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Create the page "2005 AIME II Problem 7" on this wiki! See also the search results found.
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- == Problem == ...y^6+y^4+y^2+1)(y+1) = (y^{15}+y^{14}+y^{13}+y^{12}+y^{11}+y^{10}+y^9+y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=\frac{y^{16}-1}{y-1}</cmath>2 KB (279 words) - 12:33, 27 October 2019
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- MC(90,"\mbox{semiminor axis}",7,D((0,0)--(0,3),green+linewidth(1)),E); MC("\mbox{semimajor axis}",7,D((0,0)--(5,0),red+linewidth(1)),S);5 KB (892 words) - 21:52, 1 May 2021
- ...s for any given Diophantine equations. This is known as [[Hilbert's tenth problem]]. The answer, however, is no. ...en, the theorem was finally proven by [[Andrew Wiles]] after he spent over 7 years working on the 200-page proof, and another year fixing an error in th9 KB (1,434 words) - 01:15, 4 July 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME II Problems]]1 KB (135 words) - 12:24, 22 March 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME I Problems]]1 KB (154 words) - 12:30, 22 March 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME I Problems]]1 KB (135 words) - 12:31, 22 March 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME II Problems]]1 KB (135 words) - 12:30, 22 March 2011
- {{AIME Problems|year=2006|n=I}} == Problem 1 ==7 KB (1,173 words) - 03:31, 4 January 2023
- == Problem == *Person 1: <math>\frac{9 \cdot 6 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{9}{28}</math>4 KB (628 words) - 11:28, 14 April 2024
- == Problem == ...positive integer]]s that are divisors of at least one of <math> 10^{10},15^7,18^{11}. </math>3 KB (377 words) - 18:36, 1 January 2024
- == Problem == ...at it has roots at <math>17</math> and <math>24</math>. Hence <math>P(x)-x+7=A(x-17)(x-24)</math>. In particular, this means that4 KB (642 words) - 02:14, 1 June 2024
- == Problem == {{AIME box|year=2005|n=II|num-b=5|num-a=7}}2 KB (384 words) - 00:31, 26 July 2018
- {{AIME Problems|year=2005|n=II}} == Problem 1 ==7 KB (1,119 words) - 21:12, 28 February 2020
- == Problem == ...y^6+y^4+y^2+1)(y+1) = (y^{15}+y^{14}+y^{13}+y^{12}+y^{11}+y^{10}+y^9+y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=\frac{y^{16}-1}{y-1}</cmath>2 KB (279 words) - 12:33, 27 October 2019
- == Problem == pathpen = black + linewidth(0.7);4 KB (693 words) - 13:03, 28 December 2021
- == Problem == This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</ma6 KB (1,154 words) - 03:30, 11 January 2024
- == Problem == ...A=(0,9), B=(9,9), C=(9,0), 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);13 KB (2,080 words) - 13:14, 23 July 2024
- {{AIME Problems|year=2005|n=I}} == Problem 1 ==6 KB (983 words) - 05:06, 20 February 2019
- == Problem == size(220); pointpen = black; pen d = linewidth(0.7); pathpen = d;12 KB (2,001 words) - 20:26, 23 July 2024
- == Problem == ...Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8} = 84</math>. We can then use similar triangles with triangle <math>AQC</14 KB (2,340 words) - 16:38, 21 August 2024
- {{AIME Problems|year=2004|n=II}} == Problem 1 ==9 KB (1,410 words) - 05:05, 20 February 2019
