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Page title matches

• 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

Page text matches

• 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) - 11:36, 19 May 2024
• 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

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