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- == Problem ==2 KB (307 words) - 23:58, 17 November 2024
- == Problem ==1 KB (235 words) - 13:52, 25 June 2023
- == Problem == ...ue of the apothem. We could just apply s, which is the side length in this problem, <math>\frac{5\sqrt{6}}{3}</math> into the hexagon area formula, <math>\fra1 KB (210 words) - 11:36, 2 July 2024
- ...ate|[[2006 AMC 12A Problems|2006 AMC 12A #16]] and [[2006 AMC 10A Problems/Problem 23|2006 AMC 10A #23]]}} == Problem ==2 KB (286 words) - 09:16, 19 December 2021
- == Problem == Simplify the problem - in <math>2</math> dimensions, place <math>4</math> circles in each quadra2 KB (364 words) - 03:54, 16 January 2023
- == Problem == <!-- [[Image:2006_AMC10A-16.png]] -->5 KB (811 words) - 10:44, 30 November 2024
- == Problem == ...multiples of <math>4</math> (with a few exceptions that don't affect this problem).2 KB (336 words) - 09:51, 11 May 2024
- #REDIRECT[[2005 AMC 12B Problems/Problem 12]]45 bytes (5 words) - 16:22, 12 July 2011
- ==Problem== From the problem, we have <math>10a+b-(a+b)=9a</math>2 KB (279 words) - 10:57, 17 July 2023
- == Problem == First, calculate how many of each type of problem she got right:1 KB (167 words) - 19:30, 11 January 2024
- == Problem ==1 KB (218 words) - 14:52, 19 August 2023
- == Problem == However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero.2 KB (336 words) - 12:40, 31 December 2024
- == Problem ==2 KB (228 words) - 01:01, 23 January 2023
- == Problem ==2 KB (310 words) - 10:28, 3 August 2021
- == Problem ==1 KB (166 words) - 11:20, 5 July 2013
- == Problem ==2 KB (317 words) - 09:26, 5 November 2023
- ==Problem== ...h> squares possible <math>(25</math> - <math>1\times1</math> squares <math>16</math> - <math>2\times2</math> squares <math>9</math> - <math>3\times3</mat2 KB (377 words) - 00:37, 14 August 2024
- == Problem ==1 KB (240 words) - 11:59, 30 March 2023
- == Problem ==3 KB (577 words) - 15:33, 9 October 2022
- ==Problem==3 KB (463 words) - 15:35, 15 February 2021
Page text matches
- ==Problem== &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{-15}{16}\right\rfloor\2 KB (257 words) - 09:57, 16 June 2023
- == Problem == \ = 2\int_0^1 1 + \frac{3}{8}x dx = 2(x + \frac{3}{16}x^2) \left.\right|_{\;0}^{\;1}8 KB (1,016 words) - 23:17, 30 December 2023
- *2016 Regional - Saturday 2/27/16 *2016 State Finals - Saturday 5/7/168 KB (1,182 words) - 13:26, 3 April 2024
- ==Problem== =\frac{2}{2}+\frac{4}{4}+\frac{6}{8}+\frac{8}{16}+\cdots\1 KB (193 words) - 20:13, 18 May 2021
- == Problem 38== We can determine that <math>\frac{a+b+c}{3}+d+16=a+b+c+d</math>. This, with some algebra, means that <math>\frac{1}{3}(a+b+c1 KB (200 words) - 22:35, 28 August 2020
- == Problem 46== pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16);3 KB (415 words) - 17:01, 24 May 2020
- ...[[MAML]] (Maine Association of Math Leagues) Meets. Training includes the problem set "Pete's Fabulous 42." ...ed using the top 15 winners in West Virginia State Math Field Day. Winners 16-30 are used as potential alternates for the team. West Virginia State Math22 KB (3,532 words) - 10:25, 27 September 2024
- In team rounds, teams of 4 people work fast to answer a problem in 4 minutes. However, the faster a team turns in an answer, the more point **16 points4 KB (632 words) - 17:21, 21 December 2024
- ([[2000 AMC 12/Problem 6|Source]]) ==Problem 1==4 KB (682 words) - 12:13, 8 December 2024
- * <math>16! = 20922789888000</math> \left\lfloor\frac {100}{49}\right\rfloor=14+2=16</math>10 KB (809 words) - 15:40, 17 March 2024
- ...)=a^2+16a-80=0</math>. The sum of these values of <math>a</math> is <math>-16</math>. * [[1977_Canadian_MO_Problems/Problem_1 | 1977 Canadian MO Problem 1]]4 KB (768 words) - 16:56, 24 June 2024
- ...ast two digits of <math> 7^{81}-3^{81} </math>. ([[Euler's Totient Theorem Problem 1 Solution|Solution]]) ...ast two digits of <math> 3^{3^{3^{3}}} </math>. ([[Euler's Totient Theorem Problem 2 Solution|Solution]])4 KB (569 words) - 21:34, 30 December 2024
- <cmath>16[ABCD]^2=4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2</cmath> <cmath>16[ABCD]^2=(2(ab+cd)+(a^2+b^2-c^2-d^2))(2(ab+cd)-(a^2+b^2-c^2-d^2))</cmath>3 KB (543 words) - 18:35, 29 October 2024
- ...6\}</math> we would be trying to calculate <math>\frac 3{\frac 13 + \frac 16 - \frac 12} = \frac 30</math>, which is obviously problematic. * [[2002 AMC 12A Problems/Problem 11]]1 KB (196 words) - 23:49, 5 January 2021
- An example of a classic problem is as follows: ...divisible by 6, and there are 16 such numbers. Thus, there are <math>50+33-16=\boxed{67}</math> numbers that are divisible by either 2 or 3.4 KB (635 words) - 11:19, 2 January 2022
- 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
- ...e second. For instance, one function may map 1 to 1, 2 to 4, 3 to 9, 4 to 16, and so on. This function has the rule that it takes its input value, and ([[2006 AMC 10A Problems/Problem 2|Source]])10 KB (1,761 words) - 02:16, 12 May 2023
- ...ng may lead to a quick solution is the phrase "not" or "at least" within a problem statement. ''[[2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive integers have at least one digit that is8 KB (1,192 words) - 16:20, 16 June 2023
- ...together many of the best students from around the world to learn Olympiad problem solving skills. The WOOT program includes 16 classes, problem sets, and practice AIME and olympiad-style tests. More details may be foun947 bytes (133 words) - 18:44, 3 March 2022
- ([[2006 AMC 12A Problems/Problem 16|Source]]) ([[2006 AMC 12A Problems/Problem 21|Source]])9 KB (1,510 words) - 18:56, 16 January 2025