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- == Problem == An [[infinite]] [[geometric series]] has sum 2005. A new series, obtained by squaring each term of the original series, has 13 KB (581 words) - 21:19, 22 September 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
- Consider the set of letters A, B, and C. There are <math>3!</math> different [[permutations]] of those letters. Since order doesn't ma * [[2005_AIME_II_Problems/Problem_1 | 2005 AIME II Problem 1]]4 KB (638 words) - 21:55, 5 January 2025
- D((-5,0)--(0,0)--(0,-3)); MC(90,"\mbox{semiminor axis}",7,D((0,0)--(0,3),green+linewidth(1)),E);5 KB (892 words) - 21:52, 1 May 2021
- ...hat the imaginary part of a complex number is real: for example, <math>\Im(3 + 4i) = 4</math>. So, if <math>z\in \mathbb C</math>, we can write <math>z *[[2007 AMC 12A Problems/Problem 18]]5 KB (860 words) - 15:36, 10 December 2023
- ...mon ratio <math>-1/2</math>; however, <math>1, 3, 9, -27</math> and <math>-3, 1, 5, 9, \ldots</math> are not geometric sequences, as the ratio between c * [[1965 AHSME Problems/Problem 36 | 1965 AHSME Problem 36]]4 KB (649 words) - 21:09, 19 July 2024
- ...e an integer). However, <math>17</math> will never be a multiple of <math>3</math>, hence, no solutions exist. Case 3: <math>x_0^4 \equiv 1\mod 4, y_0^4 \equiv 0\mod 4</math>, and <math>z_0^29 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 == ...that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find <math> n. </math>1 KB (239 words) - 11:54, 31 July 2023
- == Problem == *Person 1: <math>\frac{9 \cdot 6 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{9}{28}</math>4 KB (628 words) - 11:28, 14 April 2024
- == Problem == <math>15^7 = 3^7\cdot5^7</math> so <math>15^7</math> has <math>8\cdot8 = 64</math> divisor3 KB (377 words) - 18:36, 1 January 2024
- == Problem == ...th> P(17)=10 </math> and <math> P(24)=17. </math> Given that <math> P(n)=n+3 </math> has two distinct integer solutions <math> n_1 </math> and <math> n_4 KB (642 words) - 02:14, 1 June 2024
- == Problem == ...og_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </math>3 KB (547 words) - 19:15, 4 April 2024
- == Problem == ...agical. For example, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in t2 KB (384 words) - 00:31, 26 July 2018
- {{AIME Problems|year=2005|n=II}} == Problem 1 ==7 KB (1,119 words) - 21:12, 28 February 2020
- == Problem == It follows that <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = \boxed{125}</math>.2 KB (279 words) - 12:33, 27 October 2019
- == Problem == ...= (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7));4 KB (693 words) - 13:03, 28 December 2021
