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• 2010 AMC 10B
  ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2010 AMC 10B Problems/Problem 1|Problem 1]]
  2 KB (197 words) - 18:19, 6 October 2014
• 1994 AJHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1994 AJHSME Problems/Problem 1]]
  2 KB (141 words) - 23:41, 8 October 2014
• 1994 AJHSME Problems
  ==Problem 1== [[1994 AJHSME Problems/Problem 1|Solution]]
  16 KB (2,292 words) - 13:36, 19 February 2020
• 2010 AIME II Problems/Problem 3
  == Problem == ...ath>(1^{19})(2^{18})\cdots(19^1)</math> (or alternatively, <math>19! \cdot 18! \cdots 1!</math>.)
  2 KB (303 words) - 15:17, 12 May 2020
• 2010 AIME II Problems/Problem 11
  == Problem == Therefore, there are <math>3 \cdot 2 \cdot 3=18</math> cases.
  6 KB (1,057 words) - 01:58, 8 January 2023
• 2010 USAJMO Problems/Problem 1
  == Problem == ...<math>S_{18}</math> with an <math>m</math> of <math>2</math> and <math>P_{18}=26</math> with an <math>m</math> of <math>26</math>, note that their produ
  12 KB (2,338 words) - 20:30, 13 February 2024
• 2015 AMC 10B Problems/Problem 20
  ==Problem== ...d\textbf{(B) }\text{9}\qquad\textbf{(C) }\text{12}\qquad\textbf{(D) }\text{18}\qquad\textbf{(E) }\text{24}</math>
  4 KB (710 words) - 10:04, 17 August 2023
• 2010 AIME II Problems/Problem 15
  == Problem 15 == <cmath>NE=AN-AE=\frac 12\cdot AB - \frac{AC}{AC+BC}\cdot AB=\frac 5{18}</cmath>
  9 KB (1,523 words) - 15:24, 21 November 2023
• 1995 AJHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1995 AJHSME Problems/Problem 1]]
  2 KB (141 words) - 23:42, 8 October 2014
• 1995 AJHSME Problems
  ==Problem 1== [[1995 AJHSME Problems/Problem 1|Solution]]
  14 KB (2,096 words) - 18:29, 2 January 2023
• 1956 AHSME Problems/Problem 1
  ==Problem #1== <math>\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 16 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 36 \qquad\textbf{(E)}\ 64 </math>
  434 bytes (57 words) - 16:54, 14 March 2023
• 2010 AMC 12B Problems/Problem 2
  == Problem 2 == ...be <math>(8 - 2) \times (5 - 2) = 18</math>. The answer is then <math>40 - 18 = 22</math> <math>(A)</math>.
  802 bytes (110 words) - 10:58, 4 July 2013
• The Ross Program
  ...umber Theory]] and consist of daily lectures, seminar groups, and rigorous problem sets. ...iversity in an effort to emphasize development of students' creativity and problem-solving skills. In 1964, the program was moved to Ohio State University. Ro
  3 KB (474 words) - 20:48, 28 May 2013
• 2010 AMC 12B Problems/Problem 7
  == Problem == <math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qq
  2 KB (302 words) - 15:49, 23 November 2023
• 2010 AMC 12B Problems/Problem 8
  == Problem == {{AMC10 box|year=2010|num-b=16|num-a=18|ab=B}}
  2 KB (367 words) - 13:40, 11 July 2021
• 2010 AMC 12B Problems/Problem 11
  == Problem == .... This adds another 9 palindromes to the list, bringing our total to <math>18/90 = \boxed {\frac{1}{5} } = \boxed {E}</math>
  2 KB (354 words) - 19:54, 10 March 2024
• 2010 AMC 12B Problems/Problem 16
  ...12B Problems|2010 AMC 12B #16]] and [[2010 AMC 10B Problems|2010 AMC 10B #18]]}} == Problem ==
  3 KB (419 words) - 13:42, 11 July 2021
• 2010 AMC 12B Problems/Problem 17
  == Problem == <math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qq
  3 KB (440 words) - 19:34, 17 October 2022
• 2010 AMC 12B Problems/Problem 19
  == Problem == {{AMC12 box|year=2010|num-b=18|num-a=20|ab=B}}
  6 KB (990 words) - 19:14, 25 March 2024
• 2010 AMC 10A Problems/Problem 3
  == Problem 3 == \mathrm{(C)}\ 18
  2 KB (276 words) - 19:50, 5 March 2023

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