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  • <cmath>12-35-37</cmath> * [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]
    6 KB (943 words) - 09:44, 17 January 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) - 10:25, 27 September 2024
  • * <math>37! = 13763753091226345046315979581580902400000000</math> ([[2007 iTest Problems/Problem 6|Source]])
    10 KB (809 words) - 15:40, 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) - 21:11, 22 September 2024
  • == Problem 1 == [[2004 AMC 12A Problems/Problem 1|Solution]]
    13 KB (1,953 words) - 23:31, 25 January 2023
  • == Problem 1 == [[2003 AMC 12B Problems/Problem 1|Solution]]
    13 KB (1,987 words) - 17:53, 10 December 2022
  • ==Problem 1== [[1991 AJHSME Problems/Problem 1|Solution]]
    17 KB (2,246 words) - 12:37, 19 February 2020
  • == Problem 1 == [[2005 AIME II Problems/Problem 1|Solution]]
    7 KB (1,119 words) - 20:12, 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} </m
    3 KB (499 words) - 17:52, 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} =105
    2 KB (249 words) - 08:37, 23 January 2024
  • == Problem == ...e sum of the possible remainders when <math> n </math> is divided by <math>37</math>?
    2 KB (250 words) - 22:56, 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) - 12:34, 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 f
    6 KB (899 words) - 19:58, 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) - 11:57, 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) - 18:19, 8 May 2024
  • == Problem 1 == [[1994 AIME Problems/Problem 1|Solution]]
    7 KB (1,141 words) - 06:37, 7 September 2018
  • == Problem 1 == [[2000 AIME II Problems/Problem 1|Solution]]
    6 KB (947 words) - 20:11, 19 February 2019
  • == Problem == 13. <math>74 = 37 \cdot 2 => 35 + 39</math> - refuted
    8 KB (1,365 words) - 14:38, 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}=\box
    7 KB (965 words) - 22:39, 11 September 2024
  • == Problem == ...0 &\equiv 0 \pmod {37}, \ -d &\equiv -9 \pmod {37}, \ d &\equiv 9 \pmod {37}.\end{align*} Since <math>d</math> is a digit, it must be equal to <math>9<
    4 KB (584 words) - 13:38, 11 August 2024

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