{"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":"bd172cf3-9b59-5628-999c-1a6275cde8c4","task_key":"test--bd172cf3-9b59-5628-999c-1a6275cde8c4","task_revision_id":"1","upstream_id":"","short_description":"Let $x, y, z$ be real numbers satisfying…","config":"","split":"test","body":"{\"problem\":\"Let $x, y, z$ be real numbers satisfying $$\\\\frac{1}{x}+y+z=x+\\\\frac{1}{y}+z=x+y+\\\\frac{1}{z}=3$$ The sum of all possible values of $x+y+z$ can be written as $\\\\frac{m}{n}$, where $m, n$ are positive integers and $\\\\operatorname{gcd}(m, n)=1$. Find $100 m+n$.\"}","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":[]}