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Create the page "2005 AIME II/Problem 10" on this wiki! See also the search results found.
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- == Problem == {{AIME box|year=2005|n=II|num-b=9|num-a=11}}3 KB (436 words) - 20:35, 13 August 2024
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- ...d only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. ==== Problem ====9 KB (1,703 words) - 01:20, 7 December 2024
- *[[2007 AMC 12A Problems/Problem 18]] *[[1984 AIME Problems/Problem 8|1984 AIME Problem 8]]5 KB (860 words) - 15:36, 10 December 2023
- ...s for any given Diophantine equations. This is known as [[Hilbert's tenth problem]]. The answer, however, is no. <math> \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 30\qquad \mathrm{(D) \ } 90\qquad \mathrm{(E) \ } 219 KB (1,434 words) - 01:15, 4 July 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME II Problems]]1 KB (135 words) - 12:24, 22 March 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME I Problems]]1 KB (154 words) - 12:30, 22 March 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME I Problems]]1 KB (135 words) - 12:31, 22 March 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME II Problems]]1 KB (135 words) - 12:30, 22 March 2011
- {{AIME Problems|year=2006|n=I}} == Problem 1 ==7 KB (1,173 words) - 03:31, 4 January 2023
- == Problem == ...o <math>9^3=729</math>, we try 8, 9, and 10, which works, so <math>n - 3 = 10</math> and <math>n = \boxed{13}</math>.1 KB (239 words) - 11:54, 31 July 2023
- == Problem == .... A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common [[ratio]] of the original3 KB (581 words) - 21:19, 22 September 2024
- == Problem == ...r of [[positive integer]]s that are divisors of at least one of <math> 10^{10},15^7,18^{11}. </math>3 KB (377 words) - 18:36, 1 January 2024
- == Problem == ...ath> be a polynomial with integer coefficients that satisfies <math> P(17)=10 </math> and <math> P(24)=17. </math> Given that <math> P(n)=n+3 </math> has4 KB (642 words) - 02:14, 1 June 2024
- {{AIME Problems|year=2005|n=II}} == Problem 1 ==7 KB (1,119 words) - 21:12, 28 February 2020
- == Problem == <cmath>(y^{12}+y^8+y^4+1)(y^2+1)=(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)</cmath>2 KB (279 words) - 12:33, 27 October 2019
- == Problem == ...</math> The radii of <math> C_1 </math> and <math> C_2 </math> are 4 and 10, respectively, and the [[center]]s of the three circles are all [[collinear4 KB (693 words) - 13:03, 28 December 2021
- == Problem == This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</ma6 KB (1,154 words) - 03:30, 11 January 2024
- == Problem == defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-013 KB (2,080 words) - 13:14, 23 July 2024
- == Problem == ...</math>, <math>a_{m-1}a_{m-2} = 3</math>; from the recursion given in the problem <math>a_{m-p+1} = a_{m-p-1} - 3/a_{m-p}</math>, so <math>a_{m-p+1} = 3p/a_{3 KB (499 words) - 18:52, 21 November 2022
- {{AIME Problems|year=2005|n=I}} == Problem 1 ==6 KB (983 words) - 05:06, 20 February 2019
- == Problem == D(CR(A,16));D(CR(B,4));D(shift((0,12)) * yscale(3^.5 / 2) * CR(C,10), linetype("2 2") + d + red);12 KB (2,001 words) - 20:26, 23 July 2024
