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• Quaternion
  1,007 bytes (155 words) - 20:47, 14 October 2013
• 2020 AMC 10A Problems
  <cmath>\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}</cmath> What is the sum of all real numbers <math>x</math> for which <math>|x^2-12x+34|=2?</math>
  13 KB (1,968 words) - 18:32, 29 February 2024
• Asymptote: Logical Operators and Loops
  if (floor((i-j)/2)==((i-j)/2)) if (floor((i-j)/2)==((i-j)/2))
  2 KB (324 words) - 16:50, 2 October 2016
• Pigeonhole Principle/Solutions
  ...riends, then the remaining friends must have from <math>1</math> to <math>n-2</math> friends for the remaining friends not to also have no friends. By p ...same remainder when divided by 5. Since there are 5 possible remainders (0-4), by the pigeonhole principle, at least two of the integers must share the
  10 KB (1,617 words) - 01:34, 26 October 2021
• 2002 USAMO Problems/Problem 3
  ...it is necessary and sufficient that <math> P(x) = Q(x) + \prod_{i=1}^{n}(x-x_i) </math>. ...s. But a polynomial with real coefficients must have an even number of non-real roots, so <math>P(x) </math> must have <math>n </math> real roots. Sim
  4 KB (688 words) - 13:38, 4 July 2013
• 2007 AIME I Problems/Problem 10
  [[Image:AIME I 2007-10.png]] ...ath> (<math>j < k</math>) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that <math>j\ne k</math> to solve the p
  13 KB (2,328 words) - 00:12, 29 November 2023
• 2004 USAMO Problems
  ...egers <math> x, y \in S </math>, if <math> x+y \in S </math>, then <math> x-y \in S </math>. * [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamo-test.shtml 2004 USAMO Problems]
  3 KB (474 words) - 09:20, 14 May 2021
• 2007 AIME II Problems/Problem 3
  ...ince <math>\triangle ABE \cong \triangle CDF</math> and are both <math>5-12-13</math> [[right triangle]]s, we can come to the conclusion that the two ne A slightly more analytic/brute-force approach:
  6 KB (933 words) - 00:05, 8 July 2023
• 2007 AIME II Problems/Problem 13
  {{AIME box|year=2007|n=II|num-b=12|num-a=14}}
  3 KB (600 words) - 11:10, 22 January 2023
• 2007 AIME II Problems
  <div style="text-align:center;">[[Image:2007 AIME II-3.png]]</div> ..._{i+1}</math> if <math>a_{i}</math> is [[even]]. How many four-digit parity-monotonic integers are there?
  9 KB (1,435 words) - 01:45, 6 December 2021
• 2003 USAMO Problems/Problem 1
  ...-1} + a </math> is divisible by 5, which is true when <math> k \equiv -3^{n-1}a \pmod{5} </math>. Since there is an odd digit in each of the residue cl ...s after it, and all multiples of 5 end in 5. Therefore, <math> a*10^x*5^{n-x}</math> always contains a 5 as its <math> (x+1)^{st}</math> digit, and we
  4 KB (736 words) - 22:17, 3 March 2023
• 2003 USAMO Problems/Problem 3
  ...{} c_j = \cdots = c_{j+i} = \cdots = c_k = j </math> (<math> 0 \le i \le k-j</math>). ''Proof.'' Since the <math>j </math> terms <math>c_0, \ldots, c_{j-1} </math> are all less than <math>j </math>, no other terms that precede <m
  3 KB (636 words) - 13:39, 4 July 2013
• Pentagon
  1 KB (191 words) - 09:59, 6 June 2022
• 2004 IMO Shortlist Problems/N2
  ...e-i}) = p^e + \sum_{i=0}^{e-1}p^i(p^{e-i} - p^{e-i-1}) = p^e + e(p^e - p^{e-1}) </math>. </center> ...^{e_i} - e_i \cdot p_i^{e_i}]}{\prod p_i^{e_i}} = \prod \left(e_i \frac{p_i-1}{p_i} + 1 \right) </math>.
  6 KB (1,007 words) - 09:10, 29 August 2011
• Parallelepiped
  2 KB (248 words) - 20:08, 17 August 2023
• Chicken McNugget Theorem
  ...am + bn</math> for [[nonnegative]] integers <math>a, b</math> is <math>mn-m-n</math>. ...e proof is based on the fact that in each pair of the form <math>(k, mn-m-n-k)</math>, exactly one element is expressible.
  17 KB (2,748 words) - 19:22, 24 February 2024
• Dodecagon
  A '''dodecagon''' is a 12-sided [[polygon]]. The sum of its internal [[angle]]s is <math>1800^{\circ}<
  1 KB (219 words) - 13:08, 15 June 2018
• Triangular number
  <math>1 +2+3 + 4............. +(n-1)+(n)</math>. ...math>T_{n} = 1 + 2 + \ldots + (n-1) + n = (1 + 2 + \ldots + n-1) + n = T_{n-1} + n</math>.
  2 KB (275 words) - 08:39, 7 July 2021
• 2007 AMC 12A Problems/Problem 24
  ...mod{4}</math> is that <math>v_2(2j+1)=0,</math> so we must have <math>v_2(n-1)>v_2(n+1)</math> since <math>v_2(2k) \geq 1 >v_2(2j+1).</math> Therefore, {{AMC12 box|year=2007|num-b=23|num-a=25|ab=A}}
  4 KB (588 words) - 14:40, 23 August 2023
• User talk:1=2
  * [[1962 AHSME]] (Complete '''w/o solutions w/o problems 41-50''')
  13 KB (1,464 words) - 17:28, 6 January 2024

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