{"kind":"task","effective_mode":"full","benchmark":{"kind":"benchmark","effective_mode":"full","slug":"math","formal_name":"MATH","introduction":"MATH organises 12,500 high-school competition problems into seven subjects and five difficulty levels, each with a full worked solution. It underlies much of the later mathematics evaluation work; this catalogue imports the full test split: 5,000 problems across all seven subjects.","introduction_ja":"","introduction_en":"","category":"Category not supplied","task_count":null,"acquisition_status":"Acquisition status not supplied","official_url":"https://github.com/hendrycks/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":"293abe16-5b49-5749-90dd-7cf5ccbb2caf","task_key":"intermediate~5falgebra--test--293abe16-5b49-5749-90dd-7cf5ccbb2caf","task_revision_id":"1","upstream_id":"","short_description":"For each $x$ in $[0,1]$, define","config":"intermediate_algebra","split":"test","body":"{\"level\":\"Level 5\",\"problem\":\"For each $x$ in $[0,1]$, define\\n\\\\[\\\\begin{cases}  f(x) = 2x, \\\\qquad\\\\qquad \\\\mathrm{if} \\\\quad 0 \\\\leq x \\\\leq \\\\frac{1}{2};\\\\\\\\  f(x) = 2-2x, \\\\qquad \\\\mathrm{if} \\\\quad \\\\frac{1}{2} < x \\\\leq 1.  \\\\end{cases}\\\\]Let $f^{[2]}(x) = f(f(x))$, and $f^{[n + 1]}(x) = f^{[n]}(f(x))$ for each integer $n \\\\geq 2$.  Then the number of values of $x$ in $[0,1]$ for which $f^{[2005]}(x) = \\\\frac {1}{2}$ can be expressed in the form $p^a,$ where $p$ is a prime and $a$ is a positive integer.  Find $p + a.$\",\"type\":\"Intermediate Algebra\"}","display_format":"math","language":"","answer_status":"published","assets":[],"source_url":"https://github.com/hendrycks/math","history":"initial import","indexing_mode":"index","subproblems":[],"grids":[]}