AIME 2026 I · 第 9 题

Solution 1

Before we continue, let's rephrase the problem as such: There are six slots labeled "$A, B, C, D, E, F$" to put plates in (it's just easier to think about it as a row instead of a die). We know that $2, 4, 6$ must be the top plates. WLOG, let them be the top plates of slots $A, B, C$. The game mechanics make it so that if, say, plates $1, 2,$ and $5$ are in a particular slot, we must have plate $1$ underneath plate $2$ underneath plate $5$ (because she puts the stickers on from lowest number to highest).

Let's count the total number of ways we can have $2,4,6$ showing on the top. Plate $5$ has $4$ options: hide under plate $6$ or take an empty spot. Similarly, plate $3$ has $5$ options and plate $1$ has all $6$ options, making $120$ total ways. For there to be exactly one face left blank, we have three simple cases:

1. Plates $1$ and $3$ occupy the unused spaces: $6\cdot 1=6$

2. Plates $1$ and $5$ occupy the unused spaces: $6\cdot 3=18$

3. Plates $3$ and $5$ occupy the unused spaces: $6\cdot 5=30$

Hence the answer is $\frac{6+18+30}{120}=\frac{9}{20}\implies \boxed{\textbf{029}}$.

~Mathkiddie

Solution 2

Let the six faces of the die be labeled abstractly, and let $f_i$ denote the face that lands on top when sticker $i$ is placed. A sticker is visible at the end exactly when that face is never used again later.

We condition on the event that stickers $2,4,6$ are visible. This means $f_3,f_4,f_5,f_6 \neq f_2$ and $f_5,f_6 \neq f_4$, so in particular $f_2,f_4,f_6$ are all different faces.

First count all outcomes consistent with this condition. Choose $f_2$ in $6$ ways, then $f_4 \neq f_2$ in $5$ ways, then $f_6 \neq f_2,f_4$ in $4$ ways. Next choose $f_3 \neq f_2$ in $5$ ways, $f_5 \neq f_2,f_4$ in $4$ ways, and finally $f_1$ freely in $6$ ways. Thus the total number of outcomes with all even stickers visible is

$$6\cdot5\cdot4\cdot5\cdot4\cdot6=14400.$$ Now count those outcomes that leave exactly one face blank. Since $f_2,f_4,f_6$ are distinct, they occupy three faces; call them $A,B,C$. The remaining faces are $D,E,F$. To have exactly one blank face at the end, among $D,E,F$ exactly two must appear in $\{f_1,f_3,f_5\}$.

There are $6\cdot5\cdot4=120$ choices for $(A,B,C)$. Fix such a choice. Choose which of $D,E,F$ is the blank face; there are $3$ choices. Without loss of generality, suppose $F$ is blank. Then $f_3\in\{B,C,D,E\}$, $f_5\in\{C,D,E\}$, and $f_1\in\{A,B,C,D,E\}$. This gives $4\cdot3\cdot5=60$ possibilities. From these subtract the cases where $D$ is never used ($24$ cases) and where $E$ is never used ($24$ cases), and add back the cases where neither $D$ nor $E$ is used ($6$ cases). Hence the number of valid triples $(f_1,f_3,f_5)$ is

$$60-24-24+6=18.$$ Multiplying by the $3$ choices of the blank face gives $54$ favorable outcomes for each fixed $(A,B,C)$.

Therefore the number of favorable outcomes is

$$120\cdot54=6480.$$ The desired conditional probability is

$$p=\frac{6480}{14400}=\frac{9}{20}.$$ Thus $m+n=9+20=\boxed{029}.$

~Gray_Wolf

Solution 3

To solve this problem, we can count two things: how many ways there can be one blank face and all the even digits visible, and how many ways all the even digits can be visible. For simplicity, we will consider rotations and reflections to be different.

To compute the first value, we can split it into two cases; the sole covered value can either be covered by an odd sticker or by an even sticker.

Case $1$: The one covered sticker is covered by an odd sticker.

Place the odd stickers first, since they cannot cover any even stickers. There are $3$ different pairs of stickers that can work here: the $3$ covers the $1$, the $5$ covers the $1$, or the $5$ covers the $3$. In any of these cases, there are 6 ways to place the covered sticker and 5 ways to place the sticker that won't cover it. This yields $3 \cdot 6 \cdot 5 = 90$ ways to place the odd stickers. There are now $4$ open faces of the die, and $3$ even numbers to place. This can be done in $4 \cdot 3 \cdot 2 = 24$ ways. This yields a total of $90 \cdot 24 = 2160$ ways for this case.

Case $2$: The one covered sticker is covered by an even sticker.

Once again, place the odd stickers first. This time, there can be no coverups, so the amount of ways to do this is merely $6 \cdot 5 \cdot 4 = 120$. For the even stickers, we need to keep in mind that we cannot, for instance, have the $2$ cover the $3$; the covering sticker must be greater than the covered sticker. There are $6$ total pairs that work here; $2$ covers $1$, $4$ covers $1$, $6$ covers $1$, $4$ covers $3$, $6$ covers $3$, and $6$ covers $5$. In all of these cases, once the covering sticker is placed, the remaining $2$ stickers have $3$ open faces left, yielding $3 \cdot 2 = 6$ total placements. This yields a total of $120 \cdot 6 \cdot 6 = 4320$ ways for this case. Adding these two yields a total of $6480$ for the first value.

To compute the second value, go in sequential order. There are $6$ ways to place the $1$, $6$ ways to place the $2$ (since it could cover up the $1$), $5$ ways to place the $3$ (since it can't cover up the $2$), $5$ ways to place the $4$, $4$ ways to place the $5$, and $4$ ways to place the $6$, yielding a total of $6 \cdot 6 \cdot 5 \cdot 5 \cdot 4 \cdot 4 \cdot = 14400$ ways for this to be accomplished.

Finally, to obtain the requested probability, we merely divide the former by the latter; $\frac{m}{n} = \frac{6480}{14400} = \frac{9}{20}$. $m + n = \boxed{029}$.

~ARP10

Solution 4

We will first determine the probability that all even stickers are left uncovered. We can place stickers $1$ and $2$ anywhere, $3$ and $4$ cannot cover $2$, and $5$ and $6$ cannot cover $2$ or $4$. Thus, the probability is $1\cdot 1\cdot \frac{5}{6}\cdot \frac{5}{6} \cdot \frac{2}{3}\cdot \frac{2}{3}=\frac{25}{81}$.

We then split the condition that there is exactly one blank face into cases. Notice that exactly one sticker must be covered at some point.

Case 1: A sticker is covered on the second roll

Sticker $2$ covers sticker $1$, and the rest must go on blank faces, giving a probability of $\frac{1}{6}\cdot \frac{5}{6} \cdot \frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{3}= \frac{5}{324}$.

Case 2: A sticker is covered on the third roll

Sticker $3$ must go on top of sticker $1$, so the probability is $\frac{5}{6}\cdot \frac{1}{6} \cdot \frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{3}= \frac{5}{324}$.

Case 3: A sticker is covered on the fourth roll

Sticker $4$ can cover either sticker $1$ or $3$, so we get a probability of $\frac{5}{6}\cdot \frac{2}{3} \cdot \frac{1}{3}\cdot \frac{1}{2} \cdot \frac{1}{3}= \frac{5}{162}$.

Case 4: A sticker is covered on the fifth roll

Sticker $5$ can cover sticker $1$ or sticker $3$, so the probability in this case is $\frac{5}{6}\cdot \frac{2}{3} \cdot \frac{1}{2}\cdot \frac{1}{3} \cdot \frac{1}{3}= \frac{5}{162}$.

Case 5: A sticker is covered on the final roll

Sticker $6$ covers sticker $1$, $3$, or $5$, giving us $\frac{5}{6}\cdot \frac{2}{3} \cdot \frac{1}{2}\cdot \frac{1}{3} \cdot \frac{1}{2}= \frac{5}{108}$.

Adding our probabilities for all cases, the probability that all even stickers are visible and there is exactly one blank face is $\frac{5}{36}$. Dividing this by $\frac{25}{81}$ means the desired conditional probability is $\frac{405}{900}$, which simplifies to $\frac{9}{20}$, so our final answer is $9+20=\boxed{029}$.

~joiceeliu

Solution 5

We will find the probability that all even numbered stickers are visible and the probability that exactly one face has been left blank and all even numbered stickers are visible.

Notice that there are 6 ways to place 1 and 2. However, to place 3 it cannot be on top of 2, so there are 5 ways. Similarly, there are 5, 4 and 4 ways to place 4, 5 and 6. This means that there are $6*6*5*5*4*4$ ways to place the stickers such that all even numbered stickers are visible, and the probability is $\frac{6*6*5*5*4*4}{6^4}$.

Notice that for one face to be blank and all even numbered stickers to be visible, there has to be a face with two overlapping stickers such that the bottom sticker is odd. There are 6 ways to choose the blank face and 5 ways to choose the overlapping face. Notice that the number on the top sticker of the overlapping stickers has to be more that the number on the bottom sticker, so we will do casework on the bottom sticker.

Case 1. The bottom sticker is 1. Then, all arrangements of the visible stickers are possible giving $5!=120$ arrangements.

Case 2. The bottom sticker is 3. There are 4 ways to place the 2, and 3 ways to place the 1 because it cannot be on top of the 3 nor the 2. There are 3 spots remaining, so 4, 5 and 6 can be permuted in these spots to get $3!=6$ arrangements. In total, there are $4*3*6=72$ arrangements in this case.

Case 3. The bottom sticker is 5. Then, the sticker above the 5 has to be 6, and the 1, 2, 3 and 4 have to not be on the 5 or 6, giving $4!=24$ arrangements.

Thus, there are $6*5*(120+72+24)=6*5*216$ ways to place the stickers such that one face is blank and all even numbered stickers to be visible. Hence, the probability is $\frac{6*5*216}{6^4}$.

By conditional probability, the answer is the quotient of these two values, so it is

$$\frac{\frac{6*5*216}{6^4}}{\frac{6*6*5*5*4*4}{6^4}}=\frac{216}{6*5*4*4}=\frac{36}{5*4*4}=\frac{9}{20},$$ answer extraction gives $9+20=\boxed{029}$.

~steve4916

Solution 6 (5 minutes)

Consider the $6$ sides of the dice ordered from $1$ to $6$ (for simplicity, order matters here). We will count the number of cases that satisfies each condition and divide them.

To let all the even numbers visible, you'll have $6$ choices for $1$ and $2$ because you don't care about whether $1$ gets overlapped or not. You then have $5$ choices for $3$ and $4$, because you cannot put over $2$. You then have $4$ choices for $5$ and $6$, because $2$ and $4$ cannot be overlapped. This gives a total of $6\times6\times5\times5\times4\times4$ ways of doing this.

To let all the even numbers visible AND exactly $1$ blank face, we consider putting all the even numbers on the dice first without overlapping, and then insert $1$ of the odd numbers behind $1$ of the larger numbers, and then cover the remaining $3$ faces with $2$ odd numbers (all cases are valid because at most you will have an even number above an odd number which is valid for any combination). There are $6\times5\times4$ ways of choosing $3$ faces to put even numbers, $5 + 3 + 1$ ways of inserting an odd behind a larger number, and finally $3\times2$ ways of putting the remaining odd numbers, for a total of $6\times5\times4\times3\times3\times3\times2$ ways of doing this.

Therefore, the desired probability is $\frac{6\times5\times4\times3\times3\times3\times2}{6\times6\times5\times5\times4\times4}=\frac{9}{20}$, yielding an answer of $\boxed{029}$.

~metrixgo (Mistake corrected by BraveCobra22, It’s not true that any even can cover any odd, not that only even can cover odd. These too mistakes numerically cancel and still give the right answer.)

Solution 7 (Similar to solution 4)

Let's first calculate the total number of cases criteria. This shouldn't be too difficult.

There are $6$ ways to put number $1$. Then, we don't care where number $2$ goes, as it can cover $1$ or go away from $1$, so $6$ ways. Similarly, for $3$, we cannot have it cover $2$. But, it doesn't matter if it covers $1$ or not, so $5$ ways. $4$ is also $5$ ways. $5$ is then $4$ ways, and $6$ is also $4$ ways. The total number of cases is $6^2 \times 5^2 \times 4^2$.

Now, onto the criterion. We need that one face must be empty, which implies that one number must be covered. This number can be either $1, 3,$ or $5$. We could do casework, or notice something intriguing.

Take $5$. We see that the only number that can cover it (as we go in numerical order), is $6$. Then, the total number of ways is $6 \times 5 \times 4 \times 3 \times 2 \times 1$, or $6!$. (We have $6 \times 5$ and $4 \times 3$ due to the fact that none of $1$ or $3$ may be covered). The number of cases in this is $6!$.

Then, notice that if we take $3$, only $4, 5,$ or $6$ may cover it. Then, the number of cases is $6 \times 5 \times 4 \times 1 \times 3 \times 2 + 6 \times 5 \times 4 \times 3 \times 1 \times 2 + 6 \times 5 \times 4 \times 3 \times 2 \times 1$, or $3 \times 6!$. All these cases of such are just $6!$. (As an exercise to the reader, I leave it to you to prove how. Check sub-solution for answer).

Then, for $1$, we have that it may be covered by $2, 3, 4, 5,$ or $6$. Thus, we have the total number of cases as $5 \times 6!$.

Our conditional probability $p$ is then

$$\frac{1 \times 6! + 3 \times 6! + 5 \times 6!}{6 \times 6 \times 5 \times 5 \times 4 \times 4}, \text{ or } \frac{6!(9)}{6 \times 6 \times 5 \times 5 \times 4 \times 4}, \text{ or } \frac{9}{5 \times 4}, \text{ or } \frac{9}{20}.$$ Then, our answer is $9+20 = \boxed{029}$.

~Pinotation

Answer to Reader Problem

Consider a die with $n$ faces. Note that when we map a number to fill in the gaps of the others, the remaining numbers cannot map onto the other, or we will break the rule that five faces are filled. Thus, we must have that the remaining numbers fall into the open slots. However, the remaining slots for such numbers include the new number itself, so we subtract 1 from the count. Thus, we are just left with $n!$. Now, notice that if we choose an odd number $k$ to replace, any integer greater than $k$ but less than or equal to $n$ can cover $k$ up. Thus, the result is just $n - k$ slots for each of the $6!$ permutations. Thus the number of cases obtained is $(n - k)(n!)$. $\square$

~Pinotation

Solution 8

I will denote $\substack{\textstyle i \\ \textstyle j}$ as $j$ getting replaced by $i$. Note that the conditional probability theorem states that $P(A|B) = \frac{P(A \cap B)}{P(B)}$ where $A$ is the event of just one face staying blank and $B$ is the event of all the event numbers being shown. First, we will find the numerator. Note that for any $i < j$, the event of $\substack{\textstyle i \\ \textstyle j}$ happening is impossible. This is because as $i < j$, $i$ will get placed on the die before $j$ will and therefore it is impossible that $i$ can replace $j$ in this case. Knowing this, we realize that for the numerator, we need one face to stay blank and all the even numbers to show. Thus, we can break it up into the odd and even numbers namely the stickers $1, 3, 5$ and the stickers $2, 4, 6$. Either an odd sticker can replace another odd sticker or an even sticker can replace an odd sticker. We can list out the following cases knowing this

1 $\substack{\textstyle 5 \\ \textstyle 3}$ 2 4 6

1 $\substack{\textstyle 3 \\ \textstyle 5}$ 2 4 6

$\substack{\textstyle 5 \\ \textstyle 1}$ 2 4 6

$\substack{\textstyle 3 \\ \textstyle 1}$ 2 4 6

$\substack{\textstyle 1 \\ \textstyle 5}$ 2 4 6

$\substack{\textstyle 1 \\ \textstyle 3}$ 2 4 6

$\substack{\textstyle 2 \\ \textstyle 1}$ 3 5 4 6

1 $\substack{\textstyle 2 \\ \textstyle 3}$ 5 4 6

1 3 $\substack{\textstyle 2 \\ \textstyle 5}$ 4 6

$\substack{\textstyle 4 \\ \textstyle 1}$ 3 2 4 6

1 $\substack{\textstyle 4 \\ \textstyle 3}$ 5 4 6

1 3 $\substack{\textstyle 4 \\ \textstyle 5}$ 2 6

$\substack{\textstyle 6 \\ \textstyle 1}$ 3 5 2 4

1 $\substack{\textstyle 6 \\ \textstyle 3}$ 5 2 4

1 3 $\substack{\textstyle 6 \\ \textstyle 5}$ 2 4

These are all the cases of which either an odd replaces an odd or an even replaces an odd. However, recall that for any $i < j$, $i$ replacing $j$ is impossible. Therefore we can remove some cases and get our final case list

1 $\substack{\textstyle 5 \\ \textstyle 3}$ 2 4 6

$\substack{\textstyle 5 \\ \textstyle 1}$ 2 4 6

$\substack{\textstyle 3 \\ \textstyle 1}$ 2 4 6

$\substack{\textstyle 2 \\ \textstyle 1}$ 3 5 4 6

$\substack{\textstyle 4 \\ \textstyle 1}$ 3 2 4 6

1 $\substack{\textstyle 4 \\ \textstyle 3}$ 5 4 6

$\substack{\textstyle 6 \\ \textstyle 1}$ 3 5 2 4

1 $\substack{\textstyle 6 \\ \textstyle 3}$ 5 2 4

1 3 $\substack{\textstyle 6 \\ \textstyle 5}$ 2 4

This gives us $9$ cases. Now, note that for each $\substack{\textstyle i \\ \textstyle j}$, we can treat this as one block. The reason why we're doing this is so that we can count the number of permutations of each of the $9$ cases. Treating these as one block makes it easier to count the permutations. For each case, we have $6$ ways to arrange that one block and $5 \cdot 4 \cdot 3 \cdot 2$ to arrange the rest of the numbers. We can therefore write $9 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2$ total ways. Thus, $P(A \cap B) = \frac{9 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{6^6}$.

Now, finding $P(B)$ is easier. Note that since we're just counting the probability that the even numbers are showing, we can just note that this is basically finding the probability that the even numbers replace at least one of the odd numbers. Essentially, we don't care if the odd numbers get replaced by the even numbers or not. So we have $6$ ways to place the $1$. We have $6$ ways to place the $2$ since we don't care if $2$ covers the $1$ or not. For $3$, we can place it anywhere but the place where the $2$ is there so we have $5$ ways. Similarly, $5$ ways to place the $4$. Finally, there are $4$ ways to place the $5$ and $4$ ways to place the $6$ giving us $6 \cdot 6 \cdot 5 \cdot 5 \cdot 4 \cdot 4 = 6 \cdot 5 \cdot 4 \cdot 120$. Thus, $P(B) = \frac{6 \cdot 5 \cdot 4 \cdot 120}{6^6}$.

Finally, the answer is

$\frac{\frac{9 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{6^6}}{\frac{6 \cdot 5 \cdot 4 \cdot 120}{6^6}} = \frac{9}{20} \implies 9 + 20 = \boxed{29}$.

~ilikemath247365

Video Solution (Fast and Easy 🔥🚀)

2026 AIME I #9

By piacademyus.org