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• 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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