{"kind":"task","effective_mode":"full","benchmark":{"kind":"benchmark","effective_mode":"full","slug":"omni-math","formal_name":"Omni-MATH","introduction":"Omni-MATH evaluates mathematical reasoning on Olympiad-level problems. Its official dataset contains 4,428 problems accompanied by domain and difficulty information.","introduction_ja":"","introduction_en":"","category":"Category not supplied","task_count":null,"acquisition_status":"Acquisition status not supplied","official_url":"https://huggingface.co/datasets/KbsdJames/Omni-MATH","indexing_mode":"index","profile":{"resources":[],"task_format":"","scoring":"","metric":"","size":"","answer_access":"","license":"","citation":"","maintainer":"","released":"","why_hard":"","related":[],"history":"","construction":"","limitations":"","saturation":"","scores":[]}},"task_id":"aafec071-b493-5dc2-9d20-2671516c67cb","task_key":"test--aafec071-b493-5dc2-9d20-2671516c67cb","task_revision_id":"1","upstream_id":"","short_description":"Let $f(x)=x^{2}+2 x+1$. Let $g(x)=f(f(\\cdots f(x)))$, where there are $2009 f…","config":"","split":"test","body":"{\"problem\":\"Let $f(x)=x^{2}+2 x+1$. Let $g(x)=f(f(\\\\cdots f(x)))$, where there are $2009 f \\\\mathrm{~s}$ in the expression for $g(x)$. Then $g(x)$ can be written as $g(x)=x^{2^{2009}}+a_{2^{2009}-1} x^{2^{2009}-1}+\\\\cdots+a_{1} x+a_{0}$ where the $a_{i}$ are constants. Compute $a_{2^{2009}-1}$.\"}","display_format":"math","language":"","answer_status":"published","assets":[],"source_url":"https://huggingface.co/datasets/KbsdJames/Omni-MATH","history":"initial import","indexing_mode":"index","subproblems":[],"grids":[]}