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- == Problem ==549 bytes (89 words) - 15:37, 14 December 2023
- == Problem ==324 bytes (48 words) - 15:48, 14 December 2023
- == Problem ==721 bytes (120 words) - 16:02, 14 December 2023
- == Problem ==367 bytes (59 words) - 16:14, 14 December 2023
- == Problem ==495 bytes (73 words) - 16:25, 14 December 2023
- == Problem ==565 bytes (90 words) - 16:37, 14 December 2023
- == Problem ==698 bytes (121 words) - 16:53, 14 December 2023
- == Problem ==618 bytes (105 words) - 17:08, 14 December 2023
- == Problem ==2 KB (399 words) - 14:44, 15 March 2024
- == Problem ==904 bytes (172 words) - 17:36, 14 December 2023
- ==Problem==812 bytes (125 words) - 22:16, 15 December 2023
- ==Problem==1 KB (165 words) - 10:37, 24 December 2023
- ==Problem== ...se that <math>a,b,c</math> are positive reals satisfying<cmath>(a^3+4)(b^3+6)(c^3+8) = 8(a+b+c)^3.</cmath> Find the sum of all possible values of <math>353 bytes (60 words) - 22:25, 15 December 2023
- ==Problem== Now notice that there is a one-to-one correspondence between this problem and the number of ways to distribute 8 balls into 3 boxes and also 8 other6 KB (932 words) - 10:44, 10 February 2024
- == Problem == ...f}{1.1x},</math> since that's the fraction of female players now. From the problem, we are given1 KB (168 words) - 00:20, 31 December 2023
- ==Problem 6== We can change the problem to (6x5)+(2x10)+(1x25) which can be evaluated as (30)+(20)+(25) which is eq427 bytes (61 words) - 17:57, 20 January 2024
- ==Problem== ...1 to find the elements in set <math>A</math>, which are <math>(11,10,9,8,7,6,4)</math>. Add these up to get <math>\boxed{055}</math>. -westwoodmonster2 KB (391 words) - 23:57, 12 May 2024
- 58 bytes (10 words) - 01:10, 13 March 2024
- #REDIRECT [[2024 USAMO Problems/Problem 5]]43 bytes (4 words) - 11:20, 24 March 2024
- {{USAMO newbox|year=2024|num-b=5|after=Last Problem}}1,002 bytes (183 words) - 17:23, 2 June 2024
Page text matches
- == Problem == ...math>a</math> and <math>b</math> must be digits. Therefore, <math>(a,b) = (6,5)</math>, and <math>x+y+m=\boxed{\textbf{(E) }154}</math>.5 KB (845 words) - 19:23, 17 September 2023
- == Problem == ...ven) to get 123. This will get us the numbers 1,3,5...21,23(or numbers 2,4,6...22,24), which gives us 12 more numbers, and adding that to the 50 origina3 KB (517 words) - 19:15, 15 October 2023
- == Problem == {{USAMO newbox|year=2006|num-b=4|num-a=6}}7 KB (1,280 words) - 17:23, 26 March 2016
- == Problem == ...owing three numbers. <math>a_1=\frac{9}{2}, a_2=\frac{4}{3}, a_3=\frac{7}{6}</math> shows that <math>n=7</math> is possible.3 KB (486 words) - 22:43, 5 August 2014
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[2004 USAMO Problems/Problem 1 | Problem 1]]812 bytes (92 words) - 12:27, 22 March 2011
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[2005 USAMO Problems/Problem 1 | Problem 1]]548 bytes (59 words) - 18:13, 17 September 2012
- ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #3]] and [[2006 AMC 10A Problems/Problem 3|2006 AMC 10A #3]]}} == Problem ==1 KB (155 words) - 17:30, 16 December 2021
- ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #5]] and [[2006 AMC 10A Problems/Problem 5|2006 AMC 10A #5]]}} == Problem ==1 KB (176 words) - 17:57, 16 December 2021
- ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}} == Problem ==3 KB (528 words) - 18:29, 7 May 2024
- == Problem == {{AMC12 box|year=2006|ab=A|num-b=6|num-a=8}}901 bytes (144 words) - 22:46, 5 February 2016
- ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #8]] and [[2006 AMC 10A Problems/Problem 9|2008 AMC 10A #9]]}} == Problem ==3 KB (450 words) - 02:00, 13 January 2024
- == Problem == ...s (6 would too, but it is faster to just test 6). Trying 6, <math>13 \cdot 6 = 72</math>. 28 is not divisible by 3, so we know that this number is not c3 KB (429 words) - 18:14, 26 September 2020
- ==Word Problem AMC 8 Algebra Video== <math>6(0)-y=-4</math>5 KB (932 words) - 12:57, 26 July 2023
- ...of|difficulty=5-8|breakdown=<u>Individual</u>: 5 (Problem 1-5), 6 (Problem 6-10)<br><u>Team</u>: 7.5<br><u>HMIC</u>: 8}} In the Team Round, 6-8 person teams compete together on a 60 minute test. The problems are weigh4 KB (539 words) - 16:58, 19 February 2023
- **[[2007 iTest Problems/Problem 1|Problem 1]] **[[2007 iTest Problems/Problem 2|Problem 2]]3 KB (305 words) - 15:10, 5 November 2023
- == Problem 1 == [[2006 AMC 10B Problems/Problem 1|Solution]]14 KB (2,059 words) - 01:17, 30 January 2024
- == Problem 1 == [[2005 AMC 10B Problems/Problem 1|Solution]]12 KB (1,874 words) - 21:20, 23 December 2020
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AMC 10A Problems/Problem 1]]2 KB (182 words) - 18:09, 6 October 2014
- {{problem}} ...ch that <math>AD</math> bisects <math>\angle A</math>. Given that <math>AD=6,BD=4</math>, and <math>DC=3</math>, find <math>AB</math>.2 KB (306 words) - 16:11, 21 February 2023
- == Problem == {{AMC12 box|year=2000|num-b=4|num-a=6}}714 bytes (102 words) - 17:01, 29 July 2023
