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- == Problem ==3 KB (527 words) - 12:50, 5 December 2024
- ==Problem==2 KB (313 words) - 17:09, 15 March 2017
- == Problem 37 ==932 bytes (136 words) - 17:27, 19 September 2022
- == Problem ==2 KB (264 words) - 11:39, 29 July 2024
- ==Problem== Thus, our problem has been simplified to maximizing the area of the blue quadrilateral.4 KB (631 words) - 09:08, 13 April 2023
- == Problem ==1 KB (231 words) - 01:40, 16 August 2023
- ==Problem==652 bytes (105 words) - 19:47, 26 January 2021
- == Problem ==3 KB (427 words) - 19:19, 13 January 2025
- == Problem ==477 bytes (69 words) - 09:43, 24 February 2022
- == Problem ==3 KB (442 words) - 19:55, 15 June 2018
- == Problem 37==3 KB (560 words) - 10:11, 3 May 2020
- ==Problem==2 KB (291 words) - 22:26, 28 December 2023
- ==Problem==1 KB (173 words) - 02:17, 3 July 2018
- ==Problem==1 KB (225 words) - 18:53, 15 July 2018
- ...''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}3 KB (520 words) - 18:27, 14 January 2020
- ==Problem==2 KB (245 words) - 01:13, 25 January 2020
- == Problem ==753 bytes (117 words) - 16:36, 21 July 2024
- 2 KB (266 words) - 15:34, 24 November 2020
- == Problem == We use [[mass points]] for this problem. Let <math>\text{m} A</math> denote the mass of point <math>A</math>.2 KB (329 words) - 10:38, 20 July 2024
- == Problem == |before=[[1963 TMTA High School Algebra I Contest Problem 36| Problem 36]]898 bytes (122 words) - 12:36, 2 February 2021
Page text matches
- <cmath>12-35-37</cmath> * [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]6 KB (978 words) - 12:02, 6 March 2025
- ...cluding Art of Problem Solving, the focus of MATHCOUNTS is on mathematical problem solving. Students are eligible for up to three years, but cannot compete be ...ics]]. The focus of MATHCOUNTS curriculum is in developing [[mathematical problem solving]] skills.11 KB (1,540 words) - 21:44, 13 March 2025
- ...[[MAML]] (Maine Association of Math Leagues) Meets. Training includes the problem set "Pete's Fabulous 42." ...ouri sends two teams, Red and Blue, and a few alternates for a total of 36-37 students to the Iowa site ARML competition.22 KB (3,532 words) - 11:25, 27 September 2024
- * <math>37! = 13763753091226345046315979581580902400000000</math> ([[2007 iTest Problems/Problem 6|Source]])10 KB (809 words) - 16:40, 17 March 2024
- 111 has a block length of three, and factors into 37 and 3. * [[2000 AMC 8 Problems/Problem 11]]10 KB (1,572 words) - 22:11, 22 September 2024
- == Problem 1 == [[2004 AMC 12A Problems/Problem 1|Solution]]13 KB (1,953 words) - 00:31, 26 January 2023
- == Problem 1 == [[2003 AMC 12B Problems/Problem 1|Solution]]13 KB (1,987 words) - 18:53, 10 December 2022
- ==Problem 1== [[1991 AJHSME Problems/Problem 1|Solution]]17 KB (2,246 words) - 13:37, 19 February 2020
- == Problem 1 == [[2005 AIME II Problems/Problem 1|Solution]]7 KB (1,119 words) - 21:12, 28 February 2020
- == Problem == ...> a_0, a_1,\ldots,a_m </math> be a sequence of reals such that <math>a_0 = 37, a_1 = 72, a_m = 0, </math> and <math> a_{k+1} = a_{k-1} - \frac 3{a_k} </m3 KB (499 words) - 18:52, 21 November 2022
- == Problem == ...There are fifteen of these (<math>2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43</math> and <math>47</math>) so there are <math> {15 \choose 2} =1052 KB (249 words) - 09:37, 23 January 2024
- == Problem == ...e sum of the possible remainders when <math> n </math> is divided by <math>37</math>?2 KB (250 words) - 23:56, 2 December 2024
- == Problem 1 == ...is the sum of the possible remainders when <math> n </math> is divided by 37?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 == ...) = 998 < 1000</math>. However, for <math>9k \ge 963 \Longrightarrow n \le 37</math>, we can no longer apply this approach.11 KB (1,857 words) - 12:57, 18 July 2024
- == Problem == ...+3}} \implies 2^{x} < 2^{40-(6n+3)}\implies 2^{x} < 2^{37-6n} \implies x < 37 - 6n</cmath>8 KB (1,283 words) - 19:19, 8 May 2024
- == Problem 1 == [[1994 AIME Problems/Problem 1|Solution]]7 KB (1,141 words) - 07:37, 7 September 2018
- == Problem 1 == [[2000 AIME II Problems/Problem 1|Solution]]6 KB (947 words) - 21:11, 19 February 2019
- == Problem == 13. <math>74 = 37 \cdot 2 => 35 + 39</math> - refuted8 KB (1,365 words) - 15:38, 10 December 2024
- == Problem == ...+9)\right]\left[(13^2+9)(19^2+9)\right]\left[(25^2+9)(31^2+9)\right]\left[(37^2+9)(43^2+9)\right]\left[(49^2+9)(55^2+9)\right]}=\frac{61^2+9}{1^2+9}=\box7 KB (965 words) - 23:39, 11 September 2024
