2 \cdot 3 = 6$ factors). So everything else consists of $2$'s. Here, the remainder modulo $3$ is $2 \cdot (1 \cdot 1 \cdot 1) = 2$ so this isn't divisible by $3$. Similarly by Pigeonhole Principle, we could have only $1$ triple that looks like this $(1, 2, 2)$ or $4$ triples that look like this $(1, 1, 2)$. Neither of these cases will give a remainder of $0$ modulo $3$ so again we have $0$ cases for this case.

Claim Case 3: $(x, y, z) = (1, 6, 0)$

Assume this case does indeed have a positive number of ways that work. WLOG, assume $a_1 = 3$ and we will multiply the result by $7$ because of symmetry. Then, since exactly $3$ triples contain $a_1$, those triples would vanish and we would be left with $4$ triples all of which have factors that are only equal to $1$. This gives $4 \cdot (1 \cdot 1 \cdot 1) = 4$ which isn't divisible by $3$. Thus, we have $0$ cases for $a_1 = 3$ and ultimately $0 \cdot 7 = 0$ cases in total.

We have shown our claim is true. Finally, the answer is

$1 + 28 + 28 + 42 + 168 + 126 + 0 + 0 + 0 = \boxed{393}$.

~ilikemath247365