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• Law of Cosines
  draw(D--(30,4)--(34,4)--(34,0)--D); 1. If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
  6 KB (1,003 words) - 09:11, 7 June 2023
• Angle bisector
  A=(4,2); D=(3,4);
  3 KB (575 words) - 15:27, 19 March 2023
• Law of Sines
  pair A=(-1,5), B=(-4,-1), C=(4,-1), D, O; *If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
  4 KB (658 words) - 16:19, 28 April 2024
• Sequence
  ...ath>\mathbb{N}</math> corresponds to the sequence <math>X = (x_n) = (0, 1, 4, 9, 16, \ldots)</math>. ...> is called the [[limit]] of <math>(x_n)</math> and is written <math>\lim_{n \to \infty} x_n</math>. The statement that <math>(x_n)</math> converges to
  2 KB (413 words) - 21:18, 13 November 2022
• Geometric sequence
  For example, <math>1, 2, 4, 8</math> is a geometric sequence with common ratio <math>2</math> and <mat ...progression if and only if <math>a_2 / a_1 = a_3 / a_2 = \cdots = a_n / a_{n-1}</math>. A similar definition holds for infinite geometric sequences. It
  4 KB (644 words) - 12:55, 7 March 2022
• Arithmetic sequence
  ...on difference <math>-8</math>; however, <math>7, 0, 7, 14</math> and <math>4, 12, 36, 108, \ldots</math> are not arithmetic sequences, as the difference ...progression if and only if <math>a_2 - a_1 = a_3 - a_2 = \cdots = a_n - a_{n-1}</math>. A similar definition holds for infinite arithmetic sequences. It
  4 KB (736 words) - 02:00, 7 March 2024
• Fermat's Last Theorem
  ...\geq 3</math>, there are no solutions to the equation <math>a^n + b^n = c^n</math>. ...he never published it, though he did publish a proof for the case <math>n=4</math>. It seems unlikely that he would have circulated a proof for the sp
  3 KB (453 words) - 11:13, 9 June 2023
• 2005 USAMO Problems/Problem 4
  ...ece of length <math>k_i</math> from the end of leg <math>L_i \; (i = 1,2,3,4)</math> and still have a stable table? For <math>0 \le x \le n</math>, it is easy to see that the number of stable tables is <math>(x+1)^2
  7 KB (1,276 words) - 20:51, 6 January 2024
• Diophantine equation
  If <math>n>1</math>, <math>2n, n^2 - 1, n^2 + 1</math> is a Pythagorean triple. ...ny <math>m,n</math>(<math>m>n</math>), we have <math>m^2 - n^2, 2mn, m^2 + n^2</math> is a Pythagorean triple.
  9 KB (1,434 words) - 13:10, 20 February 2024
• Area
  ...s to the [[circumcenter]]. This creates a triangle that is <math>\frac{1}{n},</math> of the total area (consider the regular [[octagon]] below as an ex ...ound using [[trigonometry]] to be of length <math>\frac s2 \cot \frac{180}{n}^{\circ}</math>.
  6 KB (1,181 words) - 22:37, 22 January 2023
• Floor function
  ...r+\left\lfloor a+\frac{1}{n}\right\rfloor+\ldots+\left\lfloor a+\frac{n-1}{n}\right\rfloor</cmath> *<math>\lfloor -3.2 \rfloor = -4</math>
  3 KB (508 words) - 21:05, 26 February 2024
• Pascal's triangle
  ...he sum of the values on row <math>n</math> of Pascal's Triangle is <math>2^n</math>. ...ved from the combinatorics identity <math>{n \choose k}+{n \choose k+1} = {n+1 \choose k+1}</math>. Thus, any number in the interior of Pascal's Triang
  5 KB (838 words) - 17:20, 3 January 2023
• Newton's Sums
  Consider a polynomial <math>P(x)</math> of degree <math>n</math>, <center><math> P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0</math></center>
  4 KB (690 words) - 13:11, 20 February 2024
• 1985 IMO Problems/Problem 5
  ...ath> and <math>BC</math> again at distinct points <math>K</math> and <math>N</math> respectively. Let <math>M</math> be the point of intersection of the {{IMO box|year=1985|num-b=4|num-a=6}}
  3 KB (496 words) - 13:35, 18 January 2023
• Trigonometry
  ...=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} \cdots</cmath>
  8 KB (1,217 words) - 20:15, 7 September 2023
• Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality
  ...</cmath> where <math>n_1>1,~~0<n_2<1,~~-1<n_3<0,~~n_4<-1</math>, and <math>n</math> is the root mean power. ...1}{a}}</math> and the harmonic mean's root mean power is -1 as <math>\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}=\sqrt[-1]{\frac{x_1^{-1}+\cdots+x_a^{-
  5 KB (912 words) - 20:06, 14 March 2023
• Goldbach Conjecture
  <math>4 = 2 + 2</math> ...might expect the total number of ways to write a large even integer <math>n</math> as the sum of two odd primes to be roughly
  7 KB (1,201 words) - 16:59, 19 February 2024
• Riemann Hypothesis
  ...function, it is easy to see that <math>\zeta(s)=0</math> when <math>s=-2,-4,-6,\ldots</math>. These are called the trivial zeros. This hypothesis is o ...uld hold. The Riemann Hypothesis would also follow if <math>M(n)\le C\sqrt{n}</math> for any constant <math>C</math>.
  2 KB (425 words) - 12:01, 20 October 2016
• Quadratic residues
  ...ulo]] <math>m</math> if there is some integer <math>n</math> so that <math>n^2-a</math> is [[divisibility | divisible]] by <math>m</math>. ...p-1}{2}}</math>, so <math>\left(\frac{-1}{p}\right)=1 \iff p \equiv 1 \mod 4</math>
  5 KB (778 words) - 13:10, 29 November 2017
• Power of a Point Theorem
  Let <math>P</math> be a point, and let <math>S</math> be an <math>n</math>-sphere. Let two arbitrary lines passing through <math>P</math> inter ...th> intersect at <math>R</math>. If <math>AR:BR=1:4</math> and <math>CR:DR=4:9</math>, find the ratio <math>AB:CD</math>.
  5 KB (827 words) - 17:30, 21 February 2024
• 2004 AIME I
  * [[2004 AIME I Problems/Problem 4]] {{AIME box|year=2004|n=I|before=[[2003 AIME I]], [[2003 AIME II|II]]|after=[[2004 AIME II]]}}
  1 KB (135 words) - 18:15, 19 April 2021
• Collatz Problem
  ...sitive integer]] <math>n</math>, the sequence <math>\{n,f(n),f(f(n)),f(f(f(n))),\ldots\}</math> contains 1. This conjecture is still open. Some people h ==Properties of <math>f(n)</math> ==
  1 KB (231 words) - 19:45, 24 February 2020
• 2004 AIME II
  * [[2004 AIME II Problems/Problem 4]] {{AIME box|year=2004|n=II|before=[[2004 AIME I]]|after=[[2005 AIME I]], [[2005 AIME II|II]]}}
  1 KB (135 words) - 12:24, 22 March 2011
• 2005 AIME I
  * [[2005 AIME I Problems/Problem 4 | Problem 4]] {{AIME box|year=2005|n=I|before=[[2004 AIME I|2004 AIME I]], [[2004 AIME II|II]]|after=[[2005 AIME
  1 KB (154 words) - 12:30, 22 March 2011
• 2006 AIME I
  * [[2006 AIME I Problems/Problem 4]] {{AIME box|year=2006|n=I|before=[[2005 AIME I]], [[2005 AIME II|II]]|after=[[2006 AIME II]]}}
  1 KB (135 words) - 12:31, 22 March 2011
• 2005 AIME II
  * [[2005 AIME II Problems/Problem 4]] {{AIME box|year=2005|n=II|before=[[2005 AIME I]]|after=[[2006 AIME I]], [[2006 AIME II|II]]}}
  1 KB (135 words) - 12:30, 22 March 2011
• Mock AMC
  | n/a | n/a
  51 KB (6,175 words) - 20:58, 6 December 2023
• Bertrand's Postulate
  ...em. A more widely known version states that there is a prime between <math>n</math> and <math>2n</math>. ...closer look at the [[combinations|binomial coefficient]] <math>\binom{2n}{n}</math>. Assuming that the reader is familiar with that proof, the Bertrand
  2 KB (309 words) - 21:43, 11 January 2010
• Riemann zeta function
  <cmath>\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}= 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots</cmath>
  9 KB (1,547 words) - 03:04, 13 January 2021
• Modular arithmetic/Introduction
  draw((0,0), linewidth(4)); <math>1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, \ldots </math>
  15 KB (2,396 words) - 20:24, 21 February 2024
• Perfect square
  ...so that <math>m^2=n</math>. The first few perfect squares are <math>0, 1, 4, 9, 16, 25, 36</math>. ...th>n</math> square numbers (starting with <math>1</math>) is <math>\frac{n(n+1)(2n+1)}{6}</math>
  954 bytes (155 words) - 01:14, 29 November 2023
• Modular arithmetic/Intermediate
  ...th>n</math>), if the difference <math>{a - b}</math> is divisible by <math>n</math>. ...ntegers modulo <math>n</math>''' (usually known as "the integers mod <math>n</math>," or <math>\mathbb{Z}_n</math> for short). This structure gives us
  14 KB (2,317 words) - 19:01, 29 October 2021
• 2006 AIME I Problems
  {{AIME Problems|year=2006|n=I}} == Problem 4 ==
  7 KB (1,173 words) - 03:31, 4 January 2023
• 2006 AIME I Problems/Problem 15
  === Solution 4 === {{AIME box|year=2006|n=I|num-b=14|after=Last Question}}
  6 KB (910 words) - 19:31, 24 October 2023
• 2006 AIME I Problems/Problem 14
  ...> is not divisible by the square of any prime. Find <math> \lfloor m+\sqrt{n}\rfloor. </math> (The notation <math> \lfloor x\rfloor </math> denotes the triple M=(B+C)/2,S=(4*A+T)/5;
  6 KB (980 words) - 21:45, 31 March 2020
• 2006 AIME I Problems/Problem 13
  ...h> S_n=\sum_{k=1}^{2^{n-1}}g(2k). </math> Find the greatest integer <math> n </math> less than 1000 such that <math> S_n </math> is a [[perfect square]] .../math> but not by <math>2^n, \ldots,</math> and <math>2^{n-1}-2^{n-2} = 2^{n-2}</math> elements of <math>S</math> that are divisible by <math>2^1</math>
  10 KB (1,702 words) - 00:45, 16 November 2023
• 2006 AIME I Problems/Problem 10
  ...ough the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra {{AIME box|year=2006|n=I|num-b=9|num-a=11}}
  4 KB (731 words) - 17:59, 4 January 2022
• 2006 AIME I Problems/Problem 8
  pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1, label("$\mathcal{Q}$",(4.2,-1),NW);
  5 KB (730 words) - 15:05, 15 January 2024
• 2006 AIME I Problems/Problem 7
  for(int i=0; i<4; i=i+1) { pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10);
  4 KB (709 words) - 01:50, 10 January 2022
• 2006 AIME I Problems/Problem 5
  <math> b = 4 </math> <math>b=4</math>
  3 KB (439 words) - 18:24, 10 March 2015
• 2006 AIME I Problems/Problem 4
  ...oduct <math> 1!2!3!4!\cdots99!100!. </math> Find the remainder when <math> N </math> is divided by <math>1000</math>. ...<math>5</math> dividing it, for <math>76</math> extra; every n! for <math>n\geq 50</math> has one more in addition to that, for a total of <math>51</ma
  2 KB (278 words) - 08:33, 4 November 2022
• 2006 AIME I Problems/Problem 3
  ...math> so we take <math>N_0 = 25</math> and <math>n = 2.</math> Then <cmath>N = 7 \cdot 10^2 + 25 = \boxed{725},</cmath> and indeed, <math>725 = 29 \cdot ...ow that <math>N<1000</math> (because this is an AIME problem). Thus, <math>N</math> has <math>1,</math> <math>2</math> or <math>3</math> digits. Checkin
  4 KB (622 words) - 03:53, 10 December 2022
• 2022 AIME I Problems/Problem 1
  Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4) ...]</math> from <math>5\cdot[(1)+(2)]</math> to get <cmath>b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.</cmath>
  4 KB (670 words) - 13:03, 13 November 2023
• 2021 JMC 10
  ...bf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 7</math> ==Problem 4==
  12 KB (1,784 words) - 16:49, 1 April 2021
• 2006 AMC 12B Problems
  ...ne <math>x\spadesuit y = (x + y)(x - y)</math>. What is <math>3\spadesuit(4\spadesuit 5)</math>? == Problem 4 ==
  13 KB (2,058 words) - 12:36, 4 July 2023
• 2006 AMC 12A Problems
  == Problem 4 == [[2006 AMC 12A Problems/Problem 4|Solution]]
  15 KB (2,223 words) - 13:43, 28 December 2020
• 2005 AMC 12A Problems
  ...8 \qquad (\mathrm {B}) \ -4 \qquad (\mathrm {C})\ 2 \qquad (\mathrm {D}) \ 4 \qquad (\mathrm {E})\ 8 == Problem 4 ==
  13 KB (1,971 words) - 13:03, 19 February 2020
• 2004 AMC 12A Problems
  == Problem 4 == [[2004 AMC 12A Problems/Problem 4|Solution]]
  13 KB (1,953 words) - 00:31, 26 January 2023
• 2003 AMC 12A Problems
  Members of the Rockham Soccer League buy socks and T-shirts. Socks cost &#36;4 per pair and each T-shirt costs &#36;5 more than a pair of socks. Each memb <math> \mathrm{(A) \ } 4.5\qquad \mathrm{(B) \ } 9\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 18
  13 KB (1,955 words) - 21:06, 19 August 2023
• 2002 AMC 12A Problems
  <math>(2x+3)(x-4)+(2x+3)(x-6)=0 </math> <math> \mathrm{(A) \ } \frac{7}{2}\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 13 </
  12 KB (1,792 words) - 13:06, 19 February 2020
• 2000 AMC 12 Problems
  ...00^{4000} \qquad \textbf{(D)}\ 4,000,000^{2000} \qquad \textbf{(E)}\ 2000^{4,000,000}</math> == Problem 4 ==
  13 KB (1,948 words) - 12:26, 1 April 2022
• 2001 AMC 12 Problems
  Let <math>P(n)</math> and <math>S(n)</math> denote the product and the sum, respectively, of the digits ...example, <math>P(23) = 6</math> and <math>S(23) = 5</math>. Suppose <math>N</math> is a
  13 KB (1,957 words) - 12:53, 24 January 2024
• 2002 AMC 12B Problems
  \qquad\mathrm{(C)}\ 4 when <math>x=4</math>?
  10 KB (1,547 words) - 04:20, 9 October 2022
• 2003 AMC 12B Problems
  <cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath> == Problem 4 ==
  13 KB (1,987 words) - 18:53, 10 December 2022
• 2004 AMC 12B Problems
  == Problem 4 == [[2004 AMC 12B Problems/Problem 4|Solution]]
  13 KB (2,049 words) - 13:03, 19 February 2020
• 2005 AMC 12B Problems
  ...x</math> has the property that <math>x\%</math> of <math>x</math> is <math>4</math>. What is <math>x</math>? \mathrm{(B)}\ 4 \qquad
  12 KB (1,781 words) - 12:38, 14 July 2022
• 2006 AMC 12B Problems/Problem 14
  ...h>, <math>J</math> and <math>N</math> are all positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches? <math>253=N(4B+5J)</math>
  1 KB (227 words) - 17:21, 8 December 2013
• 2006 AMC 12B Problems/Problem 18
  ...e object only makes <math>1</math> move, it is obvious that there are only 4 possible points that the object can move to. At this point we can guess that for n moves, there are <math>(n + 1)^2</math> different endpoints. Thus, for 10 moves, there are <math>11^2
  2 KB (354 words) - 16:57, 28 December 2020
• 2006 AMC 12B Problems/Problem 20
  ...th hold at the same time if and only if <math>10^k \leq x < \frac{10^{k+1}}4</math>. ...math>k</math> the length of the interval <math>\left[ 10^k, \frac{10^{k+1}}4 \right)</math> is <math>\frac 32\cdot 10^k</math>.
  3 KB (485 words) - 14:09, 21 May 2021
• 2006 AMC 12B Problems/Problem 22
  ...divisible by <math>10</math>. What is the smallest possible value of <math>n</math>? n{5}\right\rfloor +
  5 KB (881 words) - 15:52, 23 June 2021
• 2006 AMC 12B Problems/Problem 23
  MP("B",D(B),plain.N,f); MP("B",D(B),plain.N,f);
  7 KB (1,169 words) - 14:04, 10 June 2022
• 2006 AMC 12A Problems/Problem 17
  pair f = (4.34, 74.58); label("F", f, N);
  6 KB (958 words) - 23:29, 28 September 2023
• 2006 AMC 12A Problems/Problem 23
  Given a finite sequence <math>S=(a_1,a_2,\ldots ,a_n)</math> of <math>n</math> real numbers, let <math>A(S)</math> be the sequence <math>\left(\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2}\right)</math>
  3 KB (466 words) - 22:40, 29 September 2023
• 2006 AMC 12A Problems/Problem 24
  <cmath>2(x+x^3+x^5\cdots)(1+x^2+x^4\cdots)(1+x+x^2+x^3\cdots) = \frac{2x}{(1-x)^3(1+x)^2}</cmath> ...n-1}+...+P^{n-1}x+P^n)+(x^n-Px^{n-1}+...-P^{n-1}x+P^n)</math>, where <math>n=2006</math> (we may omit the coefficients, as we are seeking for the number
  8 KB (1,332 words) - 17:37, 17 September 2023
• 2006 AMC 12A Problems/Problem 25
  ...any ways are there to choose <math>k</math> elements from an ordered <math>n</math> element [[set]] without choosing two consecutive members? ...n with <math>k</math> elements where the largest possible element is <math>n-k+1</math>, with no restriction on consecutive numbers. Since this process
  8 KB (1,405 words) - 11:52, 27 September 2022
• 2005 AMC 12A Problems/Problem 20
  ...h>, and <math>f^{[n + 1]}(x) = f^{[n]}(f(x))</math> for each integer <math>n \geq 2</math>. For how many values of <math>x</math> in <math>[0,1]</math> ...h>\frac{5}{2^{n+1}}</math>, <math>\cdots</math> ,<math>\frac{2^{n+1}-1}{2^{n+1}}</math>.
  3 KB (437 words) - 23:49, 28 September 2022
• 2005 AMC 12B Problems/Problem 12
  ...f <math>m,n,</math> and <math>p</math> is zero. What is the value of <math>n/p</math>? ...xtbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}</math>
  2 KB (317 words) - 12:27, 16 December 2021
• 2005 AMC 12B Problems/Problem 21
  ...e greatest integer <math>k</math> such that <math>7^k</math> divides <math>n</math>? ...\mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}</math>
  888 bytes (140 words) - 20:04, 24 December 2020
• 2005 AMC 12B Problems/Problem 22
  <cmath>z_{n+1} = \frac {iz_{n}}{\overline {z_{n}}},</cmath> where <math>\overline {z_{n}}</math> is the [[complex conjugate]] of <math>z_{n}</math> and <math>i^{2}=-1</math>. Suppose that <math>|z_{0}|=1</math> and
  4 KB (660 words) - 17:40, 24 January 2021
• 2005 AMC 12B Problems/Problem 24
  ...th> are relatively prime positive integers. What is the value of <math>m + n</math>? draw((0,-0.5)--(0,4),Arrows);
  4 KB (761 words) - 09:10, 1 August 2023
• 2005 AMC 12B Problems/Problem 25
  ...osing which ant moves to <math>A</math>. Hence, there are <math>2 \times 2=4</math> ways the ants can move to different points. ...ath> can actually move to four different points, there is a total of <math>4 \times 20=80</math> ways the ants can move to different points.
  10 KB (1,840 words) - 21:35, 7 September 2023
• Power of a Point Theorem/Introductory Problem 4
  ...sect at a right angle at <math>E</math> . Given that <math> BE = 16, DE = 4, </math> and <math> AD = 5 </math>, find <math> CE </math>. pair D = (0,4);
  1 KB (177 words) - 02:14, 26 November 2020
• Root (polynomial)
  ...wice, triple roots three times, and so on, there are in fact exactly <math>n</math> complex roots of <math>P(x)</math>. ...1}x^{n-1}}{c_n} + \dots + \frac{c_1x}{c_n} + \frac{c_0}{c_n} = \sum_{j=0}^{n} \frac{c_jx^j}{c_n}.</cmath>
  8 KB (1,427 words) - 21:37, 13 March 2022
• 2006 AMC 10A Problems
  == Problem 4 == [[2006 AMC 10A Problems/Problem 4|Solution]]
  13 KB (2,028 words) - 16:32, 22 March 2022
• 2006 AMC 10A Problems/Problem 12
  MP('8', (16,-4), W); MP('8', (20,-8), N);
  3 KB (424 words) - 10:14, 17 December 2021
• 2006 AMC 10A Problems/Problem 16
  D((0,0)--(4*t,0)--(2*t,8)--cycle); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N);
  5 KB (732 words) - 23:19, 19 September 2023
• 2006 AMC 10A Problems/Problem 17
  D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W);
  6 KB (1,066 words) - 00:21, 2 February 2023
• Set
  ...multiple indistinct elements, such as the following: <math>\{1,4,5,3,24,4,4,5,6,2\}</math> Such an entity is actually called a multiset. ...t of size <math>n</math> then <math>\mathcal{P}(A)</math> has size <math>2^n</math>.
  11 KB (2,021 words) - 00:00, 17 July 2011
• Newman's Tauberian Theorem
  label("\Large{$\Gamma_-$}",(-.45,.4)); Now, let <math>N(R)</math> be an upper bound for the quantity
  6 KB (1,034 words) - 07:55, 12 August 2019
• 2006 USAMO Problems
  ...that there exist integers <math>m</math> and <math>n</math> with <math>0<m<n<p</math> and ...th>N </math> but every subset of size <math>k</math> has sum at most <math>N/2</math>.
  3 KB (520 words) - 09:24, 14 May 2021
• 1991 AJHSME Problems
  <math>\frac{16+8}{4-2}=</math> <math>\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \te
  17 KB (2,246 words) - 13:37, 19 February 2020
• Olimpiada de Mayo/1997
  * [[1997 I Problems/Problem 4|Problem 4]] * [[1997 II Problems/Problem 4|Problem 4]]
  856 bytes (98 words) - 14:53, 3 July 2009
• 2022 USAJMO Problems/Problem 1
  ...<math>a_n-g_n</math> is divisible by <math>m</math> for all integers <math>n>1</math>; ...\nmid d</math> and <math>m|a+(n-1)d-gr^{n-1}</math> for all integers <math>n>1</math>.
  4 KB (792 words) - 00:29, 13 April 2024
• 2006 AIME II Problems/Problem 2
  .../math> is a positive integer. Find the number of possible values for <math>n</math>. <math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math>
  1 KB (164 words) - 14:58, 14 April 2020
• 2005 AIME II Problems/Problem 1
  ...imes the number of possible sets of 3 cards that can be drawn. Find <math> n. </math> ...<math>{n \choose 6} = \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}</math>.
  1 KB (239 words) - 11:54, 31 July 2023
• 2005 AIME II Problems/Problem 2
  ...and <math> n </math> are [[relatively prime]] [[integer]]s, find <math> m+n. </math> *Person 2: <math>\frac{6 \cdot 4 \cdot 2}{6 \cdot 5 \cdot 4} = \frac 25</math>
  4 KB (628 words) - 11:28, 14 April 2024
• 2005 AIME II Problems/Problem 3
  ...and <math> n </math> are [[relatively prime]] [[integer]]s. Find <math> m+n. </math> ...-r} = 2005</math>. Then we form a new series, <math>a^2 + a^2 r^2 + a^2 r^4 + \ldots</math>. We know this series has sum <math>20050 = \frac{a^2}{1 -
  3 KB (581 words) - 07:54, 4 November 2022
• 2005 AIME II Problems/Problem 5
  ...th power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).) {{AIME box|year=2005|n=II|num-b=4|num-a=6}}
  3 KB (547 words) - 19:15, 4 April 2024
• 2005 AIME II Problems
  {{AIME Problems|year=2005|n=II}} ...imes the number of possible sets of 3 cards that can be drawn. Find <math> n. </math>
  7 KB (1,119 words) - 21:12, 28 February 2020
• 2005 AIME II Problems/Problem 7
  Let <math> x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}. </math> Find <math>(x+1)^{48}</math ...\sqrt[2^n]{5} + 1)(\sqrt[2^n]{5} - 1) = (\sqrt[2^n]{5})^2 - 1^2 = \sqrt[2^{n-1}]{5} - 1 </math>.
  2 KB (279 words) - 12:33, 27 October 2019
• 2005 AIME II Problems/Problem 8
  ...> n </math> is not divisible by the square of any [[prime]], find <math> m+n+p. </math> pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7));
  4 KB (693 words) - 13:03, 28 December 2021
• 2005 AIME II Problems/Problem 9
  ...ath> less than or equal to <math>1000</math> is <math> (\sin t + i \cos t)^n = \sin nt + i \cos nt </math> true for all real <math> t </math>? ...</math> for all [[real number]]s <math>t</math> and all [[integer]]s <math>n</math>. So, we'd like to somehow convert our given expression into a form
  6 KB (1,154 words) - 03:30, 11 January 2024
• 2005 AIME II Problems/Problem 12
  ..., D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F);draw(G--O); dot(A^^B^^C^^D^^E^^F^^G^^ ...90); draw(A--B--C--D--A);draw(E--O--F);draw(G--O--J);draw(F--G,linetype("4 4")); dot(A^^B^^C^^D^^E^^F^^G^^J^^O); label("\(A\)",A,(-1,1));label("\(B\)",B
  13 KB (2,080 words) - 21:20, 11 December 2022
• 2005 AIME II Problems/Problem 10
  ...</math> and <math> n </math> are relatively prime integers, find <math> m+n. </math> import three; currentprojection = perspective(4,-15,4); defaultpen(linewidth(0.7));
  3 KB (436 words) - 03:10, 23 September 2020
• Euler's identity
  ...states that for any [[real number]] <math>\theta</math> and integer <math>n</math>, ...i\sin(\theta))^n = (e^{i\theta})^n = e^{in\theta} = \cos(n\theta) + i\sin(n\theta)</math>.
  3 KB (452 words) - 23:17, 4 January 2021
• 2005 AIME I Problems
  {{AIME Problems|year=2005|n=I}} ...<math> k. </math> For example, <math> S_3 </math> is the sequence <math> 1,4,7,10,\ldots. </math> For how many values of <math> k </math> does <math> S_
  6 KB (983 words) - 05:06, 20 February 2019
• 2005 AIME I Problems/Problem 2
  ...h> k</math>. For example, <math> S_3 </math> is the [[sequence]] <math> 1,4,7,10,\ldots. </math> For how many values of <math> k </math> does <math> S_ ...th>(12,167)</math>, <math>(167,12)</math>,<math>(334,6)</math>, <math>(501,4)</math>, <math>(668,3)</math>, <math>(1002,2)</math> and <math>(2004,1)</ma
  2 KB (303 words) - 01:31, 5 December 2022
• 2005 AIME I Problems/Problem 3
  ...s,, so <math>n</math> must be in the form <math>n=p\cdot q</math> or <math>n=p^3</math> for distinct [[prime number]]s <math>p</math> and <math>q</math> In the first case, the three proper divisors of <math>n</math> are <math>1</math>, <math>p</math> and <math>q</math>. Thus, we nee
  2 KB (249 words) - 09:37, 23 January 2024
• 2005 AIME I Problems/Problem 4
  ...<math>n \leq 14</math>. In fact, when <math>n = 14</math> we have <math>n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5</math>, so this number works and no larg ...</math>. The [[quadratic formula]] yields <math>r = \frac{7 \pm \sqrt{49 - 4(1)(-s^2 - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}</math>. <math>\sqrt{4s
  8 KB (1,248 words) - 11:43, 16 August 2022
• 2005 AIME I Problems/Problem 5
  Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on o ...and what coins are silver, so the solution is <math>\boxed{9\cdot \binom 8 4=630}</math>.
  5 KB (830 words) - 01:51, 1 March 2023

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