AIME 2026 I · 第 7 题

Solution 1

We note that the function must cycle groups of non-overlapping subsets of A. Since $\pi^6(a)=a$, the cycles must be factors of 6, so they can be 1, 2, 3, or 6. We can split the 6 elements of A into: one cycle of 6, two cycles of 3, three cycles of 2, six cycles of 1, one cycle of 3 and three cycles of 1, one cycle of 2 and four cycles of 1, two cycles of 2 and two cycles of 1, and one cycle of 3 and one cycle of 2 and one cycle of 1.

Case 1:

$\binom{6}{6}*\frac{6!}{6}=5!=120$

Case 2:

$\frac{\binom{6}{3}\binom{3}{3}}{2!}*(\frac{3!}{3})^2=40$

Case 3:

$\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}*(\frac{2!}{2})^3=15$

Case 4:

For six cycles of 1, there is only $1$ way.

Case 5:

$\binom{6}{3}*\frac{3!}{3}=40$

Case 6:

$\binom{6}{2}*\frac{2!}{2}=15$

Case 7:

$\frac{\binom{6}{2}\binom{4}{2}}{2!}*(\frac{2!}{2})^2=45$

Case 8:

$\binom{6}{3}\binom{3}{2}*\frac{3!}{3}*\frac{2!}{2}=120$

Summing up our cases, we get $120+40+15+1+40+15+45+120=\boxed{396}$

~Reese's Pieces

Solution 2

We have that the cycles of our function must be in lengths which are factors of 6, so that $\pi^6(a) = a$ We can then case on what cycles there are

Case 1: 1 cycle of Length 6:

We have $5!=120$ ways to do this (Just think circular rotations)

Case 2: 2 cycles of length 3:

We have $\binom63$ ways to choose what is in the first cycle, then $2! = 2$ ways to arrange it. Then 3 choose 3 times 2!. We have to divide by 2! due to symmetry. We get $\frac{20 * 2! * 1 * 2!}{2!} = 40$

Case 3: One cycle of length 3, one of length 2 and one length 1:

We have $\binom63$ ways to choose what is in the first cycle, then $2! = 2$ ways to arrange it.Then $\binom32\cdot1!$ for length 2 cycle and then 1 way for the length 1 cycle. This is $20 \cdot 2 \cdot 3 = 120$.

Case 4: 1 Cycle of Length 3 and 3 of length 1:

We have $\binom63$ ways to choose what is in the first cycle, then $2! = 2$ ways to arrange it. Then only one way for the other 3 cycles because each has to go with itself. This is $20 \cdot 2! = 40$ ways.

Case 5: 3 Cycles of Length 2:

We have $\binom62$ ways to choose what is in the first cycle, then $1! = 1$ ways to arrange it. Then $\binom42$ for the next cycle and $\binom22$ for the last. We divide by $3! = 6$ due to symmetry. We get $15 \cdot 6/6 = 15$ ways.

Case 6: 2 cycles of length 2, 2 cycles of length 1:

We have $\binom62$ ways to choose what is in the first cycle, then $1! = 1$ ways to arrange it. Then $\binom42$ for the next cycle and one way for the last two. We divide by $2! = 2$ due to symmetry. We get $15 \cdot 6/2 = 45$ ways.

Case 7: 1 cycle of length 2, 4 cycles of length 1:

We have $\binom62=15$ ways to choose what is in the first cycle, then $1! = 1$ ways to arrange it. There is only one way for the other 4 cycles. So we have $15$ ways.

Case 8: 6 cycles of length 1:

There is just one way.

Adding these up we get $\boxed{396}$

~Aarav22

~Boywithnuke (Extremely minor edit)

Solution 3

Alternatively, we can use complementary counting. Obviously it must be a permutation, giving $6!=720$ total possibilities. However, note that if there is a cycle of length 4 or 5, since they aren't a factor of 6, the problem statement will not be satisfied. This is because if there is a cycle of 4, for the numbers involved in the cycle, $\pi(\pi(\pi(\pi(a))))=a.$ Similarly for numbers involved in a cycle of 5, $\pi(\pi(\pi(\pi(\pi(a)))))=a$. We can subtract the possibilities where cycles of $\{4,1,1\},\{4,2\},\{5,1\}$ are formed. We get

$$720-\binom64\cdot3!-\binom64\cdot3!-\binom65\cdot4!=720-90-90-144=\boxed{396}$$ Dinga linga doo, edited by mathdivine

Solution 4 (Partitioning and Circular Casework)

The number of cyclic connotations is the number of partitions of $6$ using only the numbers $1, 2, 3,$ or $6$. These partitions represent the number of cycles, as partition count is cyclic.

Thus, we have 8 different cycles:

$$1+1+1+1+1+1$$ $$2+1+1+1+1$$ $$2+2+1+1$$ $$2+2+2$$ $$3+1+1+1$$ $$3+2+1$$ $$3+3$$ $$\text{and}~ 6$$ Label them top down cases $A, B, C, \dots, H$.

Case A: $1+1+1+1+1+1$

We need all cycles of $1$. This implies that $f(a)=a$ for all $6$ elements. There is $1$ mapping.

Case B: $2+1+1+1+1$

We need a cycle of $2$ and $4$ cycles of $1$. This implies that for $4$ values, we have $f(a)=a$, which is done in $\binom{6}{4} = 15$ ways. The remaining two values map on to one another, giving us $(2-1)!$ cases and the number of ways is $15 \cdot 1 = 15$.

Case C: $2+2+1+1$

We need $2$ cycles of $2$ and $2$ cycles of $1$. This implies that for $2$ values, we have $f(a)=a$, which is done in $\binom{6}{2} = 15$ ways. For the remaining two cycles, we distribute upon $4$ numbers. Then, say the cycles are in an ordered pair $((a,b),(c,d))$. Then, $((a,b),(c,d)) = ((c,d),(a,b))$, thus we choose two elements from the remaining $4$, and divide by $2$ permutations, giving us a total of $3$ different sets, and a total of $15 \cdot 3 = 45$ cases.

Case D: $2+2+2$

We have $3$ cycles of $2$. Imagine the ordered triplet $((a,b),(c,d),(e,f))$, in which each is uniquely determined from the set $\{1, 2, 3, 4, 5, 6\}$. We choose $2$ from $6$ elements, $2$ from the remaining $4$, and the last two are determined uniquely. Then, we have that all $3! = 6$ permutations of this triplet are the same. Thus, we divide by $6$ and obtain $\frac{90}{6} = 15$ cases.

Case E: $3+1+1+1$

We have $3$ one cycles and one $3$ cycle. For the first, we just determine

$$\binom{6}{3} = 20$$ cases. For the one cycle, we order the remaining as $(a,b,c)$. The orientation remains the same except when the two endpoints ($a$ and $c$) are switched. Thus, there are $(3-1)! = 2! = 2$ cases that work here, for a total

$$20 \cdot 2 = 40$$ cases.

Case F: $3+2+1$

We have one $3$ cycle, one $2$ cycle, and one $1$ cycle. There are $6$ ways to choose the $1$ cycle. From the remaining $5$, we choose $2$, giving $10$ cases. However, we must be careful to not undercount and naively divide by $2$, as when we select only one pair. For the remaining $3$, we continue as in previous case to obtain $2$ ways (as once again, we have $(a,b,c)$, and swapping $a$ and $c$ given they map to one another is the same). Thus there is $6 \cdot 10 \cdot 2 = 120$ cases.

Case G: $3+3$

Two cycles of $3$. For the first, we have $\binom{6}{3} = 20$ ways. For the second, only $2$. Then, we have $40$ cases.

Case H: $6$

One cycle of $6$. We have $(a,b,c,d,e,f)$, but swapping the endpoints are the same due to cyclic rotations. For the six elements there are $6$ cases for the two that get swapped, and thus we divide the overcount by $6$. Then, there are $(6-1)! = 5! = 120$ possibilities.

Alas, we have to add all $8$ partitions, to obtain

$$1 + 15 + 45 + 15 + 40 + 120 + 40 + 120 = 396$$ The answer is clearly $\boxed{396}$.

~Pinotation

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2026 AIME I #7

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