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  • == Problem == draw((0,0)--(18,0));
    2 KB (307 words) - 23:58, 17 November 2024
  • == Problem == ...[[User:Integralarefun|Integralarefun]] ([[User talk:Integralarefun|talk]]) 18:19, 27 September 2023 (EDT)
    2 KB (268 words) - 17:19, 27 September 2023
  • == Problem == ...in area. They also have the same altitude. If the base of the triangle is 18 inches, the median of the trapezoid is:
    878 bytes (143 words) - 19:56, 1 April 2017
  • *Two different [[prime number]]s between <math>4</math> and <math>18</math> are chosen. When their sum is subtracted from their product, which o ([[2000 AMC 12/Problem 6|Source]])
    4 KB (682 words) - 12:13, 8 December 2024
  • ...at <math>n>k</math>. The definition that <math>|N| > |K|</math> (as in our problem) is that there exists a surjective mapping from <math>N</math> to <math>K</ ...select a fifth sock without creating a pair. We may use this to prove the problem:
    11 KB (1,986 words) - 18:13, 19 June 2024
  • ...d only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. ==== Problem ====
    9 KB (1,703 words) - 00:20, 7 December 2024
  • * <math>18! = 6402373705728000</math> ([[2007 iTest Problems/Problem 6|Source]])
    10 KB (809 words) - 15:40, 17 March 2024
  • ...hat is the units digit of <math>k^2 + 2^k</math>? ([[2008 AMC 12A Problems/Problem 15]]) ...27^5=n^5</math>. Find the value of <math>{n}</math>. ([[1989 AIME Problems/Problem 9|1989 AIME, #9]])<br><br>
    16 KB (2,660 words) - 22:42, 28 August 2024
  • Other, odder inductions are possible. If a problem asks you to prove something for all integers greater than 3, you can use <m ...ernational Mathematics Olympiad | IMO]]. A good example of an upper-level problem that can be solved with induction is [http://www.artofproblemsolving.com/Fo
    5 KB (768 words) - 23:59, 28 September 2024
  • ([[2006 AMC 12A Problems/Problem 16|Source]]) ([[2006 AMC 12A Problems/Problem 21|Source]])
    9 KB (1,510 words) - 18:56, 16 January 2025
  • *[[2007 AMC 12A Problems/Problem 18]] *[[1984 AIME Problems/Problem 8|1984 AIME Problem 8]]
    5 KB (860 words) - 14:36, 10 December 2023
  • * [[2017_USAJMO_Problems/Problem_3 | 2017 USAJMO Problem 3]] * [[2016_AMC_12A_Problems/Problem_17 | 2016 AMC 12A Problem 17]] (See Solution 2)
    2 KB (318 words) - 22:35, 13 September 2024
  • ==Problem== ...and in this situation, the value of <math>I + M + O</math> would be <math>18</math>. Now, we use this process on <math>2001</math> to get <math>667 * 3
    2 KB (320 words) - 18:37, 30 November 2024
  • ...ns can significantly help in solving functional identities. Consider this problem: === Problem Examples ===
    4 KB (743 words) - 23:28, 17 November 2024
  • ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 10B Problems/Problem 1]]
    2 KB (182 words) - 20:57, 23 January 2021
  • ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 12B Problems/Problem 1 | Problem 1]]
    2 KB (210 words) - 23:06, 6 October 2014
  • * <math>91 \not\equiv 18 \pmod{6}</math> because <math>\frac{91 - 18}{6} = \frac{73}{6}</math>, which is not an integer. === Sample Problem ===
    16 KB (2,410 words) - 13:05, 3 January 2025
  • ...ngruence: <math>x \equiv 5 \pmod{21}</math>, and <math>x \equiv -3 \equiv 18 \pmod{21}</math>. (Note that two values of <math>x</math> that are congrue ...following topics expand on the flexible nature of modular arithmetic as a problem solving tool:
    14 KB (2,317 words) - 18:01, 29 October 2021
  • ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 10A Problems/Problem 1]]
    2 KB (180 words) - 17:06, 6 October 2014
  • == Problem 1 == ...<math> \overline{AC} </math> is perpendicular to <math> \overline{CD}, AB=18, BC=21, </math> and <math> CD=14. </math> Find the perimeter of <math> ABCD
    7 KB (1,173 words) - 02:31, 4 January 2023

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