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2024年 AIME I 数学邀请赛答案解析,看看你水平如何？

2024年 AIME I 数学邀请赛真题

Problem 1

Every morning Aya goes for a $9$-kilometer-long walk and stops at a coffee shop afterwards. When she walks at a constant speed of $s$ kilometers per hour, the walk takes her 4 hours, including $t$ minutes spent in the coffee shop. When she walks $s+2$ kilometers per hour, the walk takes her 2 hours and 24 minutes, including $t$ minutes spent in the coffee shop. Suppose Aya walks at $s+\frac{1}{2}$ kilometers per hour. Find the number of minutes the walk takes her, including the $t$ minutes spent in the coffee shop.

Solution 1

$\frac{9}{s} + t = 4$ in hours and $\frac{9}{s+2} + t = 2.4$ in hours.

Subtracting the second equation from the first, we get,

$\frac{9}{s} - \frac{9}{s+2} = 1.6$

Multiplying by $(s)(s+2)$, we get

$9s+18-9s=18=1.6s^{2} + 3.2s$

Multiplying by 5/2 on both sides, we get

$0 = 4s^{2} + 8s - 45$

Factoring gives us

$(2s-5)(2s+9) = 0$, of which the solution we want is $s=2.5$.

Substituting this back to the first equation, we can find that $t = 0.4$ hours.

Lastly, $s + \frac{1}{2} = 3$ kilometers per hour, so

$\frac{9}{3} + 0.4 = 3.4$ hours, or $\framebox{204}$ minutes

Solution 2

The amount of hours spent while walking on the first travel is $\frac{240-t}{60}$. Thus, we have the equation $(240-t)(s) = 540$, and by the same logic, the second equation yields $(144-t)(s+2) = 540$. We have $240s-st = 540$, and $288+144s-2t-st = 540$. We subtract the two equations to get $96s+2t-288 = 0$, so we have $48s+t = 144$, so $t = 144-48s$, and now we have $(96+48s)(s) = 540$. The numerator of $s$ must evenly divide 540, however, $s$ must be less than 3. We can guess that $s = 2.5$. Now, $2.5+0.5 = 3$. Taking $\frac{9}{3} = 3$, we find that it will take three hours for the 9 kilometers to be traveled. The t minutes spent at the coffeeshop can be written as $144-48(2.5)$, so t = 24. $180 + 24 = 204$

Problem 2

There exist real numbers $x$ and $y$, both greater than 1, such that $\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10$. Find $xy$.

Solution 1

By properties of logarithms, we can simplify the given equation to $x\log_xy=4y\log_yx=10$. Let us break this into two separate equations:

\[x\log_xy=10\]\[4y\log_yx=10.\]We multiply the two equations to get:\[4xy\left(\log_xy\log_yx\right)=100.\]

Also by properties of logarithms, we know that $\log_ab\cdot\log_ba=1$; thus, $\log_xy\cdot\log_yx=1$. Therefore, our equation simplifies to:

\[4xy=100\implies xy=\boxed{025}.\]

Solution 2

Convert the two equations into exponents:

\[x^{10}=y^x~(1)\]\[y^{10}=x^{4y}~(2).\]

Take $(1)$ to the power of $\frac{1}{x}$:

\[x^{\frac{10}{x}}=y.\]

Plug this into $(2)$:

\[x^{(\frac{10}{x})(10)}=x^{4(x^{\frac{10}{x}})}\]\[{\frac{100}{x}}={4x^{\frac{10}{x}}}\]\[{\frac{25}{x}}={x^{\frac{10}{x}}}=y,\]

So $xy=\boxed{025}$

Solution 3

Similar to solution 2, we have:

$x^{10}=y^x$ and $y^{10}=x^{4y}$

Take the tenth root of the first equation to get

$x=y^{\frac{x}{10}}$

Substitute into the second equation to get

$y^{10}=y^{\frac{4xy}{10}}$

This means that $10=\frac{4xy}{10}$, or $100=4xy$, meaning that $xy=\boxed{25}$.

Solution 4

The same with other solutions, we have obtained $x^{10}=y^x$ and $y^{10}=x^{4y}$. Then, $x^{10}y^{10}=y^xx^{4y}$. So, an obvious solution is to have $x^{10}=x^{4y}$ and $y^{10}=y^{x}$. Solving, we get $x=10$ and $y=2.5$.

Solution 5

Using the first expression, we see that $x^{10} = y^x$. Now, taking the log of both sides, we get $\log_y(x^{10}) = \log_y(y^x)$. This simplifies to $10 \log_y(x) = x$. This is still equal to the second equation in the problem statement, so $10 \log_y(x) = x = 4y \log_y(x)$. Dividing by $\log_y(x)$ on both sides, we get $x = 4y = 10$. Therefore, $x = 10$ and $y = 2.5$, so $xy = \boxed{25}$.

Solution 6

Put\[y=x^a\].We see:\[ax=10\]and\[4x^a/a=10\]which gives rise to\[xy=25\]which is the required answer.

Problem

Alice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes either $1$ token or $4$ tokens from the stack. Whoever removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ for which there exists a strategy for Bob that guarantees that Bob will win the game regardless of Alice's play.

Solution 1

Let's first try some experimentation. Alice obviously wins if there is one coin. She will just take it and win. If there are 2 remaining, then Alice will take one and then Bob will take one, so Bob wins. If there are $3$, Alice will take $1$, Bob will take one, and Alice will take the final one. If there are $4$, Alice will just remove all $4$ at once. If there are $5$, no matter what Alice does, Bob can take the final coins in one try. Notice that Alice wins if there are $1$, $3$, or $4$ coins left. Bob wins if there are $2$ or $5$ coins left.

After some thought, you may realize that there is a strategy for Bob. If there is n is a multiple of $5$, then Bob will win. The reason for this is the following: Let's say there are a multiple of $5$ coins remaining in the stack. If Alice takes $1$, Bob will take $4$, and there will still be a multiple of $5$. If Alice takes $4$, Bob will take $1$, and there will still be a multiple of $5$. This process will continue until you get $0$ coins left. For example, let's say there are $205$ coins. No matter what Alice does, Bob can simply just do the complement. After each of them make a turn, there will always be a multiple of $5$ left. This will continue until there are $5$ coins left, and Bob will end up winning.

After some more experimentation, you'll realize that any number that is congruent to $2$ mod $5$ will also work. This is because Bob can do the same strategy, and when there are $2$ coins left, Alice is forced to take $1$ and Bob takes the final coin. For example, let's say there are $72$ coins. If Alice takes $1$, Bob will take $4$. If Alice takes $4$, Bob will take $1$. So after they each make a turn, the number will always be equal to $2$ mod $5$. Eventually, there will be only $2$ coins remaining, and we've established that Alice will simply take $1$ and Bob will take the final coin.

So we have to find the number of numbers less than or equal to $2024$ that are either congruent to $0$ mod $5$ or $2$ mod $5$. There are $404$ numbers in the first category: $5, 10, 15, \dots, 2020$. For the second category, there are $405$ numbers. $2, 7, 12, 17, \dots, 2022$. So the answer is $404 + 405 = \boxed{809}$

Solution 2

Let's use winning and losing positions, where $W$ marks a win for Alice.

$1$ coin: $W$

$2$ coins: $L$

$3$ coins: $W$

$4$ coins: $W$

$5$ coins: $L$

$6$ coin: $W$

$7$ coins: $L$

$8$ coins: $W$

$9$ coins: $W$

$10$ coins: $L$

$11$ coin: $W$

$12$ coins: $L$

$13$ coins: $W$

$14$ coins: $W$

$15$ coins: $L$

We can see that losing positions occur when $n$ is congruent to $0, 2 \mod{5}$ and winning positions occur otherwise. As $n$ ranges from $1$ to $2020$, $\frac{2}{5}$ of these values are losing positions where Bob will win. As $n$ ranges from $2021$ to $2024$, $2022$ is the only value where Bob will win. Thus, the answer is $2020\times\frac{2}{5}+1=\boxed{809}$

Solution 3

Denote by $A_i$ and $B_i$ Alice's or Bob's $i$th moves, respectively.

Case 1: $n \equiv 0 \pmod{5}$.

Bob can always take the strategy that $B_i = 5 - A_i$. This guarantees him to win.

In this case, the number of $n$ is $\left\lfloor \frac{2024}{5} \right\rfloor = 404$.

Case 2: $n \equiv 1 \pmod{5}$.

In this case, consider Alice's following strategy: $A_1 = 1$ and $A_i = 5 - B_{i-1}$ for $i \geq 2$. Thus, under Alice's this strategy, Bob has no way to win.

Case 3: $n \equiv 4 \pmod{5}$.

In this case, consider Alice's following strategy: $A_1 = 4$ and $A_i = 5 - B_{i-1}$ for $i \geq 2$. Thus, under Alice's this strategy, Bob has no way to win.

Case 4: $n \equiv 2 \pmod{5}$.

Bob can always take the strategy that $B_i = 5 - A_i$. Therefore, after the $\left\lfloor \frac{n}{5} \right\rfloor$th turn, there are two tokens leftover. Therefore, Alice must take 1 in the next turn that leaves the last token on the table. Therefore, Bob can take the last token to win the game. This guarantees him to win.

In this case, the number of $n$ is $\left\lfloor \frac{2024 - 2}{5} \right\rfloor +1 = 405$.

Case 5: $n \equiv 3 \pmod{5}$.

Consider Alice's following strategy: $A_1 = 1$ and $A_i = 5 - B_{i-1}$ for $i \geq 2$. By doing so, there will finally be 2 tokens on the table and Bob moves first. Because Bob has the only choice of taking 1 token, Alice can take the last token and win the game.

Therefore, in this case, under Alice's this strategy, Bob has no way to win.

Putting all cases together, the answer is $404 + 405 = \boxed{\textbf{(809) }}$.

Solution 4 (Grundy Values)

Since the game Alice and Bob play is impartial (the only difference between player 1 and player 2 is that player 1 goes first (note that games like chess are not impartial because each player can only move their own pieces)), we can use the Sprague-Grundy Theorem to solve this problem. We will use induction to calculate the Grundy Values for this game.

We claim that heaps of size congruent to $0,2 \bmod{5}$ will be in outcome class $\mathcal{P}$ (win for player 2 = Bob), and heaps of size equivalent to $1,3,4 \bmod{5}$ will be in outcome class $\mathcal{N}$ (win for player 1 = Alice). Note that the mex (minimal excludant) of a set of nonnegative integers is the least nonnegative integer not in the set. e.g. mex$(1, 2, 3) = 0$ and mex$(0, 1, 2, 4) = 3$.

$\text{heap}(0) = \{\} = *\text{mex}(\emptyset) = 0$

$\text{heap}(1) = \{0\} = *\text{mex}(0) = *$

$\text{heap}(2) = \{*\} = *\text{mex}(1) = 0$

$\text{heap}(3) = \{0\} = *\text{mex}(0) = *$

$\text{heap}(4) = \{0, *\} = *\text{mex}(0, 1) = *2$

$\text{heap}(5) = \{*, *2\} = *\text{mex}(1, 2) = 0$

$\text{heap}(6) = \{0, 0\} = *\text{mex}(0, 0) = *$

$\text{heap}(7) = \{*, *\} = *\text{mex}(1, 1) = 0$

$\text{heap}(8) = \{*2, 0\} = *\text{mex}(0, 2) = *$

$\text{heap}(9) = \{0, *\} = *\text{mex}(0, 1) = *2$

$\text{heap}(10) = \{*, *2\} = *\text{mex}(1, 2) = 0$

We have proven the base case. We will now prove the inductive hypothesis: If $n \equiv 0 \bmod{5}$, $\text{heap}(n) = 0$, $\text{heap}(n+1) = *$, $\text{heap}(n+2) = 0$, $\text{heap}(n+3) = *$, and $\text{heap}(n+4) = *2$, then $\text{heap}(n+5) = 0$, $\text{heap}(n+6) = *$, $\text{heap}(n+7) = 0$, $\text{heap}(n+8) = *$, and $\text{heap}(n+9) = *2$.$\text{heap}(n+5) = \{\text{heap}(n+1), \text{heap}(n+4)\} = \{*, *2\} = *\text{mex}(1, 2) = 0$$\text{heap}(n+6) = \{\text{heap}(n+2), \text{heap}(n+5)\} = \{0, 0\} = *\text{mex}(0, 0) = *$$\text{heap}(n+7) = \{\text{heap}(n+3), \text{heap}(n+6)\} = \{*, *\} = *\text{mex}(1, 1) = 0$$\text{heap}(n+8) = \{\text{heap}(n+4), \text{heap}(n+7)\} = \{*2, 0\} = *\text{mex}(2, 1) = *$$\text{heap}(n+9) = \{\text{heap}(n+5), \text{heap}(n+8)\} = \{0, *\} = *\text{mex}(0, 1) = *2$

We have proven the inductive hypothesis. QED.

There are $2020*\frac{2}{5}=808$ positive integers congruent to $0,2 \bmod{5}$ between 1 and 2020, and 1 such integer between 2021 and 2024. $808 + 1 = \boxed{809}$.

Solution 5 (no modular arithmetic)

We start with $n$ as some of the smaller values. After seeing the first 4 where Bob wins automatically, with trial and error we see that $2, 5, 7,$ and $10$ are spaced alternating in between 2 and 3 apart. This can also be proven with modular arithmetic, but this is an easier solution for some people. We split them into 2 different sets with common difference 5: {2,7,12 ...} and {5,10,15...}. Counting up all the numbers in each set can be done as follows:

Set 1 ${2,7,12...}$

$2024-2=2022$ (because the first term is two)

$\lfloor \frac{2024}{5} \rfloor = 404$

Set 2 ${5,10,15}$

$\lfloor \frac{2024}{5} \rfloor = 404$

And because we forgot 2022 we add 1 more.

$404+404+1=809$

Problem 3

Jen enters a lottery by picking $4$ distinct numbers from $S=\{1,2,3,\cdots,9,10\}.$ $4$ numbers are randomly chosen from $S.$ She wins a prize if at least two of her numbers were $2$ of the randomly chosen numbers, and wins the grand prize if all four of her numbers were the randomly chosen numbers. The probability of her winning the grand prize given that she won a prize is $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

This is a conditional probability problem. Bayes' Theorem states that\[P(A|B)=\dfrac{P(B|A)\cdot P(A)}{P(B)}\]

in other words, the probability of $A$ given $B$ is equal to the probability of $B$ given $A$ times the probability of $A$ divided by the probability of $B$. In our case, $A$ represents the probability of winning the grand prize, and $B$ represents the probability of winning a prize. Clearly, $P(B|A)=1$, since by winning the grand prize you automatically win a prize. Thus, we want to find $\dfrac{P(A)}{P(B)}$.

Let us calculate the probability of winning a prize. We do this through casework: how many of Jen's drawn numbers match the lottery's drawn numbers?

To win a prize, Jen must draw at least $2$ numbers identical to the lottery. Thus, our cases are drawing $2$, $3$, or $4$ numbers identical.

Let us first calculate the number of ways to draw exactly $2$ identical numbers to the lottery. Let Jen choose the numbers $a$, $b$, $c$, and $d$; we have $\dbinom42$ ways to choose which $2$ of these $4$ numbers are identical to the lottery. We have now determined $2$ of the $4$ numbers drawn in the lottery; since the other $2$ numbers Jen chose can not be chosen by the lottery, the lottery now has $10-2-2=6$ numbers to choose the last $2$ numbers from. Thus, this case is $\dbinom62$, so this case yields $\dbinom42\dbinom62=6\cdot15=90$ possibilities.

Next, let us calculate the number of ways to draw exactly $3$ identical numbers to the lottery. Again, let Jen choose $a$, $b$, $c$, and $d$. This time, we have $\dbinom43$ ways to choose the identical numbers and again $6$ numbers left for the lottery to choose from; however, since $3$ of the lottery's numbers have already been determined, the lottery only needs to choose $1$ more number, so this is $\dbinom61$. This case yields $\dbinom43\dbinom61=4\cdot6=24$.

Finally, let us calculate the number of ways to all $4$ numbers matching. There is actually just one way for this to happen.

In total, we have $90+24+1=115$ ways to win a prize. The lottery has $\dbinom{10}4=210$ possible combinations to draw, so the probability of winning a prize is $\dfrac{115}{210}$. There is actually no need to simplify it or even evaluate $\dbinom{10}4$ or actually even know that it has to be $\dbinom{10}4$; it suffices to call it $a$ or some other variable, as it will cancel out later. However, let us just go through with this. The probability of winning a prize is $\dfrac{115}{210}$. Note that the probability of winning a grand prize is just matching all $4$ numbers, which we already calculated to have $1$ possibility and thus have probability $\dfrac1{210}$. Thus, our answer is $\dfrac{\frac1{210}}{\frac{115}{210}}=\dfrac1{115}$. Therefore, our answer is $1+115=\boxed{116}$.

Shortcut

One may also use complimentary counting as a shortcut to calculate the probability of winning a prize, in which the cases are that either one number is shared or no numbers are shared. There are $4 \cdot { {10 - 4} \choose {4-1}} = 4 \cdot 20 = 80$ ways to choose the former and ${{10-4} \choose 4} = 15$ ways for the latter. Therefore, there are $95$ ways to NOT choose a prize, so there are $210-95 = 115$ ways to choose a prize, and the answer follows.

Solution 2

For getting all $4$ right, there is only $1$ way.

For getting $3$ right, there is $\dbinom43$ multiplied by $\dbinom61$ = $24$ ways.

For getting $2$ right, there is $\dbinom42$ multiplied by $\dbinom62$ = $90$ ways.

$\frac{1}{1+24+90}$ = $\frac{1}{115}$

Therefore, the answer is $1+115 = \boxed{116}$

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2024年 AIME I 数学邀请赛真题和答案解析

2024年 AIME I 数学邀请赛真题

问题 1

每天早上，Aya 都会步行一9 美元公里，然后在咖啡店结束。有一天，她以$s 美元每小时公里的速度行走，步行需要4 美元几个小时，包括$t 美元在咖啡店的几分钟。另一个早上，她以$s+2美元每小时公里的速度步行，步行需要2 美元数小时和24 美元分钟，包括$t 美元在咖啡店的几分钟。今天早上，如果她以$s+\frac12$每小时公里的速度步行，步行需要多少分钟，包括在咖啡店的$t 美元几分钟？

问题 2

实数$x 美元和$y 美元$x，y>1$满足 $\log_x（y^x）=\log_y（x^{4y}）=10.$ 的值是什么$xy 美元？

问题 3

Alice 和 Bob 玩以下游戏。一堆$n$代币摆在他们面前。玩家轮流，Alice 先走。在每个回合中，玩家从堆栈中移除1 美元一个或多个4 美元代币。移除最后一个令牌的玩家获胜。找到小于或等于的$n$正整数的数量，2024 美元以便有一种策略可以保证 Bob 获胜，而不管 Alice 的移动如何。

问题 4

Jen 通过从$S=\{1,2,3，\cdots，9,10\}.$4 美元随机选择的数字中选择4 美元不同的数字来参加抽奖 如果她的数字中至少有两个是2 美元随机选择的数字，$S.$她就会中奖，如果她的所有四个数字都是随机选择的数字，她就会赢得大奖。鉴于她中奖，她赢得大奖的概率是 $\tfrac{m}{n}$ 和 $m 美元 $n$ 是相对素数正整数。查找 $m+n$.

问题 5

矩形 $ABCD 美元 和 $EFGH 美元 绘制$D，E，C，F$为共线。此外，$A，D，H，G$所有 S 都位于一个圆圈上。如果 $BC=16，$ $AB=107，美元 $FG=17，美元 的长度$EF=184，$是多少 $CE 美元？

[asy] import graph;单位尺寸（0.1 厘米）;对 A = （0,0）;对 B = （70,0）;对 C = （70,16）;对 D = （0,16）;对 E = （3,16）;对 F = （90,16）;对 G = （90,33）;对 H = （3,33）;dot（A^^B^^C^^D^^E^^F^^G^^H）;标签（“$A$”， A， S）;标签（“$B$”， B， S）;标签（“$C$”， C， N）;标签（“$D$”， D， N）;标签（“$E$”， E， S）;标签（“$F$”， F， S）;标签（“$G$”， G， N）;标签（“$H$”， H， N）;绘制（E--D--A--B--C--E--H--G--F--C）;[/亚西]

问题 6

考虑沿着$8\times 8$网格上从左下角到右上角的线条的长度路径16 美元。找到正好改变方向四次的此类路径的数量，如下例所示。

[ASY] 尺寸（7.5cm）;usepackage（“抖音”）;label（“\begin{tikzpicture}[scale=.4]\draw（0,0）grid（8,8）;\draw[线宽=2，red]（0,0）--（2,0）--（2,3）--（5,3）--（5,8）--（8,8）;\end{tikzpicture}”，origin）;label（“\begin{tikzpicture}[scale=.4]\draw（0,0）grid（8,8）;\draw[线宽=2，red]（0,0）--（0,3）--（3,3）--（3,5）--（8,5）--（8,8）;\end{tikzpicture}”，E）;[/亚西]

问题 7

求 的最大可能的实部 ，其中 $z$ 是一个复数。$|z|=4$\[（75+117i）z+\frac{96+144i}{z}\]

问题 8

34 美元半径为 8 个圆可以相切放置，以便这些圆按顺序彼此相切，第一个圆与 $\overline{BC}$ $\overline{AC}$相$\三角形 ABC$切$\overline{AB}$，最后一个圆与 相切，如图所示。同样， 2024 美元 1 美元半径圆可以$\overline{BC}$以相同的方式相切放置。的$\三角形 ABC$半径可以表示为 $\frac{m}{n}$，其中 $m 美元 和 $n$ 是相对素数的正整数。查找 $m+n$.

对 A = （2,1）;对 B = （0,0）;对 C = （3,0）;点（A^^B^^C）;标签（“$A$”， A， N）;标签（“$B$”， B， S）;标签（“$C$”， C， S）;draw（A--B--C--循环）;for（real i=0.62; i<2.7; i+=0.29）{ draw（circle（（i，0.145）， 0.145））; }[/亚西]

问题 9

设 $ABCD 美元 是一个菱形，其顶点都位于双曲线上$\tfrac{x^2}{20}-\tfrac{y^2}{24}=1$，并且按该顺序排列。如果它的对角线在原点相交，则找到小于所有菱形 $BD^2$ 的最大数字$ABCD 美元。

问题 10

设 $ABC$ 是一个内接在 circle $\omega$中的三角形。设切线 to $\omega$ at $B 美元 和 $C$ intersect at 点 $D 美元，并设 $\overline{AD}$ $\omega$ intersect at $P 美元。如果 $AB=5 美元， $BC=9$， 和 $AC=10 美元， $AP 美元 可以写成形式 $\frac{m}{n}$，其中 $m 美元 和 $n$ 是相对质数。查找 $m + n$.

问题 11

正八边形的每个顶点都以相等的概率独立地着色为红色或蓝色。然后可以旋转八边形以使所有蓝色顶点都位于最初存在红色顶点的位置的概率为 $\tfrac{m}{n}$，其中 $m 美元 和 $n$ 是相对素数正整数。什么是 $m+n$？

问题 12

Define $f（x）=||x|-\tfrac{1}{2}|$ 和 $g（x）=||x|-\tfrac{1}{4}|$.求 的图形的交集数\[y=4 g（f（\sin （2 \pi x））） \quad\text{ 和 }\quad x=4 g（f（\cos （3 \pi y）））.\]

问题 13

设 $p$ 为存在可被 $p^{2}$整除的正整数$n$$n^{4}+1$的最小素数。求可被 整除的最小正整数$m 美元$m^{4}+1$。$p^{2}$

问题 14

设 $ABCD 美元 为四面体，使得 $AB = CD = \sqrt{41}$、 $AC = BD = \sqrt{80}$和 $BC = AD = \sqrt{89}$。四面体内部存在一个点$I 美元，使得到$I 美元四面体的每个面的距离都相等。这个距离可以写成 $\frac{m \sqrt{n}}{p}$的形式，当 $m 美元， $n$， 和 $p$ 是正整数，$m 美元并且$p$是相对素数，并且$n$不能被任何素数的平方整除。查找 $m+n+p$.

问题 15

设 $\mathcal{B}$ 为具有表面积54 美元和体积 23 美元的矩形框的集合。设 $r 美元 是可以包含每个矩形框的最小球体的半径，这些矩形框是 的$\mathcal{B}$元素。的值$r^2$可以写为 $\frac{p}{q}$，其中 $p$ 和 $q$ 是相对素数的正整数。查找 $p+q$.

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