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- == Problem == We can find the value of <math>a_{9}</math> by its bounds using three conditions:3 KB (535 words) - 22:25, 5 March 2025
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- ..., then adding together the totals of each part. Casework is a very general problem-solving approach, and as such has wide applicability. '''Solution''': We divide the problem into cases, based on how long the word is.5 KB (709 words) - 17:40, 24 September 2024
- *[[2007 AMC 12A Problems/Problem 18]] *[[1984 AIME Problems/Problem 8|1984 AIME Problem 8]]5 KB (860 words) - 15:36, 10 December 2023
- ...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 == ...of potential values of <math>n</math> is the number of multiples of <math>9</math> from <math>0</math> to <math>1000</math>, or <math>112</math>.11 KB (1,857 words) - 12:57, 18 July 2024
- == Problem == ...m from <math> S, </math> the [[probability]] that it is divisible by <math>9</math> is <math> p/q, </math> where <math> p </math> and <math> q </math> a8 KB (1,283 words) - 19:19, 8 May 2024
- == Problem == We can find the value of <math>a_{9}</math> by its bounds using three conditions:3 KB (535 words) - 22:25, 5 March 2025
- == Problem == ...tive integer divisors of <math>2004^{2004}</math> are divisible by exactly 2004 positive integers?2 KB (359 words) - 19:58, 24 December 2024
- == Problem == BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \9 KB (1,500 words) - 20:06, 8 October 2024
- == Problem == ...al. Also note some other conditions we have picked up in the course of the problem, namely that <math>b_1</math> is divisible by <math>8</math>, <math>b_2</ma6 KB (950 words) - 14:18, 15 January 2024
- ==Problem== From the initial problem statement, we have <math>1000w\cdot\frac{1}{4}t=\frac{1}{4}</math>.4 KB (592 words) - 19:02, 26 September 2020
- == Problem == ...B, ABBA, BABA, BBAA, ABBB, BABB, BBAB, BBBA</math>). Thus, we have <math>{9 \choose 2} \cdot 22 = 36\cdot22 = 792</math> numbers of this form.3 KB (508 words) - 01:16, 19 January 2024
- == Problem == ...h>. So the probability that they both pick two red candies is <math>\frac{9}{38} \cdot \frac{28}{153} = \frac{14}{323}</math>. The same calculation wo2 KB (330 words) - 13:42, 1 January 2015
- {{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
- == 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
