{"status": "success", "data": {"description_md": "An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\\pi$? \n\n<center>\n<img class=\"problem-image\" height=\"252\" src=\"https://latex.artofproblemsolving.com/a/a/3/aa3c94ca7814105299a9086afbfcb5b4ecaf2609.png\" width=\"252\"/>\n</center><br>\n\n$\\textbf{(A) } 46 \\qquad \\textbf{(B) } 48 \\qquad \\textbf{(C) } 56 \\qquad \\textbf{(D) } 60 \\qquad \\textbf{(E) } 64$", "description_html": "<p>An even number of circles are nested, starting with a radius of <span class=\"katex--inline\">1</span> and increasing by <span class=\"katex--inline\">1</span> each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius <span class=\"katex--inline\">2</span> but outside the circle of radius <span class=\"katex--inline\">1.</span> An example showing <span class=\"katex--inline\">8</span> circles is displayed below. What is the least number of circles needed to make the total shaded area at least <span class=\"katex--inline\">2023\\pi</span>?</p>&#10;<center>&#10;<img class=\"problem-image\" height=\"252\" src=\"https://latex.artofproblemsolving.com/a/a/3/aa3c94ca7814105299a9086afbfcb5b4ecaf2609.png\" width=\"252\"/>&#10;</center><br/>&#10;<p><span class=\"katex--inline\">\\textbf{(A) } 46 \\qquad \\textbf{(B) } 48 \\qquad \\textbf{(C) } 56 \\qquad \\textbf{(D) } 60 \\qquad \\textbf{(E) } 64</span></p>&#10;", "hints_md": "", "hints_html": "", "editorial_md": "", "editorial_html": "", "flag_hint": "", "point_value": 3, "problem_name": "2023 AMC 10A Problem 15", "can_next": true, "can_prev": true, "nxt": "/problem/23_amc10A_p16", "prev": "/problem/23_amc10A_p14"}}