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- ==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
- ==Problem== Find the least odd prime factor of <math>2019^8+1</math>.8 KB (1,264 words) - 01:22, 7 March 2024
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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
- ...eorems concerning [[polygon]]s, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local [[coordinate syst ...school students made it popular. The technique greatly simplifies certain problems.5 KB (812 words) - 15:43, 1 March 2025
- ...into anything. Using that fact, you can use the Games theorem to solve any problem. 14. Gmaas killed Thanos with a swat of Gmaas's tail. Gmaas barfed and created69 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== ...s C)}{\cos B\cos C}=\frac{16\cdot 21}{(2/\sqrt{35})(3/\sqrt{40})}=560\sqrt{14}.</cmath>7 KB (1,129 words) - 16:27, 6 January 2025
- ==Problem== {{AIME box|year=2019|n=II|num-b=12|num-a=14}}7 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
- {{AIME Problems|year=2019|n=II}} ==Problem 1==7 KB (1,254 words) - 14:45, 21 August 2023
- ==Problem== ...frac{1}{2}a+\frac{1}{54}a=\frac{56}{3}</cmath> This means that <math>\frac{14}{27}a=\frac{56}{3}</math>, so <cmath>a=36</cmath> <cmath>b=36^{\frac{54}{367 KB (1,165 words) - 15:57, 30 December 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 == ...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
- ==Problem== Find the least odd prime factor of <math>2019^8+1</math>.8 KB (1,264 words) - 01:22, 7 March 2024
- ==Problem== Back to our problem. Assume <math>XP=x</math>, <math>PY=y</math> and <math>x<y</math>. Then we13 KB (2,252 words) - 13:39, 5 January 2025
- ...hing. Using that fact, you can use the Almighty Gmaas theorem to solve any problem. 14. Almighty Gmaas killed Thanos with a swat of Almighty Gmaas's tail. Almi99 KB (14,096 words) - 23:49, 19 February 2025
- {{AIME Problems|year=2020|n=I}} ==Problem 1==7 KB (1,257 words) - 17:51, 5 February 2022
