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- == Problem == Now, consider the strip of length <math>1024</math>. The problem asks for <math>s_{941, 10}</math>. We can derive some useful recurrences f6 KB (899 words) - 20:58, 12 May 2022
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- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME I Problems]]1 KB (135 words) - 18:15, 19 April 2021
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME II Problems]]1 KB (135 words) - 12:24, 22 March 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME I Problems]]1 KB (154 words) - 12:30, 22 March 2011
- {{AIME Problems|year=2005|n=I}} == Problem 1 ==6 KB (983 words) - 05:06, 20 February 2019
- {{AIME Problems|year=2004|n=I}} == Problem 1 ==9 KB (1,434 words) - 13:34, 29 December 2021
- == Problem == Now, consider the strip of length <math>1024</math>. The problem asks for <math>s_{941, 10}</math>. We can derive some useful recurrences f6 KB (899 words) - 20:58, 12 May 2022
- == Problem == ...n't need to be nearly as rigorous). A more natural manner of attacking the problem is to think of the process in reverse, namely seeing that <math>n \equiv 111 KB (1,857 words) - 12:57, 18 July 2024
- == Problem == ...AC \parallel DE, \angle ABC=120^\circ, AB=3, BC=5, </math> and <math>DE = 15. </math> Given that the [[ratio]] between the area of triangle <math> ABC <3 KB (486 words) - 22:15, 7 April 2023
- == Problem == ...d power, we will also satisfy the congruence. Thus, <math>2^{3}, 2^{9}, 2^{15},</math> etc. work; or, <cmath>y-x \equiv 3 \pmod 6</cmath>8 KB (1,283 words) - 19:19, 8 May 2024
- == Problem == pair A=(0,0),B=(0,25),C=(70/3,25),D=(70/3,0),E=(0,8),F=(70/3,22),G=(15,0);9 KB (1,500 words) - 20:06, 8 October 2024
- ==Problem== ...math>\frac{250}{800}(60)=\frac{150}{8}</math>. The train then has <math>60-15-\frac{50}{3}-\frac{150}{8}=230/24</math> minutes left to travel 250 miles,4 KB (592 words) - 19:02, 26 September 2020
- {{AIME Problems|year=2004|n=II}} == Problem 1 ==9 KB (1,410 words) - 05:05, 20 February 2019
- {{AIME Problems|year=2003|n=II}} == Problem 1 ==7 KB (1,127 words) - 09:02, 11 July 2023
- * [[2019 AMC 8 Problems/Problem 24]] * [[2016 AMC 10A Problems/Problem 19]]5 KB (812 words) - 15:43, 1 March 2025
- == Problem 1 == ...largest integer <math>k</math> such that <math>2004^k</math> divides <math>2004!</math>.6 KB (1,052 words) - 13:52, 9 June 2020
- ==Problem== Hence, <cmath>N = \frac{2005 \cdot 2004 \cdot 2003}{3 \cdot 2\cdot 1} \equiv \boxed{010} (\mathrm{mod} \hskip .2cm7 KB (1,187 words) - 05:58, 3 February 2025
- [[2019 AIME II Problems/Problem 15]] Solution 5 [[2023 USAJMO Problems/Problem 6]] Solution 110 KB (1,116 words) - 12:37, 11 June 2024