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values | answer
stringclasses 10
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stringclasses 9
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stringclasses 10
values | solution
stringclasses 5
values | dataset
stringclasses 2
values | aggregated_pass_rate
float64 0.8
1
| accessor_pass@k
dict |
|---|---|---|---|---|---|---|---|---|
amc23_68
|
50.0
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_12
|
68
|
For complex number $u = a+bi$ and $v = c+di$ (where $i=\sqrt{-1}$), define the binary operation
$u \otimes v = ac + bdi$
Suppose $z$ is a complex number such that $z\otimes z = z^{2}+40$. What is $|z|^2$?
|
amc23
| 1 |
{
"OpenThinker3-7B": {
"K": 64,
"pass@K": 1
}
}
|
|
amc23_70
|
5.0
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_14
|
70
|
For how many ordered pairs $(a,b)$ of integers does the polynomial $x^3+ax^2+bx+6$ have $3$ distinct integer roots?
|
amc23
| 0.796875 |
{
"OpenThinker3-7B": {
"K": 64,
"pass@K": 0.796875
}
}
|
|
aime24_71
|
294
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_6
|
71
|
Consider the paths of length $16$ that follow the lines from the lower left corner to the upper right corner on an $8\times 8$ grid. Find the number of such paths that change direction exactly four times, as in the examples shown below.
|
We divide the path into eight “$R$” movements and eight “$U$” movements. Five sections of alternative $RURUR$ or $URURU$ are necessary in order to make four “turns.” We use the first case and multiply by $2$.
For $U$, we have seven ordered pairs of positive integers $(a,b)$ such that $a+b=8$.
For $R$, we subtract $1$ from each section (to make the minimum stars of each section $0$) and we use Stars and Bars to get ${7 \choose 5}=21$.
Thus our answer is $7\cdot21\cdot2=\boxed{294}$.
~eevee9406
Draw a few examples of the path. However, notice one thing in common - if the path starts going up, there will be 3 "segments" where the path goes up, and two horizontal "segments." Similarly, if the path starts going horizontally, we will have three horizontal segments and two vertical segments. Those two cases are symmetric, so we only need to consider one. If our path starts going up, by stars and bars, we can have $\binom{7}{2}$ ways to split the 8 up's into 3 lengths, and have $\binom{7}{1}$ to split the 8-horizontals into 2 lengths. We multiply them together, and multiply by 2 for symmetry, giving us $2*\binom{7}{2}*\binom{7}{1}=294.$
~nathan27 (original by alexanderruan)
Notice that the $RURUR$ case and the $URURU$ case is symmetrical. WLOG, let's consider the RURUR case.
Now notice that there is a one-to-one correspondence between this problem and the number of ways to distribute 8 balls into 3 boxes and also 8 other balls into 2 other boxes, such that each box has a nonzero amount of balls.
There are ${8+2-3 \choose 2}$ ways for the first part, and ${8+1-2 \choose 1}$ ways for the second part, by stars and bars.
The answer is $2\cdot {7 \choose 2} \cdot {7 \choose 1} = \boxed{294}$.
~northstar47
Feel free to edit this solution
Starting at the origin, you can either first go up or to the right. If you go up first, you will end on the side opposite to it (the right side) and if you go right first, you will end up on the top. It can then be observed that if you choose the turning points in the middle $7 \times 7$ grid, that will automatically determine your start and ending points. For example, in the diagram if you choose the point $(3,2)$ and $(5,3)$, you must first move three up or two right, determining your first point, and move 5 up or 3 right, determining your final point. Knowing this is helpful because if we first move anywhere horizontally, we have $7$ points on each column to choose from and starting from left to right, we have $6,5,4,3,2,1$ points on that row to choose from. This gives us $7(6)+7(5)+7(4)+7(3)+7(2)+7(1)$ which simplifies to $7\cdot21$. The vertical case is symmetrical so we have $7\cdot21\cdot2 = \boxed{294}$
~KEVIN_LIU
As in Solution 1, there are two cases: $RURUR$ or $URURU$. We will work with the first case and multiply by $2$ at the end. We use stars and bars; we can treat the $R$s as the stars and the $U$s as the bars. However, we must also use stars and bars on the $U$s to see how many different patterns of bars we can create for the reds. We must have $1$ bar in $8$ blacks, so we use stars and bars on the equation \[x + y = 8\]. However, each divider must have at least one black in it, so we do the change of variable $x' = x-1$ and $y' = x-1$. Our equation becomes \[x' + y' = 6\]. By stars and bars, this equation has $\binom{6 + 2 - 1}{1} = 7$ valid solutions. Now, we use stars and bars on the reds. We must distribute two bars amongst the reds, so we apply stars and bars to \[x + y + z = 8\]. Since each group must have one red, we again do a change of variables with $x' = x-1$, $y' = y-1$, and $z' = z-1$. We are now working on the equation \[x' + y' + z' = 5\]. By stars and bars, this has $\binom{5 + 3 - 1}{2} = 21$ solutions. The number of valid paths in this case is the number of ways to create the bars times the number of valid arrangements of the stars given fixed bars, which equals $21 \cdot 7 = 147$. We must multiply by two to account for both cases, so our final answer is $147 \cdot 2 = \boxed{294}$.
~ [cxsmi](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Cxsmi)
|
aime24
| 1 |
{
"OpenThinker3-7B": {
"K": 64,
"pass@K": 1
}
}
|
amc23_48
|
7.0
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_14
|
48
|
How many complex numbers satisfy the equation $z^5=\overline{z}$, where $\overline{z}$ is the conjugate of the complex number $z$?
|
amc23
| 1 |
{
"OpenThinker3-7B": {
"K": 64,
"pass@K": 1
}
}
|
|
amc23_81
|
144.0
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_8
|
81
|
How many nonempty subsets $B$ of ${0, 1, 2, 3, \cdots, 12}$ have the property that the number of elements in $B$ is equal to the least element of $B$? For example, $B = {4, 6, 8, 11}$ satisfies the condition.
|
amc23
| 0.984375 |
{
"OpenThinker3-7B": {
"K": 64,
"pass@K": 0.984375
}
}
|
|
amc23_77
|
1625.0
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_4
|
77
|
Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?
|
amc23
| 1 |
{
"OpenThinker3-7B": {
"K": 64,
"pass@K": 1
}
}
|
|
amc23_78
|
4.0
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_5
|
78
|
You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?
|
amc23
| 0.890625 |
{
"OpenThinker3-7B": {
"K": 64,
"pass@K": 0.890625
}
}
|
|
aime24_68
|
809
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_3
|
68
|
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.
|
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 $404 + 405 = \boxed{809}$
~lprado
We will use winning and losing positions, where a $W$ marks when Alice wins and an $L$ marks when Bob wins.
$1$ 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. In other words, there will be $2$ losing positions out of every $5$ consecutive values of n. 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 $2020\times\frac{2}{5}+1=\boxed{809}$
~alexanderruan
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 $404 + 405 = \boxed{\textbf{(809) }}$.
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. $808 + 1 = \boxed{809}$.
~numerophile
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.
$404+404+1=809$
-Multpi12
(Edits would be appreciated)
LaTexed by BossLu99
|
aime24
| 0.953125 |
{
"OpenThinker3-7B": {
"K": 64,
"pass@K": 0.953125
}
}
|
aime24_69
|
116
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_4
|
69
|
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 $m+n$.
|
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 $1+115=\boxed{116}$.
~Technodoggo
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 $1+115 = \boxed{116}$
~e___
|
aime24
| 1 |
{
"OpenThinker3-7B": {
"K": 64,
"pass@K": 1
}
}
|
aime24_84
|
033
|
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_4
|
84
|
Let $x,y$ and $z$ be positive real numbers that satisfy the following system of equations:
\[\log_2\left({x \over yz}\right) = {1 \over 2}\]\[\log_2\left({y \over xz}\right) = {1 \over 3}\]\[\log_2\left({z \over xy}\right) = {1 \over 4}\]
Then the value of $\left|\log_2(x^4y^3z^2)\right|$ is $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
|
Denote $\log_2(x) = a$, $\log_2(y) = b$, and $\log_2(z) = c$.
Then, we have:
$a-b-c = \frac{1}{2}$
$-a+b-c = \frac{1}{3}$
$-a-b+c = \frac{1}{4}$
Now, we can solve to get $a = \frac{-7}{24}, b = \frac{-9}{24}, c = \frac{-5}{12}$. Plugging these values in, we obtain $|4a + 3b + 2c| = \frac{25}{8} \implies \boxed{033}$. ~akliu
$\log_2(y/xz) + \log_2(z/xy) = \log_2(1/x^2) = -2\log_2(x) = \frac{7}{12}$
$\log_2(x/yz) + \log_2(z/xy) = \log_2(1/y^2) = -2\log_2(y) = \frac{3}{4}$
$\log_2(x/yz) + \log_2(y/xz) = \log_2(1/z^2) = -2\log_2(z) = \frac{5}{6}$
$\log_2(x) = -\frac{7}{24}$
$\log_2(y) = -\frac{3}{8}$
$\log_2(z) = -\frac{5}{12}$
$4\log_2(x) + 3\log_2(y) + 2\log_2(z) = -25/8$
$25 + 8 = \boxed{033}$
~Callisto531
Adding all three equations, $\log_2(\frac{1}{xyz}) = \frac{1}{2}+\frac{1}{3}+\frac{1}{4} = \frac{13}{12}$. Subtracting this from every equation, we have: \[2\log_2x = -\frac{7}{12},\] \[2\log_2y = -\frac{3}{4},\] \[2\log_2z = -\frac{5}{6}\] Our desired quantity is the absolute value of $4\log_2x+3\log_2y+2\log_2z = 2(\frac{7}{12})+3/2(\frac{3}{4})+\frac{5}{6} = \frac{25}{8}$, so our answer is $25+8 = \boxed{033}$.
~Spoirvfimidf
|
aime24
| 1 |
{
"OpenThinker3-7B": {
"K": 64,
"pass@K": 1
}
}
|
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