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- ==Problem== {{AIME box|year=2019|n=II|num-b=12|num-a=14}}7 KB (1,051 words) - 20:45, 27 January 2024
- == Problem == ...d to </math>BX<math> and </math>XD<math>, which are crucial lengths in the problem. Suppose </math>BX = r, XD = s<math> for simplicity. We have:10 KB (1,622 words) - 16:51, 28 February 2025
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- ...n McNugget Theorem''' (or '''Postage Stamp Problem''' or '''Frobenius Coin Problem''') states that for any two [[relatively prime]] [[positive integer]]s <mat ...t Theorem has also been called the Frobenius Coin Problem or the Frobenius Problem, after German mathematician Ferdinand Frobenius inquired about the largest17 KB (2,823 words) - 23:06, 15 November 2024
- It is used to solve problems of the form: how many ways can one distribute <math>k</math> indistinguisha ...ach urn, then there would be <math>{n \choose k}</math> possibilities; the problem is that you can repeat urns, so this does not work.<math>n</math> and then5 KB (795 words) - 17:39, 31 December 2024
- ...into anything. Using that fact, you can use the Games theorem to solve any problem. 13. Gmaas was the first person to use the Infinity Gauntlet. EDIT: Gmaas creat69 KB (11,805 words) - 20:49, 18 December 2019
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2018 AIME II Problems|Entire Test]]1 KB (133 words) - 18:13, 18 March 2020
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2019 AIME I Problems|Entire Test]]1 KB (133 words) - 17:41, 29 March 2019
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2019 AIME II Problems|Entire Test]]1 KB (133 words) - 15:43, 22 March 2019
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2020 AIME I Problems|Entire Test]]1 KB (133 words) - 18:11, 18 March 2020
- {{AIME Problems|year=2018|n=II}} ==Problem 1==9 KB (1,385 words) - 00:26, 21 January 2024
- Here are the problems from the 2019 AMC 10C, a mock contest created by the AoPS user fidgetboss_4000. ==Problem 1==12 KB (1,917 words) - 12:14, 29 November 2021
- ==Problem== [[File:2019 AIME II 7.png|450px|right]]7 KB (1,053 words) - 14:58, 14 January 2024
- ==Problem== ...pretty. Let <math>S</math> be the sum of positive integers less than <math>2019</math> that are <math>20</math>-pretty. Find <math>\tfrac{S}{20}</math>.3 KB (474 words) - 01:38, 22 December 2024
- ==Problem== Obviously <math>n\le 90</math>. We see that the problem's condition is equivalent to: 96 is the smallest number that can be formed9 KB (1,543 words) - 19:48, 3 January 2025
- {{AIME Problems|year=2019|n=I}} ==Problem 1==8 KB (1,331 words) - 06:57, 4 January 2021
- {{AIME Problems|year=2019|n=II}} ==Problem 1==7 KB (1,254 words) - 14:45, 21 August 2023
- ==Problem== pair L2 = (21/2,-13/8);9 KB (1,508 words) - 14:02, 7 September 2024
- ==Problem== Another way to solve this problem is to do casework on all the perfect squares from <math>1^2</math> to <math20 KB (3,220 words) - 03:24, 14 August 2024
- ==Problem== ...}+z\right)^6=2z^6 + \frac{15z^4}{2} + \frac{15z^2}{8} + \frac{1}{32}=\frac{13}{54}.</cmath>10 KB (1,878 words) - 13:19, 1 February 2024
- ==Problem== * <math>13^2</math> has two possibilities: <math>168</math> and <math>169</math> or <m7 KB (1,208 words) - 09:21, 4 October 2022
- ==Problem== Before we start thinking about the problem, let’s draw it out;25 KB (4,645 words) - 22:46, 30 January 2025
- ==Problem== ...sier with the intuition that <math>(z-20)</math> must be a divisor for the problem to lead anywhere. Now we know <math>(z+1)(z-19)\in i\mathbb{R}</math> so us8 KB (1,534 words) - 13:05, 5 August 2024
