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- ==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== ...9</math> and <math>10</math>. <math>\boxed{(8,9)}</math> and <math>\boxed{(9,10)}</math> both work.7 KB (1,208 words) - 09:21, 4 October 2022
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- ==Problem== ...ty that <math>B-J \ge 2</math> is <math>1-\frac{1}{2}-\frac{1}{20} = \frac{9}{20} \implies \boxed{029}</math>5 KB (831 words) - 18:47, 29 January 2025
- ...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
- ...into anything. Using that fact, you can use the Games theorem to solve any problem. 9. He hates contemporary music. He only like Renaissance music, Baroque music69 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== draw(J+9/8*(K-J)--K+9/8*(J-K),dashed);7 KB (1,053 words) - 14:58, 14 January 2024
- ==Problem== <cmath>\cos C=\cos\left(\tfrac 12 (\pi-Q)\right)=\sin\tfrac 12 Q=\sqrt{\frac{9}{40}}.</cmath>7 KB (1,129 words) - 16:27, 6 January 2025
- ==Problem== ...ing <math>2019</math>, and <math>f\left(\tfrac{1+\sqrt{3}i}{2}\right)=2015+2019\sqrt{3}i</math>. Find the remainder when <math>f(1)</math> is divided by <m4 KB (706 words) - 22:18, 28 December 2023
- ==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== ...nds will converge to such a value (since it is unique, as indicated in the problem). Let's keep track of the lower bound.8 KB (1,172 words) - 18:21, 8 August 2024
- ==Problem== ...e <math>ABC</math> has side lengths <math>AB=7, BC=8, </math> and <math>CA=9.</math> Circle <math>\omega_1</math> passes through <math>B</math> and is t14 KB (2,229 words) - 14:57, 27 December 2024
- ==Problem== ...r arc <math>\widehat{A_3A_4}</math> of the circle has area <math>\tfrac{1}{9}.</math> There is a positive integer <math>n</math> such that the area of t7 KB (1,051 words) - 20:45, 27 January 2024
- {{AIME Problems|year=2019|n=I}} ==Problem 1==8 KB (1,331 words) - 06:57, 4 January 2021
- ==Problem== Consider the integer <cmath>N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.</c3 KB (433 words) - 07:57, 9 February 2023
- ==Problem== ...own above. Note that <math>m\angle KPL = 90^{\circ}</math> as given in the problem. Since <math>\angle KPL \cong \angle KLN</math> and <math>\angle PKL \cong11 KB (1,782 words) - 23:29, 29 January 2025
- {{AIME Problems|year=2019|n=II}} ==Problem 1==7 KB (1,254 words) - 14:45, 21 August 2023