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• 2006 AIME I Problems/Problem 15
  6 KB (910 words) - 19:31, 24 October 2023
• 2006 AMC 12B Problems/Problem 15
  3 KB (458 words) - 16:40, 6 October 2019
• 2006 AMC 12A Problems/Problem 15
  919 bytes (138 words) - 12:45, 4 August 2017
• 2005 AMC 12B Problems/Problem 15
  2 KB (411 words) - 21:02, 21 December 2020
• 2006 AMC 10A Problems/Problem 15
  3 KB (532 words) - 17:49, 13 August 2023
• 2005 AIME I Problems/Problem 15
  </asy></center><!-- Asymptote replacement for Image:2005_I_AIME-15.png by azjps -->
  5 KB (906 words) - 23:15, 6 January 2024
• 2005 AIME II Problems/Problem 15
  D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP("y=ax",(14,14 * (69/100)^.5),E),EndArrow [[Image:2005_AIME_II_-15.png||center|800px]]
  12 KB (2,000 words) - 13:17, 28 December 2020
• 2004 AIME I Problems/Problem 15
  15&2^{13}-4\cdot 7\\ 17&2^{15}-16\cdot 9\\
  9 KB (1,491 words) - 01:23, 26 December 2022
• 2004 AIME II Problems/Problem 15
  <cmath>s_{13, 4} = 2s_{15 - 13, 3} + 1 = 2s_{2, 3}+1</cmath> (46,15)&64&16\\
  6 KB (899 words) - 20:58, 12 May 2022
• 1983 AIME Problems/Problem 15
  <!-- [[Image:1983_AIME-15.png|200px]] --> <cmath>0 \geq f(3) = 25 - 15\cos \alpha - 20 \sin \alpha</cmath>
  19 KB (3,221 words) - 01:05, 7 February 2023
• 1984 AIME Problems/Problem 15
  \frac{x^2}{15}+\frac{y^2}{7}-\frac{z^2}{9}-\frac{w^2}{33}=1\\ \frac{x^2}{63}+\frac{y^2}{55}+\frac{z^2}{39}+\frac{w^2}{15}=1\\
  6 KB (1,051 words) - 04:52, 8 May 2024
• 1985 AIME Problems/Problem 15
  [[Image:AIME 1985 Problem 15.png]]
  2 KB (245 words) - 22:44, 4 March 2024
• 1986 AIME Problems/Problem 15
  <math>2a^2 + 5b^2 = - \frac {15}{2}ab \ \ \ \ (2)</math></div>
  11 KB (1,722 words) - 09:49, 13 September 2023
• 1987 AIME Problems/Problem 15
  5 KB (838 words) - 18:05, 19 February 2022
• 1988 AIME Problems/Problem 15
  7 KB (1,186 words) - 10:16, 4 June 2023
• 1989 AIME Problems/Problem 15
  ...ac{[APC] + [BPC]}{[APB]} = 3</math>, so <math>CP = \dfrac{3}{4} \cdot CF = 15</math>. ...ed out, so <math>w_C = 1</math> and <math>w_F = 3</math>. Thus, <math>CP = 15</math> and <math>PF = 5</math>.
  13 KB (2,091 words) - 00:20, 26 October 2023
• 1990 AIME Problems/Problem 15
  4 KB (644 words) - 16:24, 28 May 2023
• 1991 AIME Problems/Problem 15
  4 KB (658 words) - 16:58, 10 November 2023
• 2006 AMC 10B Problems/Problem 15
  3 KB (445 words) - 22:01, 20 August 2022
• 1992 AIME Problems/Problem 15
  2 KB (358 words) - 01:54, 2 October 2020
• 1993 AIME Problems/Problem 15
  3 KB (449 words) - 21:39, 21 September 2023
• 1994 AIME Problems/Problem 15
  .../2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue);
  4 KB (717 words) - 22:20, 3 June 2021
• 1995 AIME Problems/Problem 15
  ...ve. This is an infinite geometric series whose sum is <math>\frac{3/64}{1-(15/32)}=\frac{3}{34}</math>, so the answer is <math>\boxed{037}</math>. ...e roll <math>4</math> consecutive <tt>H</tt>'s, and there is a <math>\frac{15}{16}</math> probability we roll a <tt>T</tt>. Thus,
  6 KB (979 words) - 13:20, 11 April 2022
• 1996 AIME Problems/Problem 15
  ...\Longrightarrow \cos 2\theta = \frac{\sqrt{3}}{2} \Longrightarrow \theta = 15^{\circ}</math>. The answer is <math>\lfloor 1000r \rfloor = \left\lfloor 10 ...cdot 4}}{8} = \frac {\sqrt {6} \pm \sqrt {2}}{4}</math>, so <math>\theta = 15^{\circ}</math> (as the other roots are too large to make sense in context).
  5 KB (710 words) - 21:04, 14 September 2020
• 1997 AIME Problems/Problem 15
  4 KB (609 words) - 22:49, 17 July 2023
• 1998 AIME Problems/Problem 15
  9 KB (1,671 words) - 22:10, 15 March 2024
• 1999 AIME Problems/Problem 15
  ...has pairwise parallel planar and oppositely equal length (<math>4\sqrt{13},15,17</math>) edges and can be inscribed in a parallelepiped (rectangular box) <math>q^2+r^2=15^2</math>
  7 KB (1,169 words) - 15:28, 13 May 2024
• 2000 AIME I Problems/Problem 15
  Round 7: <math>b</math> to <math>b</math>, <math>15</math> to right, <math>16</math> left in deck, <math>n = -2 + 8k</math>, be ...ans our sieving process will return to normal after Round 7, with <math>31-15=16</math> cards remaining. Since <math>16</math> is a perfect power of <mat
  15 KB (2,673 words) - 19:16, 6 January 2024
• 2001 AIME I Problems/Problem 15
  11 KB (1,837 words) - 18:53, 22 January 2024
• 2002 AIME I Problems/Problem 15
  7 KB (1,181 words) - 20:32, 8 January 2024
• 2003 AIME I Problems/Problem 15
  8 KB (1,382 words) - 14:23, 29 December 2022
• 2003 AIME II Problems/Problem 15
  4 KB (675 words) - 17:23, 30 July 2022
• 2002 AIME II Problems/Problem 15
  7 KB (1,182 words) - 09:56, 7 February 2022
• 2001 AIME II Problems/Problem 15
  4 KB (518 words) - 15:01, 31 December 2021
• 2000 AIME II Problems/Problem 15
  3 KB (469 words) - 21:14, 7 July 2022
• 2005 AMC 10B Problems/Problem 15
  45 bytes (5 words) - 16:48, 16 July 2011
• Mock AIME 1 2006-2007 Problems/Problem 15
  3 KB (414 words) - 13:45, 19 February 2016
• Mock AIME 2 2006-2007 Problems/Problem 15
  4 KB (739 words) - 17:04, 24 November 2023
• 2005 AMC 10A Problems/Problem 15
  2 KB (306 words) - 19:29, 13 December 2021
• 2004 AMC 10A Problems/Problem 15
  3 KB (457 words) - 02:16, 5 July 2021
• 2003 AMC 10A Problems/Problem 15
  2 KB (261 words) - 14:34, 17 August 2023
• 2004 AMC 12A Problems/Problem 15
  {{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #15]] and [[2004 AMC 10A Problems/Problem 17|2004 AMC 10A #17]]}}
  2 KB (309 words) - 22:27, 15 August 2023
• 2003 AMC 12A Problems/Problem 15
  {{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #15]] and [[2003 AMC 10A Problems|2003 AMC 10A #19]]}}
  3 KB (380 words) - 21:53, 19 March 2022
• Mock AIME 4 2006-2007 Problems/Problem 15
  3 KB (532 words) - 20:29, 31 August 2020
• Mock AIME 3 Pre 2005 Problems/Problem 15
  2 KB (312 words) - 10:38, 4 April 2012
• 2007 AIME I Problems/Problem 15
  [[Image:AIME I 2007-15.png]]
  4 KB (673 words) - 22:14, 6 August 2022
• 2007 AIME II Problems/Problem 15
  ...triangle <math>ABC</math> are <math>13,</math> <math>14,</math> and <math>15,</math> the radius of <math>\omega</math> can be represented in the form <m B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C);
  11 KB (2,099 words) - 17:51, 4 January 2024
• 2007 AMC 12A Problems/Problem 15
  1 KB (152 words) - 14:52, 7 August 2017
• 2005 AMC 12A Problems/Problem 15
  14 KB (1,970 words) - 17:02, 18 August 2023
• 2006 AIME II Problems/Problem 15
  Thus, <math>A = \frac{1}{15\sqrt{7}}</math> and <math>x + y + z = 30A = \frac2{\sqrt{7}}</math> and the
  4 KB (725 words) - 17:18, 27 June 2021
• 2000 AMC 12 Problems/Problem 15
  {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #15]] and [[2000 AMC 10 Problems|2000 AMC 10 #24]]}}
  3 KB (441 words) - 21:11, 29 April 2023
• 2002 AMC 12B Problems/Problem 15
  2 KB (395 words) - 15:50, 3 April 2022
• 2007 AMC 10A Problems/Problem 15
  2 KB (326 words) - 10:29, 4 June 2021
• 2008 AMC 12A Problems/Problem 15
  {{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #15]] and [[2008 AMC 10A Problems/Problem 24|2008 AMC 10A #24]]}}
  4 KB (547 words) - 04:19, 30 September 2023
• 2007 AMC 12B Problems/Problem 15
  2 KB (250 words) - 15:41, 27 July 2021
• 2008 AIME I Problems/Problem 15
  6 KB (1,041 words) - 00:54, 1 February 2024
• Mock AIME 1 2007-2008 Problems/Problem 15
  2 KB (276 words) - 16:27, 26 December 2015
• 2008 AIME II Problems/Problem 15
  4 KB (628 words) - 16:23, 2 January 2024
• 2008 AMC 10B Problems/Problem 15
  1,016 bytes (166 words) - 11:39, 8 October 2023
• 2008 AMC 10A Problems/Problem 15
  6 KB (1,033 words) - 15:19, 1 July 2021
• 1995 AHSME Problems/Problem 15
  1 KB (183 words) - 21:54, 2 August 2016
• 2002 AMC 10A Problems/Problem 15
  993 bytes (147 words) - 01:50, 26 January 2018
• 2002 AMC 10B Problems/Problem 15
  45 bytes (5 words) - 18:14, 28 July 2011
• 2000 AMC 10 Problems/Problem 15
  45 bytes (4 words) - 23:54, 26 November 2011
• 1985 AJHSME Problems/Problem 15
  3 KB (473 words) - 08:14, 13 January 2023
• 1986 AJHSME Problems/Problem 15
  1 KB (155 words) - 15:32, 9 November 2017
• 2002 AMC 12A Problems/Problem 15
  {{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #15]] and [[2002 AMC 10A Problems|2002 AMC 10A #21]]}} \text{(E) }15
  3 KB (519 words) - 21:34, 23 June 2023
• 2004 AMC 12B Problems/Problem 15
  {{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #15]] and [[2004 AMC 10B Problems|2004 AMC 10B #17]]}}
  2 KB (326 words) - 09:47, 17 October 2020
• 2004 AMC 10B Problems/Problem 15
  <math> \textbf{(A)}\ \textdollar 1.15\qquad\textbf{(B)}\ \textdollar 1.20\qquad\textbf{(C)}\ \textdollar 1.25\qqu .../math>. Her coins are worth <math>200-5n=\boxed{\mathrm{(A)\ }\textdollar1.15}</math>.
  1 KB (220 words) - 00:53, 24 July 2014
• 2009 AMC 12A Problems/Problem 15
  4 KB (634 words) - 16:34, 3 December 2020
• 2001 AMC 12 Problems/Problem 15
  1 KB (225 words) - 15:26, 30 November 2021
• 2009 AMC 10B Problems/Problem 15
  {{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #15]] and [[2009 AMC 12B Problems|2009 AMC 12B #8]]}}
  3 KB (545 words) - 13:37, 3 September 2023
• 2009 AMC 12B Problems/Problem 15
  2 KB (336 words) - 16:58, 27 December 2020
• 1987 AJHSME Problems/Problem 15
  731 bytes (110 words) - 16:40, 30 July 2023
• 2009 AIME I Problems/Problem 15
  .../math> and <math>COP</math>, with <math>BO=CO=7</math> and <math>OP=7 \tan 15=7(2-\sqrt{3})=14-7\sqrt3</math>. Then, the area of [<math>\triangle BPC</ma
  6 KB (1,048 words) - 19:35, 2 January 2023
• 1988 AJHSME Problems/Problem 15
  721 bytes (100 words) - 23:55, 4 July 2013
• 2009 AIME II Problems/Problem 15
  11 KB (1,849 words) - 19:43, 2 January 2023
• 1989 AJHSME Problems/Problem 15
  1 KB (194 words) - 00:03, 5 July 2013
• 1990 AJHSME Problems/Problem 15
  863 bytes (124 words) - 00:06, 5 July 2013
• 1991 AJHSME Problems/Problem 15
  draw((1,0)--(1+9*sqrt(3)/2,9/2)--(1+9*sqrt(3)/2,15/2)--(1+5*sqrt(3)/2,11/2)--(1+5*sqrt(3)/2,9/2)--(1+2*sqrt(3),4)--(1+2*sqrt(3 draw((1+9*sqrt(3)/2,15/2)--(9*sqrt(3)/2,15/2)--(5*sqrt(3)/2,11/2)--(5*sqrt(3)/2,5));
  1 KB (197 words) - 00:07, 5 July 2013
• 2010 AMC 12A Problems/Problem 15
  <math>\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{
  2 KB (255 words) - 21:47, 3 July 2013
• 2010 AIME I Problems/Problem 15
  ...BC}</math> with <math>AB = 12</math>, <math>BC = 13</math>, and <math>AC = 15</math>, let <math>M</math> be a point on <math>\overline{AC}</math> such th draw((0,0)--(15,0));
  14 KB (2,210 words) - 13:14, 11 January 2024
• 2007 AMC 8 Problems/Problem 15
  1 KB (187 words) - 17:18, 3 November 2023
• 2010 AMC 10A Problems/Problem 15
  46 bytes (5 words) - 13:25, 26 May 2020
• Northeastern WOOTers Mock AIME I Problems/Problem 15
  == Problem 15 ==
  1 KB (226 words) - 21:35, 7 August 2021
• 2010 AIME II Problems/Problem 15
  == Problem 15 == ...math>ABC</math>, <math>AC = 13</math>, <math>BC = 14</math>, and <math>AB=15</math>. Points <math>M</math> and <math>D</math> lie on <math>AC</math> wit
  9 KB (1,523 words) - 15:24, 21 November 2023
• 2010 AMC 12B Problems/Problem 15
  ...ath>i^x</math> only cycles as <math>1, i, -1, -i</math>). So we have <math>15\cdot 1\cdot 1+20\cdot 1\cdot 1=35</math> ordered triples.
  2 KB (360 words) - 17:29, 26 May 2023
• 1999 AMC 8 Problems/Problem 15
  3 KB (516 words) - 14:50, 21 December 2022
• 2010 AMC 10B Problems/Problem 15
  1 KB (210 words) - 02:44, 26 September 2020
• 2011 AMC 12A Problems/Problem 15
  2 KB (376 words) - 23:14, 5 January 2024
• 2011 AMC 10A Problems/Problem 15
  == Problem 15 ==
  2 KB (264 words) - 15:31, 3 September 2022
• 2001 AMC 10 Problems/Problem 15
  ...e street at an angle. The length of the curb between the stripes is <math> 15 </math> feet and each stripe is <math> 50 </math> feet long. Find the dista ...)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25 </math>
  2 KB (253 words) - 11:57, 20 October 2020
• 2011 AIME I Problems/Problem 15
  8 KB (1,302 words) - 04:07, 24 July 2023
• 2011 AIME II Problems/Problem 15
  ...ber <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfl P(15) &= 171 \\
  8 KB (1,273 words) - 14:03, 7 January 2023
• 2001 AMC 8 Problems/Problem 15
  922 bytes (135 words) - 00:38, 5 July 2013
• 2011 AMC 10B Problems/Problem 15
  2 KB (287 words) - 01:48, 26 June 2022
• 2008 AMC 12B Problems/Problem 15
  3 KB (505 words) - 22:58, 7 October 2021
• 1999 AHSME Problems/Problem 15
  931 bytes (144 words) - 19:36, 1 May 2023
• 2011 AMC 12B Problems/Problem 15
  ==Problem 15==
  2 KB (284 words) - 20:54, 29 May 2023
• 2007 AMC 10B Problems/Problem 15
  ...12B Problems|2007 AMC 12B #11]] and [[2007 AMC 10B Problems|2007 AMC 10B #15]]}}
  997 bytes (137 words) - 12:22, 4 July 2013
• 1998 AHSME Problems/Problem 15
  764 bytes (123 words) - 14:29, 5 July 2013
• 1998 AJHSME Problems/Problem 15
  Problems 15, 16, and 17 all refer to the following: ==Problem 15==
  1 KB (173 words) - 20:55, 11 May 2021
• 2003 AMC 10B Problems/Problem 15
  2 KB (267 words) - 23:23, 7 September 2023
• AoPSWik:Problem of the Day/June 15, 2011
  #REDIRECT [[AoPSWiki:Problem of the Day/June 15, 2011]]
  55 bytes (7 words) - 18:20, 14 June 2011
• 1984 AHSME Problems/Problem 15
  ==Problem 15==
  3 KB (493 words) - 18:16, 4 June 2021
• 2002 AMC 8 Problems/Problem 15
  1 KB (197 words) - 04:47, 25 November 2019
• 2000 AMC 8 Problems/Problem 15
  <math> \text{(A)}\ 12\qquad\text{(B)}\ 13\qquad\text{(C)}\ 15\qquad\text{(D)}\ 18\qquad\text{(E)}\ 21 </math> <math>15</math>, thus the answer is <math>\boxed{C}</math>
  2 KB (325 words) - 20:56, 6 November 2013
• 1997 AJHSME Problems/Problem 15
  3 KB (473 words) - 10:58, 27 June 2023
• 1996 AJHSME Problems/Problem 15
  1 KB (200 words) - 11:29, 27 June 2023
• 1997 AHSME Problems/Problem 15
  2 KB (406 words) - 20:44, 15 February 2024
• 2008 AMC 8 Problems/Problem 15
  1 KB (190 words) - 19:24, 8 August 2021
• 2009 AMC 8 Problems/Problem 15
  1 KB (162 words) - 19:53, 20 August 2020
• 1994 AJHSME Problems/Problem 15
  label("$\cdots$",(6.5,0.15),S);
  2 KB (282 words) - 11:52, 27 June 2023
• 1996 AHSME Problems/Problem 15
  3 KB (537 words) - 14:07, 5 July 2013
• 1992 AJHSME Problems/Problem 15
  634 bytes (95 words) - 00:09, 5 July 2013
• 1993 AJHSME Problems/Problem 15
  732 bytes (114 words) - 00:11, 5 July 2013
• 1995 AJHSME Problems/Problem 15
  577 bytes (81 words) - 00:23, 5 July 2013
• 2007 Alabama ARML TST Problems/Problem 15
  2 KB (389 words) - 18:12, 21 March 2018
• 1985 AHSME Problems/Problem 15
  1,012 bytes (163 words) - 21:35, 19 March 2024
• 1950 AHSME Problems/Problem 15
  2 KB (255 words) - 08:55, 1 September 2021
• 2003 AMC 8 Problems/Problem 15
  currentprojection=orthographic(3/4,8/15,7/15);
  1 KB (201 words) - 19:56, 15 April 2023
• 2006 AMC 8 Problems/Problem 15
  Problems 14, 15 and 16 involve Mrs. Reed's English assignment. <math>75x = 45(760)</math> Divide both sides by <math>15</math> to make it easier to solve
  2 KB (323 words) - 22:54, 6 January 2023
• 2011 AMC 8 Problems/Problem 15
  846 bytes (112 words) - 15:09, 17 December 2023
• 2012 AMC 10A Problems/Problem 15
  6 KB (935 words) - 23:41, 13 September 2023
• 2012 AMC 12A Problems/Problem 15
  46 bytes (5 words) - 14:35, 12 February 2012
• 1991 AHSME Problems/Problem 15
  <math>\text{(A) } 15\quad
  879 bytes (143 words) - 17:38, 23 February 2018
• 1988 AHSME Problems/Problem 15
  759 bytes (128 words) - 18:23, 26 February 2018
• 1989 AHSME Problems/Problem 15
  2 KB (275 words) - 20:37, 24 March 2023
• 2012 AMC 12B Problems/Problem 15
  2 KB (259 words) - 13:42, 2 September 2020
• 2012 AMC 10B Problems/Problem 15
  ...s 2.5 wins, so the maximum number of tied teams is five. Here's a chart of 15 games where five teams each have 3 wins: Note that the total number of matches is 15, and if 4 teams tie for the most wins then they can tie for 3 wins each, bu
  2 KB (307 words) - 20:04, 13 July 2021
• 2012 AIME I Problems/Problem 15
  3 KB (591 words) - 20:41, 24 January 2021
• 2007 iTest Problems/Problem 15
  3 KB (539 words) - 21:01, 29 July 2018
• 2012 AIME II Problems/Problem 15
  == Problem 15 == ...BD=\tfrac{35}{8}</math>, and use Stewart's Theorem to find <math>AD=\tfrac{15}{8}</math>. Use Power of Point <math>D</math> to find <math>DE=\tfrac{49}{8
  13 KB (2,298 words) - 12:56, 10 September 2023
• Mock AIME 2 2006-2007/Problem 15
  #REDIRECT [[Mock AIME 2 2006-2007 Problems/Problem 15]]
  55 bytes (6 words) - 15:31, 3 April 2012
• Mock AIME 1 2006-2007/Problem 15
  #REDIRECT [[Mock AIME 1 2006-2007 Problems/Problem 15]]
  55 bytes (6 words) - 15:50, 3 April 2012
• Mock AIME 3 Pre 2005/Problem 15
  #REDIRECT [[Mock AIME 3 Pre 2005 Problems/Problem 15]]
  54 bytes (6 words) - 19:42, 9 April 2012
• Mock AIME II 2012 Problems/Problem 15
  ...h>. Now, substitute this into <math>5b_2+b_1+6=0</math> to give us <math>-15+6+b_1=0</math> or <math>b_1=9</math>. Therefore <math>f(x)=-3x^2+9x</math> ...ly. Note that <math>155^2=\underline{15*16},2,5=24025</math> (where <math>15*16, 2</math>, and <math>5</math> are digits of <math>155^2</math>), therefo
  4 KB (660 words) - 15:55, 8 March 2015
• Mock AIME I 2012 Problems/Problem 15
  15 & 3 & 8\\
  3 KB (488 words) - 20:05, 10 March 2015
• KGS math club/solution 11 15
  630 bytes (106 words) - 06:41, 8 April 2012
• 1951 AHSME Problems/Problem 15
  1 KB (191 words) - 06:08, 6 April 2024
• 1966 AHSME Problems/Problem 15
  498 bytes (87 words) - 22:11, 14 January 2018
• 1974 AHSME Problems/Problem 15
  651 bytes (101 words) - 12:43, 5 July 2013
• 2010 AMC 8 Problems/Problem 15
  ...f gumdrops. <math>30\%</math> are blue, <math>20\%</math> are brown, <math>15\%</math> are red, <math>10\%</math> are yellow, and other <math>30</math> g ...s. If we replace half of the blue gumdrops with brown gumdrops, then <math>15\%</math> of the jar's gumdrops are brown. <math>\dfrac{35}{100} \cdot 120=4
  990 bytes (148 words) - 20:55, 22 January 2023
• 2012 AMC 8 Problems/Problem 15
  1 KB (180 words) - 19:00, 15 April 2023
• 2004 AMC 8 Problems/Problem 15
  1 KB (215 words) - 20:20, 12 October 2020
• 2005 AMC 8 Problems/Problem 15
  952 bytes (155 words) - 00:11, 2 July 2023
• 2013 AMC 12A Problems/Problem 15
  3 KB (441 words) - 10:10, 4 August 2020
• 2013 AMC 10A Problems/Problem 15
  Two sides of a triangle have lengths <math>10</math> and <math>15</math>. The length of the altitude to the third side is the average of the ...e perpendicular to that altitude will be between <math>10</math> and <math>15</math>. The only answer choice that meets this requirement is <math>\boxed{
  2 KB (407 words) - 01:12, 22 September 2022
• 2013 AMC 10B Problems/Problem 15
  2 KB (240 words) - 12:18, 18 October 2022
• 2013 AMC 12B Problems/Problem 15
  {{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #15]] and [[2013 AMC 10B Problems|2013 AMC 10B #20]]}}
  2 KB (280 words) - 16:28, 25 December 2023
• 2013 AIME I Problems/Problem 15
  ...ution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}</math>. When <math>x=5</math>, <math>B=6, 9, ..., 93\Rightarrow 15</math> triples..
  4 KB (661 words) - 23:14, 26 May 2023
• 2013 AIME II Problems/Problem 15
  ==Problem 15== \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\
  8 KB (1,411 words) - 23:48, 11 January 2023
• 2003 AMC 12B Problems/Problem 15
  46 bytes (5 words) - 01:01, 5 January 2014
• 2014 AIME II Problems/Problem 15
  <math>x_6 = 15</math> <math>x_{14} = 105</math> <math>x_7 = 30</math> <math>x_{15} = 210</math>
  4 KB (770 words) - 17:44, 11 October 2023
• 1967 AHSME Problems/Problem 15
  1 KB (231 words) - 01:39, 16 August 2023
• 1962 AHSME Problems/Problem 15
  1 KB (245 words) - 08:45, 30 January 2018
• 2013 AMC 8 Problems/Problem 15
  2 KB (295 words) - 03:58, 29 December 2022
• 1952 AHSME Problems/Problem 15
  1 KB (169 words) - 23:17, 2 January 2014
• 2014 AMC 10A Problems/Problem 15
  ...12A Problems|2014 AMC 12A #11]] and [[2014 AMC 10A Problems|2014 AMC 10A #15]]}} ...hour late if he continues at this speed. He increases his speed by <math>15</math> miles per hour for the rest of the way to the airport and arrives <m
  3 KB (409 words) - 04:18, 20 June 2022
• 1994 AHSME Problems/Problem 15
  1 KB (214 words) - 02:53, 28 May 2021
• 2014 AMC 12A Problems/Problem 15
  4 KB (612 words) - 22:42, 2 August 2021
• 2014 AMC 10B Problems/Problem 15
  3 KB (536 words) - 21:02, 31 October 2022
• 2014 AMC 12B Problems/Problem 15
  1 KB (203 words) - 23:18, 3 January 2023
• 2014 AIME I Problems/Problem 15
  == Problem 15 == ...theorem, we have <math>\frac{AF}{AB} = \frac{CF}{CB}</math>, so <math>AF = 15/7</math> and <math>CF = 20/7</math>.
  10 KB (1,643 words) - 22:30, 28 January 2024
• 1993 AHSME Problems/Problem 15
  897 bytes (140 words) - 13:39, 6 November 2020
• 1992 AHSME Problems/Problem 15
  821 bytes (134 words) - 02:46, 20 February 2018
• 1968 AHSME Problems/Problem 15
  <math>\text{(A) } 15\quad
  1 KB (195 words) - 09:24, 1 January 2024
• 1990 AHSME Problems/Problem 15
  643 bytes (92 words) - 05:37, 4 February 2016
• 1969 AHSME Problems/Problem 15
  2 KB (250 words) - 18:29, 21 June 2018
• 1970 AHSME Problems/Problem 15
  1 KB (226 words) - 22:26, 13 July 2019
• 1980 AHSME Problems/Problem 15
  983 bytes (156 words) - 13:30, 9 March 2020
• 1958 AHSME Problems/Problem 15
  {{AHSME 50p box|year=1958|num-b=13|num-a=15}}
  444 bytes (62 words) - 06:13, 3 October 2014
• Mock AIME 1 Pre 2005 Problems/Problem 15
  {{Mock AIME box|year=Pre 2005|n=1|num-b=14|num-a=15|source=14769}}
  2 KB (340 words) - 01:44, 3 March 2020
• Mock AIME 5 2005-2006 Problems/Problem 15
  {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=14|num-a=15}}
  955 bytes (157 words) - 21:20, 8 October 2014
• 2005 CEMC Gauss (Grade 7) Problems/Problem 15
  ...ext{(B)}\ 20 \qquad \text{(C)}\ 16 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 15</math>
  2 KB (349 words) - 02:41, 23 October 2014
• 1987 AHSME Problems/Problem 15
  699 bytes (107 words) - 12:36, 31 March 2018
• 1986 AHSME Problems/Problem 15
  1 KB (185 words) - 18:43, 1 April 2018
• 2014 AMC 8 Problems/Problem 15
  2 KB (280 words) - 11:10, 2 July 2023
• 2015 AMC 10A Problems/Problem 15
  2 KB (372 words) - 14:45, 27 September 2020
• 2015 AMC 12A Problems/Problem 15
  3 KB (411 words) - 15:53, 23 June 2022
• 2015 AMC 12B Problems/Problem 15
  4 KB (658 words) - 22:12, 21 November 2023
• 2015 AMC 10B Problems/Problem 15
  2 KB (276 words) - 11:02, 20 November 2023
• 2016 AMC 10A Problems/Problem 15
  1 KB (211 words) - 14:53, 25 June 2023
• 2015 AIME I Problems/Problem 15
  [[File:2015 AIME I 15.png|400px|right]] <cmath> A = \frac{15}{4} \int_0^4 \sqrt{ 64 - y^2 }dy </cmath>
  9 KB (1,407 words) - 19:37, 17 February 2024
• 2015 AIME II Problems/Problem 15
  ...;O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=f ...label("$L$",(24/15,0.2));label("$n$",(-0.8,-0.12));label("$p$",((29/15,-48/15)));label("$\mathcal{P}$",(-1.6,1.1));label("$\mathcal{Q}$",(6,4));
  31 KB (5,086 words) - 19:15, 20 December 2023
• 2015 AMC 8 Problems/Problem 15
  2 KB (333 words) - 01:41, 15 January 2024
• 2016 AMC 12A Problems/Problem 15
  8 KB (1,255 words) - 09:05, 5 September 2022
• 2016 AMC 10B Problems/Problem 15
  2 KB (378 words) - 21:13, 18 June 2022
• 2016 AMC 12B Problems/Problem 15
  3 KB (554 words) - 01:25, 4 August 2023
• 2016 AIME I Problems/Problem 15
  pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1];
  14 KB (2,427 words) - 17:12, 8 January 2024
• 2016 AIME II Problems/Problem 15
  2 KB (277 words) - 20:02, 19 April 2023
• 1975 AHSME Problems/Problem 15
  887 bytes (148 words) - 23:42, 24 June 2019
• 1964 AHSME Problems/Problem 15
  == Problem 15==
  2 KB (260 words) - 21:01, 23 July 2019
• 1954 AHSME Problems/Problem 15
  == Problem 15==
  441 bytes (54 words) - 01:26, 28 February 2020
• 1965 AHSME Problems/Problem 15
  298 bytes (49 words) - 11:12, 10 June 2016
• 1977 AHSME Problems/Problem 15
  1 KB (200 words) - 21:30, 1 May 2024
• 2016 AMC 8 Problems/Problem 15
  1 KB (156 words) - 22:58, 17 May 2024
• 1979 AHSME Problems/Problem 15
  == Problem 15 ==
  1 KB (240 words) - 13:21, 6 January 2017
• 2017 AMC 10A Problems/Problem 15
  ...al length of <math>2-0=2</math>. Therefore, the probability is <math>1.5/2=15/20=\boxed{\frac{3}{4} \space \text{(C)}}.</math>
  4 KB (533 words) - 19:01, 15 March 2024
• 2017 AMC 12A Problems/Problem 15
  Notice that <math>\frac{5\pi}{4}<\frac{3.15*5}{4}<4<\frac{3\pi}{2}, f(\frac{5\pi}{4})=3-\frac{3\sqrt{2}}{2}>0 .</math>
  3 KB (564 words) - 14:12, 23 October 2021
• 2017 AMC 10B Problems/Problem 15
  ...xtbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{42}{25}\qquad\textbf{(C)}\ \frac{28}{15}\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{54}{25}</math>
  4 KB (656 words) - 01:20, 4 December 2023
• 2017 AMC 12B Problems/Problem 15
  9 KB (1,416 words) - 14:30, 23 November 2023
• 2017 AIME I Problems/Problem 15
  ==Problem 15== [[File:2017 AIME I 15.png|530px|right]]
  22 KB (3,622 words) - 17:11, 6 January 2024
• 2017 AIME II Problems/Problem 15
  [[File:2017 AIME II 15.png|300px|right]]
  6 KB (971 words) - 02:08, 22 January 2024
• 1953 AHSME Problems/Problem 15
  ==Problem 15==
  1 KB (211 words) - 14:41, 27 January 2020
• 2017 AMC 8 Problems/Problem 15
  4 KB (613 words) - 11:36, 24 December 2023
• 2008 iTest Problems/Problem 15
  1,002 bytes (137 words) - 13:25, 22 June 2018
• 2018 AMC 10A Problems/Problem 15
  label("$A$", (-8.125,-10.15), S); label("$B$", (8.125,-10.15), S);
  2 KB (342 words) - 17:10, 15 October 2023
• 2018 AMC 12A Problems/Problem 15
  46 bytes (5 words) - 16:05, 8 February 2018
• 2018 AMC 12B Problems/Problem 15
  ...e a multiple of <math>3</math>, but does not contain 3. We have <math>(12, 15, 18, 21, 24, 27, 42, 45, 48, 51, 54, 57, 60, 66, 69, 72, 75, 78, 81, 84, 87 ...3) = 15</math> cases. For <math>\underline{3AB}</math>, we also have <math>15</math> cases, but when <math>B=3, 9</math>, <math>A</math> can equal <math>
  8 KB (1,244 words) - 08:19, 30 May 2023
• 2018 AMC 10B Problems/Problem 15
  ...12B Problems|2018 AMC 12B #11]] and [[2018 AMC 10B Problems|2018 AMC 10B #15]]}}
  6 KB (974 words) - 00:28, 30 May 2023
• 1960 AHSME Problems/Problem 15
  2 KB (276 words) - 19:01, 17 May 2018
• 1961 AHSME Problems/Problem 15
  3 KB (490 words) - 11:22, 16 September 2022
• 1963 AHSME Problems/Problem 15
  1 KB (215 words) - 16:52, 7 June 2018
• 2018 AIME I Problems/Problem 15
  ==Problem 15== [[File:2018 AIME I 15.png|900px]]
  7 KB (1,148 words) - 23:33, 6 January 2024
• 2018 AIME II Problems/Problem 15
  9 KB (1,607 words) - 10:27, 4 February 2022
• 1973 AHSME Problems/Problem 15
  2 KB (306 words) - 14:00, 20 February 2020
• 1983 AHSME Problems/Problem 15
  1 KB (166 words) - 00:54, 20 February 2019
• 2018 AMC 8 Problems/Problem 15
  3 KB (454 words) - 10:15, 10 April 2024
• 2006 iTest Problems/Problem 15
  1 KB (197 words) - 23:31, 27 November 2018
• Mock AIME I 2015 Problems/Problem 15
  2 bytes (1 word) - 02:02, 7 December 2019
• 2019 AIME II Problems/Problem 15
  ...ath>Y</math>. Suppose <math>XP=10</math>, <math>PQ=25</math>, and <math>QY=15</math>. The value of <math>AB\cdot AC</math> can be written in the form <ma label("$15$", Q--Y, SW);
  7 KB (1,115 words) - 03:11, 7 January 2024
• 2019 AMC 12A Problems/Problem 15
  ...ndex.php/Perfect_square perfect squares] up to <math>14^2</math> (as <math>15^2</math> is larger than the largest possible sum of <math>x</math> and <mat
  4 KB (686 words) - 04:22, 13 November 2022
• 2019 AMC 10A Problems/Problem 15
  {{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #15]] and [[2019 AMC 12A Problems|2019 AMC 12A #9]]}} ...cursive formula, we find <math>a_3=\frac{3}{11}</math>, <math>a_4=\frac{3}{15}</math>, and so on. It appears that <math>a_n=\frac{3}{4n-1}</math>, for al
  4 KB (687 words) - 08:11, 20 November 2023
• 2019 AMC 10B Problems/Problem 15
  4 KB (704 words) - 23:59, 22 March 2023
• 2019 AMC 12B Problems/Problem 15
  46 bytes (5 words) - 13:31, 14 February 2019
• 2019 AIME I Problems/Problem 15
  By Power of a Point, <math>PX\cdot PY=PA\cdot PB=15</math>. Since <math>PX+PY=XY=11</math> and <math>XQ=11/2</math>, <cmath>XP= .../math>. Then we have <math>AP\cdot PB=XP\cdot PY</math>, that is, <math>xy=15</math>. Also, <math>XP+PY=x+y=XY=11</math>. Solve these above, we have <mat
  13 KB (2,252 words) - 11:32, 1 February 2024
• 2006 SMT/Team Problems/Problem 15
  3 KB (445 words) - 18:37, 14 January 2020
• 2020 AIME I Problems/Problem 15
  A = (-15, 27); H = (-15, 13);
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• 2019 Mock AMC 10B Problems/Problem 15
  283 bytes (49 words) - 00:23, 14 December 2019
• 2019 AMC 8 Problems/Problem 15
  ==Problem 15==
  3 KB (496 words) - 00:28, 6 May 2024
• 2020 AMC 10A Problems/Problem 15
  2 KB (261 words) - 20:04, 3 December 2023
• 2020 AMC 12A Problems/Problem 15
  4 KB (696 words) - 12:38, 13 September 2021
• 2020 AMC 10B Problems/Problem 15
  ...torname{lcm}(3,5)=15</math> before erasing. So, we first group <math>\frac{15}{5}=3</math> copies of the current cycle into one, then erase: As a quick confirmation, one cycle should have length <math>15\cdot\left(1-\frac{1}{3}\right)=10</math> at this point.</li><p>
  10 KB (1,471 words) - 13:57, 30 October 2023
• 2020 AMC 12B Problems/Problem 15
  46 bytes (5 words) - 21:15, 12 February 2020
• 2023 AIME I Problems/Problem 15
  ...satisfy (7) and <math>a^2 + b^2 < 1000</math>, we have <math>1 \leq b \leq 15</math>.
  8 KB (1,299 words) - 17:37, 3 June 2023
• 2021 JMPSC Accuracy Problems/Problem 15
  2 KB (304 words) - 01:19, 12 July 2021
• 2021 AMC 10A Problems/Problem 15
  4 KB (577 words) - 19:18, 28 October 2022
• 1959 AHSME Problems/Problem 15
  ...the legs. One of the acute angles of the triangle is: <math>\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qqua ...rect answer, one of the acute angles of the triangle will measure to <math>15</math> degrees. This implies that the other acute angle of the triangle wou
  2 KB (263 words) - 11:33, 12 August 2020
• 2020 AIME II Problems/Problem 15
  ...aw segments <math>XM</math>, and <math>YM</math>. We have <math>MT=3\sqrt{15}</math>. ...larly, Ptolemy's theorem in <math>YTMC</math> gives<cmath>16TX=11TY+3\sqrt{15}CY</cmath> while Pythagoras' theorem in <math>\triangle CYT</math> gives <m
  7 KB (1,221 words) - 16:46, 29 January 2023
• 2021 AIME I Problems/Problem 15
  10 KB (1,742 words) - 02:31, 13 November 2023
• 2019 AMC 10C Problems/Problem 15
  625 bytes (105 words) - 10:53, 6 August 2020
• 1955 AHSME Problems/Problem 15
  == Problem 15==
  1 KB (177 words) - 10:44, 15 February 2021
• 2022 AIME I Problems/Problem 15
  so <math>(\alpha, \beta, \gamma) = (15/2^{\circ}, 45/2^{\circ}, 75/2^{\circ})</math>. Hence, <cmath> abc = (1-2\sin^2(\alpha))(1-2\sin^2(\beta))(1-2\sin^2(\gamma))=\cos(15^{\circ})\cos(45^{\circ})\cos(75^{\circ})=\frac{\sqrt{2}}{8}, </cmath>
  15 KB (2,208 words) - 01:25, 1 February 2024
• 2020 CIME I Problems/Problem 15
  ==Problem 15==
  523 bytes (80 words) - 11:58, 1 September 2020
• 2021 Fall AMC 10A Problems/Problem 15
  7 KB (1,026 words) - 13:43, 5 May 2024
• 2021 Fall AMC 12A Problems/Problem 15
  2 KB (311 words) - 19:01, 24 May 2023
• 2021 AMC 12A Problems/Problem 15
  4 & 0 & \tbinom{6}{4}\tbinom{8}{0} & 15 \\ [1ex] 4 & 8 & \tbinom{6}{4}\tbinom{8}{8} & 15 \\ [1ex]
  7 KB (1,152 words) - 14:13, 29 February 2024
• 2021 AMC 12B Problems/Problem 15
  ...20|2021 AMC 10B #20]] and [[2021 AMC 12B Problems#Problem 15|2021 AMC 12B #15]]}}
  4 KB (537 words) - 13:49, 25 February 2024
• 2019 CIME I Problems/Problem 15
  2 KB (264 words) - 17:57, 4 October 2020
• 2020 AMC 8 Problems/Problem 15
  Suppose <math>15\%</math> of <math>x</math> equals <math>20\%</math> of <math>y.</math> What ...g the given condition by <math>5</math> shows that <math>y</math> is <math>15 \cdot 5 = \boxed{\textbf{(C) }75}</math> percent of <math>x</math>.
  3 KB (455 words) - 16:34, 26 January 2024
• 2021 AMC 10B Problems/Problem 15
  5 KB (806 words) - 00:13, 20 October 2023
• 1981 AHSME Problems/Problem 15
  331 bytes (50 words) - 19:09, 23 October 2021
• 1972 AHSME Problems/Problem 15
  929 bytes (153 words) - 10:07, 29 January 2021
• 1998 CEMC Gauss (Grade 7) Problems/Problem 15
  760 bytes (133 words) - 15:40, 29 January 2021
• 1985 AJHSME Problem 15
  611 bytes (86 words) - 21:51, 31 January 2021
• 1963 TMTA High School Algebra I Contest Problem 15
  679 bytes (90 words) - 09:58, 2 February 2021
• 1978 AHSME Problems/Problem 15
  == Problem 15 ==
  1 KB (207 words) - 21:10, 13 February 2021
• 1956 AHSME Problems/Problem 15
  The root(s) of <math>\frac {15}{x^2 - 4} - \frac {2}{x - 2} = 1</math> is (are): Use a common denominator: <math>\frac{15 - 2(x+2)}{x^2-4} = 1</math>. After moving the <math>1</math> to the left si
  645 bytes (100 words) - 20:07, 12 February 2021
• 2021 AIME II Problems/Problem 15
  ...l(\phantom{ }\underbrace{(k+2)^2-2q}_{n}\phantom{ }\bigr)&=k+2q+2. \hspace{15.5mm}(4) \\ ...th the least <math>k</math> and <math>m</math> are <math>(2,5), (5,10), (8,15),</math> and <math>(11,20).</math>
  14 KB (2,569 words) - 09:28, 28 March 2024
• 2002 AMC 10P Problems/Problem 15
  ...8 \quad \textbf{(C)} \; 12 \quad \textbf{(D)} \; 13 \quad \textbf{(E)} \; 15 </math>
  900 bytes (122 words) - 11:24, 9 March 2021
• 1957 AHSME Problems/Problem 15
  760 bytes (111 words) - 23:16, 27 March 2021
• 2021 JMC 10 Problems/Problem 15
  1 KB (250 words) - 16:19, 1 April 2021
• 2021 April MIMC 10 Problems/Problem 15
  806 bytes (122 words) - 13:44, 26 April 2021
• 1982 AHSME Problems/Problem 15
  \text{(D) }\text{ between 15 and 16}\qquad ...s between 4 and 5. <math>y=11</math>. So <math>x+y</math> is between <math>15</math> and <math>16</math>. Select <math>\boxed{D}</math>.
  1 KB (227 words) - 22:41, 16 August 2022
• 2013 Mock AIME I Problems/Problem 15
  9 bytes (1 word) - 13:08, 29 September 2021
• 2022 AIME II Problems/Problem 15
  ...<math>D</math>, as shown. Suppose that <math>AB = 2</math>, <math>O_1O_2 = 15</math>, <math>CD = 16</math>, and <math>ABO_1CDO_2</math> is a convex hexag point O1=(0,0),O2=(15,0),B=9*dir(30);
  14 KB (2,217 words) - 00:28, 29 June 2023
• 2021 JMPSC Sprint Problems/Problem 15
  984 bytes (129 words) - 20:54, 7 September 2021
• 2021 JMPSC Invitationals Problems/Problem 15
  2 KB (342 words) - 00:02, 12 July 2021
• 1976 AHSME Problems/Problem 15
  == Problem 15 == \textbf{(B) }15\qquad
  1 KB (198 words) - 14:46, 20 July 2021
• 2021 Fall AMC 12B Problems/Problem 15
  ...shapes like <math>\triangle ABC</math>. Then <math>\angle BAC=360^\circ/24=15^\circ</math>, and <math>\angle EAC = 45^\circ</math>, so <math>\angle{EAB}
  4 KB (686 words) - 18:38, 12 July 2023
• 2021 Fall AMC 10B Problems/Problem 15
  11 KB (1,733 words) - 23:06, 13 May 2024
• 2022 AMC 8 Problems/Problem 15
  7 KB (914 words) - 14:49, 23 November 2023
• 2022 AMC 12A Problems/Problem 15
  46 bytes (5 words) - 01:36, 12 November 2022
• 2022 AMC 10A Problems/Problem 15
  ...rilateral <math>ABCD</math> with side lengths <math>AB=7, BC=24, CD=20, DA=15</math> is inscribed in a circle. The area interior to the circle but exteri D = intersectionpoints(Circle(A,15),Circle(C,20))[1];
  5 KB (768 words) - 21:00, 21 February 2024
• 2022 AMC 10B Problems/Problem 15
  2 KB (354 words) - 13:34, 2 April 2024
• 2022 AMC 12B Problems/Problem 15
  46 bytes (5 words) - 00:14, 5 January 2023
• 2023 AMC 8 Problems/Problem 15
  6 KB (1,028 words) - 17:51, 11 May 2024
• 2023 AIME II Problems/Problem 15
  If \({\rm Rem} \left( n , 11 \right) = 8\), then \(k_n = 15\) and \(b_n = 167\). The first few values of <math>b_n</math> are <math>11, 17, 20, 10, 5, 14, 7, 15, 19, 21,</math> and <math>22</math>. We notice that <math>b_{12} = b_1 = 11
  7 KB (1,144 words) - 08:13, 1 February 2024
• Mock AIME 3 2006-2007 Problems/Problem 15
  210 bytes (41 words) - 09:51, 18 June 2023
• 2023 AMC 10A Problems/Problem 15
  4th area = <math>64\pi-49\pi = 15\pi</math>
  6 KB (967 words) - 07:01, 28 January 2024
• 2023 AMC 12A Problems/Problem 15
  draw((-50,15)--(50,15)); draw((50,15)--(50,-15));
  7 KB (1,074 words) - 21:22, 20 November 2023
• 2023 AMC 10B Problems/Problem 15
  <math>(2^{15}) \times (3^{14}) \times ((2^2)^{13}) \times (5^{12}) \times ((2 \times 3)^ ...immediately eliminate B, D, and E since 13 only appears in <math>13!, 14!, 15, 16!</math>, so <math>13\cdot 13\cdot 13\cdot 13</math> is a perfect square
  10 KB (1,510 words) - 14:27, 29 March 2024
• 2023 AMC 12B Problems/Problem 15
  ...18|2023 AMC 10B #18]] and [[2023 AMC 12B Problems/Problem 15|2023 AMC 12B #15]]}} ...<math>c</math> are positive integers such that<cmath>\frac{a}{14}+\frac{b}{15}=\frac{c}{210}.</cmath>Which of the following statements are necessarily tr
  7 KB (1,115 words) - 13:34, 21 April 2024
• Mock AIME 6 2006-2007 Problems/Problem 15
  751 bytes (134 words) - 10:27, 23 November 2023
• 2024 AIME I Problems/Problem 15
  10 KB (1,554 words) - 22:26, 13 April 2024
• 2021 WSMO Team Round/Problem 15
  335 bytes (52 words) - 13:01, 14 December 2023
• 2022 SSMO Team Round Problems/Problem 15
  859 bytes (143 words) - 13:07, 14 December 2023
• 2021 WSMO Team Round Problems/Problem 15
  361 bytes (55 words) - 22:16, 15 December 2023
• 2023 SSMO Team Round Problems/Problem 15
  2 KB (177 words) - 22:24, 15 December 2023
• 2002 AMC 12P Problems/Problem 15
  1 KB (191 words) - 19:16, 10 March 2024
• 2024 AMC 8 Problems/Problem 15
  ==Problem 15==
  4 KB (559 words) - 01:46, 18 April 2024
• 2024 AIME II Problems/Problem 15
  Case 2: <math>k = \tan 15^\circ</math>, <math>\tan 45^\circ</math>, <math>\tan 75^\circ</math>. So we only count for <math>k = \tan 15^\circ</math>.
  8 KB (1,395 words) - 17:26, 9 February 2024
• Mock AIME 2 2010 Problems/Problem 15
  3 bytes (2 words) - 23:06, 14 April 2024
• 2021 CIME I Problems/Problem 15
  0 bytes (0 words) - 23:09, 28 January 2024
• Mock AIME 5 Pre 2005 Problems/Problem 15
  280 bytes (34 words) - 00:11, 15 February 2024

Page text matches

• Pythagorean Theorem
  <cmath>8-15-17</cmath>
  5 KB (886 words) - 13:51, 15 May 2024
• 2005 AMC 12A Problems/Problem 16
  {{AMC12 box|year=2005|num-b=15|num-a=17|ab=A}}
  2 KB (307 words) - 15:30, 30 March 2024
• 2016 AMC 10A Problems/Problem 2
  ...an rewrite <math>10^{x}\cdot 100^{2x}=1000^{5}</math> as <math>10^{5x}=10^{15}</math>: 10^{5x} & =10^{15}
  1 KB (190 words) - 10:58, 16 June 2023
• 2016 AMC 10A Problems/Problem 4
  &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{-15}{16}\right\rfloor\\
  2 KB (257 words) - 10:57, 16 June 2023
• 2016 AMC 10A Problems/Problem 11
  label("$1$",(15/2,0),dir(270)); label("$1$",(15/2,0),dir(270));
  8 KB (1,016 words) - 00:17, 31 December 2023
• 2016 AMC 10A Problems/Problem 16
  {{AMC10 box|year=2016|ab=A|num-b=15|num-a=17}}
  1 KB (235 words) - 14:52, 25 June 2023
• MathILy (serious mathematics infused with levity)
  ...age or mathematical development, had its first year in 2015, and grew from 15 attendees in 2015 to 31 attendees in 2023.
  5 KB (706 words) - 23:49, 29 January 2024
• ICTM Math Contest
  The contest is given in-school May 1-15. Each school sets its own testing date and time within these guidelines.
  8 KB (1,182 words) - 14:26, 3 April 2024
• Math books
  ...roblems from the [[UK Junior Mathematical Olympiad]], for students ages 11-15.
  24 KB (3,177 words) - 12:53, 20 February 2024
• William Lowell Putnam Mathematical Competition
  **School receives &#36;15,000
  4 KB (623 words) - 13:11, 20 February 2024
• AMC historical results
  *Achievement Roll: Awarded to students in 6th or younger who score 15 points or higher on the AMC 8. Awarded to students in 8th or younger who sc *AIME floor: 81 (top ~15%)
  17 KB (1,921 words) - 20:53, 10 May 2024
• Inequality
  \\x+5-13+4x+20&\ge 3x+15
  12 KB (1,798 words) - 16:20, 14 March 2023
• American Invitational Mathematics Examination
  ...>: 3<br><u>Problem 6-10</u>: 4<br><u>Problem 10-12</u>: 5<br><u>Problem 12-15</u>: 6}} ...hile correct answers receive one point of credit, making the maximum score 15. Problems generally increase in difficulty as the exam progresses - the fi
  8 KB (1,057 words) - 12:02, 25 February 2024
• Math competition books
  ...roblems from the [[UK Junior Mathematical Olympiad]], for students ages 11-15.
  2 KB (302 words) - 16:45, 3 October 2019
• American Regions Mathematics League
  See the [[How to join an ARML team]] wiki page for more info. Teams have 15 students.
  2 KB (267 words) - 17:06, 7 March 2020
• Joining an ARML team
  ...The three top teams usually all place in the top 20, often even in the top 15 or 10. ...or is selected, although it does happen. Then the coaches select the best 15 young students (10th grade and below) to comprise a C team -- a team specif
  21 KB (3,500 words) - 18:41, 23 April 2024
• Stanford University Mathematics Camp
  Students must be current 10-11th graders who will be 15-17 years old on the first day of the program. The cost is \$6500.
  1 KB (166 words) - 17:54, 10 June 2016
• Simon's Favorite Factoring Trick
  ...the equations, we have the solutions as <math>\boxed{(5, 25), (24, 6), (6, 15), (14, 7), (8, 10), (9, 9)}</math>.
  7 KB (1,107 words) - 07:35, 26 March 2024
• Pigeonhole Principle
  # Prove that having 100 whole numbers, one can choose 15 of them so that the difference of any two is divisible by 7. ([[Pigeonhole
  11 KB (1,985 words) - 21:03, 5 August 2023
• Principle of Inclusion-Exclusion
  15 take algebra, language arts, social studies, and history. 15 take algebra, language arts, biology, and history.
  9 KB (1,703 words) - 07:25, 24 March 2024
• Factorial
  * <math>15! = 1307674368000</math>
  10 KB (809 words) - 16:40, 17 March 2024
• 2006 AIME II
  * [[2006 AIME II Problems/Problem 15]]
  1 KB (133 words) - 12:32, 22 March 2011
• Fermat's Little Theorem
  ...he units digit of <math>k^2 + 2^k</math>? ([[2008 AMC 12A Problems/Problem 15]])
  16 KB (2,658 words) - 16:02, 8 May 2024
• Alabama ARML
  ...9, Alabama did not send a team, but Grissom High in Alabama sent a team of 15 students to compete.
  2 KB (258 words) - 00:50, 28 December 2021
• LaTeX
  <cmath>1+2+3+4+5+\sin \pi = \frac{5\cdot 6}{2}+0=15.</cmath>
  1 KB (164 words) - 19:09, 14 February 2024
• North Carolina MathCounts
  * 2013 - Nikhil Reddy (15), Angela Deng (40), Lloyd Liu (42), Andrew Zhang, Coach: Michael Pillsbury
  4 KB (582 words) - 21:40, 14 May 2024
• Orthic triangle
  ...Y=15</math>. Compute the product <math>AB\cdot AC</math>. (AIME II, 2019, 15)
  8 KB (1,408 words) - 11:54, 8 December 2021
• Recursion
  * [[2004_AIME_I_Problems#Problem_15| 2004 AIME I Problem 15]]
  2 KB (316 words) - 16:03, 1 January 2024
• Combinatorial identity
  ....com/wiki/index.php/2021_AMC_12A_Problems/Problem_15 2021 AMC 12A Problem 15]
  12 KB (1,993 words) - 23:49, 19 April 2024
• Composite number
  4 6 8 9 10 12 14 15 16 18 20 21 22 24 25 26 27 28 30 32 33 34 35 36 38 39 40 42 44 45 46 48 49 ⇒ 13,15,17...... all are sum of 2 composites Hence, any odd positive ≥ 13 can be
  6 KB (350 words) - 12:58, 26 September 2023
• Circle
  ...cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",
  9 KB (1,581 words) - 18:59, 9 May 2024
• Ellipse
  ...prime integers, find <math> p+q. </math> ([[2005 AIME II Problems/Problem 15|Source]])
  5 KB (892 words) - 21:52, 1 May 2021
• Multiple
  ...nite]] number of multiples. As an example, some of the multiples of 15 are 15, 30, 45, 60, and 75.
  860 bytes (142 words) - 22:51, 26 January 2021
• Least common multiple
  ...multiply the results together. For example, to find the LCM of 8, 12 and 15, write: <math>15 = 3^1\cdot 5^1</math>
  2 KB (383 words) - 10:49, 4 September 2022
• Math Bee
  ...tes, while the second phase has 20 free-response questions to be solved in 15 minutes.
  1 KB (197 words) - 10:59, 14 April 2024
• Math Day at the Beach
  This round lasts 45 minutes and consists of 15 multiple-choice questions. Scoring consists of:
  4 KB (644 words) - 12:56, 29 March 2017
• Writing scholarships
  ...uch as creative writing or journalism. Apply for this USD 500 award by May 15 of each year, including a 400-600 word short story in prose or in script.
  7 KB (1,039 words) - 18:45, 18 January 2024
• National Scholarships
  * [[Best Buy @15 Scholarship]] of <dollar/>1,000 for students in grades 9-12. [https://www.a
  4 KB (538 words) - 00:48, 28 January 2024
• Law of Cosines
  label("d",(15,0),(0,-1)); \qquad \mathrm{(B) \ } 8/\sqrt{15}
  6 KB (1,003 words) - 09:11, 7 June 2023
• Chicago ARML
  The [[Chicago ARML]] team consists of four groups of approximately 15 high school students. The four groups are:
  2 KB (227 words) - 11:47, 4 December 2023
• Law of Sines
  \qquad \mathrm{(B) \ } 8/\sqrt{15} ([[Mock AIME 4 2006-2007 Problems/Problem 15|Source]])
  4 KB (658 words) - 16:19, 28 April 2024
• Homothety
  pair A=(15,15),B=(30,15),C=(30,30),D=(15,30),a=(60,60),b=(120,60),c=(120,120),d=(60,120); ...triangle <math>ABC</math> are <math>13,</math> <math>14,</math> and <math>15,</math> the radius of <math>\omega</math> can be represented in the form <m
  3 KB (532 words) - 01:11, 11 January 2021
• Pascal's triangle
  ...aring above it. For example, <math>{5 \choose 1}+{5 \choose 2} = 5 + 10 = 15 = {6 \choose 2}</math>. This property allows the easy creation of the firs
  5 KB (838 words) - 17:20, 3 January 2023
• 2006 AMC 10B
  * [[2006 AMC 10B Problems/Problem 15]]
  2 KB (182 words) - 21:57, 23 January 2021
• Mathematical Olympiad Summer Program
  Top non-senior USAMO finishers: In addition to the winners, the next 15 or so non-senior non-Canadian finishers are invited to attend MOP. This gro ...ree instructional sessions: 8:30 AM - 10:00 AM, 10:15 AM - 11:45 AM, and 1:15 PM - 2:45 PM. Classes usually consist of a lecture followed by a problem se
  6 KB (936 words) - 10:37, 27 November 2023
• Power of a Point Theorem
  * [[2005 AIME I Problems/Problem 15|2005 AIME I Problem 15]]
  5 KB (827 words) - 17:30, 21 February 2024
• 2004 AIME I
  * [[2004 AIME I Problems/Problem 15]]
  1 KB (135 words) - 18:15, 19 April 2021
• 2004 AIME II
  * [[2004 AIME II Problems/Problem 15]]
  1 KB (135 words) - 12:24, 22 March 2011
• 2005 AIME I
  * [[2005 AIME I Problems/Problem 15 | Problem 15]]
  1 KB (154 words) - 12:30, 22 March 2011
• 2006 AIME I
  * [[2006 AIME I Problems/Problem 15]]
  1 KB (135 words) - 12:31, 22 March 2011
• 2005 AIME II
  * [[2005 AIME II Problems/Problem 15]]
  1 KB (135 words) - 12:30, 22 March 2011
• 2006 AMC 12B
  * [[2006 AMC 12B Problems/Problem 15 | Problem 15]]
  2 KB (210 words) - 00:06, 7 October 2014
• Rhombus
  * [[2006_AMC_10B_Problems/Problem_15 | 2006 AMC 10B Problem 15]]
  3 KB (490 words) - 15:30, 22 February 2024
• Mock AMC
  [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9574 11-15]
  51 KB (6,175 words) - 20:58, 6 December 2023
• Modular arithmetic/Intermediate
  <math>x^2 - 2x - 15 \equiv 0 \pmod{21}</math>. However, since <math>15 \equiv 99 \pmod{21}</math>, the original congruence is equivalent to
  14 KB (2,317 words) - 19:01, 29 October 2021
• 2006 AMC 10A
  * [[2006 AMC 10A Problems/Problem 15]]
  2 KB (180 words) - 18:06, 6 October 2014
• 2006 AIME I Problems
  The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, </mat == Problem 15 ==
  7 KB (1,173 words) - 03:31, 4 January 2023
• 2006 AIME I Problems/Problem 14
  {{AIME box|year=2006|n=I|num-b=13|num-a=15}}
  6 KB (980 words) - 21:45, 31 March 2020
• 2006 AIME I Problems/Problem 13
  ...>k=6</math>, then <math>n<1000</math> implies that <math>\frac{n+1}{64}\le 15</math>, so <math>n+1=64,64\cdot 3^2</math>.
  10 KB (1,702 words) - 00:45, 16 November 2023
• 2006 AIME I Problems/Problem 5
  The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, </mat ...h> a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</cmath>
  3 KB (439 words) - 18:24, 10 March 2015
• 2006 AIME I Problems/Problem 4
  ...> have a factor of <math>10</math>. <math>86</math> have a factor of <math>15</math>. And so on. This gives us an initial count of <math>96 + 91 + 86 + \
  2 KB (278 words) - 08:33, 4 November 2022
• 2006 AIME I Problems/Problem 3
  ...ber after digit removal could be 10,20,30) or <math>c=5.</math> Testing 5, 15, and 25 as the numbers after removal we find that our answer is clearly <ma
  4 KB (622 words) - 03:53, 10 December 2022
• 2021 JMC 10
  ...f{(A) } 5 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 15 \qquad\textbf{(E) } 16</math> ...e written with digits in reverse order. A citizen in Malachar writes <math>15\cdot 73.</math> What does this Malacharian write as the answer?
  12 KB (1,784 words) - 16:49, 1 April 2021
• 2006 AMC 12B Problems
  \text {(A) } 5 \qquad \text {(B) } 10 \qquad \text {(C) } 15 \qquad \text {(D) } 20 \qquad \text {(E) } 25 ...three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?
  13 KB (2,058 words) - 12:36, 4 July 2023
• 2006 AMC 12A
  * [[2006 AMC 12A Problems/Problem 15]]
  1 KB (168 words) - 21:51, 6 October 2014
• 2004 AMC 12A
  * [[2004 AMC 12A Problems/Problem 15]]
  2 KB (186 words) - 17:35, 16 December 2019
• 2004 AMC 12B
  * [[2004 AMC 12B Problems/Problem 15]]
  2 KB (181 words) - 21:40, 6 October 2014
• 2005 AMC 12A
  ** [[2005 AMC 12A Problems/Problem 15|Problem 15]]
  2 KB (202 words) - 21:30, 6 October 2014
• 2005 AMC 12B
  * [[2005 AMC 12B Problems/Problem 15|Problem 15]]
  2 KB (206 words) - 23:23, 21 June 2021
• 2000 AMC 12
  * [[2000 AMC 12 Problems/Problem 15]]
  1 KB (126 words) - 13:28, 20 February 2020
• 2001 AMC 12
  * [[2001 AMC 12 Problems/Problem 15]]
  1 KB (127 words) - 21:36, 6 October 2014
• 2002 AMC 12A
  * [[2002 AMC 12A Problems/Problem 15]]
  1 KB (158 words) - 21:33, 6 October 2014
• 2003 AMC 12A
  * [[2003 AMC 12A Problems/Problem 15]]
  1 KB (162 words) - 21:52, 6 October 2014
• 2002 AMC 12B
  * [[2002 AMC 12B Problems/Problem 15]]
  1 KB (154 words) - 00:32, 7 October 2014
• 2003 AMC 12B
  * [[2003 AMC 12B Problems/Problem 15]]
  1 KB (160 words) - 20:46, 1 February 2016
• 2006 AMC 12A Problems
  <math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 20\qquad \mathrm{(D) \ } 24\ label("$y$",(15,-4),N);
  15 KB (2,223 words) - 13:43, 28 December 2020
• 2005 AMC 12A Problems
  == Problem 15 == [[2005 AMC 12A Problems/Problem 15|Solution]]
  13 KB (1,971 words) - 13:03, 19 February 2020
• 2004 AMC 12A Problems
  ...Players <math>A</math>, <math>B</math> and <math>C</math> start with <math>15</math>, <math>14</math> and <math>13</math> tokens, respectively. How many == Problem 15 ==
  13 KB (1,953 words) - 00:31, 26 January 2023
• 2003 AMC 12A Problems
  A solid box is <math>15</math> cm by <math>10</math> cm by <math>8</math> cm. A new solid is formed == Problem 15 ==
  13 KB (1,955 words) - 21:06, 19 August 2023
• 2002 AMC 12A Problems
  <math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 34\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 51\ ...xt{(A)}\ 12 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 14 \qquad \text{(D)}\ 15 \qquad \text{(E)} 16</math>
  12 KB (1,792 words) - 13:06, 19 February 2020
• 2000 AMC 12 Problems
  == Problem 15 == [[2000 AMC 12 Problems/Problem 15|Solution]]
  13 KB (1,948 words) - 12:26, 1 April 2022
• 2001 AMC 12 Problems
  ...ee numbers is <math>10</math> more than the least of the numbers and <math>15</math> == Problem 15 ==
  13 KB (1,957 words) - 12:53, 24 January 2024
• 2002 AMC 12B Problems
  == Problem 15 == [[2002 AMC 12B Problems/Problem 15|Solution]]
  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> ...{(A) } 3 \qquad \text {(B) } 5 \qquad \text {(C) } 11 \qquad \text {(D) } 15 \qquad \text {(E) } 165
  13 KB (1,987 words) - 18:53, 10 December 2022
• 2004 AMC 12B Problems
  ...{B}) 6 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 12 \qquad (\mathrm {E}) 15</math> <math>(\mathrm {A}) 13\qquad (\mathrm {B}) 14 \qquad (\mathrm {C}) 15 \qquad (\mathrm {D}) 16 \qquad (\mathrm {E}) 17</math>
  13 KB (2,049 words) - 13:03, 19 February 2020
• 2005 AMC 12B Problems
  ...math>80</math> points, <math>20\%</math> got <math>85</math> points, <math>15\%</math> got <math>90</math> points, and the rest got <math>95</math> point == Problem 15 ==
  12 KB (1,781 words) - 12:38, 14 July 2022
• 2006 AMC 12B Problems/Problem 4
  \text {(A) } 5 \qquad \text {(B) } 10 \qquad \text {(C) } 15 \qquad \text {(D) } 20 \qquad \text {(E) } 25
  1 KB (152 words) - 16:11, 8 December 2013
• 2006 AMC 12B Problems/Problem 10
  ...three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? ...third side has length <math>15</math>, and so the perimeter is <math>21+7+15=43 \Rightarrow \boxed{\text {(A)}}</math>.
  977 bytes (156 words) - 13:57, 19 January 2021
• 2006 AMC 12B Problems/Problem 14
  ...)=23</math>. The problem asks for the total cost of jam, or <math>N(5J)=11(15)=165</math> cents, or <math>1.65\implies\mathrm{(D)}</math> {{AMC12 box|year=2006|ab=B|num-b=13|num-a=15}}
  1 KB (227 words) - 17:21, 8 December 2013
• 2006 AMC 12B Problems/Problem 16
  {{AMC12 box|year=2006|ab=B|num-b=15|num-a=17}}
  1 KB (203 words) - 16:36, 18 September 2023
• 2006 AMC 12B Problems/Problem 20
  \mathrm{(D)}\ \frac 15 .../(1-r)</math> for the sum of an infinite geometric sequence, we get <math>(15/100)/(1-(1/10)) = \boxed{\frac 16}</math>.
  3 KB (485 words) - 14:09, 21 May 2021
• 2006 AMC 12A Problems/Problem 14
  {{AMC12 box|year=2006|ab=A|num-b=13|num-a=15}}
  3 KB (442 words) - 03:13, 8 August 2022
• 2006 AMC 12A Problems/Problem 16
  {{AMC12 box|year=2006|ab=A|num-b=15|num-a=17}}
  2 KB (286 words) - 10:16, 19 December 2021
• 2006 AMC 12A Problems/Problem 22
  ...e OAP = OAB - 60^{\circ} = \frac{180^{\circ}-30^{\circ}}{2} - 60^{\circ} = 15^{\circ}</math>. Since <math>OA</math> is a radius and <math>OP</math> can b ...>, then we get a [[right triangle]]. Using simple trigonometry, <math>\cos 15^{\circ} = \frac{\frac{r}{2}}{2\sqrt{3}} = \frac{r}{4\sqrt{3}}</math>.
  2 KB (343 words) - 15:39, 14 June 2023
• 2006 AMC 12A Problems/Problem 25
  ...-[[empty set | empty]] [[subset]]s <math>S</math> of <math>\{1,2,3,\ldots ,15\}</math> have the following two properties? <math>{15 \choose 1} + {14 \choose 2} + {13 \choose 3} + ... + {9 \choose 7} + {8 \ch
  8 KB (1,405 words) - 11:52, 27 September 2022
• 2005 AMC 12B Problems/Problem 6
  ...(x)\left( \frac{-1}{7} \right) \rightarrow 64=49+x^2+2x \rightarrow x^2+2x-15=0</math>. The only positive value for <math>x</math> is <math>3</math>, whi \implies k^2+2k-15=0
  2 KB (299 words) - 15:29, 5 July 2022
• 2005 AMC 12B Problems/Problem 9
  ...math>80</math> points, <math>20\%</math> got <math>85</math> points, <math>15\%</math> got <math>90</math> points, and the rest got <math>95</math> point The mean is equal to <math>10\%\cdot70 + 25\%\cdot80 + 20\%\cdot85 + 15\%\cdot90 + 30\%\cdot95 = 86</math>.
  2 KB (280 words) - 15:35, 16 December 2021
• 2005 AMC 12B Problems/Problem 11
  ...12B Problems|2005 AMC 12B #11]] and [[2005 AMC 10B Problems|2005 AMC 10B #15]]}} ...th>6</math> bills left. <math>\dbinom{6}{2} = \dfrac{6\times5}{2\times1} = 15</math> ways. However, you counted the case when you have <math>2</math> ten
  4 KB (607 words) - 21:01, 20 May 2023
• 2005 AMC 12B Problems/Problem 12
  {{AMC10 box|year=2005|ab=B|num-b=15|num-a=17}}
  2 KB (317 words) - 12:27, 16 December 2021
• 2005 AMC 12B Problems/Problem 14
  {{AMC12 box|year=2005|ab=B|num-b=13|num-a=15}}
  2 KB (278 words) - 21:12, 24 December 2020
• 2005 AMC 12B Problems/Problem 16
  {{AMC12 box|year=2005|ab=B|num-b=15|num-a=17}}
  2 KB (364 words) - 04:54, 16 January 2023
• 2005 AMC 12B Problems/Problem 18
  xaxis(-1,15,Ticks(f, 2.0)); yaxis(-1,15,Ticks(f, 2.0));
  2 KB (262 words) - 21:20, 21 December 2020
• 2005 AMC 12B Problems/Problem 19
  ...b=4</math> then <math>(a+b)+(a-b)=11+4</math>, <math>2a=15</math>, <math>a=15/2</math>, which is not a digit. Hence the only possible value for <math>a-b
  2 KB (283 words) - 20:02, 24 December 2020
• 2005 AMC 12B Problems/Problem 23
  \textbf{(A)}\ \frac {15}{2} \qquad \textbf{(C)}\ 15 \qquad
  5 KB (786 words) - 16:49, 31 January 2023
• 2005 AMC 12B Problems/Problem 24
  <math> \mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)
  4 KB (761 words) - 09:10, 1 August 2023
• 2006 AMC 10A Problems
  <math>\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18\qquad\mathrm{(C)}\ 20\qquad\mathrm{(D)}\ 24\qquad\mat label("$y$",(15,-4),N);
  13 KB (2,028 words) - 16:32, 22 March 2022
• 2006 AMC 12A Problems/Problem 12
  {{AMC10 box|year=2006|ab=A|num-b=13|num-a=15}}
  2 KB (292 words) - 11:56, 17 December 2021
• 2006 AMC 10A Problems/Problem 16
  <math>\textbf{(A) } \frac{35}{2}\qquad\textbf{(B) } 15\sqrt{2}\qquad\textbf{(C) } \frac{64}{3}\qquad\textbf{(D) } 16\sqrt{2}\qquad {{AMC10 box|year=2006|num-b=15|num-a=17|ab=A}}
  5 KB (732 words) - 23:19, 19 September 2023
• 1991 AJHSME Problems
  <math>\text{(A)}\ 5 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 25</math> fill((15,3)--(16,3)--(16,2)--(15,2)--cycle,black); fill((14,2)--(15,2)--(15,1)--(14,1)--cycle,black);
  17 KB (2,246 words) - 13:37, 19 February 2020
• 2005 AIME II Problems/Problem 4
  ...[[positive integer]]s that are divisors of at least one of <math> 10^{10},15^7,18^{11}. </math> <math>15^7 = 3^7\cdot5^7</math> so <math>15^7</math> has <math>8\cdot8 = 64</math> divisors.
  3 KB (377 words) - 18:36, 1 January 2024
• 2005 AIME II Problems
  ...r of positive integers that are divisors of at least one of <math> 10^{10},15^7,18^{11}. </math> In triangle <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \ov
  7 KB (1,119 words) - 21:12, 28 February 2020
• 2005 AIME II Problems/Problem 7
  <cmath>(y^{14}+y^{12}+y^{10}+y^8+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)=\f
  2 KB (279 words) - 12:33, 27 October 2019
• 2005 AIME II Problems/Problem 10
  import three; currentprojection = perspective(4,-15,4); defaultpen(linewidth(0.7));
  3 KB (436 words) - 03:10, 23 September 2020
• 2005 AIME I Problems
  == Problem 15 == [[2005 AIME I Problems/Problem 15|Solution]]
  6 KB (983 words) - 05:06, 20 February 2019
• 2005 AIME I Problems/Problem 3
  ...9, 23, 29, 31, 37, 41, 43</math> and <math>47</math>) so there are <math> {15 \choose 2} =105</math> ways to choose a pair of primes from the list and th
  2 KB (249 words) - 09:37, 23 January 2024
• 2005 AIME I Problems/Problem 4
  15 & 330 & no\\ \hline
  8 KB (1,248 words) - 11:43, 16 August 2022
• 2005 AIME I Problems/Problem 10
  pair A=(15,32), B=(12,19), C=(23,20), M=B/2+C/2, P=(17,22); ...}p = \frac{107 \cdot 11 - 337}{11}</math>. Solving produces that <math>p = 15</math>. [[Substitution|Substituting]] backwards yields that <math>q = 32</m
  5 KB (852 words) - 21:23, 4 October 2023
• 2005 AIME I Problems/Problem 12
  ...<math>S(4), S(5), \ldots, S(8)</math> are even, and <math>S(9), \ldots, S(15)</math> are odd, and so on.
  4 KB (647 words) - 02:29, 4 May 2021
• 2005 AIME I Problems/Problem 13
  label("$(13,15,18)$", (4,5), N); label("$(18,15,13)$", (5,4), N);
  5 KB (897 words) - 00:21, 29 July 2022
• 2005 AIME I Problems/Problem 14
  {{AIME box|year=2005|n=I|num-b=13|num-a=15}}
  3 KB (561 words) - 14:11, 18 February 2018
• 2005 AIME II Problems/Problem 15
  D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP("y=ax",(14,14 * (69/100)^.5),E),EndArrow [[Image:2005_AIME_II_-15.png||center|800px]]
  12 KB (2,000 words) - 13:17, 28 December 2020
• 2005 AIME II Problems/Problem 14
  In [[triangle]] <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \ov pair C = rotate(15,A)*(A+dir(-50));
  13 KB (2,129 words) - 18:56, 1 January 2024
• 2004 AIME I Problems/Problem 15
  15&2^{13}-4\cdot 7\\ 17&2^{15}-16\cdot 9\\
  9 KB (1,491 words) - 01:23, 26 December 2022
• 2004 AIME I Problems/Problem 14
  currentprojection = perspective(-2,-50,15); size(200); {{AIME box|year=2004|n=I|num-b=13|num-a=15}}
  4 KB (729 words) - 01:00, 27 November 2022
• 2004 AIME I Problems/Problem 11
  ...ath>\frac{72}{5x^2} - 1 = \frac{27}{x^3} - 1</math> and so <math>x = \frac{15}{8}</math>. Then <math>k = \frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\p ...knowledge, we obtain <math>\ell = 5</math> and lateral area <math>A_\ell = 15\pi</math>. The area of the base is <math>A_B = 3^2\pi = 9\pi</math>.
  5 KB (839 words) - 22:12, 16 December 2015
• 2004 AIME I Problems/Problem 10
  A [[circle]] of [[radius]] 1 is randomly placed in a 15-by-36 [[rectangle]] <math> ABCD </math> so that the circle lies completely ...h>[A'B'C'] = [ABC] \cdot \left(\frac{r_{A'B'C'}}{r_{ABC}}\right)^2 = \frac{15 \times 36}{2} \cdot \frac{25}{36} = \frac{375}{2}</math>.
  5 KB (836 words) - 07:53, 15 October 2023
• 2004 AIME I Problems/Problem 5
  \frac{a}{q} < \frac{160}{300} = \frac{8}{15} a + b < \frac{8}{15}q + (350 - \frac{7}{10}q)
  3 KB (436 words) - 18:31, 9 January 2024
• 2004 AIME I Problems/Problem 7
  ...of <math>x</math> in <math>P(x)</math> is <math>-1 + 2 - 3 + \ldots + 14 - 15 = -8</math>, so <math>P(x) = 1 -8x + Cx^2 + Q(x)</math>, where <math>Q(x)</ ...roduct of each pair of distinct terms, or <math>C = \sum_{1 \le i \neq j}^{15} S_iS_j</math>. Also, we know that
  5 KB (833 words) - 19:43, 1 October 2023
• 2004 AIME I Problems/Problem 8
  ...ath>. Thus <math>\phi(1000) = 1000\left(1 - \frac 12\right)\left(1 - \frac 15\right) = 400</math>, and the answer is <math>\frac{400}{2} - 1 = 199</math>
  4 KB (620 words) - 21:26, 5 June 2021
• 2004 AIME I Problems
  A circle of radius 1 is randomly placed in a 15-by-36 rectangle <math> ABCD </math> so that the circle lies completely with == Problem 15 ==
  9 KB (1,434 words) - 13:34, 29 December 2021
• 2004 AIME II Problems/Problem 15
  <cmath>s_{13, 4} = 2s_{15 - 13, 3} + 1 = 2s_{2, 3}+1</cmath> (46,15)&64&16\\
  6 KB (899 words) - 20:58, 12 May 2022
• 2004 AIME II Problems/Problem 14
  {{AIME box|year=2004|n=II|num-b=13|num-a=15}}
  11 KB (1,857 words) - 21:55, 19 June 2023
• 2004 AIME II Problems/Problem 13
  ...AC \parallel DE, \angle ABC=120^\circ, AB=3, BC=5, </math> and <math>DE = 15. </math> Given that the [[ratio]] between the area of triangle <math> ABC < pair D=(0,0), E=(15,0), F=IP(CR(D, 75/7), CR(E, 45/7)), A=D+ (5+(75/7))/(75/7) * (F-D), C = E+
  3 KB (486 words) - 22:15, 7 April 2023
• 2004 AIME II Problems/Problem 10
  ...d power, we will also satisfy the congruence. Thus, <math>2^{3}, 2^{9}, 2^{15},</math> etc. work; or, <cmath>y-x \equiv 3 \pmod 6</cmath> ...l and error soon gives us <math>2^9=512</math>. Notice a pattern? Trying 2^15 = 32768 also works. You could go on, but basically all powers of two 3 mod
  8 KB (1,283 words) - 19:19, 8 May 2024
• 2004 AIME II Problems/Problem 7
  pair A=(0,0),B=(0,25),C=(70/3,25),D=(70/3,0),E=(0,8),F=(70/3,22),G=(15,0); ...= B'E</math> (by SAS). By the [[Pythagorean Theorem]], we have <math>AB' = 15</math>. Similarly, from <math>BF = B'F</math>, we have
  9 KB (1,501 words) - 05:34, 30 October 2023
• 2004 AIME II Problems/Problem 5
  ...math>\frac{250}{800}(60)=\frac{150}{8}</math>. The train then has <math>60-15-\frac{50}{3}-\frac{150}{8}=230/24</math> minutes left to travel 250 miles,
  4 KB (592 words) - 19:02, 26 September 2020
• 2004 AIME II Problems
  ...|| AD, AC || DE, \angle ABC=120^\circ, AB=3, BC=5, </math> and <math>DE = 15. </math> Given that the ratio between the area of triangle <math> ABC </mat == Problem 15 ==
  9 KB (1,410 words) - 05:05, 20 February 2019
• 2000 AIME II
  * [[2000 AIME II Problems/Problem 15|Problem 15]]
  1 KB (139 words) - 08:41, 7 September 2011
• 2001 AIME I
  * [[2001 AIME I Problems/Problem 15|Problem 15]]
  1 KB (139 words) - 08:41, 7 September 2011
• 2000 AIME I
  * [[2000 AIME I Problems/Problem 15|Problem 15]]
  1 KB (135 words) - 18:05, 30 May 2015
• 1999 AIME
  * [[1999 AIME Problems/Problem 15|Problem 15]]
  1 KB (118 words) - 08:41, 7 September 2011
• 1998 AIME
  * [[1998 AIME Problems/Problem 15|Problem 15]]
  1 KB (114 words) - 08:39, 7 September 2011
• 1997 AIME
  * [[1997 AIME Problems/Problem 15|Problem 15]]
  1 KB (114 words) - 08:39, 7 September 2011
• 1996 AIME
  * [[1996 AIME Problems/Problem 15|Problem 15]]
  1 KB (114 words) - 08:39, 7 September 2011
• 1995 AIME
  * [[1995 AIME Problems/Problem 15|Problem 15]]
  1 KB (114 words) - 08:38, 7 September 2011
• 1994 AIME
  * [[1994 AIME Problems/Problem 15|Problem 15]]
  1 KB (114 words) - 08:43, 7 September 2011
• 1983 AIME
  * [[1983 AIME Problems/Problem 15|Problem 15]]
  1 KB (114 words) - 20:35, 31 October 2020
• 1983 AIME Problems
  Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>0 < p < 15</math>. Determine the [[minimum]] value taken by <math>f(x)</math> for <mat == Problem 15 ==
  7 KB (1,104 words) - 12:53, 6 July 2022
• 1984 AIME Problems
  ...</math> is either <math>8</math> or <math>0</math>. Compute <math>\frac{n}{15}</math>. MP("t_1",(A+C+(C-A)*3/4+(A-C)*5/12)/2+(0,0.15),ESE);
  6 KB (933 words) - 01:15, 19 June 2022
• 1986 AIME Problems
  ...ments. Let <math>S</math> be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of <math>S</math> have the same sum. What ...ree HT, four TH, and five TT subsequences. How many different sequences of 15 coin tosses will contain exactly two HH, three HT, four TH, and five TT sub
  5 KB (847 words) - 15:48, 21 August 2023
• 1987 AIME Problems
  ...or which there is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>? == Problem 15 ==
  6 KB (869 words) - 15:34, 22 August 2023
• 1988 AIME Problems
  == Problem 15 == [[1988 AIME Problems/Problem 15|Solution]]
  6 KB (902 words) - 08:57, 19 June 2021
• 1989 AIME Problems
  == Problem 15 == [[1989 AIME Problems/Problem 15|Solution]]
  7 KB (1,045 words) - 20:47, 14 December 2023
• 1990 AIME Problems
  A triangle has vertices <math>P_{}^{}=(-8,5)</math>, <math>Q_{}^{}=(-15,-19)</math>, and <math>R_{}^{}=(1,-7)</math>. The equation of the bisector == Problem 15 ==
  6 KB (870 words) - 10:14, 19 June 2021
• 1991 AIME Problems
  ...<math>\overline{DA}</math>, respectively. It is given that <math>PB^{}_{}=15</math>, <math>BQ^{}_{}=20</math>, <math>PR^{}_{}=30</math>, and <math>QS^{} == Problem 15 ==
  7 KB (1,106 words) - 22:05, 7 June 2021
• 1992 AIME Problems
  \text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1 == Problem 15 ==
  8 KB (1,117 words) - 05:32, 11 November 2023
• 1993 AIME Problems
  ...n{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\ :(a) the winner caught <math>15</math> fish;
  8 KB (1,275 words) - 06:55, 2 September 2021
• 1994 AIME Problems
  The increasing sequence <math>3, 15, 24, 48, \ldots\,</math> consists of those positive multiples of 3 that are == Problem 15 ==
  7 KB (1,141 words) - 07:37, 7 September 2018
• 1995 AIME Problems
  == Problem 15 == [[1995 AIME Problems/Problem 15|Solution]]
  6 KB (1,000 words) - 00:25, 27 March 2024
• 1996 AIME Problems
  ...>, <math>AB=\sqrt{30}</math>, <math>AC=\sqrt{6}</math>, and <math>BC=\sqrt{15}</math>. There is a point <math>D</math> for which <math>\overline{AD}</mat == Problem 15 ==
  6 KB (931 words) - 17:49, 21 December 2018
• 1997 AIME Problems
  == Problem 15 == [[1997 AIME Problems/Problem 15|Solution]]
  7 KB (1,098 words) - 17:08, 25 June 2020
• 1998 AIME Problems
  ...th>\overline{CD},</math> respectively, so that <math>AP = 5, PB = 15, BQ = 15,</math> and <math>CR = 10.</math> What is the [[area]] of the [[polygon]] == Problem 15 ==
  7 KB (1,084 words) - 02:01, 28 November 2023
• 1999 AIME Problems
  ...sides of the triangle have lengths <math>AB=13, BC=14,</math> and <math>CA=15,</math> and the tangent of angle <math>PAB</math> is <math>m/n,</math> wher == Problem 15 ==
  7 KB (1,094 words) - 13:39, 16 August 2020
• 2000 AIME I Problems
  == Problem 15 == [[2000 AIME I Problems/Problem 15|Solution]]
  7 KB (1,204 words) - 03:40, 4 January 2023
• 2001 AIME I Problems
  In triangle <math>ABC</math>, <math>AB=13</math>, <math>BC=15</math> and <math>CA=17</math>. Point <math>D</math> is on <math>\overline{A == Problem 15 ==
  7 KB (1,212 words) - 22:16, 17 December 2023
• 2002 AIME I Problems
  == Problem 15 == [[2002 AIME I Problems/Problem 15|Solution]]
  8 KB (1,374 words) - 21:09, 27 July 2023
• 2003 AIME I Problems
  == Problem 15 == [[2003 AIME I Problems/Problem 15|Solution]]
  6 KB (965 words) - 16:36, 8 September 2019
• 2000 AIME II Problems
  Given that <center><math>\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10! ...</math>. It is given that <math>AB=13</math>, <math>BC=14</math>, <math>CA=15</math>, and that the distance from <math>O</math> to triangle <math>ABC</ma
  6 KB (947 words) - 21:11, 19 February 2019
• 2001 AIME II Problems
  == Problem 15 == [[2001 AIME II Problems/Problem 15|Solution]]
  8 KB (1,282 words) - 21:12, 19 February 2019
• 2002 AIME II Problems
  == Problem 15 == [[2002 AIME II Problems/Problem 15|Solution]]
  7 KB (1,177 words) - 15:42, 11 August 2023
• 2003 AIME II Problems
  ...e <math>ABC,</math> <math>AB = 13,</math> <math>BC = 14,</math> <math>AC = 15,</math> and point <math>G</math> is the intersection of the medians. Points ...ame side of line <math>AB</math> as <math>C</math> so that <math>AD = BD = 15.</math> Given that the area of triangle <math>CDM</math> may be expressed a
  7 KB (1,127 words) - 09:02, 11 July 2023
• 1983 AIME Problems/Problem 2
  Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>0 < p < 15</math>. Determine the [[minimum]] value taken by <math>f(x)</math> for <mat ...that <math>|x-p|=x-p</math>, <math>|x-15|=15-x</math>, and <math>|x-p-15|=15+p-x</math>.
  1 KB (184 words) - 20:16, 14 January 2023
• 1983 AIME Problems/Problem 3
  ...for <math>x^2+18x+30</math>, so that the equation becomes <math>y=2\sqrt{y+15}</math>. ...> or <math>y=-6</math>. The second root is extraneous since <math>2\sqrt{y+15}</math> is always non-negative (and moreover, plugging in <math>y=-6</math>
  3 KB (532 words) - 05:18, 21 July 2022
• 1983 AIME Problems/Problem 14
  pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); ...pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--a--m--n--cycle); draw(
  13 KB (2,149 words) - 18:44, 5 February 2024
• 1983 AIME Problems/Problem 15
  <!-- [[Image:1983_AIME-15.png|200px]] --> <cmath>0 \geq f(3) = 25 - 15\cos \alpha - 20 \sin \alpha</cmath>
  19 KB (3,221 words) - 01:05, 7 February 2023
• 1984 AIME Problems/Problem 2
  ...</math> is either <math>8</math> or <math>0</math>. Compute <math>\frac{n}{15}</math>. Any multiple of 15 is a multiple of 5 and a multiple of 3.
  1 KB (187 words) - 20:05, 29 May 2021
• 1984 AIME Problems/Problem 5
  ...ath>. This means that <math>\log ab^3 = 15\log 2 \Longrightarrow ab^3 = 2^{15}</math> and that <math>\log a^3 b = 21\log 2 \Longrightarrow a^3 b = 2^{21}
  6 KB (863 words) - 16:10, 16 May 2024
• 1984 AIME Problems/Problem 9
  ...>AB</math> has length 3 cm. The area of [[face]] <math>ABC</math> is <math>15\mbox{cm}^2</math> and the area of face <math>ABD</math> is <math>12 \mbox { ...ghtanglemark(D,Da,Db));draw(rightanglemark(A,Db,D));draw(anglemark(Da,Db,D,15));
  6 KB (947 words) - 20:44, 26 November 2021
• 1984 AIME Problems/Problem 10
  ...symbol{\cdots}&\boldsymbol{12}&\boldsymbol{13}&\boldsymbol{14}&\boldsymbol{15}&\boldsymbol{16}&\boldsymbol{17}&\boldsymbol{18}&\boldsymbol{19}&\boldsymbo
  7 KB (1,163 words) - 23:53, 28 March 2022
• 1984 AIME Problems/Problem 12
  ...the double counted values. This means that <math>x = 0, \pm 5, \pm 10, \pm 15... \pm 1000</math> so the answer is 400 + 1 = <math>\boxed{401}</math>
  3 KB (588 words) - 14:37, 22 July 2020
• 1984 AIME Problems/Problem 14
  ...or <math>4 \bmod{6}</math>. Notice that the numbers <math>9</math>, <math>15</math>, <math>21</math>, ... , and in general <math>9 + 6n</math> for nonne ...which yields <math>n=34=9+25</math> which does not work). Thus <math>n-9,n-15,n-21,n-27,</math> and <math>n-33</math> form a prime quintuplet. However, o
  8 KB (1,346 words) - 01:16, 9 January 2024
• 1984 AIME Problems/Problem 15
  \frac{x^2}{15}+\frac{y^2}{7}-\frac{z^2}{9}-\frac{w^2}{33}=1\\ \frac{x^2}{63}+\frac{y^2}{55}+\frac{z^2}{39}+\frac{w^2}{15}=1\\
  6 KB (1,051 words) - 04:52, 8 May 2024
• 1985 AIME Problems/Problem 14
  ...t have <math>n > 10</math>, so <math>n = 15</math> and the answer is <math>15 + 10 = \boxed{25}</math>. ..., then the strongest <math>16</math> people get a total of <math>16*10-145=15</math> playing against the weakest <math>10</math> who gained <math>45</mat
  5 KB (772 words) - 22:14, 18 June 2020
• 1985 AIME Problems/Problem 11
  ...th>d = \sqrt{(9 - 49)^2 + (20 + 55)^2} = \sqrt{40^2 + 75^2} = 5\sqrt{8^2 + 15^2} = 5\cdot 17 = 85</math>.
  5 KB (932 words) - 17:00, 1 September 2020
• 1985 AIME Problems/Problem 10
  <math>\frac{15}{24} \to 11</math> <math>\frac{20}{24}\to 15</math>
  12 KB (1,859 words) - 18:16, 28 March 2022
• 1985 AIME Problems/Problem 9
  real r = 8/15^0.5, a = 57.91, b = 93.135; real r = 8/15^0.5, a = 57.91, b = 93.135;
  5 KB (763 words) - 16:20, 28 September 2019
• 1985 AIME Problems/Problem 6
  <cmath>12+16\cdot \frac ab + 5\cdot \frac ba = \frac ba\cdot 15</cmath>
  5 KB (789 words) - 03:09, 23 January 2023
• 1986 AIME Problems/Problem 15
  <math>2a^2 + 5b^2 = - \frac {15}{2}ab \ \ \ \ (2)</math></div>
  11 KB (1,722 words) - 09:49, 13 September 2023
• 1986 AIME Problems/Problem 14
  ...<math>2\sqrt{5}</math>, <math>\frac{30}{\sqrt{13}}</math>, and <math>\frac{15}{\sqrt{10}}</math>. Determine the [[volume]] of <math>P</math>. <cmath>\frac {hl}{\sqrt {h^2 + l^2}} = \frac {15}{\sqrt {10}}</cmath>
  2 KB (346 words) - 13:13, 22 July 2020
• 1986 AIME Problems/Problem 13
  ...H</tt>, and five <tt>TT</tt> subsequences. How many different sequences of 15 coin tosses will contain exactly two <tt>HH</tt>, three <tt>HT</tt>, four <
  4 KB (772 words) - 21:09, 7 May 2024
• 1986 AIME Problems/Problem 12
  ...ments. Let <math>S</math> be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of <math>S</math> have the same sum. What ...> must have more than 4 elements, otherwise its sum would be at most <math>15+14+13+12=54</math>.
  2 KB (364 words) - 19:41, 1 September 2020
• 1986 AIME Problems/Problem 11
  ...ath>1</math>). By the [[Binomial Theorem]], this is <math>(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}</math>. ...y^2</math> term is <math>{2\choose 0} + {3\choose 1} + \cdots + {17\choose 15}</math>. This is actually the sum of the first 16 triangular numbers, which
  6 KB (872 words) - 16:51, 9 June 2023
• 1986 AIME Problems/Problem 10
  ...ath>a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14</math> so <math>a+b+c\geq 15</math>. Checking each of the multiples of <math>222</math> from <math>15\cdot222</math> to <math>18\cdot222</math> by subtracting <math>N</math> fro
  3 KB (565 words) - 16:51, 1 October 2023
• 1986 AIME Problems/Problem 9
  By similar triangles, <math>BE'=\frac{d}{510}\cdot450=\frac{15}{17}d</math> and <math>EC=\frac{d}{425}\cdot450=\frac{18}{17}d</math>. Sinc ...510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}</math>.
  11 KB (1,850 words) - 18:07, 11 October 2023
• 1986 AIME Problems/Problem 8
  ...\log 1 + \log 2 + \log 4 + \log 5 +\ldots + \log 1000000 = \log (2^05^0)(2^15^0)(2^25^0)\cdots (2^65^6).</cmath> Each power of <math>2</math> appears <ma
  3 KB (487 words) - 20:52, 16 September 2020
• 1986 AIME Problems/Problem 2
  ...right)^2}{2\cdot 2\sqrt5\cdot2\sqrt6}=\frac{16}{8\sqrt{30}}=\sqrt{\frac{2}{15}},</cmath> so <math>\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}.</math>
  3 KB (460 words) - 00:44, 5 February 2022
• 1987 AIME Problems/Problem 14
  {{AIME box|year=1987|num-b=13|num-a=15}}
  7 KB (965 words) - 10:42, 12 April 2024
• 1987 AIME Problems/Problem 10
  ...ance at rate <math>r</math> from the escalator, while Bob is getting <math>15</math> seconds of help at rate <math>r</math>. Solving for <math>r</math>,
  7 KB (1,187 words) - 16:21, 27 January 2024
• 1987 AIME Problems/Problem 8
  ...or which there is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>? <cmath>\begin{array}{ccccc}\frac{15}{8} &>& \frac{k + n}{n} &>& \frac{13}{7}\\
  2 KB (393 words) - 16:59, 16 December 2020
• 1987 AIME Problems/Problem 4
  ...h>|x - 60| = 0</math>, which is when <math>x = 60</math> and <math>y = \pm 15</math>. Since the graph is [[symmetry|symmetric]] about the y-axis, we just ...using the [[Shoelace Theorem]], we get <math>2 \cdot \frac{1}{2}(80 - 48)(15) = \boxed{480}</math>.
  2 KB (371 words) - 17:25, 13 February 2024
• 1987 AIME Problems/Problem 3
  ...6,\ 2^3 = 8,\ 2 \cdot 5 = 10,</math> <math>\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,\ 3 \cdot 7 = 21,</math> <math>\ 2 \cdot 11 = 22,\ 2 \cdot 13 = 26,</math>
  3 KB (511 words) - 09:29, 9 January 2023
• 1988 AIME Problems/Problem 14
  {{AIME box|year=1988|num-b=13|num-a=15}}
  4 KB (700 words) - 17:21, 3 May 2021
• 1988 AIME Problems/Problem 13
  ...+ F_{15}) + 1 &\Longrightarrow (aF_{17} + bF_{16})\cdot x + (aF_{16} + bF_{15} + 1) = 0,\ x\not\in Q \\ &\Longrightarrow aF_{17} + bF_{16} = 0 \text{ and } aF_{16} + bF_{15} + 1 = 0 \\
  10 KB (1,585 words) - 03:58, 1 May 2023
• 1988 AIME Problems/Problem 11
  <math>504 = 3m + 15</math>
  2 KB (422 words) - 00:22, 6 September 2020
• 1988 AIME Problems/Problem 6
  ...use the arithmetic progression from left to right has difference <math>x - 15</math>. Therefore, we have <math>x = 50</math>, and because the desired ast
  5 KB (878 words) - 23:06, 20 November 2023
• 1989 AIME Problems/Problem 15
  ...ac{[APC] + [BPC]}{[APB]} = 3</math>, so <math>CP = \dfrac{3}{4} \cdot CF = 15</math>. ...ed out, so <math>w_C = 1</math> and <math>w_F = 3</math>. Thus, <math>CP = 15</math> and <math>PF = 5</math>.
  13 KB (2,091 words) - 00:20, 26 October 2023
• 1989 AIME Problems/Problem 14
  {{AIME box|year=1989|num-b=13|num-a=15}}
  2 KB (408 words) - 17:28, 16 September 2023
• 1989 AIME Problems/Problem 11
  <math>n = 4: 5000+15*116 = 6740</math>
  5 KB (851 words) - 18:01, 28 December 2022
• 1989 AIME Problems/Problem 6
  label("$P$",(6,15),N); label("$X$",(12.5,15),N);
  6 KB (980 words) - 15:08, 14 May 2024
• 1989 AIME Problems/Problem 4
  <math>m = 15</math> gives a solution for k. <math>10 + 5a = 15^3</math>
  3 KB (552 words) - 12:41, 3 March 2024
• 1990 AIME Problems/Problem 14
  {{AIME box|year=1990|num-b=13|num-a=15}}
  7 KB (1,086 words) - 08:16, 29 July 2023
• 1990 AIME Problems/Problem 12
  ...th of each of the 12 sides is <math>2 \cdot 12\sin 15</math>. <math>24\sin 15 = 24\sin (45 - 30) = 24\frac{\sqrt{6} - \sqrt{2}}{4} = 6(\sqrt{6} - \sqrt{2
  6 KB (906 words) - 13:23, 5 September 2021
• 1990 AIME Problems/Problem 7
  ...gle]] has [[vertex|vertices]] <math>P_{}^{}=(-8,5)</math>, <math>Q_{}^{}=(-15,-19)</math>, and <math>R_{}^{}=(1,-7)</math>. The [[equation]] of the [[ang pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17);
  8 KB (1,319 words) - 11:34, 22 November 2023
• 1990 AIME Problems/Problem 5
  The [[prime factorization]] of <math>75 = 3^15^2 = (2+1)(4+1)(4+1)</math>. For <math>n</math> to have exactly <math>75</ma
  1 KB (175 words) - 03:45, 21 January 2023
• 2006 AMC 10B Problems/Problem 10
  ...times as long as a second side, and the length of the third side is <math>15</math>. What is the greatest possible perimeter of the triangle? The lengths of the sides are: <math>x</math>, <math>3x</math>, and <math>15</math>.
  900 bytes (132 words) - 13:57, 26 January 2022
• 2006 AMC 10B Problems/Problem 14
  {{AMC10 box|year=2006|ab=B|num-b=13|num-a=15}}
  2 KB (264 words) - 21:10, 19 September 2023
• 1991 AIME Problems/Problem 14
  {{AIME box|year=1991|num-b=13|num-a=15}}
  2 KB (284 words) - 03:56, 23 January 2023
• 1991 AIME Problems/Problem 12
  ...<math>\overline{DA}</math>, respectively. It is given that <math>PB^{}_{}=15</math>, <math>BQ^{}_{}=20</math>, <math>PR^{}_{}=30</math>, and <math>QS^{} ...",P,N);label("\(Q\)",Q,E);label("\(R\)",R,SW);label("\(S\)",S,W); label("\(15\)",B/2+P/2,N);label("\(20\)",B/2+Q/2,E);label("\(O\)",O,SW); </asy></center
  8 KB (1,270 words) - 23:36, 27 August 2023
• 2006 AMC 10B Problems/Problem 16
  {{AMC10 box|year=2006|ab=B|num-b=15|num-a=17}}
  2 KB (336 words) - 10:51, 11 May 2024
• 1991 AIME Problems/Problem 11
  ...</math> and <math>m \angle MOA = 15^\circ</math>. Thus <math>AM = (1) \tan{15^\circ} = 2 - \sqrt {3}</math>, which is the radius of one of the circles. T Note that it is very useful to understand the side lengths of a 15-75-90 triangle, as these triangles often appear on higher level math contes
  4 KB (740 words) - 19:33, 28 December 2022
• 1991 AIME Problems/Problem 10
  bab & 2 & 4 & 15 \\
  5 KB (813 words) - 06:10, 25 February 2024
• 1991 AIME Problems/Problem 9
  ...only the [[positive]] root will work, so the value of <math>y = \frac{29}{15}</math> and <math>m + n = \boxed{044}</math>. <cmath>\csc x + \cot x = \frac {841}{435} = \frac {29}{15},</cmath>
  10 KB (1,590 words) - 14:04, 20 January 2023
• 1992 AIME Problems/Problem 14
  {{AIME box|year=1992|num-b=13|num-a=15}}
  4 KB (667 words) - 01:26, 16 August 2023
• 1992 AIME Problems/Problem 4
  \text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1
  3 KB (476 words) - 14:13, 20 April 2024
• 2006 AMC 10B Problems/Problem 23
  <math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{( label("$7$",(1.45,0.15));
  5 KB (861 words) - 00:53, 25 November 2023
• 2006 AMC 10B Problems/Problem 24
  pair top=X+15*dir(X--A), bottom=X+15*dir(X--B);
  4 KB (558 words) - 14:38, 6 April 2024
• 1993 AIME Problems/Problem 14
  {{AIME box|year=1993|num-b=13|num-a=15}}
  3 KB (601 words) - 09:25, 19 November 2023
• 1993 AIME Problems/Problem 13
  ...ac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}</math>. Therefore, <math>t = CD = CX\cdot\tan(\angle BXA) = 100 \cdot \frac{8}{15} = \frac{160}{3}</math>, and the answer is <math>\boxed{163}</math>.
  8 KB (1,231 words) - 20:06, 26 November 2023
• 1993 AIME Problems/Problem 3
  ...n{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\ :(a) the winner caught <math>15</math> fish;
  2 KB (364 words) - 00:05, 9 July 2022
• 1994 AIME Problems/Problem 15
  .../2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue);
  4 KB (717 words) - 22:20, 3 June 2021
• 1994 AIME Problems/Problem 14
  {{AIME box|year=1994|num-b=13|num-a=15}}
  2 KB (303 words) - 00:03, 28 December 2017
• 1994 AIME Problems/Problem 11
  ..., and <math>15</math>'s. Because <math>0</math>, <math>6</math>, and <math>15</math> are all multiples of <math>3</math>, the change will always be a mul
  4 KB (645 words) - 15:12, 15 July 2019
• 1994 AIME Problems/Problem 6
  draw((-15,-10)--(15,-10)); draw((-15,10)--(15,10));
  4 KB (721 words) - 16:14, 8 March 2021
• 1994 AIME Problems/Problem 1
  The increasing [[sequence]] <math>3, 15, 24, 48, \ldots\,</math> consists of those [[positive]] multiples of 3 that
  946 bytes (139 words) - 21:05, 1 September 2023
• 1995 AIME Problems/Problem 15
  ...ve. This is an infinite geometric series whose sum is <math>\frac{3/64}{1-(15/32)}=\frac{3}{34}</math>, so the answer is <math>\boxed{037}</math>. ...e roll <math>4</math> consecutive <tt>H</tt>'s, and there is a <math>\frac{15}{16}</math> probability we roll a <tt>T</tt>. Thus,
  6 KB (979 words) - 13:20, 11 April 2022
• 1995 AIME Problems/Problem 14
  {{AIME box|year=1995|num-b=13|num-a=15}}
  3 KB (484 words) - 13:11, 14 January 2023
• 1995 AIME Problems/Problem 8
  For <math>y=3</math>, we have <math>15,27,\cdots ,99</math>, or <math>8</math> cases.
  4 KB (646 words) - 17:37, 1 January 2024
• 1996 AIME Problems/Problem 15
  ...\Longrightarrow \cos 2\theta = \frac{\sqrt{3}}{2} \Longrightarrow \theta = 15^{\circ}</math>. The answer is <math>\lfloor 1000r \rfloor = \left\lfloor 10 ...cdot 4}}{8} = \frac {\sqrt {6} \pm \sqrt {2}}{4}</math>, so <math>\theta = 15^{\circ}</math> (as the other roots are too large to make sense in context).
  5 KB (710 words) - 21:04, 14 September 2020
• 1996 AIME Problems/Problem 14
  {{AIME box|year=1996|num-b=13|num-a=15}}
  5 KB (923 words) - 21:21, 22 September 2023
• 1996 AIME Problems/Problem 13
  ...>, <math>AB=\sqrt{30}</math>, <math>AC=\sqrt{6}</math>, and <math>BC=\sqrt{15}</math>. There is a point <math>D</math> for which <math>\overline{AD}</mat pair B=(0,0), C=(15^.5, 0), A=IP(CR(B,30^.5),CR(C,6^.5)), E=(B+C)/2, D=foot(B,A,E);
  3 KB (521 words) - 01:18, 25 February 2016
• 1996 AIME Problems/Problem 12
  <math>\sum\limits_{k = 1}^{9}\sum\limits_{j = 1}^{k}j = 45+36+28+21+15+10+6+3+1 = 165</math>.
  5 KB (879 words) - 11:23, 5 September 2021
• 1996 AIME Problems/Problem 11
  Hence, the answer is <math>\frac{18\pi}{15}+\frac{5\pi}{15}=\frac{23\pi}{15}\cdot\frac{180}{\pi}=\boxed{276}</math>
  6 KB (1,022 words) - 20:23, 17 April 2021
• 1996 AIME Problems/Problem 6
  <math>\frac{5}{16}+\frac{5}{16}-\frac{5}{32}=\frac{15}{32}</math>
  3 KB (461 words) - 00:33, 16 May 2024
• 1997 AIME Problems/Problem 14
  {{AIME box|year=1997|num-b=13|num-a=15}}
  5 KB (874 words) - 22:30, 1 April 2022
• 1997 AIME Problems/Problem 9
  ...+ 6\cdot 3^5\sqrt{5} + 15\cdot 3^4 \cdot 5 + 20\cdot 3^3 \cdot 5\sqrt{5} + 15 \cdot 3^2 \cdot 25 + 6 \cdot 3 \cdot 25\sqrt{5} + 5^3\right)</math> <math>=
  4 KB (586 words) - 21:53, 30 December 2023
• 1998 AIME Problems/Problem 14
  {{AIME box|year=1998|num-b=13|num-a=15}}
  2 KB (390 words) - 21:05, 29 May 2023
• 1998 AIME Problems/Problem 11
  ...th>\overline{CD},</math> respectively, so that <math>AP = 5, PB = 15, BQ = 15,</math> and <math>CR = 10.</math> What is the area of the polygon that is triple P=(5,0,0),Q=(20,0,15),R=(20,10,20),Pa=(15,20,20),Qa=(0,20,5),Ra=(0,10,0);
  7 KB (1,084 words) - 11:48, 13 August 2023
• 1998 AIME Problems/Problem 9
  real m=60-12*sqrt(15); <math>60 - m = 12\sqrt{15}</math><br />
  4 KB (624 words) - 18:34, 18 February 2018
• 1998 AIME Problems/Problem 2
  <math>\sum_{i=1}^{15} i=\frac{(15)(16)}{2}</math> ordered pairs. For <math>x > 15</math>, <math>y</math> must follow <math>x < y\le 30</math>. Hence, there a
  6 KB (913 words) - 16:34, 6 August 2020
• 1999 AIME Problems/Problem 15
  ...has pairwise parallel planar and oppositely equal length (<math>4\sqrt{13},15,17</math>) edges and can be inscribed in a parallelepiped (rectangular box) <math>q^2+r^2=15^2</math>
  7 KB (1,169 words) - 15:28, 13 May 2024
• 1999 AIME Problems/Problem 14
  ...sides of the triangle have lengths <math>AB=13, BC=14,</math> and <math>CA=15,</math> and the [[tangent]] of angle <math>PAB</math> is <math>m/n,</math> pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14));
  7 KB (1,184 words) - 13:25, 22 December 2022
• 1999 AIME Problems/Problem 11
  &= 0.5(\cos 0 - \cos 10 + \cos 5 - \cos 15 + \cos 10 \ldots + \cos 165 - \cos 175+ \cos 170 - \cos 180)
  4 KB (614 words) - 04:38, 8 December 2023
• 2000 AIME I Problems/Problem 15
  Round 7: <math>b</math> to <math>b</math>, <math>15</math> to right, <math>16</math> left in deck, <math>n = -2 + 8k</math>, be ...ans our sieving process will return to normal after Round 7, with <math>31-15=16</math> cards remaining. Since <math>16</math> is a perfect power of <mat
  15 KB (2,673 words) - 19:16, 6 January 2024
• 2000 AIME I Problems/Problem 14
  ...20*dir(75),p); MP("x",P,17*dir(245),p); MP("2x",Q,15*dir(70),p); MP("2x",A,15*dir(-90),p); MP("2y",P,2*left,p); MP("3x",P,10*dir(-95),p); MP("x+y",C,5*di {{AIME box|year=2000|n=I|num-b=13|num-a=15|t=721509}}
  8 KB (1,275 words) - 03:04, 27 February 2022
• 2000 AIME I Problems/Problem 11
  Add all powers of 2: 15 Or, <math>156*1875+15*1248</math> <math>=311220</math>
  4 KB (667 words) - 13:58, 31 July 2020
• 2000 AIME I Problems/Problem 5
  ...>5|a,b</math>, so there are 2 pairs of <math>a</math> and <math>b: (20,5),(15,10)</math>. ...ack in both. Thus, <math>P(\text{both white}) = \frac{1}{10}\cdot \frac{6}{15} = \frac{1}{25}</math>. <math>m + n = 26</math>.
  7 KB (1,011 words) - 20:09, 4 January 2024
• 2000 AIME I Problems/Problem 2
  ...,Arrows(5)); D((0,-15)--(0,15),linewidth(0.6),Arrows(5)); D((-15,-15)--(15,15),linewidth(0.6),Arrows(5));
  3 KB (434 words) - 22:43, 16 May 2021
• 2001 AIME I Problems/Problem 14
  n&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\\hline Our sub-cases are still the same. However, our equations become <math>2x+3y=15,16,17.</math> Computing yields <math>28+37+49=114</math> sequences.
  13 KB (2,298 words) - 19:46, 9 July 2020
• 2005 AMC 10B
  *[[2005 AMC 10B Problems/Problem 15]]
  1 KB (165 words) - 12:40, 14 August 2020
• 2001 AIME I Problems/Problem 9
  In [[triangle]] <math>ABC</math>, <math>AB=13</math>, <math>BC=15</math> and <math>CA=17</math>. Point <math>D</math> is on <math>\overline{A pair A=(0,0),B=(13,0),C=IP(CR(A,17),CR(B,15)), D=A+p*(B-A), E=B+q*(C-B), F=C+r*(A-C);
  4 KB (673 words) - 20:15, 21 February 2024
• 2001 AIME I Problems/Problem 6
  ...that the sum is equal to <math>\sum_{i = 0}^{5}{i+2\choose2}(6 - i) = 6 + 15 + 24 + 30 + 30 + 21 = 126</math>. The requested probability is <math>\frac{ #Now, for the first case, there are <math>{6\choose4} = 15</math> ways for this. We do not have to consider the order because the comb
  11 KB (1,729 words) - 20:50, 28 November 2023
• 2001 AIME I Problems/Problem 1
  ...ity (again, higher ones give <math>b > 9</math>), giving us a sum of <math>15</math>. ...= 0</math> as we have been doing so far, then the sum is <math>495 + 120 + 15 = \boxed{630}</math>.
  4 KB (687 words) - 18:37, 27 November 2022
• 2002 AIME I Problems/Problem 14
  {{AIME box|year=2002|n=I|num-b=13|num-a=15}}
  2 KB (267 words) - 19:18, 21 June 2021
• 2002 AIME I Problems/Problem 7
  <cmath>(10^{2002} + 1)^{\frac {10}7} = 10^{2860}+\dfrac{10}{7}10^{858}+\dfrac{15}{49}10^{-1144}+\cdots</cmath>
  2 KB (316 words) - 19:54, 4 July 2013
• 2002 AIME I Problems/Problem 6
  ...25}(x_2)=(3+\sqrt{5})+(3-\sqrt{5})=6</math> <math>\Rightarrow x_1x_2=225^6=15^{12}</math>. Thus, <math>\log_{30}\left(x_1y_1x_2y_2\right) = \log_{30}\left(15^{12}\cdot2^{12} \right) = \log_{30}\left(30^{12} \right) = \boxed{012}</mat
  1 KB (194 words) - 19:55, 23 April 2016
• 2002 AIME I Problems/Problem 5
  ...es. So in total, we have overcounted by <math>9+6=15</math>, and <math>198-15=\fbox{183}</math>. ...ing we counted them 6 times. Therefore, our answer is <math>198-3(6-1)=198-15=\boxed{183}.</math>
  1 KB (220 words) - 20:50, 12 November 2022
• 2003 AIME I Problems/Problem 14
  {{AIME box|year=2003|n=I|num-b=13|num-a=15}}
  3 KB (477 words) - 14:23, 4 January 2024
• 2003 AIME I Problems/Problem 8
  ...3ad + 4d^2 = 30a + 30d</math>, which upon rearranging yields <math>2d(2d - 15) = 3a(10 - d)</math>. ...th>a = 18</math>. Alternatively, note that <math>3|2d</math> or <math>3|2d-15</math> implies that <math>3|d</math>, so only <math>9</math> may work. Henc
  5 KB (921 words) - 23:21, 22 January 2023
• 2003 AIME I Problems/Problem 7
  pair O=(0,0),A=(-15,0),B=(-6,0),C=(15,0),D=(0,8); MP("6",B/2); MP("15",C/2); MP("9",(A+B)/2);
  3 KB (490 words) - 18:13, 13 February 2021
• 2003 AIME II Problems/Problem 14
  {{AIME box|year=2003|n=II|num-b=13|num-a=15}}
  9 KB (1,461 words) - 15:09, 18 August 2023
• 2003 AIME II Problems/Problem 11
  ...ame side of line <math>AB</math> as <math>C</math> so that <math>AD = BD = 15.</math> Given that the area of triangle <math>CDM</math> may be expressed a <math>DM=\sqrt{15^2-\left(\frac{25} {2}\right)^2}=\frac{5} {2} \sqrt{11}.</math>
  5 KB (772 words) - 19:47, 1 August 2023
• 2003 AIME II Problems/Problem 8
  <math>21*1848-35*1716+15*1440=1848+20*132+(20*1716-35*1716+15*1716)+15*(-276)</math> <math>=1848+5*132+15(132-276)</math>
  5 KB (793 words) - 15:18, 14 July 2023
• 2003 AIME II Problems/Problem 6
  ...e <math>ABC,</math> <math>AB = 13,</math> <math>BC = 14,</math> <math>AC = 15,</math> and point <math>G</math> is the intersection of the medians. Points ...-14-15</math> triangle is a <math>5-12-13</math> triangle and a <math>9-12-15</math> triangle "glued" together on the <math>12</math> side, <math>[ABC]=\
  5 KB (787 words) - 17:38, 30 July 2022
• 2002 AIME II Problems/Problem 14
  {{AIME box|year=2002|n=II|num-b=13|num-a=15}}
  4 KB (658 words) - 19:15, 19 December 2021
• 2002 AIME II Problems/Problem 13
  [[WLOG]], let <math>W_C=15</math>. <math>W_P=W_C+W_X=15+11=26</math>.
  6 KB (935 words) - 13:23, 3 September 2021
• 2002 AIME II Problems/Problem 12
  ...7}>.4</math>. All the rest work. Therefore there are <math>3\cdot5=\textbf{15}</math> possibilities here. Taking all these cases into account, we find that there are <math>4+15+4=23</math> ways to have <math>a_{10} = .4</math> and <math>a_n\leq .4</mat
  7 KB (1,127 words) - 13:34, 19 June 2022
• 2002 AIME II Problems/Problem 7
  Thus, <math>k \equiv 0, 15 \pmod{16}</math>. <math>k \equiv 15 \pmod{16}</math>
  3 KB (403 words) - 12:10, 9 September 2023
• 2001 AIME II Problems/Problem 14
  ...>, we see that the solutions common to both equations have arguments <math>15^\circ , 105^\circ, 195^\circ, </math> and <math>\ 285^\circ</math>. We can {{AIME box|year=2001|n=II|num-b=13|num-a=15}}
  2 KB (380 words) - 15:03, 22 July 2018
• 2001 AIME II Problems/Problem 13
  ...> use Law of Cosines on <math>\triangle ABD</math> to find <math>AD=2\sqrt{15}</math>
  4 KB (743 words) - 03:32, 23 January 2023
• 2001 AIME II Problems/Problem 10
  ...next 90 numbers (6 each), so our total is <math>4\cdot 16 + 6 \cdot \frac{15 \cdot 16}{2} = \boxed{784}</math>.
  4 KB (549 words) - 23:16, 19 January 2024
• 2001 AIME II Problems/Problem 7
  ..., 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(CR(D(O1), 30)); D(CR(D(O2), 15)); D(CR(D(O3), 10));
  7 KB (1,112 words) - 02:15, 26 December 2022
• 2001 AIME II Problems/Problem 6
  ...r corresponding side lengths, so <math>\frac{[EFGH]}{[ABCD]} = \left(\frac 15\right)^2 = \frac{1}{25}</math>, and the answer is <math>10n + m = \boxed{25
  4 KB (772 words) - 19:31, 6 December 2023
• 2001 AIME II Problems/Problem 4
  D((0,0)--(10/(15/8),10),EndArrow(6)); D((0,0)--(13,13 * 3/10),EndArrow(6));
  2 KB (240 words) - 20:34, 4 July 2013
• 2000 AIME II Problems/Problem 14
  ...}\right)-\left(1+\sum_{k=1}^{32m-1} {k\cdot k!}\right) = \sum_{k=32m}^{32m+15}k\cdot k!.</math> &=16! +\sum_{m=1}^{62}\sum_{k=32m}^{32m+15}k\cdot k!
  7 KB (1,131 words) - 14:49, 6 April 2023
• 2000 AIME II Problems/Problem 12
  ...</math>. It is given that <math>AB=13</math>, <math>BC=14</math>, <math>CA=15</math>, and that the distance from <math>O</math> to <math>\triangle ABC</m ...ternatively, a <math>13-14-15</math> triangle may be split into <math>9-12-15</math> and <math>5-12-13</math> [[right triangle]]s):
  3 KB (532 words) - 13:14, 22 August 2020
• 2000 AIME II Problems/Problem 7
  Given that <center><math>\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10! <cmath>\frac {19!}{2!17!}+\frac {19!}{3!16!}+\frac {19!}{4!15!}+\frac {19!}{5!14!}+\frac {19!}{6!13!}+\frac {19!}{7!12!}+\frac {19!}{8!11
  2 KB (281 words) - 12:09, 5 April 2024
• 2005 AMC 10B Problems/Problem 14
  {{AMC10 box|year=2005|ab=B|num-b=13|num-a=15}}
  5 KB (882 words) - 22:12, 30 April 2024
• 2005 AMC 10B Problems/Problem 24
  ...>. Testing <math>a-b=4</math> gives us <math>2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{7}{2}</math>, which is impossible, as <math>a</math> and <math
  5 KB (845 words) - 19:23, 17 September 2023
• 2006 AMC 12A Problems/Problem 1
  ...cost <math>8\cdot 2=16</math> dollars. In total, the purchase costs <math>15+16=\boxed{\textbf{(A) }31}</math> dollars.
  846 bytes (115 words) - 17:20, 16 December 2021
• 2006 AMC 12A Problems/Problem 3
  <math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\tex
  1 KB (155 words) - 17:30, 16 December 2021
• 2006 AMC 12A Problems/Problem 6
  label("$y$",(15,-4),N); label("$y$",(15,-4),N);
  3 KB (528 words) - 18:29, 7 May 2024
• 2006 AMC 12A Problems/Problem 8
  ...[[set]]s of two or more consecutive positive integers have a sum of <math>15</math>? ...then their average (which could be a fraction) must be a divisor of <math>15</math>. If the number of integers in the list is odd, then the average must
  3 KB (450 words) - 02:00, 13 January 2024
• 2006 AMC 12A Problems/Problem 9
  <math>\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 18\qquad\mathrm{(E)}\ 20</math>
  3 KB (429 words) - 18:14, 26 September 2020
• Inradius
  *[[2007 AIME II Problems/Problem 15]]
  2 KB (336 words) - 00:44, 23 April 2024
• 2007 iTest
  **[[2007 iTest Problems/Problem 15|Problem 15]]
  3 KB (305 words) - 15:10, 5 November 2023
• Minnesota ARML
  ...p high school mathematics students in Minnesota. Each year, four teams of 15 students are selected out of a group of approximately 75-80 trainees. The
  4 KB (680 words) - 16:45, 10 June 2015
• 2006 AMC 10B Problems
  ...three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? == Problem 15 ==
  14 KB (2,059 words) - 01:17, 30 January 2024
• 2005 AMC 10B Problems
  == Problem 15 == [[2005 AMC 10B Problems/Problem 15|Solution]]
  12 KB (1,874 words) - 21:20, 23 December 2020
• 2005 AMC 10A
  * [[2005 AMC 10A Problems/Problem 15]]
  2 KB (182 words) - 18:09, 6 October 2014
• Areteem Camp
  University of Chicago August 2 thru August 15.
  2 KB (370 words) - 19:48, 6 January 2015
• New York State Math League
  The format of NYSML is identical to that of [[ARML]], with 15-member teams competing in Individual, Team, Relay, and Power rounds, althou
  1 KB (201 words) - 14:39, 25 June 2023
• Sigma notation
  .... This sum is written in summation notation as <math>\sum_{k=1}^5 5k=5+10+15+20+25</math>. In this case, 1 is the lower limit of summation, the number
  2 KB (335 words) - 17:17, 8 February 2024
• University of South Carolina High School Math Contest/1993 Exam/Problem 29
  \qquad \mathrm{(B) \ } 8/\sqrt{15} ...frac{3}{4}\sqrt{15})} = \frac{6\cdot\frac{4}{3}}{\sqrt{15}} = \frac8{\sqrt{15}} \Longrightarrow \mathrm{(B)}</math>.
  2 KB (219 words) - 09:57, 31 August 2012
• University of South Carolina High School Math Contest/1993 Exam/Problems
  == Problem 15 == [[University of South Carolina High School Math Contest/1993 Exam/Problem 15|Solution]]
  14 KB (2,102 words) - 22:03, 26 October 2018
• Mock AIME 2 Pre 2005
  * [[Mock AIME 2 Pre 2005 Problems/Problem 15|Problem 15]]
  2 KB (181 words) - 10:58, 18 March 2015
• Mock AIME 7 Pre 2005
  * [[Mock AIME 7 Pre 2005 Problems/Problem 15|Problem 15]]
  1 KB (146 words) - 16:33, 14 October 2022
• Mock AIME 1 2005-2006
  * [[Mock AIME 1 2005-2006/Problem 15|Problem 15]]
  1 KB (135 words) - 17:41, 21 January 2017
• Mock AIME 1 2006-2007
  ** [[Mock AIME 1 2006-2007 Problems/Problem 15|Problem 15]]
  1 KB (155 words) - 16:06, 3 April 2012
• Mock AIME 2 2006-2007
  * [[Mock AIME 2 2006-2007 Problems/Problem 15|Problem 15]]
  1 KB (145 words) - 10:55, 4 April 2012
• Mock AIME 1 2006-2007 Problems/Problem 1
  ...or do a bit of casework to find that the only solution is <math>x = 8, y = 15</math>. In the second case, we have <math>17^2 = y^2 - x^2 = (y - x)(y + x ...se, this value may be either <math>\frac {17}{8}</math> or <math>\frac{17}{15}</math>. In the second, it may be either <math>\frac{145}{144}</math> or <
  2 KB (329 words) - 15:53, 3 April 2012
• Mock AIME 1 2006-2007 Problems/Problem 7
  ...[CFD]}=\frac{\frac{5x}{9}}{\frac{2x}{27}}=\frac{5\cdot 27}{9\cdot 2}=\frac{15}{2}</cmath> ...[CFD]}=\frac{[AEF]}{[CFE]}\cdot\frac{[CFE]}{[CFD]}=5\cdot\frac{3}{2}=\frac{15}{2}=\frac{m}{n}\implies m+n=\boxed{17}.</cmath>
  3 KB (518 words) - 16:54, 25 November 2015
• Mock AIME 1 2006-2007 Problems/Problem 14
  *[[Mock AIME 1 2006-2007 Problems/Problem 15 | Next Problem]]
  3 KB (541 words) - 17:32, 22 November 2023
• Mock AIME 1 2006-2007/Problems
  ==Problem 15== [[Mock AIME 1 2006-2007 Problems/Problem 15|Solution]]
  8 KB (1,355 words) - 14:54, 21 August 2020
• Mock AIME 2 2006-2007 Problems/Problem 6
  If <math>\tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>0^\circ \le ...irc\cos 35^\circ}=\frac{\sin 15^{\circ}(\cos 10^\circ-\cos 60^\circ)}{\cos 15^\circ(\cos 10^\circ+\cos 60^\circ)}</cmath>
  1 KB (157 words) - 10:51, 4 April 2012
• Mock AIME 2 2006-2007 Problems/Problem 14
  {{Mock AIME box|year=2006-2007|n=2|num-b=13|num-a=15}}
  2 KB (284 words) - 10:53, 4 April 2012
• Mock AIME 2 2006-2007 Problems
  If <math>\tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>0^\circ \le == Problem 15 ==
  5 KB (848 words) - 23:49, 25 February 2017
• Base numbers/Conversion
  ...t. The standard notation is to use the letters a=10,b=11,c=12,d=13,e=14,f=15.
  7 KB (1,177 words) - 15:56, 18 April 2020
• 1959 IMO Problems/Problem 1
  ...rac{46}{31} </math>. Again, both results are 1 off from being multiples of 15. ...ime number is always being produced, and even with larger values, like say 15, implementing it in gives us <math> \frac{319}{199} </math>, we keep ending
  5 KB (767 words) - 10:59, 23 July 2023
• 1971 Canadian MO Problems/Problem 1
  <math>(EP)(CE)=(BE)(ED)</math> and <math>2r-1=15.</math> Hence, <math>r=8.</math>
  680 bytes (114 words) - 21:38, 9 July 2019
• 1959 IMO Problems/Problem 4
  ...riangle), and <math>75^\circ</math> angle is constructed by constructing a 15 degree angle on top of the 60 degree point.
  6 KB (939 words) - 17:31, 15 July 2023
• 1995 AHSME Problems/Problem 8
  ...{16}{3} } \qquad \mathrm{(C) \ \frac {20}{3} } \qquad \mathrm{(D) \ \frac {15}{2} } \qquad \mathrm{(E) \ 8 } </math> ...(0,6), B = (8,0), C= (0,0), D = (8/3,4), E = (8/3,0), F = (0, 3), G = (38/15,1.6);
  1 KB (199 words) - 13:58, 5 July 2013
• Congruent (geometry)
  X=(10,0); Y=(13,2); Z=(15,0); X=(10,0); Y=(13,2); Z=(15,0);
  10 KB (1,655 words) - 21:43, 24 March 2022
• 2005 AMC 10A Problems/Problem 5
  ...would get <math>\left\lfloor\frac{8+7}{5}\right\rfloor = \left\lfloor\frac{15}{5}\right\rfloor=3</math> free windows.
  1 KB (182 words) - 03:42, 29 April 2023
• 2005 AMC 10A Problems/Problem 14
  {{AMC10 box|year=2005|ab=A|num-b=13|num-a=15}}
  4 KB (694 words) - 19:25, 13 December 2021
• 2005 AMC 10A Problems/Problem 16
  {{AMC10 box|year=2005|ab=A|num-b=15|num-a=17}}
  2 KB (279 words) - 11:57, 17 July 2023
• Perfect cube
  * [[2005_AMC_10A_Problems/Problem_15 | 2005 AMC 10A Problem 15]]
  704 bytes (91 words) - 14:12, 24 August 2023
• System of equations
  ...Thus <center><math> 4(a+b+c+d+e) = 60 \Leftrightarrow a + b + c + d + e = 15. </math></center> ...uation from the second equation leaves <math>d</math> on the LHS and <math>15-8=7</math> on the RHS. And thus we continue on in this way to find that <m
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• MathPath
  | 12:15 - 1:15 || Lunch | 3:15 - 4:30 || Breakout Session
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• 1970 Canadian MO Problems
  draw((16/5,12/5)--(16/5-.2,12/5-.15)--(16/5-.2+.15,12/5-.15-.2)--(16/5+.15,12/5-.2)--cycle,black);
  4 KB (604 words) - 04:32, 8 October 2014
• Reduced fraction
  ...15 and 27 are divisible by 3. So in order to reduce, we write <math>\frac{15}{27} = \frac{3 \cdot 5}{3\cdot 9} = \frac5 9</math>, and 5 and 9 are relati
  1 KB (225 words) - 13:22, 7 March 2021
• Mock MathCounts
  * <b>Add variety.</b> Nobody wants to see a sprint round with 15 counting problems or 20 algebra problems. Generally, algebra and geometry s
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• 2007 iTest Problems
  ===Problem 15=== [[2007 iTest Problems/Problem 15|Solution]]
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• 2005 Alabama ARML TST
  ...to an easy [[AIME]], Alabaman scores average about 5 right, while the top 15 people usually answer 8 or more correctly. The first link contains the full * [[2005 Alabama ARML TST Problems/Problem 15]]
  1 KB (170 words) - 01:01, 19 June 2018
• 2006 iTest
  *[[2006 iTest Problems/Problem 15|Problem 15]]
  3 KB (320 words) - 09:56, 23 April 2024
• Areteem Institute Summer Math Programs
  University of Chicago August 2 thru August 15.
  2 KB (373 words) - 19:58, 6 January 2015
• 2005 AMC 10A Problems
  == Problem 15 == [[2005 AMC 10A Problems/Problem 15|Solution]]
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• Division
  For instance: <math>15 \div 2.5 = 150 \div 25 = 6.</math>
  2 KB (259 words) - 09:52, 23 January 2020
• 2004 AMC 12A Problems/Problem 7
  ...layers <math>A</math>, <math>B</math>, and <math>C</math> start with <math>15</math>, <math>14</math>, and <math>13</math> tokens, respectively. How man First round: <math>15,14,13</math> (given)
  4 KB (588 words) - 16:51, 24 March 2023
• 2003 AMC 10A
  * [[2003 AMC 10A Problems/Problem 15]]
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• 2003 AMC 10A Problems
  A solid box is <math>15</math> cm by <math>10</math> cm by <math>8</math> cm. A new solid is formed <math> \mathrm{(A) \ } 12\qquad \mathrm{(B) \ } 15\qquad \mathrm{(C) \ } 18\qquad \mathrm{(D) \ } 21\qquad \mathrm{(E) \ } 24
  13 KB (1,900 words) - 22:27, 6 January 2021
• 1999 AMC 8 Problems/Problem 16
  {{AMC8 box|year=1999|num-b=15|num-a=17}}
  1 KB (167 words) - 20:30, 11 January 2024
• 2003 AMC 10A Problems/Problem 14
  <math> \mathrm{(A) \ } 12\qquad \mathrm{(B) \ } 15\qquad \mathrm{(C) \ } 18\qquad \mathrm{(D) \ } 21\qquad \mathrm{(E) \ } 24 {{AMC10 box|year=2003|ab=A|num-b=13|num-a=15}}
  2 KB (336 words) - 15:49, 19 August 2023
• 2003 AMC 10A Problems/Problem 16
  {{AMC10 box|year=2003|ab=A|num-b=15|num-a=17}}
  1 KB (218 words) - 15:52, 19 August 2023
• 2003 AMC 10A Problems/Problem 19
  #REDIRECT[[2003 AMC 12A Problems/Problem 15]]
  45 bytes (5 words) - 18:31, 31 July 2011
• 1997 AJHSME Problems/Problem 24
  ...{1}{2}\pi\cdot3^2 - \frac{1}{2}\pi\cdot2^2 =\frac{\pi}{2}(5^2 + 3^2 - 2^2)=15\pi</math> ...\pi - 15\pi = 10\pi</math>.Therefore the ratio of shaded:unshaded is <math>15\pi : 10\pi =\boxed{ \text{(C)}\ 3:2}</math>.
  2 KB (394 words) - 17:05, 20 October 2023
• 2004 AMC 10A
  * [[2004 AMC 10A Problems/Problem 15]]
  2 KB (182 words) - 01:29, 7 October 2014
• 2003 AMC 10A Problems/Problem 21
  <math>3*5=15</math> cases for case 2 10+15+3= <math>28</math> total
  7 KB (994 words) - 17:51, 11 April 2024
• 2003 AMC 10A Problems/Problem 22
  ...lugging this into our equation for line <math>GA</math> gives us <math>G=(-15,20)</math>, so <math>GF= \boxed{\mathrm{(B)}\ 20}</math>
  9 KB (1,446 words) - 22:48, 8 May 2024
• 2004 AMC 10A Problems
  ...ns. Players <math>A</math>, <math>B</math>, and <math>C</math> start with 15, 14, and 13 tokens, respectively. How many rounds will there be in the gam == Problem 15 ==
  15 KB (2,092 words) - 20:32, 15 April 2024
• 2004 AMC 12A Problems/Problem 15
  {{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #15]] and [[2004 AMC 10A Problems/Problem 17|2004 AMC 10A #17]]}}
  2 KB (309 words) - 22:27, 15 August 2023
• 2001 AMC 8
  ** [[2001 AMC 8 Problems/Problem 15]]
  1 KB (138 words) - 10:26, 22 August 2013
• 2001 AMC 8 Problems
  <math>\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24</math> <math>\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72</math>
  13 KB (1,994 words) - 13:04, 18 February 2024
• 1985 AIME Problems
  ==Problem 15== [[Image:AIME 1985 Problem 15.png]]
  7 KB (1,071 words) - 19:24, 23 February 2024
• Pythagorean triple
  <math>(8, 15, 17)</math> <nowiki>*</nowiki> <math>(9, 12, 15)</math>
  4 KB (684 words) - 16:45, 1 August 2020
• 2003 AMC 12A Problems/Problem 3
  ...g tag, for PoTW on the Wiki front page--><onlyinclude>A solid box is <math>15</math> cm by <math>10</math> cm by <math>8</math> cm. A new solid is formed The volume of the original box is <math>15\cdot10\cdot8=1200.</math>
  1 KB (183 words) - 15:36, 19 August 2023
• 2003 AMC 12A Problems/Problem 8
  This is not too bad with casework. Notice that <math>1*60=2*30=3*20=4*15=5*12=6*10=60</math>. Hence, <math>60</math> has <math>12</math> factors, of
  2 KB (326 words) - 15:40, 19 August 2023
• 2003 AMC 12A Problems/Problem 15
  {{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #15]] and [[2003 AMC 10A Problems|2003 AMC 10A #19]]}}
  3 KB (380 words) - 21:53, 19 March 2022
• 1987 IMO Problems/Problem 4
  ...|<font color="#FF 69 B4">Maha</font><font color="#FF00FF">maya</font>]] 21:15, 21 May 2012 (EDT)
  5 KB (923 words) - 19:51, 21 January 2024
• Minkowski Inequality
  ...ofproblemsolving.com/Forum/viewtopic.php?p=432791#432791 AIME 1991 Problem 15]
  1 KB (240 words) - 16:49, 29 December 2021
• Georgia Tech High School Math Competition
  ...n as Georgia Tech). The 2014 competition will occur on Saturday, February 15, 2014. ...ation is online at [http://hsmc.gatech.edu]. The cost is <nowiki>$</nowiki>15 per contestant, which includes lunch and a T-shirt for the contestant. One
  3 KB (475 words) - 21:51, 31 December 2013
• Fundamental Theorem of Arithmetic
  ...e, most integers have many factorizations into 2 parts: <math>30 = 2 \cdot 15 = 3 \cdot 10 = 5 \cdot 6</math>. Thus, the Fundamental Theorem of Arithmet
  2 KB (376 words) - 23:28, 4 August 2022
• 2020 AMC 10A Problems
  ...7, a,</math> and <math>b</math> have an average (arithmetic mean) of <math>15</math>. What is the average of <math>a</math> and <math>b</math>? <math>\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60</math>
  13 KB (1,968 words) - 18:32, 29 February 2024
• Asymptote: Useful commands and their Output
  path p=(0,0)..(20,15)..(40,-5)..(50,0);
  6 KB (871 words) - 21:14, 12 June 2023
• Pigeonhole Principle/Solutions
  ...al [[Modular arithmetic/Introduction |modulo]] 7 residues. To avoid having 15 with the same residue, 14 numbers with different modulo 7 residues can be p ...2}\approx 1.414</math>, we get <math>\frac{\sqrt{29-14.14}}{2}=\frac{\sqrt{15.86}}{2}<\frac{4}{2}=2</math>.
  10 KB (1,617 words) - 01:34, 26 October 2021
• 2021 Fall AMC 10B
  **[[2021 Fall AMC 10B Problems/Problem 15|Problem 15]]
  2 KB (205 words) - 10:53, 1 December 2021
• 2007 AIME II Problems/Problem 9
  ...h>, then if <math>BY = b</math> we have a system of equations. <math>a+b = 15, b+41-a = 52</math>. We can then solve for <math>a</math>, and since <math>
  5 KB (818 words) - 11:05, 7 June 2022
• 2007 AIME II Problems/Problem 10
  **Probability: <math>{6\choose4}/64 = \frac{15}{64}</math> <math>=\frac{(1)(64)+(6)(64)+(15)(32)+(20)(16)+(15)(8)+(6)(4)+(1)(2)}{(64)(64)}</math>
  8 KB (1,367 words) - 11:48, 23 October 2022
• 2005 AMC 10A Problems/Problem 23
  O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.14213562373095 draw(Circle((0,0),15));
  6 KB (1,045 words) - 09:46, 4 April 2023
• Mock AIME 1 Pre 2005 Problems
  Let <math>N</math> denote the number of permutations of the <math>15</math>-character string <math>AAAABBBBBCCCCCC</math> such that == Problem 15 ==
  6 KB (1,100 words) - 22:35, 9 January 2016
• Mock AIME 3 Pre 2005 Problems
  ...<math>D</math> such that <math>\angle{BAE}</math> is right. If <math>BD = 15, DE = 2,</math> and <math>BC = 16</math>, then <math>CD</math> can be expre ==Problem 15==
  7 KB (1,135 words) - 23:53, 24 March 2019
• Mock AIME 3 Pre 2005 Problems/Problem 2
  ...alent to <math>8</math> dividers, and there are <math>{8 + 7 \choose 7} = {15 \choose 7} = 6435 \equiv \boxed{435} \pmod{1000}</math>.
  950 bytes (137 words) - 10:16, 29 November 2019
• Mock AIME 3 Pre 2005 Problems/Problem 8
  which equals either <math>-1,1,15</math>. ...math> then note that <math>\pm 1 \pm 2 \pm 3 \pm 4 \pm 5</math> is at most 15 so we must have <math>\pm 1 \pm 2 \pm 3 \pm 4 \pm 5=9</math>, which forces
  3 KB (520 words) - 12:55, 11 January 2019
• Mock AIME 3 Pre 2005 Problems/Problem 9
  ...<math>D</math> such that <math>\angle{BAE}</math> is right. If <math>BD = 15, DE = 2,</math> and <math>BC = 16</math>, then <math>CD</math> can be expre ...5</math> and <math>DP=6.5.</math> Then the mass of <math>P</math> is <math>15</math> and the mass of <math>D</math> is <math>8.5</math> and the mass of <
  2 KB (278 words) - 16:32, 27 December 2019
• Mock AIME 3 Pre 2005 Problems/Problem 11
  ...the second equation and rearranging variables gives <math>45x=8\cdot 15+bx=15^2\implies x=5</math>. Back-substitution yields <math>AC=21</math> and conse
  2 KB (325 words) - 15:32, 22 March 2015
• Mock AIME 3 Pre 2005 Problems/Problem 12
  ...ence then repeats itself. We hence find that there are a total of <math>11*15 - 1</math> or <math>\boxed{164}</math> numbers that satisfy the inequality.
  714 bytes (105 words) - 23:59, 24 April 2013
• Mock AIME 3 Pre 2005 Problems/Problem 14
  {{Mock AIME box|year=Pre 2005|n=3|num-b=13|num-a=15}}
  3 KB (563 words) - 02:05, 25 November 2023
• Mock AIME 3 Pre 2005 Answer Key
  15. 041
  124 bytes (0 words) - 15:02, 3 April 2012
• 2007 AIME I Problems
  ...ath>, and ending with <math>39</math>. For example, <math>0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39</math> is a move sequence. How many move sequences are possible f == Problem 15 ==
  7 KB (1,218 words) - 15:28, 11 July 2022
• 2007 AIME I Problems/Problem 3
  ...43bi - b^3i</math>, so <math>b(b^2 - 225) = 0</math> and <math>b = -15, 0, 15</math>. Since <math>b > 0</math>, the solution is <math>\boxed{015}</math>.
  1,003 bytes (163 words) - 15:34, 18 February 2017
• 2007 AIME I Problems/Problem 8
  The first two equations show that <math>m - n = 29 - \frac{43}{2} = \frac{15}{2}</math>. The last two equations show that <math>\frac{m}{n} = -2</math>.
  4 KB (728 words) - 00:11, 29 November 2023
• 2007 AIME I Problems/Problem 10
  ...e arranged in <math>{2\choose1} = 2</math> positions; this totals to <math>15 \cdot 6 \cdot 2 = 180</math>. Now, the third and fourth columns have a fixe
  13 KB (2,328 words) - 00:12, 29 November 2023
• 2007 AIME I Problems/Problem 6
  ..., beginning with 0 and ending with 39. For example, <math>0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39</math> is a move sequence. How many move sequences are possible f **There are <math>1 \cdot \left(\frac{24-15}{3} + 1\right) = 4</math> ways to reach <math>26</math>.
  10 KB (1,519 words) - 00:11, 29 November 2023
• 2007 AIME I Problems/Problem 9
  ...7}</math>. Thus, <math>\frac rx = \sqrt{\frac{1 + \frac{15}{17}}{1 - \frac{15}{17}}}</math>, and <math>x = \frac{r}{4}</math>. :<math>r\left(\frac{2}{\frac{17}{17} + \frac{8}{17} + \frac{15}{17}} + 2\right) = 34</math>
  11 KB (1,851 words) - 12:31, 21 December 2021
• 2007 AIME I Problems/Problem 12
  var theta=15; ...MA("45^\circ",A,extension(A,C,Bp,Cp),Bp,3); MA("15^\circ",C,Y,Cp,8); MA("15^\circ",C,A,Cp,9);
  10 KB (1,458 words) - 20:50, 3 November 2023
• 2007 AIME I
  * [[2007 AIME I Problems/Problem 15]]
  1 KB (135 words) - 12:32, 22 March 2011
• 2007 AIME I Problems/Problem 14
  {{AIME box|year=2007|n=I|num-b=13|num-a=15}}
  13 KB (2,185 words) - 23:30, 5 December 2022
• 2007 AIME II
  * [[2007 AIME II Problems/Problem 15]]
  1 KB (135 words) - 12:34, 22 March 2011
• 2007 AIME II Problems/Problem 15
  ...triangle <math>ABC</math> are <math>13,</math> <math>14,</math> and <math>15,</math> the radius of <math>\omega</math> can be represented in the form <m B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C);
  11 KB (2,099 words) - 17:51, 4 January 2024
• 2007 AIME II Problems/Problem 14
  {{AIME box|year=2007|n=II|num-b=13|num-a=15}}
  7 KB (1,335 words) - 17:44, 25 January 2022
• 2007 AIME II Problems
  == Problem 15 == ...triangle <math>ABC</math> are <math>13,</math> <math>14,</math> and <math>15,</math> the radius of <math>\omega</math> can be represented in the form <m
  9 KB (1,435 words) - 01:45, 6 December 2021
• 2007 AMC 12A
  * [[2007 AMC 12A Problems/Problem 15]]
  1 KB (168 words) - 21:50, 6 October 2014
• 2007 AMC 12A Problems
  ...\ 2 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 10 \qquad \textbf{(D)} \ 15 \qquad \textbf{(E)} \ 20</math> ==Problem 15==
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• Pascal Triangle Related Problems
  6: 1 6 15 20 15 6 1
  4 KB (513 words) - 20:18, 3 January 2023
• 2007 AMC 12B Problems/Problem 5
  <math>\mathrm {(A)}\ 13 \qquad \mathrm {(B)}\ 14 \qquad \mathrm {(C)}\ 15 \qquad \mathrm {(D)}\ 16 \qquad \mathrm {(E)} 17</math>
  864 bytes (121 words) - 10:50, 4 July 2013
• 1951 AHSME
  ** [[1951 AHSME Problems/Problem 15|Problem 15]]
  3 KB (258 words) - 14:25, 20 February 2020
• LaTeX:Examples
  ==Solution for USAMTS Problem 4/4/15== ...Abel. [http://aops-cdn.artofproblemsolving.com/LaTeX/Examples/usamts%204-4-15.tex Click here for the source file]. This file gives an example of writing
  5 KB (687 words) - 02:03, 4 February 2020
• Chicken McNugget Theorem
  \color{magenta}\cancel{11} & \color{blue}\cancel{12} & 13 & 14 & \cancel{15} & \color{blue}\cancel{16} & 17 & 18 & \cancel{19} & \color{blue}\cancel{20 ...66</math> (Source: [[2015 AMC 10B Problems/Problem 15|AMC 10B 2015 Problem 15]])
  17 KB (2,748 words) - 19:22, 24 February 2024
• Triangular number
  <math>1, 3, 6, 10, 15, 21</math>.
  2 KB (275 words) - 08:39, 7 July 2021
• 2007 AMC 12A Problems/Problem 1
  ...m{(A)}\ 2\qquad \mathrm{(B)}\ 5\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 15\qquad \mathrm{(E)}\ 20</math>
  911 bytes (127 words) - 22:13, 22 March 2021
• 2007 AMC 12A Problems/Problem 10
  {{AMC10 box|year=2007|ab=A|num-b=13|num-a=15}}
  2 KB (231 words) - 14:02, 3 June 2021
• 2007 AMC 12A Problems/Problem 12
  {{AMC10 box|year=2007|ab=A|num-b=15|num-a=17}}
  3 KB (445 words) - 08:59, 24 March 2023
• 2007 AMC 12A Problems/Problem 14
  {{AMC12 box|year=2007|ab=A|num-b=13|num-a=15}}
  2 KB (278 words) - 02:10, 16 February 2021
• 2007 AMC 12A Problems/Problem 22
  S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\ S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\
  15 KB (2,558 words) - 19:33, 4 February 2024
• 2007 AMC 12A Problems/Problem 16
  {{AMC12 box|year=2007|ab=A|num-b=15|num-a=17}}
  2 KB (336 words) - 05:01, 4 November 2022
• 2007 AMC 12A Problems/Problem 25
  ...</math> sets in this category. Adding them up, there are <math>3 + 6 + 6 = 15</math> spacy subsets with <math>4</math> elements. Adding these all up, we have a total of <math>1 + 12 + 45 + 56 + 15 = \boxed{\mathrm{(E)}\ 129}</math> spacy subsets. ~[[User:emerald_block|eme
  9 KB (1,461 words) - 23:07, 27 January 2024
• 2005 AMC 12A Problems/Problem 4
  ...s. Together, they receive a discount of <math>100 \cdot \left\lfloor \frac{15}{5} \right\rfloor = 300</math>, so they save <math>300-200=100\ \mathrm{(A)
  1 KB (158 words) - 13:05, 7 August 2019
• 2005 AMC 12A Problems/Problem 14
  {{AMC12 box|year=2005|num-b=13|num-a=15|ab=A}}
  2 KB (245 words) - 20:07, 4 March 2024
• 2005 AMC 12A Problems/Problem 17
  draw((15,0)--(10,10),Arrow); draw((15.5,0)--(30,10),Arrow);
  2 KB (215 words) - 13:56, 19 January 2021
• 2009 AIME II
  ** [[2009 AIME II Problems/Problem 15|Problem 15]]
  1 KB (156 words) - 17:59, 20 June 2020
• 2006 AIME II Problems/Problem 3
  This is the same as the number of multiples of 3 in the sequence 3, 9, 15, 21, ..., 192, 195. There are clearly 33 terms in this sequence.
  4 KB (562 words) - 18:37, 30 October 2020
• 2006 AIME II Problems/Problem 6
  ...ees, so <math>\angle BAE = \frac{90 - 60}{2} = 15</math>. Thus, <math>\cos 15 = \cos (45 - 30) = \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{1}{AE}</math>, so ...note that <cmath>AB = 1 = \overline{AA'} + \overline{A'B} = \frac{x}{\tan(15)} + x</cmath>
  7 KB (1,067 words) - 12:23, 8 April 2024
• 2006 AIME II Problems/Problem 14
  {{AIME box|year=2006|n=II|num-b=13|num-a=15}}
  2 KB (263 words) - 23:32, 28 February 2021
• 2006 AIME II Problems/Problem 5
  ...ll these probabilities from <math>\frac{47}{288}</math> leaves <math>\frac{15}{288}=\frac{5}{96}</math> chance of getting a 1 on die <math>A</math> and a <cmath>32+16(1-6x)(6x+1)=47 \Longrightarrow 16(1-36x^2)=15</cmath>
  5 KB (712 words) - 12:10, 5 November 2023
• 2006 AIME II Problems/Problem 9
  ...le O_1r_1r</math> by the [[Pythagorean theorem]] yields <math>r_1r = \sqrt{15}</math>. On <math>\mathcal{C}_3</math>, we can do the same thing to get <ma ...rmined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</math>.
  3 KB (553 words) - 10:45, 26 August 2015
• 2006 AIME II Problems/Problem 13
  The total number of integers <math>N</math> is <math>5 + 10 = \boxed{15}</math>.
  4 KB (675 words) - 10:40, 14 July 2022
• 2006 AIME II Problems/Problem 15
  Thus, <math>A = \frac{1}{15\sqrt{7}}</math> and <math>x + y + z = 30A = \frac2{\sqrt{7}}</math> and the
  4 KB (725 words) - 17:18, 27 June 2021
• 2006 AIME II Problems
  == Problem 15 == [[2006 AIME II Problems/Problem 15|Solution]]
  8 KB (1,350 words) - 12:00, 4 December 2022
• 2004 AMC 12B Problems/Problem 1
  ...{B}) 6 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 12 \qquad (\mathrm {E}) 15</math>
  1 KB (170 words) - 19:21, 22 October 2022
• Primary Mathematics World Contest
  *Individual Round: The individual round consists of 15 problems, numbered I1-I15.
  1 KB (183 words) - 13:57, 15 October 2018
• 2005 PMWC Problems
  9+10+11+12 &=& 13+14+15\end{eqnarray*}</cmath> (e.g. the last number of the third row is <math>15</math>).
  9 KB (1,449 words) - 20:49, 2 October 2020
• 1998 PMWC Problems
  ...math>\frac{1*2*3+2*4*6+3*6*9+4*8*12+5*10*15}{1*3*5+2*6*10+3*9*15+4*12*20+5*15*25}</math> ...ing together nine squares with sides equal to <math>1, 4, 7, 8, 9, 10, 14, 15,</math> and <math>18</math>. What is the sum of the areas of the squares o
  11 KB (1,738 words) - 19:25, 10 March 2015
• 1999 PMWC Problems
  ...},{"1","2","3","4","5","6","7"},{"13","12","11","10","9","8",""},{"","14","15","16","17","18","19"}, {"","..","..","..","..","20",""},{"","..","..","..",
  6 KB (703 words) - 21:21, 21 April 2014
• 2000 PMWC Problems
  ...math> cubes. A number of the smaller cubes are removed by punching out the 15 designated columns from front to back, top to bottom, and side to side. Fin
  11 KB (1,713 words) - 22:47, 13 July 2023
• 2002 PMWC Problems
  defaultpen(fontsize(15));
  5 KB (725 words) - 16:07, 23 April 2014
• 2003 PMWC Problems
  draw((15,3.46)--(14,1.73)--(13,3.46)--cycle); draw((14,1.73)--(16,1.73)--(15,3.46)--cycle);
  7 KB (918 words) - 16:15, 22 April 2014
• 2005 PMWC Problems/Problem T6
  9+10+11+12 &=& 13+14+15\end{eqnarray*}</cmath> ...the <math>80</math>th row (e.g. the last number of the third row is <math>15</math>).
  882 bytes (140 words) - 19:07, 10 March 2015
• 1997 PMWC Problems/Problem I1
  Evaluate <math>29 \frac{27}{28} \times 27\frac{14}{15}</math>. <math>\frac{29 \cdot 28 + 27}{28} \cdot \frac{27 \cdot 15 + 14}{15} = \frac{839 \cdot 419}{420} = \frac{839(420 - 1)}{420} = 839 - \frac{839}{
  558 bytes (62 words) - 10:47, 22 May 2014
• 1997 PMWC Problems/Problem I6
  <math>6s-6l=-15</math> <math>s=\dfrac{15}{2}</math>
  660 bytes (105 words) - 14:23, 20 April 2014
• 1997 PMWC Problems/Problem I7
  ...g = 12 \Longrightarrow g = 15</math>. So there are 12 boys, and <math>12 + 15 = 27</math> students in the class.
  750 bytes (111 words) - 14:23, 20 April 2014
• 1997 PMWC Problems/Problem I9
  <cmath>x = \frac{15}{2}</cmath> There are <math>1000 \cdot \frac{15}{2} = 7500</math> millilitres of the acid.
  947 bytes (136 words) - 14:26, 20 April 2014
• 1997 PMWC Problems/Problem I11
  <math>6750= 5lw = \dfrac{15}{2}w^2</math>
  944 bytes (154 words) - 12:44, 13 August 2014
• 1997 PMWC Problems/Problem T2
  &+&13\left(\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)+15\left(\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\\
  1 KB (139 words) - 19:07, 10 March 2015
• 2004 AMC 10A Problems/Problem 20
  E = (0,Tan(15)); F = (1 - Tan(15),1);
  4 KB (710 words) - 02:47, 18 April 2024
• 2007 AMC 12B
  * [[2007 AMC 12B Problems/Problem 15]]
  1 KB (168 words) - 21:22, 6 October 2014
• 2007 AMC 12B Problems
  <math>\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17</math> ==Problem 15==
  12 KB (1,814 words) - 12:58, 19 February 2020
• 2002 AMC 12A Problems/Problem 2
  <math> \textbf{(A) } 15\qquad \textbf{(B) } 34\qquad \textbf{(C) } 43\qquad \textbf{(D) } 51\qquad ...138-3}{9}=\frac{135}{9}=15</math>. Our answer is <math>\boxed{\textbf{(A) }15}</math>.
  1 KB (163 words) - 12:46, 8 November 2021
• 1999 AHSME Problems
  == Problem 15 == [[1999 AHSME Problems/Problem 15|Solution]]
  13 KB (1,945 words) - 18:28, 19 June 2023
• Lexington Math Tournament
  The Theme Round consists of a total of 15 problems, which are divided between 3 themes with 5 problems each. The prob ...his round, 4-6 person teams work together to solve a test consisting of 10-15 problems in 60 minutes.
  3 KB (484 words) - 20:04, 12 March 2024
• 2016 AMC 10A Problems/Problem 21
  #redirect [[2016 AMC 12A Problems/Problem 15]]
  46 bytes (5 words) - 15:33, 26 January 2021
• Ceva's theorem/Problems
  ...AC</math>, and <math>BC</math> have lengths <math>13, 14</math>, and <math>15</math>, respectively. If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <mat ...> and <math>DC = y</math>, then <math>10x = 40y</math>, and <math>{x + y = 15}</math>. From this, we find <math>x = 12</math> and <math>y = 3</math>.
  530 bytes (82 words) - 16:16, 9 May 2021
• Summation
  **<math>\sum_{i=5}^{15} i + 1</math>
  3 KB (482 words) - 16:39, 8 October 2023
• Concurrence/Problems
  ...apply a bit of [[number theory]] to find that the answer is a <math>13-12-15</math> triangle:
  3 KB (458 words) - 15:44, 1 December 2015
• Median
  Find the median of <math>\{3, 4, 5, 15, 9\}</math>.
  2 KB (370 words) - 15:30, 12 November 2023
• 2004 AMC 12A Problems/Problem 11
  {{AMC10 box|year=2004|ab=A|num-b=13|num-a=15}}
  2 KB (310 words) - 17:39, 31 December 2022
• 2004 AMC 12A Problems/Problem 14
  {{AMC12 box|year=2004|ab=A|num-b=13|num-a=15}}
  4 KB (689 words) - 03:35, 16 January 2023
• 2004 AMC 12A Problems/Problem 16
  {{AMC12 box|year=2004|ab=A|num-b=15|num-a=17}}
  2 KB (228 words) - 02:01, 23 January 2023
• 2005 iTest
  *[[2005 iTest Problems/Problem 15|Problem 15]]
  3 KB (283 words) - 02:37, 24 January 2024
• 2008 AMC 12B Problems/Problem 23
  ...\text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15</math>
  5 KB (814 words) - 18:02, 17 January 2023
• 2000 AMC 12 Problems/Problem 11
  ...AMC 12 Problems|2000 AMC 12 #11]] and [[2000 AMC 10 Problems|2000 AMC 10 #15]]}}
  2 KB (405 words) - 21:48, 21 April 2024
• 2000 AMC 12 Problems/Problem 14
  {{AMC12 box|year=2000|num-b=13|num-a=15}}
  2 KB (272 words) - 18:18, 22 April 2023
• 2000 AMC 12 Problems/Problem 15
  {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #15]] and [[2000 AMC 10 Problems|2000 AMC 10 #24]]}}
  3 KB (441 words) - 21:11, 29 April 2023
• 2000 AMC 12 Problems/Problem 16
  ...top to bottom, is <math>1,2,\ldots,13,</math>, the second column <math>14,15,\ldots,26</math> and so on across the board, some squares have the same num {{AMC12 box|year=2000|num-b=15|num-a=17}}
  2 KB (310 words) - 11:28, 3 August 2021
• 2000 AMC 12 Problems/Problem 18
  ...\cdot 7=97^{\text{th}}</math> day of year <math>N</math> and the <math>200-15\cdot 7=95^{\text{th}}</math> day of year <math>N+1</math> are Tuesdays. If
  2 KB (382 words) - 19:20, 12 May 2023
• 2000 AMC 12 Problems/Problem 19
  ...] <math>ABC</math>, <math>AB = 13</math>, <math>BC = 14</math>, <math>AC = 15</math>. Let <math>D</math> denote the [[midpoint]] of <math>\overline{BC}</ ...ABC</math> to solve for <math>BE</math>. Solving <math>\frac{13}{BE}=\frac{15}{14-BE}</math>, we get that <math>BE=\frac{13}{2}</math>. <math>D</math> is
  3 KB (547 words) - 17:37, 17 February 2024
• 2007 AMC 10A
  ** [[2007 AMC 10A Problems/Problem 15]]
  2 KB (182 words) - 03:21, 31 December 2019
• 2007 AMC 10A Problems/Problem 17
  <math>\text{(A)}\ 15 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 60 \qquad \t ...h>3^2 \cdot 5 = 45</math>, which makes <math>n = \sqrt[3]{3^3 \cdot 5^3} = 15</math>. The minimum possible value for the sum of <math>m</math> and <math>
  1 KB (201 words) - 08:04, 11 February 2023
• 2007 AMC 10A Problems
  ...m{(A)}\ 2\qquad \mathrm{(B)}\ 5\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 15\qquad \mathrm{(E)}\ 20</math> A school store sells 7 pencils and 8 notebooks for <math>\$4.15</math>. It also sells 5 pencils and 3 notebooks for <math>\$1.77</math>. Ho
  13 KB (2,058 words) - 17:54, 29 March 2024
• 2007 AMC 10A Problems/Problem 12
  ...h> B </math> takes 4, where there are <math> \binom{6}{2} </math> = <math> 15 </math> ways to arrange the tourists. ...h> B </math> takes 2, where there are <math> \binom{6}{2} </math> = <math> 15 </math> ways to arrange the tourists.
  3 KB (411 words) - 18:07, 14 March 2023
• 1995 AHSME
  * [[1995 AHSME Problems/Problem 15|Problem 15]]
  2 KB (174 words) - 00:30, 2 October 2014
• 1995 AHSME Problems
  ...{16}{3} } \qquad \mathrm{(C) \ \frac {20}{3} } \qquad \mathrm{(D) \ \frac {15}{2} } \qquad \mathrm{(E) \ 8 } </math> == Problem 15 ==
  17 KB (2,387 words) - 22:44, 26 May 2021
• 1995 AHSME Problems/Problem 23
  The sides of a triangle have lengths <math>11,15,</math> and <math>k</math>, where <math>k</math> is an integer. For how man ...e is either opposite side <math>15</math> or side <math>k</math>. If <math>15</math> is the largest side,
  1 KB (183 words) - 14:05, 5 July 2013
• 1997 PMWC Problems
  Evaluate <math>29 \dfrac{27}{28} \times 27 \frac{14}{15}</math> draw((0.95,-0.15)--(1.09,0));
  15 KB (2,057 words) - 19:13, 10 March 2015
• 1995 AHSME Problems/Problem 14
  {{AHSME box|year=1995|num-b=13|num-a=15}}
  1 KB (202 words) - 18:05, 6 April 2016
• 1951 AHSME Problems
  At <math> 2: 15</math> o'clock, the hour and minute hands of a clock form an angle of: == Problem 15 ==
  23 KB (3,641 words) - 22:23, 3 November 2023
• 1951 AHSME Problems/Problem 16
  {{AHSME 50p box|year=1951|num-b=15|num-a=17}}
  1 KB (166 words) - 12:20, 5 July 2013
• 1990 USAMO Problems/Problem 3
  ...\le 17</math> such that <math>n+k</math> is relatively prime to <math>n+k+15</math>, <math>n+k+1</math> is relatively prime to <math>n+k+14</math>. ...h>n+b \equiv n+a \pmod{15}</math>, <math>n+b</math> is relatively prime to 15. But
  3 KB (460 words) - 19:14, 18 July 2016
• Algebraic manipulation
  \frac{x^2 - 2x}{x-5} &= \frac{15}{x-5} \\ x^2 - 2x &= 15 \\
  4 KB (562 words) - 18:49, 8 November 2020
• 1959 AHSME
  * [[1959 AHSME Problems/Problem 15|Problem 15]]
  3 KB (257 words) - 14:19, 20 February 2020
• 1959 AHSME Problems
  == Problem 15== <math>\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qqua
  22 KB (3,345 words) - 20:12, 15 February 2023
• 1966 AHSME Problems
  Given that the ratio of <math>3x - 4</math> to <math>y + 15</math> is constant, and <math>y = 3</math> when <math>x = 2</math>, then, w ...\qquad \text{(B)} \ 21 \qquad \text{(C)} \ \sqrt {389} \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{undetermined}</math>
  19 KB (3,159 words) - 22:10, 11 March 2024
• 1966 AHSME
  ** [[1966 AHSME Problems/Problem 15|Problem 15]]
  2 KB (217 words) - 14:15, 20 February 2020
• 1966 AHSME Problems/Problem 1
  Given that the [[ratio]] of <math>3x - 4</math> to <math>y + 15</math> is constant, and <math>y = 3</math> when <math>x = 2</math>, then, w ...Then <math>k = \frac{3(2)-4}{(3)+15} = \frac{1}{9} = \frac{3x - 4}{(12) + 15}</math>. Solving gives <math>3x - 4 = 3 \Longrightarrow x = \frac 73 \Right
  712 bytes (99 words) - 12:39, 5 July 2013
• 2002 AMC 12B Problems/Problem 16
  {{AMC12 box|year=2002|ab=B|num-b=15|num-a=17}}
  2 KB (317 words) - 10:26, 5 November 2023
• 2002 AMC 12B Problems/Problem 14
  {{AMC12 box|year=2002|ab=B|num-b=13|num-a=15}}
  2 KB (282 words) - 14:04, 12 July 2021
• 2002 AMC 12B Problems/Problem 11
  ...12B Problems|2002 AMC 12B #11]] and [[2002 AMC 10B Problems|2002 AMC 10B #15]]}}
  3 KB (470 words) - 11:57, 10 August 2022
• 2002 AMC 12B Problems/Problem 12
  This leaves <math>n\in\{12,14,15,16,18\}</math>, and a quick substitution shows that out of these only <math {{AMC10 box|year=2002|ab=B|num-b=15|num-a=17}}
  4 KB (579 words) - 05:54, 17 October 2023
• Construction
  15. Construct a regular hexagon inside a given circle. 23. Construct <math>15^\circ, 30^\circ, 45^\circ, 60^\circ, 75^\circ</math> angles. Hence or other
  3 KB (443 words) - 20:52, 28 August 2014
• 2007 AMC 8 Problems
  <math>\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36</math> ...him <math>\textdollar 35</math> and his cousin gives him <math>\textdollar 15</math>. He earns
  12 KB (1,800 words) - 20:01, 8 May 2023
• 2001 AMC 12 Problems/Problem 4
  ...ee numbers is <math>10</math> more than the least of the numbers and <math>15</math> and the greatest is <math>m + 15</math>. The middle of the three numbers is the median, <math>5</math>. So
  2 KB (311 words) - 21:53, 10 February 2024
• 2004 AMC 10A Problems/Problem 16
  <math> \mathrm{(A) \ } 12 \qquad \mathrm{(B) \ } 15 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 19\qquad \mathrm{(E) \ } {{AMC10 box|year=2004|ab=A|num-b=15|num-a=17}}
  2 KB (339 words) - 00:25, 14 February 2024
• 2002 AMC 12B Problems/Problem 8
  ...ve Mondays, there are only three possibilities for their dates: <math>(1,8,15,22,29)</math>, <math>(2,9,16,23,30)</math>, and <math>(3,10,17,24,31)</math
  2 KB (329 words) - 15:48, 14 October 2023
• 2003 AMC 12B Problems/Problem 22
  pair O = (15*15/17,8*15/17), C = (17,0), D = (0,0), P = (25.6,19.2), Q = (25.6, 18.5); ...rac{1}{2} AO \cdot BO = 60 = \frac{1}{2} ON \cdot AB = \frac{\sqrt{8^2 + 15^2}}{2} \cdot ON \Longrightarrow ON = \frac{120}{17} \approx 7.06 \Rightarro
  4 KB (551 words) - 14:17, 23 June 2022
• 2003 AMC 12B Problems/Problem 1
  <cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath> 3+(-6+9)+(-12+15)+(-18+21)&=3*4\\
  970 bytes (134 words) - 00:09, 14 September 2015
• 2004 AMC 10A Problems/Problem 19
  draw((0, 15)--(3, 12)^^(0, 16)--(3, 13)); draw((0,-1)--(0,15), dashed);
  2 KB (220 words) - 14:19, 21 April 2021
• 2007 AMC 12B Problems/Problem 11
  #REDIRECT [[2007 AMC 10B Problems/Problem 15]]
  46 bytes (5 words) - 16:22, 5 June 2011
• 2004 AMC 12B Problems/Problem 16
  {{AMC12 box|year=2004|ab=B|num-b=15|num-a=17}}
  1 KB (240 words) - 12:59, 30 March 2023
• 2008 AMC 12A
  **[[2008 AMC 12A Problems/Problem 15|Problem 15]]
  2 KB (193 words) - 21:49, 6 October 2014
• 2008 AMC 12B
  **[[2008 AMC 12B Problems/Problem 15|Problem 15]]
  2 KB (193 words) - 00:11, 7 October 2014
• 2008 AMC 10A
  **[[2008 AMC 10A Problems/Problem 15|Problem 15]]
  2 KB (195 words) - 18:08, 28 June 2021
• 2008 AMC 10B
  **[[2008 AMC 10B Problems/Problem 15|Problem 15]]
  2 KB (188 words) - 18:44, 6 October 2014
• 2002 AMC 10A Problems/Problem 25
  \frac {DE}{DE + 5} = \frac {3}{4} & \Longrightarrow DE = 15 \end{align*} So the sides of <math>\triangle CDE</math> are <math>15,36,39</math>, which we recognize to be a <math>5 - 12 - 13</math> [[right t
  8 KB (1,308 words) - 07:05, 19 December 2022
• 2002 AMC 10A Problems/Problem 22
  15&3&6\\
  3 KB (430 words) - 23:13, 13 September 2023
• 2008 AMC 10A Problems
  ...a new computer at two different stores. Store <math>A</math> offers <math>15\%</math> off the sticker price followed by a <math>\textdollar90</math> reb ==Problem 15==
  14 KB (2,138 words) - 15:08, 18 February 2023
• Partition (combinatorics)
  ...11</math>. One could continue this computation to find that <math>P(7) = 15</math>, <math>P(8) = 22</math>, <math>P(9) = 30</math>, <math>P(10) = 42</m
  10 KB (1,508 words) - 14:24, 17 September 2017
• 2008 AMC 12A Problems
  ...ff the same sticker price with no rebate. Heather saves <math> \textdollar 15</math> by buying the computer at store <math>A</math> instead of store <mat ==Problem 15==
  13 KB (2,025 words) - 13:56, 2 February 2021
• 2008 AMC 12A Problems/Problem 6
  ...\%</math> off the same sticker price with no rebate. Heather saves <math>\$15</math> by buying the computer at store <math>A</math> instead of store <mat Heather saves <math>\$15</math> at store <math>A</math>, so <math>0.85x-90+15=0.75x</math>.
  2 KB (240 words) - 19:53, 4 June 2021
• 2008 AMC 12A Problems/Problem 19
  Since there are <math>15</math> terms in <math>A</math>, there are <math>15^2 = 225</math> ways to choose one term from each <math>A</math>. The produc = 15^2 - 1
  5 KB (895 words) - 22:54, 9 January 2021
• 2008 AMC 12A Problems/Problem 16
  ...th>\log(a^3b^7)</math>, <math>\log(a^5b^{12})</math>, and <math>\log(a^8b^{15})</math> are the first three terms of an [[arithmetic sequence]], and the < ...ssion]], then <math>a^3b^7</math>, <math>a^5b^{12}</math>, and <math>a^8b^{15}</math> are in [[geometric progression]]. Therefore,
  3 KB (577 words) - 16:33, 9 October 2022
• 2008 AMC 12A Problems/Problem 13
  label("\(30^{\circ}\)",(0.65,0.15),O); {{AMC10 box|year=2008|ab=A|num-b=15|num-a=17}}
  4 KB (630 words) - 20:32, 4 June 2021
• 2008 AMC 12A Problems/Problem 7
  It will take <math>\frac{1}{4}</math> of an hour or <math>15</math> minutes to get to shore. ...<math>30</math> gallons of water can enter the boat, only <math>\frac{30}{15}=2</math> net gallons can enter the boat per minute.
  2 KB (313 words) - 18:08, 4 June 2021

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