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- == Problem ==6 KB (980 words) - 20:45, 31 March 2020
- == Problem == ...B=2</math> and <math>J=3</math>, then <math>4B+5J=4(2)+5(3)=23</math>. The problem asks for the total cost of jam, or <math>N(5J)=11(15)=165</math> cents, or1 KB (227 words) - 16:21, 8 December 2013
- ...ate|[[2006 AMC 12A Problems|2006 AMC 12A #14]] and [[2006 AMC 10A Problems/Problem 22|2006 AMC 10A #22]]}} == Problem ==3 KB (442 words) - 02:13, 8 August 2022
- == Problem ==2 KB (278 words) - 20:12, 24 December 2020
- #REDIRECT [[2006 AMC 12A Problems/Problem 12]]46 bytes (5 words) - 09:56, 20 February 2016
- #REDIRECT [[2006 AMC 12A Problems/Problem 14]]46 bytes (5 words) - 13:46, 14 January 2016
- == Problem == [[Image:2005_I_AIME-14.png|center]]3 KB (561 words) - 13:11, 18 February 2018
- == Problem == In [[triangle]] <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <ma14 KB (2,340 words) - 15:38, 21 August 2024
- == Problem ==4 KB (729 words) - 00:00, 27 November 2022
- == Problem == ...n't need to be nearly as rigorous). A more natural manner of attacking the problem is to think of the process in reverse, namely seeing that <math>n \equiv 111 KB (1,857 words) - 11:57, 18 July 2024
- == Problem == draw(circumcircle((14,0),p,r));14 KB (2,351 words) - 20:06, 8 December 2024
- == Problem == ...s 190 or lower. We use the fact that sufficiently high multiples of 6, 10, 14, 22, etc. can be represented as <math>n+n</math>. We bash each case until w8 KB (1,365 words) - 14:38, 10 December 2024
- == Problem == ...alf of their points by playing the <math>n-10</math> stronger players. The problem also tells us that the <math>n-10</math> people who aren't part of the lose5 KB (772 words) - 21:14, 18 June 2020
- == Problem ==2 KB (346 words) - 12:13, 22 July 2020
- == Problem == ~ pi_is_3.147 KB (965 words) - 22:39, 11 September 2024
- == Problem ==4 KB (700 words) - 16:21, 3 May 2021
- == Problem ==2 KB (410 words) - 00:37, 25 August 2024
- == Problem ==7 KB (1,086 words) - 07:16, 29 July 2023
- == Problem ==2 KB (264 words) - 20:10, 19 September 2023
- == Problem == </asy></center><!-- asy replaced Image:AIME 1991 Solution 14.png by minsoens -->4 KB (547 words) - 03:46, 1 December 2024
Page text matches
- == Problem == ...minus the second largest rectangle, which is <math>(5x+20) - (3x+6) = 2x + 14</math>.2 KB (335 words) - 04:52, 18 December 2024
- == Problem == {{AMC10 box|year=2016|ab=A|num-b=12|num-a=14}}2 KB (402 words) - 13:54, 25 June 2023
- == Problem == {{AHSME 50p box|year=1953|num-b=12|num-a=14}}878 bytes (143 words) - 19:56, 1 April 2017
- * <math>14! = 87178291200</math> ...9277696409600000000000000</math> (Note: this number is 82 digits long with 14 terminal zeroes!)10 KB (809 words) - 15:40, 17 March 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME II Problems/Problem 1]]1 KB (133 words) - 11:32, 22 March 2011
- ===2023 AIME I Problem 5=== ([[2023 AIME I Problems/Problem 5|Source]])6 KB (922 words) - 16:34, 13 January 2025
- * [[1991 AIME Problems/Problem 12]] * [[2001 AIME I Problems/Problem 13]]1 KB (179 words) - 18:41, 3 January 2025
- ...e that works. 12348 - 28 ==> 12320 ==> 1232 +28 ==> 1260 ==> 126 + 14 ==> 14 YAY! - 01 + 10 - 23 + 14 ← last block is always two digits and positive10 KB (1,572 words) - 21:11, 22 September 2024
- This is a problem where constructive counting is not the simplest way to proceed. This next e ...proceed with the construction. If we were to go like before and break the problem down by each box, we'd get a fairly messy solution.13 KB (2,018 words) - 14:31, 10 January 2025
- ...0</math>, dividing both sides by <math>\gcd(93,42) = 3</math> gives <math>14 \cdot 93 - 31 \cdot 42 = 0</math>. We can add <math>k</math> times this eq * [[2020 AMC 10A Problems/Problem 24]]6 KB (923 words) - 16:39, 30 September 2024
- * [[Mock_AIME_2_2006-2007_Problems#Problem_8 | Mock AIME 2 2006-2007 Problem 8]] ([[number theory]]) *[[1994_AIME_Problems/Problem 9|1994 AIME Problem 9]]2 KB (316 words) - 15:03, 1 January 2024
- *The ellipse with axis lengths <math>14</math> and <math>16</math> has the general equation of <math>\frac{x^2}{a^2 ...latively prime integers, find <math> p+q. </math> ([[2005 AIME II Problems/Problem 15|Source]])5 KB (892 words) - 20:52, 1 May 2021
- ([[University of South Carolina High School Math Contest/1993 Exam/Problem 29|Source]]) ([[2006 AIME I Problems/Problem 14|Source]])6 KB (1,003 words) - 23:02, 19 May 2024
- *[[2007 AMC 12A Problems/Problem 18]] *[[1984 AIME Problems/Problem 8|1984 AIME Problem 8]]5 KB (860 words) - 14:36, 10 December 2023
- ...c sequence with common difference <math>-8</math>; however, <math>7, 0, 7, 14</math> and <math>4, 12, 36, 108, \ldots</math> are not arithmetic sequences * [[2005_AMC_10A_Problems/Problem_17 | 2005 AMC 10A Problem 17]]4 KB (736 words) - 01:00, 7 March 2024
- == Practice Problem == ...ath>. If the sides of triangle <math>ABC</math> are <math>13,</math> <math>14,</math> and <math>15,</math> the radius of <math>\omega</math> can be repre3 KB (533 words) - 12:51, 2 September 2024
- ...ns can significantly help in solving functional identities. Consider this problem: === Problem Examples ===4 KB (743 words) - 23:28, 17 November 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 10B Problems/Problem 1]]2 KB (182 words) - 20:57, 23 January 2021
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME I Problems/Problem 1]]1 KB (135 words) - 17:15, 19 April 2021
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME II Problems/Problem 1]]1 KB (135 words) - 11:24, 22 March 2011