{"status": "success", "data": {"description_md": "According to the standard convention for exponentiation, \n$2^{2^{2^2}} = 2^{\\left(2^{\\left(2^2\\right)}\\right)} = 2^{16} = 65,536$.\nIf the order in which the exponentiations are performed is changed, how many <u>other</u> values are possible?\n\n$\\text{(A)}\\ 0 \\qquad \\text{(B)}\\ 1 \\qquad \\text{(C)}\\ 2 \\qquad \\text{(D)}\\ 3 \\qquad \\text{(E)}\\ 4$", "description_html": "<p>According to the standard convention for exponentiation,<br/>\n <span class=\"katex--inline\">2^{2^{2^2}} = 2^{\\left(2^{\\left(2^2\\right)}\\right)} = 2^{16} = 65,536</span> .<br/>\nIf the order in which the exponentiations are performed is changed, how many <u>other</u> values are possible?</p>\n<p> <span class=\"katex--inline\">\\text{(A)}\\ 0 \\qquad \\text{(B)}\\ 1 \\qquad \\text{(C)}\\ 2 \\qquad \\text{(D)}\\ 3 \\qquad \\text{(E)}\\ 4</span> </p>\n<hr><p>Full credit goes to <a href=\"https://maa.org/\">MAA</a> for authoring these problems. These problems were taken on the <a href=\"https://artofproblemsolving.com/\">AOPS</a> website.</p>", "hints_md": "", "hints_html": "", "editorial_md": "", "editorial_html": "", "flag_hint": "", "point_value": 1, "problem_name": "2002 AMC 10A Problem 3", "can_next": true, "can_prev": true, "nxt": "/problem/02_amc10A_p04", "prev": "/problem/02_amc10A_p02"}}