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• 2020 USAMO Problems
  ...th> be an odd prime. An integer <math>x</math> is called a <i>quadratic non-residue</i> if <math>p</math> does not divide <math>x - t^2</math> for any i ...p</math>, and both <math>a</math> and <math>4 - a</math> are quadratic non-residues. Calculate the remainder when the product of the elements of <math>
  4 KB (614 words) - 13:47, 22 November 2023
• 1979 IMO Problems
  ...ove that:<cmath> a_{2n-1}=0, \quad a_{2n}={(2+\sqrt2)^{n-1} - (2-\sqrt2)^{n-1} \over\sqrt2}. </cmath>
  3 KB (448 words) - 11:07, 14 June 2024
• Taylor polynomial
  ...f(a) + f'(a)(x - a) + \frac{f''(a)(x-a)^2}{2} + \dots + \frac{f^{(n)}(a)(x-a)^n}{n!}.</cmath> In the formula above, <math>f^{(k)}</math> denotes the [[ Taylor polynomials are often used to approximate non-polynomial functions that cannot be calculated exactly, such as [[Trigonomet
  7 KB (1,280 words) - 12:39, 9 December 2022
• 2020 USOJMO Problems/Problem 5
  ...th>(i,j)</math> satisfying <math>1\leq i<j\leq 100</math> and <math>|a_ib_j-a_jb_i|=1</math>. Determine the largest possible value of <math>N</math> ove ...h>(i, j)</math> good if <math>1\leq i < j \leq 100</math> and <math>|a_ib_j-a_jb_i|=1</math>. Note that we can reorder the pairs <math>(a_1, b_1), (a_2,
  8 KB (1,486 words) - 19:16, 6 October 2023
• 2020 USOJMO Problems/Problem 4
  ...>y+x+\angle GBE=\frac{\pi}{2}</math>, or <math>\angle GBE = \frac{\pi}{2}-x-y</math>. ...= \pi - \angle BEG - \angle GBE = \pi - (\frac{\pi}{2}-y)-(\frac{\pi}{2}-x-y)=x+2y</math>. But <math>\angle FGB = \angle BGE</math>, so <math>\angle FG
  6 KB (1,131 words) - 19:15, 6 October 2023
• 1989 IMO Problems/Problem 3
  ...f <math>k\ge \sqrt{2n}+\frac 12</math>, then <math>k(k-1)\ge 2n-\frac 14>2n-2</math>, contradicting the last inequality in <math>(*)</math>, so we're do ...ity). So now <math>\dbinom{k}{2}\le E(\dbinom{f}{2})\le (2/n)((n-1)(n)/2)=n-1</math>, which is what we want.
  3 KB (580 words) - 11:56, 30 January 2021
• 1995 IMO Problems/Problem 5
  1 KB (214 words) - 20:19, 5 July 2020
• 2020 USOJMO Problems
  ...led diagram. Failure to meet this requirement will result in an automatic 1-point deduction. ...th>(i,j)</math> satisfying <math>1\leq i<j\leq 100</math> and <math>|a_ib_j-a_jb_i|=1</math>. Determine the largest possible value of <math>N</math> ove
  5 KB (799 words) - 16:46, 5 August 2023
• 2021 AIME I Problems/Problem 3
  ...ers <math>a > b \ge 0</math> (note that if <math>a=b</math>, then <math>2^a-2^b = 0</math>). We first observe <math>a</math> must be at most 10; if <mat ...lts in a different positive integer <math>n</math>. Then there are <math>55-5 = \boxed{050}</math> integers which can be expressed as a difference of tw
  7 KB (1,174 words) - 08:21, 13 May 2023
• 2021 AIME I Problems/Problem 10
  ...th> and <math>m+18k</math>, then <math>d\mid18k+18</math>, so <math>d\mid{m-18}</math>. This follows from the Euclidean Algorithm. &\implies d\mid m-18 \text{ and } d\mid k'+1.
  6 KB (1,015 words) - 16:30, 26 January 2024
• 2021 AIME I Problems/Problem 12
  <ol style="margin-left: 1.5em;"> [[File:AIME-I-2021-12a.png| 600px |center]]
  8 KB (1,309 words) - 11:57, 3 December 2023
• 2021 AIME I Problems/Problem 14
  Then <math>\sigma(a^n)-1 = \sum_{i=1}^n a^i = a\left(\frac{a^n - 1}{a-1}\right)</math>. In order for this expression to be divisible by <math>2021 ...</math> by LTE, we have <math>v_{43}(a)+v_{43}{(a-1)}+v_{43}{(n)}-v_{43}{(a-1)} \geq 1,</math> which simplifies to <math>v_{43}(n) \geq 1,</math> which
  13 KB (2,185 words) - 02:28, 13 November 2023
• Pólya Enumeration Theorem
  2 KB (308 words) - 19:51, 24 November 2021
• 2021 AIME I Problems
  Call a three-term strictly increasing arithmetic sequence of integers special if the sum ...k</math> such that the two parabolas<cmath>y=x^2-k~~\text{and}~~x=2(y-20)^2-k</cmath>intersect in four distinct points, and these four points lie on a c
  7 KB (1,182 words) - 14:54, 13 March 2023
• 2022 AIME I Problems/Problem 3
  P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C);
  24 KB (3,832 words) - 20:59, 2 March 2024
• 2022 AIME I Problems/Problem 12
  ...= \binom{4}{2}</math> - let's prove that <math>S_n = n \cdot \binom{2n-2}{n-1}</math> (note that you can assume this and answer the problem if you're ru ...bsets, so the total number of ways to choose the subsets are <math>\binom{n-1}{l}^2</math>.
  9 KB (1,471 words) - 16:41, 1 February 2024
• 2022 AIME II Problems
  ...at<cmath>\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.</cmath>Find the least possible value of <math>a+b.</math> ...p of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to <math>1000</math> cents,
  8 KB (1,397 words) - 09:18, 15 August 2022
• 2022 AMC 12B Problems
  Define <math>x\diamond y</math> to be <math>|x-y|</math> for all real numbers <math>x</math> and <math>y.</math> What is th Let <math> f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n </math>, where <math>i = \sqrt{-1}</math>. What is <
  15 KB (2,233 words) - 13:02, 10 November 2023
• 2021 Fall AMC 10B Problems
  <cmath>x(x-y)+y(y-z)+z(z-x) = 1?</cmath> <math>\textbf{(B)} \: x=y-1</math> and <math>y=z-1</math>
  16 KB (2,450 words) - 09:34, 9 June 2024
• 2021 Fall AMC 10A Problems
  What is the value of <math>\frac{(2112-2021)^2}{169}</math>? The six-digit number <math>\underline{2}\,\underline{0}\,\underline{2}\,\underline{1
  14 KB (2,162 words) - 21:33, 2 November 2023

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