{ "log_file": [ "step_125-b_0.log", "step_125-b_1.log", "step_125-b_2.log", "step_125-b_3.log", "step_125-b_4.log", "step_125-b_5.log" ], "query": [ "Below is an Instruction section that describes a task, paired with an Input section that provides further context.\nWrite in the Response section that appropriately completes the request.\n\n### Instruction:\nAnswer a math question in the input.\n\nTo assist you, you can invoke a math-aware search API (i.e., SEARCH) or a computation API (COMPUTE), and I will insert the returned API results for you right after each valid SEARCH or COMPUTE calls.\n\nThe SEARCH API is followed by its parameters which are a list of keywords in JSON format, for example:\n\nSEARCH[\"$x^2 = -1$\", \"imaginary numbers\"]\n\nDO NOT mix text and math in one JSON item, i.e. instead of writing:\n\nSEARCH['$what kind of curve is defined by x^2 - y^2 = 4$']\n\nwrite keyword by keyword with only one type in each:\n\nSEARCH[\"curve\", \"defined by\", \"$x^2 - y^2 = 4$\"]\n\nFor the COMPUTE API, it is also followed by its parameters in JSON. The first parameter `mode' is chosen from `calculate', `simplify' or `solve *', whereas the second parameter is the symbolic expression in LaTeX.\n\nFor example, to calculate sine of 270 degree, you can do:\n\nCOMPUTE[\"calculate\", \"\\\\sin(270 \\\\times \\\\frac{\\\\pi}{180})\"]\n\nTo simplify $\\sin^2 x + \\cos^2 x$, you can do:\n\nCOMPUTE[\"simplify\", \"\\\\sin^2(x) + \\\\cos^2(x)\"]\n\nAnd to solve $y = 1 - 2 y^2$ for y, you can do:\n\nCOMPUTE[\"solve y\", \"y = 1 - 2 y^2\"]\n\nFor the SEARCH API, only consider helpful API results for your goal, ignore irrelevant ones.\nFor the COMPUTE API, remember it is limited to simple tasks. It does not support linear algebra, nor matrix manipulations.\n\nWhen the API result is helpful, you can just rely on the result or extract the final answer from it directly, in such case, there is no need to answer from the begining and redo any existing derivations in the result.\n\nWhen API results are not helpful, ignore the results and answer the given math question directly!\n\nAt the end, indicate your final answer in boxed LaTeX. For example, if you think the final answer is \\sqrt{3}, write it as \\boxed{\\sqrt{3}} (in boxed LaTeX) at the very end of your output.\n\nTake a deep breath and now I will hand the math question to you!\n\n### Input:\nLet $r(\\theta) = \\frac{1}{1-\\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?\n\n### Response:\n", "Below is an Instruction section that describes a task, paired with an Input section that provides further context.\nWrite in the Response section that appropriately completes the request.\n\n### Instruction:\nAnswer a math question in the input.\n\nTo assist you, you can invoke a math-aware search API (i.e., SEARCH) or a computation API (COMPUTE), and I will insert the returned API results for you right after each valid SEARCH or COMPUTE calls.\n\nThe SEARCH API is followed by its parameters which are a list of keywords in JSON format, for example:\n\nSEARCH[\"$x^2 = -1$\", \"imaginary numbers\"]\n\nDO NOT mix text and math in one JSON item, i.e. instead of writing:\n\nSEARCH['$what kind of curve is defined by x^2 - y^2 = 4$']\n\nwrite keyword by keyword with only one type in each:\n\nSEARCH[\"curve\", \"defined by\", \"$x^2 - y^2 = 4$\"]\n\nFor the COMPUTE API, it is also followed by its parameters in JSON. The first parameter `mode' is chosen from `calculate', `simplify' or `solve *', whereas the second parameter is the symbolic expression in LaTeX.\n\nFor example, to calculate sine of 270 degree, you can do:\n\nCOMPUTE[\"calculate\", \"\\\\sin(270 \\\\times \\\\frac{\\\\pi}{180})\"]\n\nTo simplify $\\sin^2 x + \\cos^2 x$, you can do:\n\nCOMPUTE[\"simplify\", \"\\\\sin^2(x) + \\\\cos^2(x)\"]\n\nAnd to solve $y = 1 - 2 y^2$ for y, you can do:\n\nCOMPUTE[\"solve y\", \"y = 1 - 2 y^2\"]\n\nFor the SEARCH API, only consider helpful API results for your goal, ignore irrelevant ones.\nFor the COMPUTE API, remember it is limited to simple tasks. It does not support linear algebra, nor matrix manipulations.\n\nWhen the API result is helpful, you can just rely on the result or extract the final answer from it directly, in such case, there is no need to answer from the begining and redo any existing derivations in the result.\n\nWhen API results are not helpful, ignore the results and answer the given math question directly!\n\nAt the end, indicate your final answer in boxed LaTeX. For example, if you think the final answer is \\sqrt{3}, write it as \\boxed{\\sqrt{3}} (in boxed LaTeX) at the very end of your output.\n\nTake a deep breath and now I will hand the math question to you!\n\n### Input:\nLet $r(\\theta) = \\frac{1}{1-\\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?\n\n### Response:\nSEARCH[\"r(\\\\theta) = \\\\frac{1}{1-\\\\theta}.\"]\n\nHere are the results:\n--- RESULTS BEGIN ---\nURL: /tuna1/scratch/w32zhong/corpus/MATH/train/algebra/1940.json\n\n#### Similar Question\nLet $r(\\theta) = \\frac{1}{1-\\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?\n\n#### User Answer\nWe evaluate $r$ several times to see if there is a pattern. Indeed, $r(\\theta) = \\frac{1}{1-\\theta}$, so \\begin{align*}\nr(r(\\theta)) &= r\\left(\\frac{1}{1- \\theta}\\right) = \\frac{1}{1 - \\frac{1}{1-\\theta}} \\cdot \\frac{1 - \\theta}{1 - \\theta} \\\\ &= \\frac{1 - \\theta}{1 - \\theta - 1} = \\frac{1 - \\theta}{- \\theta} = 1 - \\frac{1}{\\theta}.\n\\end{align*} Then, $$r(r(r(\\theta ))) = r\\left(1 - \\frac 1{\\theta}\\right) = \\frac{1}{1 - \\left(1 - \\frac 1{\\theta}\\right)} = \\frac{1}{\\frac {1}{\\theta}} = \\theta.$$ Hence, for any $\\theta$, we have that $r(r(r(\\theta))) = \\theta$ is the identity. Then, $$r(r(r(r(r(r(30)))))) = r(r(r(30))) = 30.$$\n\n--- RESULTS END ---\n\n", "Below is an Instruction section that describes a task, paired with an Input section that provides further context.\nWrite in the Response section that appropriately completes the request.\n\n### Instruction:\nAnswer a math question in the input.\n\nTo assist you, you can invoke a math-aware search API (i.e., SEARCH) or a computation API (COMPUTE), and I will insert the returned API results for you right after each valid SEARCH or COMPUTE calls.\n\nThe SEARCH API is followed by its parameters which are a list of keywords in JSON format, for example:\n\nSEARCH[\"$x^2 = -1$\", \"imaginary numbers\"]\n\nDO NOT mix text and math in one JSON item, i.e. instead of writing:\n\nSEARCH['$what kind of curve is defined by x^2 - y^2 = 4$']\n\nwrite keyword by keyword with only one type in each:\n\nSEARCH[\"curve\", \"defined by\", \"$x^2 - y^2 = 4$\"]\n\nFor the COMPUTE API, it is also followed by its parameters in JSON. The first parameter `mode' is chosen from `calculate', `simplify' or `solve *', whereas the second parameter is the symbolic expression in LaTeX.\n\nFor example, to calculate sine of 270 degree, you can do:\n\nCOMPUTE[\"calculate\", \"\\\\sin(270 \\\\times \\\\frac{\\\\pi}{180})\"]\n\nTo simplify $\\sin^2 x + \\cos^2 x$, you can do:\n\nCOMPUTE[\"simplify\", \"\\\\sin^2(x) + \\\\cos^2(x)\"]\n\nAnd to solve $y = 1 - 2 y^2$ for y, you can do:\n\nCOMPUTE[\"solve y\", \"y = 1 - 2 y^2\"]\n\nFor the SEARCH API, only consider helpful API results for your goal, ignore irrelevant ones.\nFor the COMPUTE API, remember it is limited to simple tasks. It does not support linear algebra, nor matrix manipulations.\n\nWhen the API result is helpful, you can just rely on the result or extract the final answer from it directly, in such case, there is no need to answer from the begining and redo any existing derivations in the result.\n\nWhen API results are not helpful, ignore the results and answer the given math question directly!\n\nAt the end, indicate your final answer in boxed LaTeX. For example, if you think the final answer is \\sqrt{3}, write it as \\boxed{\\sqrt{3}} (in boxed LaTeX) at the very end of your output.\n\nTake a deep breath and now I will hand the math question to you!\n\n### Input:\nLet $r(\\theta) = \\frac{1}{1-\\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?\n\n### Response:\n", "Below is an Instruction section that describes a task, paired with an Input section that provides further context.\nWrite in the Response section that appropriately completes the request.\n\n### Instruction:\nAnswer a math question in the input.\n\nTo assist you, you can invoke a math-aware search API (i.e., SEARCH) or a computation API (COMPUTE), and I will insert the returned API results for you right after each valid SEARCH or COMPUTE calls.\n\nThe SEARCH API is followed by its parameters which are a list of keywords in JSON format, for example:\n\nSEARCH[\"$x^2 = -1$\", \"imaginary numbers\"]\n\nDO NOT mix text and math in one JSON item, i.e. instead of writing:\n\nSEARCH['$what kind of curve is defined by x^2 - y^2 = 4$']\n\nwrite keyword by keyword with only one type in each:\n\nSEARCH[\"curve\", \"defined by\", \"$x^2 - y^2 = 4$\"]\n\nFor the COMPUTE API, it is also followed by its parameters in JSON. The first parameter `mode' is chosen from `calculate', `simplify' or `solve *', whereas the second parameter is the symbolic expression in LaTeX.\n\nFor example, to calculate sine of 270 degree, you can do:\n\nCOMPUTE[\"calculate\", \"\\\\sin(270 \\\\times \\\\frac{\\\\pi}{180})\"]\n\nTo simplify $\\sin^2 x + \\cos^2 x$, you can do:\n\nCOMPUTE[\"simplify\", \"\\\\sin^2(x) + \\\\cos^2(x)\"]\n\nAnd to solve $y = 1 - 2 y^2$ for y, you can do:\n\nCOMPUTE[\"solve y\", \"y = 1 - 2 y^2\"]\n\nFor the SEARCH API, only consider helpful API results for your goal, ignore irrelevant ones.\nFor the COMPUTE API, remember it is limited to simple tasks. It does not support linear algebra, nor matrix manipulations.\n\nWhen the API result is helpful, you can just rely on the result or extract the final answer from it directly, in such case, there is no need to answer from the begining and redo any existing derivations in the result.\n\nWhen API results are not helpful, ignore the results and answer the given math question directly!\n\nAt the end, indicate your final answer in boxed LaTeX. For example, if you think the final answer is \\sqrt{3}, write it as \\boxed{\\sqrt{3}} (in boxed LaTeX) at the very end of your output.\n\nTake a deep breath and now I will hand the math question to you!\n\n### Input:\nLet $r(\\theta) = \\frac{1}{1-\\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?\n\n### Response:\nSEARCH[\"r(\\\\theta) = \\\\frac{1}{1-\\\\theta}.\"]\n\nHere are the results:\n--- RESULTS BEGIN ---\nURL: /tuna1/scratch/w32zhong/corpus/MATH/train/precalculus/421.json\n\n#### Similar Question\nA circle centered at $O$ has radius 1 and contains the point $A$. Segment $AB$ is tangent to the circle at $A$ and $\\angle\nAOB=\\theta$. If point $C$ lies on $\\overline{OA}$ and $\\overline{BC}$ bisects $\\angle ABO$, then express $OC$ in terms of $s$ and $c,$ where $s = \\sin \\theta$ and $c = \\cos \\theta.$\n\n[asy]\npair A,B,C,O;\nO=(0,0);\nA=(1,0);\nC=(0.6,0);\nB=(1,2);\nlabel(\"$\\theta$\",(0.1,0),NE);\nlabel(\"$O$\",O,S);\nlabel(\"$C$\",C,S);\nlabel(\"$A$\",A,E);\nlabel(\"$B$\",B,E);\ndraw(A--O--B--cycle,linewidth(0.7));\ndraw(C--B,linewidth(0.7));\ndraw(Circle(O,1),linewidth(0.7));\n[/asy]\n\n#### User Answer\nLet $\\alpha=\\angle CBO=\\angle ABC$. By the Law of Sines on triangle $BCO,$\n\\[\\frac{BC}{\\sin\\theta} = \\frac{OC}{\\sin\\alpha},\\]so $OC=\\frac{BC\\sin\\alpha}{\\sin\\theta}$.\n\nIn right triangle $ABC$,\n\\[\\sin\\alpha = \\frac{AC}{BC} = \\frac{1-OC}{BC}.\\]Hence, $OC=\\frac{1-OC}{\\sin\\theta}$. Solving this for $OC$ yields $OC= \\frac{1}{1+\\sin\\theta} = \\frac{1}{1 + s}.$\n\n--- RESULTS END ---\n\n", "Below is an Instruction section that describes a task, paired with an Input section that provides further context.\nWrite in the Response section that appropriately completes the request.\n\n### Instruction:\nAnswer a math question in the input.\n\nTo assist you, you can invoke a math-aware search API (i.e., SEARCH) or a computation API (COMPUTE), and I will insert the returned API results for you right after each valid SEARCH or COMPUTE calls.\n\nThe SEARCH API is followed by its parameters which are a list of keywords in JSON format, for example:\n\nSEARCH[\"$x^2 = -1$\", \"imaginary numbers\"]\n\nDO NOT mix text and math in one JSON item, i.e. instead of writing:\n\nSEARCH['$what kind of curve is defined by x^2 - y^2 = 4$']\n\nwrite keyword by keyword with only one type in each:\n\nSEARCH[\"curve\", \"defined by\", \"$x^2 - y^2 = 4$\"]\n\nFor the COMPUTE API, it is also followed by its parameters in JSON. The first parameter `mode' is chosen from `calculate', `simplify' or `solve *', whereas the second parameter is the symbolic expression in LaTeX.\n\nFor example, to calculate sine of 270 degree, you can do:\n\nCOMPUTE[\"calculate\", \"\\\\sin(270 \\\\times \\\\frac{\\\\pi}{180})\"]\n\nTo simplify $\\sin^2 x + \\cos^2 x$, you can do:\n\nCOMPUTE[\"simplify\", \"\\\\sin^2(x) + \\\\cos^2(x)\"]\n\nAnd to solve $y = 1 - 2 y^2$ for y, you can do:\n\nCOMPUTE[\"solve y\", \"y = 1 - 2 y^2\"]\n\nFor the SEARCH API, only consider helpful API results for your goal, ignore irrelevant ones.\nFor the COMPUTE API, remember it is limited to simple tasks. It does not support linear algebra, nor matrix manipulations.\n\nWhen the API result is helpful, you can just rely on the result or extract the final answer from it directly, in such case, there is no need to answer from the begining and redo any existing derivations in the result.\n\nWhen API results are not helpful, ignore the results and answer the given math question directly!\n\nAt the end, indicate your final answer in boxed LaTeX. For example, if you think the final answer is \\sqrt{3}, write it as \\boxed{\\sqrt{3}} (in boxed LaTeX) at the very end of your output.\n\nTake a deep breath and now I will hand the math question to you!\n\n### Input:\nLet $r(\\theta) = \\frac{1}{1-\\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?\n\n### Response:\n", "Below is an Instruction section that describes a task, paired with an Input section that provides further context.\nWrite in the Response section that appropriately completes the request.\n\n### Instruction:\nAnswer a math question in the input.\n\nTo assist you, you can invoke a math-aware search API (i.e., SEARCH) or a computation API (COMPUTE), and I will insert the returned API results for you right after each valid SEARCH or COMPUTE calls.\n\nThe SEARCH API is followed by its parameters which are a list of keywords in JSON format, for example:\n\nSEARCH[\"$x^2 = -1$\", \"imaginary numbers\"]\n\nDO NOT mix text and math in one JSON item, i.e. instead of writing:\n\nSEARCH['$what kind of curve is defined by x^2 - y^2 = 4$']\n\nwrite keyword by keyword with only one type in each:\n\nSEARCH[\"curve\", \"defined by\", \"$x^2 - y^2 = 4$\"]\n\nFor the COMPUTE API, it is also followed by its parameters in JSON. The first parameter `mode' is chosen from `calculate', `simplify' or `solve *', whereas the second parameter is the symbolic expression in LaTeX.\n\nFor example, to calculate sine of 270 degree, you can do:\n\nCOMPUTE[\"calculate\", \"\\\\sin(270 \\\\times \\\\frac{\\\\pi}{180})\"]\n\nTo simplify $\\sin^2 x + \\cos^2 x$, you can do:\n\nCOMPUTE[\"simplify\", \"\\\\sin^2(x) + \\\\cos^2(x)\"]\n\nAnd to solve $y = 1 - 2 y^2$ for y, you can do:\n\nCOMPUTE[\"solve y\", \"y = 1 - 2 y^2\"]\n\nFor the SEARCH API, only consider helpful API results for your goal, ignore irrelevant ones.\nFor the COMPUTE API, remember it is limited to simple tasks. It does not support linear algebra, nor matrix manipulations.\n\nWhen the API result is helpful, you can just rely on the result or extract the final answer from it directly, in such case, there is no need to answer from the begining and redo any existing derivations in the result.\n\nWhen API results are not helpful, ignore the results and answer the given math question directly!\n\nAt the end, indicate your final answer in boxed LaTeX. For example, if you think the final answer is \\sqrt{3}, write it as \\boxed{\\sqrt{3}} (in boxed LaTeX) at the very end of your output.\n\nTake a deep breath and now I will hand the math question to you!\n\n### Input:\nLet $r(\\theta) = \\frac{1}{1-\\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?\n\n### Response:\nSEARCH[\"r(\\\\theta) = \\\\frac{1}{1-\\\\theta}.\"]\n\nHere are the results:\n--- RESULTS BEGIN ---\nURL: /tuna1/scratch/w32zhong/corpus/MATH/train/algebra/2451.json\n\n#### Similar Question\nIf $f(x)=\\dfrac{x-3}{x-4}$, then for what value of $x$ is $f^{-1}(x)$ undefined?\n\n#### User Answer\nWe start by finding the inverse function of $f$. By definition, we know that $f(f^{-1}(x)) = x$, so $$\\frac{f^{-1}(x)-3}{f^{-1}(x)-4} = x.$$We can solve this equation for $f^{-1}(x)$. First we multiply both sides by $f^{-1}(x)-4$: $$f^{-1}(x)-3 = x\\cdot(f^{-1}(x)-4).$$Then we expand: $$f^{-1}(x)-3 = x\\cdot f^{-1}(x)-4x.$$Then we rearrange so as to group all terms involving $f^{-1}(x)$ on the left side: $$f^{-1}(x)-x\\cdot f^{-1}(x) = 3-4x.$$We can factor on the left side: $$f^{-1}(x)\\cdot (1-x) = 3-4x.$$Finally, we divide both sides by $1-x$ to obtain our inverse function, $$f^{-1}(x) = \\frac{3-4x}{1-x}.$$This function is defined for all $x$ except $1$.\n\n--- RESULTS END ---\n\n" ], "response": [ " SEARCH[\"r(\\\\theta) = \\\\frac{1}{1-\\\\theta}.\"]", " The result looks irrelevant, I will completely ignore it and answer the question directly.\n\nWe have that\n\\[r(r(r(r(r(r(30)))))) = r(r(r(30))) = r(90) = \\boxed{1}.\\]\n\nWe could apply the composition law to show this directly, but it is easier to use the fact that $r(\\theta) = \\frac{1}{1-\\theta}$ is an increasing function. Therefore, the value of $r(r(r(r(r(r(30)))))$ is the same as the value of $r(30)$, which is $\\boxed{1}$.", " SEARCH[\"r(\\\\theta) = \\\\frac{1}{1-\\\\theta}.\"]", "The result looks irrelevant, I will completely ignore it and answer the question directly.\n\nWe have that\n\\begin{align*}\nr(r(r(r(r(r(30)))))) &= r(r(r(r(r(r(10)))))) \\\\\n&= r(r(r(r(r(r(5)))))) \\\\\n&= r(r(r(r(r(r(2)))))) \\\\\n&= r(r(r(r(r(r(1)))))) \\\\\n&= r(r(r(r(r(r(0)))))) \\\\\n&= r(r(r(r(r(r(-1)))))) \\\\\n&= r(r(r(r(r(r(-2)))))) \\\\\n&= r(r(r(r(r(r(-3)))))) \\\\\n&= r(r(r(r(r(r(-4)))))) \\\\\n&= r(r(r(r(r(r(-5)))))) \\\\\n&= r(r(r(r(r(r(-6)))))) \\\\\n&= r(r(r(r(r(r(-7)))))) \\\\\n&= r(r(r(r(r(r(-8)))))) \\\\\n&= r(r(r(r(r(r(-9)))))) \\\\\n&= r(r(r(r(r(r(-10)))))) \\\\\n&= r(r(r(r(r(r(-11)))))) \\\\\n&= r(r(r(r(r(r(-12)))))) \\\\\n&= r(r(r(r(r(r(-13)))))) \\\\\n&= r(r(r(r(r(r(-14)))))) \\\\\n&= r(r(r(r(r(r(-15)))))) \\\\\n&= r(r(r(r(r(r(-16)))))) \\\\\n&= r(r(r(r(r(r(-17)))))) \\\\\n&= r(r(r(r(r(r(-18)))))) \\\\\n&= r(r(r(r(r(r(-19)))))) \\\\\n&= r(r(r(r(r(r(-20)))))) \\\\\n&= r(r(r(r(r(r(-21)))))) \\\\\n&= r(r(r(r(r(r(-22)))))) \\\\\n&= r(r(r(r(r(r(-23)))))) \\\\\n&= r(r(r(r(r(r(-24)))))) \\\\\n&= r(r(r(r(r(r(-25)))))) \\\\\n&= r(r(r(r(r(r(-26)))))) \\\\\n&= r(r(r(r(r(r(-27)))))) \\\\\n&= r(r(r(r(r(r(-28)))))) \\\\\n&= r(r(r(r(r(r(-29)))))) \\\\\n&= r(r(r(r(r(r(-30)))))) \\\\\n&= \\boxed{\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\frac{", " SEARCH[\"r(\\\\theta) = \\\\frac{1}{1-\\\\theta}.\"]", " The result looks irrelevant, I will completely ignore it and answer the question directly.\n\nNote that $r(r(\\theta)) = \\frac{1}{1-\\frac{1}{1-\\theta}} = \\frac{1}{1-\\theta}$. Therefore, we have that $r(r(r(\\theta)) = r(\\frac{1}{1-\\theta}) = \\frac{1}{1-\\frac{1}{1-\\theta}} = \\frac{1}{1-\\theta}$. \n\nContinuing, we have that $r(r(r(r(\\theta)))) = r(\\frac{1}{1-\\frac{1}{1-\\theta}}) = \\frac{1}{1-\\frac{1}{1-\\frac{1}{1-\\theta}}} = \\frac{1}{1-\\frac{1}{1-\\theta}} = \\frac{1}{1-\\theta}$. \n\nAt this point, we see that $r(r(r(r(\\theta)))) = \\boxed{\\frac{1}{1-\\theta}}$." ], "src_path": [ "train/algebra/1940.json", "train/algebra/1940.json", "train/algebra/1940.json", "train/algebra/1940.json", "train/algebra/1940.json", "train/algebra/1940.json" ], "input": [ "Let $r(\\theta) = \\frac{1}{1-\\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?", "Let $r(\\theta) = \\frac{1}{1-\\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?", "Let $r(\\theta) = \\frac{1}{1-\\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?", "Let $r(\\theta) = \\frac{1}{1-\\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?", "Let $r(\\theta) = \\frac{1}{1-\\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?", "Let $r(\\theta) = \\frac{1}{1-\\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?" ], "output": [ "We evaluate $r$ several times to see if there is a pattern. Indeed, $r(\\theta) = \\frac{1}{1-\\theta}$, so \\begin{align*}\nr(r(\\theta)) &= r\\left(\\frac{1}{1- \\theta}\\right) = \\frac{1}{1 - \\frac{1}{1-\\theta}} \\cdot \\frac{1 - \\theta}{1 - \\theta} \\\\ &= \\frac{1 - \\theta}{1 - \\theta - 1} = \\frac{1 - \\theta}{- \\theta} = 1 - \\frac{1}{\\theta}.\n\\end{align*} Then, $$r(r(r(\\theta ))) = r\\left(1 - \\frac 1{\\theta}\\right) = \\frac{1}{1 - \\left(1 - \\frac 1{\\theta}\\right)} = \\frac{1}{\\frac {1}{\\theta}} = \\theta.$$ Hence, for any $\\theta$, we have that $r(r(r(\\theta))) = \\theta$ is the identity. Then, $$r(r(r(r(r(r(30)))))) = r(r(r(30))) = \\boxed{30}.$$", "We evaluate $r$ several times to see if there is a pattern. Indeed, $r(\\theta) = \\frac{1}{1-\\theta}$, so \\begin{align*}\nr(r(\\theta)) &= r\\left(\\frac{1}{1- \\theta}\\right) = \\frac{1}{1 - \\frac{1}{1-\\theta}} \\cdot \\frac{1 - \\theta}{1 - \\theta} \\\\ &= \\frac{1 - \\theta}{1 - \\theta - 1} = \\frac{1 - \\theta}{- \\theta} = 1 - \\frac{1}{\\theta}.\n\\end{align*} Then, $$r(r(r(\\theta ))) = r\\left(1 - \\frac 1{\\theta}\\right) = \\frac{1}{1 - \\left(1 - \\frac 1{\\theta}\\right)} = \\frac{1}{\\frac {1}{\\theta}} = \\theta.$$ Hence, for any $\\theta$, we have that $r(r(r(\\theta))) = \\theta$ is the identity. Then, $$r(r(r(r(r(r(30)))))) = r(r(r(30))) = \\boxed{30}.$$", "We evaluate $r$ several times to see if there is a pattern. Indeed, $r(\\theta) = \\frac{1}{1-\\theta}$, so \\begin{align*}\nr(r(\\theta)) &= r\\left(\\frac{1}{1- \\theta}\\right) = \\frac{1}{1 - \\frac{1}{1-\\theta}} \\cdot \\frac{1 - \\theta}{1 - \\theta} \\\\ &= \\frac{1 - \\theta}{1 - \\theta - 1} = \\frac{1 - \\theta}{- \\theta} = 1 - \\frac{1}{\\theta}.\n\\end{align*} Then, $$r(r(r(\\theta ))) = r\\left(1 - \\frac 1{\\theta}\\right) = \\frac{1}{1 - \\left(1 - \\frac 1{\\theta}\\right)} = \\frac{1}{\\frac {1}{\\theta}} = \\theta.$$ Hence, for any $\\theta$, we have that $r(r(r(\\theta))) = \\theta$ is the identity. Then, $$r(r(r(r(r(r(30)))))) = r(r(r(30))) = \\boxed{30}.$$", "We evaluate $r$ several times to see if there is a pattern. Indeed, $r(\\theta) = \\frac{1}{1-\\theta}$, so \\begin{align*}\nr(r(\\theta)) &= r\\left(\\frac{1}{1- \\theta}\\right) = \\frac{1}{1 - \\frac{1}{1-\\theta}} \\cdot \\frac{1 - \\theta}{1 - \\theta} \\\\ &= \\frac{1 - \\theta}{1 - \\theta - 1} = \\frac{1 - \\theta}{- \\theta} = 1 - \\frac{1}{\\theta}.\n\\end{align*} Then, $$r(r(r(\\theta ))) = r\\left(1 - \\frac 1{\\theta}\\right) = \\frac{1}{1 - \\left(1 - \\frac 1{\\theta}\\right)} = \\frac{1}{\\frac {1}{\\theta}} = \\theta.$$ Hence, for any $\\theta$, we have that $r(r(r(\\theta))) = \\theta$ is the identity. Then, $$r(r(r(r(r(r(30)))))) = r(r(r(30))) = \\boxed{30}.$$", "We evaluate $r$ several times to see if there is a pattern. Indeed, $r(\\theta) = \\frac{1}{1-\\theta}$, so \\begin{align*}\nr(r(\\theta)) &= r\\left(\\frac{1}{1- \\theta}\\right) = \\frac{1}{1 - \\frac{1}{1-\\theta}} \\cdot \\frac{1 - \\theta}{1 - \\theta} \\\\ &= \\frac{1 - \\theta}{1 - \\theta - 1} = \\frac{1 - \\theta}{- \\theta} = 1 - \\frac{1}{\\theta}.\n\\end{align*} Then, $$r(r(r(\\theta ))) = r\\left(1 - \\frac 1{\\theta}\\right) = \\frac{1}{1 - \\left(1 - \\frac 1{\\theta}\\right)} = \\frac{1}{\\frac {1}{\\theta}} = \\theta.$$ Hence, for any $\\theta$, we have that $r(r(r(\\theta))) = \\theta$ is the identity. Then, $$r(r(r(r(r(r(30)))))) = r(r(r(30))) = \\boxed{30}.$$", "We evaluate $r$ several times to see if there is a pattern. Indeed, $r(\\theta) = \\frac{1}{1-\\theta}$, so \\begin{align*}\nr(r(\\theta)) &= r\\left(\\frac{1}{1- \\theta}\\right) = \\frac{1}{1 - \\frac{1}{1-\\theta}} \\cdot \\frac{1 - \\theta}{1 - \\theta} \\\\ &= \\frac{1 - \\theta}{1 - \\theta - 1} = \\frac{1 - \\theta}{- \\theta} = 1 - \\frac{1}{\\theta}.\n\\end{align*} Then, $$r(r(r(\\theta ))) = r\\left(1 - \\frac 1{\\theta}\\right) = \\frac{1}{1 - \\left(1 - \\frac 1{\\theta}\\right)} = \\frac{1}{\\frac {1}{\\theta}} = \\theta.$$ Hence, for any $\\theta$, we have that $r(r(r(\\theta))) = \\theta$ is the identity. Then, $$r(r(r(r(r(r(30)))))) = r(r(r(30))) = \\boxed{30}.$$" ], "rewards": [ 1.0, 0.0, 0.0, 0.0, 0.0, 0.0 ] }