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- == Problem ==2 KB (354 words) - 16:57, 28 December 2020
- == Problem == ...by the commutative property(this is the same as the equation given in the problem. We are just rearranging). So we can set <math>\frac{1}{x} = x</math> which2 KB (334 words) - 18:34, 18 September 2020
- == Problem ==2 KB (262 words) - 21:20, 21 December 2020
- == Problem ==2 KB (254 words) - 14:39, 5 April 2024
- == Problem ==1 KB (158 words) - 01:33, 29 May 2023
- == Problem ==1 KB (195 words) - 15:33, 16 December 2021
- == Problem == ...y of South Carolina High School Math Contest/1993 Exam/Problem 17|Previous Problem]]2 KB (331 words) - 00:37, 26 January 2023
- ==Problem==2 KB (260 words) - 17:42, 7 July 2023
- == Problem == The problem is asking for <math>\frac{1}{a}+\frac{1}{b}= \frac{a+b}{ab}</math>3 KB (458 words) - 13:41, 26 August 2023
- == Problem ==3 KB (426 words) - 18:20, 18 July 2022
- == Problem ==2 KB (319 words) - 00:37, 25 March 2024
- == Problem ==2 KB (277 words) - 18:15, 25 November 2020
- ...ate|[[2004 AMC 12A Problems|2004 AMC 12A #18]] and [[2004 AMC 10A Problems/Problem 22|2004 AMC 10A #22]]}} == Problem ==5 KB (738 words) - 13:11, 27 March 2023
- {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #18]] and [[2000 AMC 10 Problems|2000 AMC 10 #25]]}} == Problem ==2 KB (409 words) - 19:33, 25 May 2024
- == Problem == [[Image:2002_12B_AMC-18.png]]3 KB (376 words) - 19:16, 20 August 2019
- ==Problem== ...alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Prob6 KB (867 words) - 00:17, 20 May 2023
- ==Problem==2 KB (302 words) - 04:51, 16 January 2023
- ==Problem== ...}\ 10 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 21</math>5 KB (758 words) - 16:35, 15 February 2021
- ==Problem== You can solve this problem by setting up a simple equation with the Pythagorean Theorem. The hypotenus2 KB (385 words) - 14:17, 4 June 2021
- ==Problem==875 bytes (139 words) - 20:19, 23 March 2023
- == Problem == From the problem, we know that8 KB (1,339 words) - 14:15, 1 August 2022
- == Problem == A simpler way to tackle this problem without all that modding is to keep the equation as:6 KB (914 words) - 11:07, 7 September 2023
- ==Problem==1 KB (167 words) - 13:59, 5 July 2013
- ==Problem==1 KB (214 words) - 12:01, 2 February 2015
- == Problem ==999 bytes (153 words) - 20:43, 28 May 2023
- == Problem ==1 KB (187 words) - 14:29, 5 July 2013
- == Problem ==2 KB (270 words) - 14:35, 5 July 2013
- ==Problem==2 KB (274 words) - 10:26, 8 November 2021
- ==Problem==1 KB (235 words) - 09:03, 22 January 2023
- ==Problem==2 KB (270 words) - 18:54, 28 December 2023
- == Problem ==2 KB (461 words) - 16:29, 27 August 2016
- #REDIRECT[[2002 AMC 12B Problems/Problem 14]]45 bytes (5 words) - 16:15, 29 July 2011
- == Problem == ...> and <math>\sin(\angle E) = \frac{12}{20}</math>, so <math>96 = 18 + 18 + 18 + x</math>.6 KB (904 words) - 12:54, 22 October 2023
- {{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #18]] and [[2009 AMC 10A Problems|2009 AMC 10A #25]]}} == Problem ==6 KB (1,012 words) - 19:16, 14 September 2022
- == Problem == Hence the answer is <math>\frac{36}{70}=\frac{18}{35}</math>. We know this is a little bit larger than <math>\frac 12</math>3 KB (402 words) - 10:29, 2 August 2021
- == Problem ==5 KB (822 words) - 01:35, 7 February 2024
- == Problem == \text{(B) }183 KB (485 words) - 03:13, 1 September 2023
- == Problem ==6 KB (930 words) - 22:14, 18 January 2024
- ==Problem==1 KB (185 words) - 19:11, 26 August 2016
- ==Problem==1 KB (170 words) - 23:56, 4 July 2013
- ==Problem==2 KB (267 words) - 04:54, 23 June 2022
- ==Problem==1 KB (195 words) - 13:25, 28 December 2021
- ==Problem==2 KB (266 words) - 00:07, 5 July 2013
- == Problem == ...h>6</math> points during any one of its paths. Therefore we can divide the problem into <math>3</math> cases, focusing on <math>1</math> quadrant; then multip5 KB (910 words) - 01:40, 2 February 2021
- 46 bytes (5 words) - 13:27, 26 May 2020
- == Problem == ...find the average arc length where the third jump could land to satisfy the problem. To do this, I can apply average function value with our old buddy calculus6 KB (1,105 words) - 13:39, 9 January 2024
- ==Problem==1,000 bytes (149 words) - 05:43, 31 December 2022
- #redirect [[2010 AMC 12B Problems/Problem 16]]46 bytes (5 words) - 20:45, 26 May 2020
- == Problem ==2 KB (306 words) - 21:50, 2 November 2021
- #redirect [[2011 AMC 12A Problems/Problem 11]]46 bytes (5 words) - 19:19, 27 June 2020
- == Problem==5 KB (782 words) - 14:29, 1 April 2024
- #redirect [[2001 AMC 12 Problems/Problem 10]]45 bytes (4 words) - 02:23, 5 December 2019
- ==Problem==3 KB (508 words) - 19:26, 7 August 2023
- ==Problem== <math>\text{(A)}\ \dfrac{1}{36} \qquad \text{(B)}\ \dfrac{1}{18} \qquad \text{(C)}\ \dfrac{1}{6} \qquad \text{(D)}\ \dfrac{11}{36} \qquad \2 KB (333 words) - 22:55, 17 October 2023
- ==Problem==2 KB (278 words) - 17:21, 20 February 2020
- ...the distance from <math>X</math> to <math>CD</math>.<noinclude>[[category:Problem of the Day]]</noinclude>433 bytes (76 words) - 18:06, 17 June 2011
- ==Problem== {{:AoPSWiki:Problem of the Day/June 18, 2011}}2 KB (287 words) - 21:50, 17 June 2011
- ==Problem== ...le bounded by the x-axis, the y-axis, and the line <math>x+y=2</math>. The problem is asking for <math>x</math>, which is just the inradius. The inradius is <3 KB (446 words) - 18:21, 4 June 2021
- #REDIRECT [[2003 AMC 12B Problems/Problem 12]]46 bytes (5 words) - 00:56, 5 January 2014
- ...ath> units. What is the total area of both triangles?<noinclude>[[category:Problem of the Day]]</noinclude>235 bytes (36 words) - 20:05, 17 July 2011
- ==Problem== {{:AoPSWiki:Problem of the Day/July 18, 2011}}767 bytes (125 words) - 13:59, 10 July 2016
- ==Problem==2 KB (258 words) - 20:01, 15 April 2023
- ==Problem== ...qrt{2^2 + 1^2} = \sqrt{5}</math>. This does not need to be found for this problem, as you can do a one-to-one correspondance with three of the four sides of3 KB (516 words) - 20:05, 15 April 2023
- ==Problem==2 KB (296 words) - 02:00, 28 February 2022
- ==Problem==918 bytes (130 words) - 19:49, 31 October 2016
- ==Problem==893 bytes (115 words) - 13:31, 25 January 2024
- ==Problem== <math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20 </math>2 KB (364 words) - 14:13, 5 July 2013
- == Problem == draw((-18,1)--(-12, 1), EndArrow);2 KB (210 words) - 13:37, 19 October 2020
- ==Problem==2 KB (279 words) - 19:37, 15 April 2023
- ==Problem== draw((17,2)--(18,8)--(22,8)--(23,2));2 KB (367 words) - 13:30, 30 October 2016
- {{:AoPSWiki:Problem of the Day/August 18, 2011}}48 bytes (6 words) - 04:43, 18 August 2011
- ==Problem==2 KB (415 words) - 14:43, 5 June 2016
- ==Problem==666 bytes (101 words) - 05:42, 31 August 2015
- ==Problem==2 KB (263 words) - 01:05, 11 November 2019
- ==Problem==895 bytes (142 words) - 12:53, 12 November 2017
- ==Problem== <asy>/* AMC8 2003 #18 Problem */2 KB (250 words) - 22:17, 5 January 2024
- <noinclude>[[Category: Problem of the Day]]<noinclude>236 bytes (44 words) - 08:02, 18 September 2011
- 248 bytes (42 words) - 13:32, 18 September 2011
- ==Problem== Six bags of marbles contain <math> 18, 19, 21, 23, 25 </math> and <math> 34 </math> marbles, respectively. One ba2 KB (241 words) - 22:50, 19 March 2024
- == Problem ==863 bytes (137 words) - 22:19, 6 February 2023
- ==Problem==2 KB (376 words) - 15:08, 17 December 2023
- ==Problem==962 bytes (148 words) - 02:21, 7 January 2020
- ...12A Problems|2012 AMC 12A #14]] and [[2012 AMC 10A Problems|2012 AMC 10A #18]]}} == Problem 14 ==5 KB (775 words) - 22:33, 22 October 2023
- == Problem == We can use mass points and Stewart's to solve this problem. Because we are looking at the Incenter we then label <math>A</math> with a4 KB (717 words) - 19:07, 28 July 2021
- 1 KB (212 words) - 01:19, 16 February 2012
- == Problem ==991 bytes (164 words) - 13:54, 25 February 2018
- ==Problem==2 KB (379 words) - 14:00, 22 August 2022
- == Problem ==1 KB (196 words) - 11:56, 19 March 2017
- ==Problem==2 KB (301 words) - 09:04, 10 March 2023
- == Problem == {{AHSME box|year=1966|num-b=17|num-a=18}}492 bytes (69 words) - 03:33, 15 February 2019
- ==Problem==1 KB (153 words) - 12:43, 5 July 2013
- ==Problem== The problem states that the answer cannot be a perfect square or have prime factors les1 KB (158 words) - 13:58, 10 November 2023
- == Problem 18 == This problem is worded awkwardly. More simply, it asks: “How many ways can you order n4 KB (715 words) - 00:50, 27 December 2022
- ==Problem==2 KB (262 words) - 11:53, 23 March 2020
- ==Problem==861 bytes (126 words) - 04:07, 29 December 2022
- == Problem==4 KB (645 words) - 03:39, 28 December 2022
- ==Problem==4 KB (592 words) - 22:19, 2 November 2023
- ==Problem==3 KB (480 words) - 22:23, 26 March 2023
- ==Problem== https://youtu.be/I6JgSPbL6gY ~Math Problem Solving Skills3 KB (494 words) - 20:15, 15 June 2022
- == Problem ==749 bytes (129 words) - 18:53, 11 October 2016
- == Problem ==953 bytes (152 words) - 01:40, 16 August 2023
- ==Problem== {{AHSME 40p box|year=1962|before=Problem 17|num-a=19}}902 bytes (147 words) - 22:17, 3 October 2014
- ==Problem==3 KB (412 words) - 22:30, 18 December 2023
- == Problem==485 bytes (86 words) - 01:57, 3 January 2014
- ==Problem==5 KB (791 words) - 03:18, 20 June 2022
- ==Problem== ...f{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18 </math>1 KB (170 words) - 03:08, 28 May 2021
- ==Problem==2 KB (410 words) - 21:18, 31 May 2020
- ==Problem== ...roblem is quite similar to [[2004_AMC_12A_Problems/Problem_16|2004 AMC 12A Problem 16]].4 KB (703 words) - 19:04, 10 July 2021
- ==Problem==2 KB (369 words) - 17:44, 30 January 2021
- {{duplicate|[[2014 AMC 12B Problems|2014 AMC 12B #18]] and [[2014 AMC 10B Problems|2014 AMC 10B #24]]}} ==Problem==2 KB (331 words) - 04:43, 12 January 2021
- == Problem == Day 18: Al works; Barb rests2 KB (294 words) - 16:49, 9 September 2020
- == Problem ==1,008 bytes (157 words) - 03:02, 20 February 2018
- == Problem ==1 KB (181 words) - 01:52, 16 August 2023
- == Problem ==987 bytes (146 words) - 11:52, 4 February 2016
- == Problem ==1 KB (180 words) - 20:08, 23 February 2024
- == Problem ==608 bytes (84 words) - 16:23, 2 July 2016
- == Problem ==830 bytes (142 words) - 20:45, 18 June 2021
- == Problem ==880 bytes (138 words) - 01:40, 22 December 2015
- == Problem == <math>\text{(F) }18\qquad2 KB (393 words) - 17:01, 10 June 2018
- ==Problem==1 KB (232 words) - 14:03, 27 February 2018
- ==Problem==2 KB (340 words) - 18:23, 28 June 2015
- == Problem ==1 KB (237 words) - 11:45, 23 October 2014
- ==Problem==844 bytes (131 words) - 18:31, 12 October 2023
- ==Problem==2 KB (245 words) - 11:20, 2 July 2023
- ==Problem==1 KB (232 words) - 23:57, 22 September 2021
- ==Problem== <math> \textbf{(A) }17\qquad\textbf{(B) }18\qquad\textbf{(C) }19\qquad\textbf{(D) }20\qquad\textbf{(E) }21 </math>3 KB (455 words) - 07:19, 31 March 2023
- ==Problem== ...{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18</math>7 KB (1,193 words) - 14:25, 25 July 2022
- ==Problem== This problem becomes simple once we recognize that the domain of the function is <math>\2 KB (407 words) - 03:03, 3 August 2021
- ==Problem== We can simplify the problem first, then apply reasoning to the original problem. Let's say that there are <math>8</math> coins. Shaded coins flip heads, an3 KB (479 words) - 13:54, 27 August 2021
- ==Problem== ...the numbers on each face must be 18, because <math>\frac{1+2+\cdots+8}{2}=18</math>.4 KB (769 words) - 12:31, 7 November 2022
- ==Problem== {{AMC8 box|year=2015|num-b=18|after=Last Problem}}5 KB (641 words) - 10:28, 13 January 2024
- #REDIRECT [[2016 AMC 10A Problems/Problem 22]]46 bytes (5 words) - 13:21, 4 February 2016
- ==Problem==5 KB (813 words) - 16:55, 9 June 2023
- ==Problem==1 KB (188 words) - 22:57, 9 January 2024
- == Problem 18==1 KB (228 words) - 12:25, 8 May 2020
- == Problem 18==496 bytes (72 words) - 01:28, 28 February 2020
- == Problem 18 ==2 KB (294 words) - 08:42, 15 April 2016
- == Problem 18 == ...2</math> and <math>DQ=r-2,</math> so we have the equation <math>(r+2)(r-2)=18.</math> Solving gives <math>r=\sqrt{22}.</math>2 KB (361 words) - 08:05, 9 April 2023
- == Problem 18 ==500 bytes (77 words) - 13:05, 22 November 2016
- ==Problem==1 KB (215 words) - 23:06, 17 May 2024
- == Problem 18 ==1 KB (147 words) - 16:07, 8 January 2017
- ==Problem==3 KB (538 words) - 04:25, 21 January 2023
- ==Problem==3 KB (497 words) - 19:06, 19 December 2023
- == Problem ==7 KB (1,057 words) - 23:27, 27 August 2022
- ==Problem==7 KB (886 words) - 04:01, 23 January 2023
- ==Problem 18==421 bytes (63 words) - 21:42, 1 April 2017
- 145 bytes (26 words) - 23:51, 2 July 2017
- == Problem 18==1 KB (210 words) - 18:43, 20 October 2018
- ==Problem== ...e, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of <math>\triangle BDA</math> is <math>\frac{5\cdot 12}{2}</math>2 KB (337 words) - 17:37, 21 January 2024
- ...AMC 10A Problems/Problem 18|2018 AMC 10A #18]] and [[2018 AMC 12A Problems/Problem 13|2018 AMC 12A #13]]}} ==Problem==10 KB (1,531 words) - 17:00, 18 October 2023
- 45 bytes (5 words) - 15:30, 8 February 2018
- 45 bytes (5 words) - 14:52, 16 February 2018
- ==Problem== ...<math>B_2</math>, <math>C_1</math>, and <math>C_2</math>. We can split our problem into two cases:7 KB (1,281 words) - 17:24, 8 January 2024
- ==Problem==902 bytes (132 words) - 22:14, 13 January 2023
- == Problem ==910 bytes (127 words) - 11:03, 21 May 2018
- ==Problem==2 KB (304 words) - 09:29, 23 June 2022
- ==Problem==1 KB (202 words) - 14:46, 14 January 2024
- ==Problem==1 KB (199 words) - 20:12, 18 January 2024
- ==Problem== ...''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}2 KB (252 words) - 12:04, 27 November 2018
- 45 bytes (5 words) - 16:52, 9 February 2019
- {{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #18]] and [[2019 AMC 12A Problems|2019 AMC 12A #11]]}} ==Problem==4 KB (594 words) - 22:15, 21 November 2023
- ==Problem== ...<math>Q</math>, with <math>P</math> being closer to home. As given in the problem statement, the distances of the points <math>P</math> and <math>Q</math> fr7 KB (1,145 words) - 18:55, 12 January 2024
- ==Problem==5 KB (799 words) - 19:30, 12 November 2022
- ==Problem 18== ==Video Solution by Math-X (First understand the problem!!!)==3 KB (430 words) - 16:05, 30 December 2023
- .../wiki/index.php/2003_AMC_10A_Problems/Problem_25 because they are the same problem.146 bytes (22 words) - 22:04, 27 November 2019
- == Problem == This problem is similar to 2007 AMC10A Problem 16. View it here: https://artofproblemsolving.com/wiki/index.php/2007_AMC_16 KB (1,044 words) - 13:50, 4 April 2024
- 45 bytes (5 words) - 07:56, 1 February 2020
- {{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #18]] and [[2020 AMC 12B Problems|2020 AMC 12B #16]]}} ==Problem==11 KB (1,928 words) - 22:40, 12 November 2023
- #redirect [[2020 AMC 10B Problems/Problem 21]]46 bytes (5 words) - 11:00, 9 May 2021
- == Problem 18 ==1 KB (165 words) - 17:20, 19 January 2021
- == Problem 18==732 bytes (116 words) - 14:12, 20 July 2020
- == Problem 18== ^The problem pretends to have two solutions.846 bytes (130 words) - 17:55, 2 August 2020
- ==Problem== ...because if we do, the two would have to be adjacent in some way, which the problem disallows. So, there are <math>{4 \choose 2} - 2 = 4</math> ways to choose11 KB (1,867 words) - 12:30, 25 May 2024
- ...C 10A #21]] and [[2021 Fall AMC 10A Problems#Problem 18|2021 Fall AMC 12A #18]]}} ==Problem==5 KB (784 words) - 12:12, 10 November 2023
- ...18|2021 AMC 10A #18]] and [[2021 AMC 12A Problems/Problem 18|2021 AMC 12A #18]]}} ==Problem==9 KB (1,403 words) - 18:30, 23 October 2022
- ==Problem== We can use a result from the Art of Problem Solving <i>Introduction to Algebra</i> book Sidenote: for a semicircle with6 KB (954 words) - 16:35, 26 January 2024
- 18 bytes (3 words) - 18:02, 18 November 2020
- #redirect [[2021 AMC 12A Problems/Problem 18]]46 bytes (5 words) - 14:11, 11 February 2021
- ==Problem== Note that the problem is basically asking us to find the probability that in some permutation of8 KB (1,296 words) - 17:48, 10 November 2023
- == Problem ==698 bytes (103 words) - 16:36, 29 January 2021
- == Problem ==880 bytes (126 words) - 16:03, 4 July 2023
- == Problem == |before=[[1963 TMTA High School Algebra I Contest Problem 17| Problem 17]]1 KB (160 words) - 10:24, 2 February 2021
- ==Problem== By the equation given in the problem5 KB (885 words) - 08:48, 21 October 2023
- == Problem ==531 bytes (66 words) - 19:40, 12 February 2021
- ==Problem 18==927 bytes (152 words) - 16:05, 1 April 2021
- 0 bytes (0 words) - 13:53, 26 April 2021
- 1 KB (223 words) - 13:54, 26 April 2021
- ==Problem==961 bytes (143 words) - 21:09, 13 July 2022
- ==Problem==974 bytes (152 words) - 17:15, 11 July 2021
- ==Problem==5 KB (733 words) - 10:36, 5 November 2022
- ==Problem== ...\cdot (2+\sqrt{3}))^2 = 3\sqrt{2}^2 \implies x^2+x^2 \cdot (7+4\sqrt{3}) = 18</math>18 KB (3,011 words) - 22:05, 26 September 2023
File:Intro to Geometry- Week 18, Problem 9 Answer(1).jpeg A picture of my work for problem 9 in week 18.(2,199 × 1,784 (540 KB)) - 02:00, 16 January 2022- ==Problem== ==Video Solution by Math-X (First understand the problem!!!)==2 KB (365 words) - 14:53, 23 November 2023
- ==Problem:== The answer to this problem is the number of intersections between the graph of <math>f(x) = \sin x</ma1 KB (178 words) - 14:01, 10 November 2022
- ==Problem==1 KB (214 words) - 19:40, 7 March 2022
- ...18|2022 AMC 10A #18]] and [[2022 AMC 12A Problems/Problem 18|2022 AMC 12A #18]]}} ==Problem==4 KB (643 words) - 14:39, 20 May 2024
- #redirect [[2022 AMC 10A Problems/Problem 18]]46 bytes (5 words) - 05:17, 19 November 2022
- ==Problem== ...ber of options that force <math>x = y = z = 0</math> is <math>3 \cdot 3! = 18</math>.13 KB (2,072 words) - 22:10, 5 July 2023
- #REDIRECT [[2022 AMC 10B Problems/Problem 19]] {{AMC12 box|year=2022|ab=B|num-b=18|num-a=20}}124 bytes (16 words) - 18:20, 18 November 2022
- ==Problem== ...th>4</math> as <math>2023 + 4(3) = 2035</math>. So now, we have solved the problem as we just go <math>\frac{2035}{5} = 407</math> hops right, and just do 4 m3 KB (489 words) - 02:00, 1 February 2024
- 271 bytes (53 words) - 01:01, 27 October 2023
- ...22|2023 AMC 10A #22]] and [[2023 AMC 12A Problems/Problem 18|2023 AMC 12A #18]]}} ==Problem==4 KB (527 words) - 23:54, 20 April 2024
- ==Problem== ...ath> be the number of vertices with <math>3</math> edges (this is what the problem asks for) and <math>B</math> be the number of vertices with <math>4</math>7 KB (1,160 words) - 03:46, 31 May 2024
- #redirect[[2023 AMC 12B Problems/Problem 15]]45 bytes (5 words) - 20:45, 15 November 2023
- ==Problem== ...s average on all the quizzes she took during the second semester was <math>18</math> points higher than her average for the first semester and was again6 KB (964 words) - 20:27, 23 May 2024
- == Problem ==966 bytes (157 words) - 02:40, 31 December 2023
- ==Problem== ...s is just a quick method if time is short or you do not know how to do the problem and want to guess at it.5 KB (869 words) - 22:19, 26 May 2024
Page text matches
- == Problem == draw((0,0)--(18,0));2 KB (307 words) - 15:30, 30 March 2024
- == Problem == ...[[User:Integralarefun|Integralarefun]] ([[User talk:Integralarefun|talk]]) 18:19, 27 September 2023 (EDT)2 KB (268 words) - 18:19, 27 September 2023
- == Problem == ...in area. They also have the same altitude. If the base of the triangle is 18 inches, the median of the trapezoid is:878 bytes (143 words) - 20:56, 1 April 2017
- ...ecause this keeps showing up in number theory problems. Let's look at this problem below: ...u through the thinking behind SFFT). Now we use factor pairs to solve this problem.7 KB (1,129 words) - 17:57, 24 May 2024
- ...at <math>n>k</math>. The definition that <math>|N| > |K|</math> (as in our problem) is that there exists a surjective mapping from <math>N</math> to <math>K</ ...select a fifth sock without creating a pair. We may use this to prove the problem:11 KB (1,985 words) - 21:03, 5 August 2023
- ...d only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. ==== Problem ====9 KB (1,703 words) - 07:25, 24 March 2024
- * <math>18! = 6402373705728000</math> ([[2007 iTest Problems/Problem 6|Source]])10 KB (809 words) - 16:40, 17 March 2024
- ...hat is the units digit of <math>k^2 + 2^k</math>? ([[2008 AMC 12A Problems/Problem 15]]) ...27^5=n^5</math>. Find the value of <math>{n}</math>. ([[1989 AIME Problems/Problem 9|1989 AIME, #9]])<br><br>16 KB (2,658 words) - 16:02, 8 May 2024
- Other, odder inductions are possible. If a problem asks you to prove something for all integers greater than 3, you can use <m ...ernational Mathematics Olympiad | IMO]]. A good example of an upper-level problem that can be solved with induction is [http://www.artofproblemsolving.com/Fo5 KB (768 words) - 20:45, 1 September 2022
- *[[2007 AMC 12A Problems/Problem 18]] *[[1984 AIME Problems/Problem 8|1984 AIME Problem 8]]5 KB (860 words) - 15:36, 10 December 2023
- * [[2017_USAJMO_Problems/Problem_3 | 2017 USAJMO Problem 3]] * [[2016_AMC_12A_Problems/Problem_17 | 2016 AMC 12A Problem 17]] (See Solution 2)2 KB (280 words) - 15:30, 22 February 2024
- ==Problem== ...and in this situation, the value of <math>I + M + O</math> would be <math>18</math>. Now, we use this process on <math>2001</math> to get <math>667 * 32 KB (276 words) - 05:25, 9 December 2023
- ...ns can significantly help in solving functional identities. Consider this problem: === Problem Examples ===2 KB (361 words) - 14:40, 24 August 2021
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 10B Problems/Problem 1]]2 KB (182 words) - 21:57, 23 January 2021
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 12B Problems/Problem 1 | Problem 1]]2 KB (210 words) - 00:06, 7 October 2014
- * <math>91 \not\equiv 18 \pmod{6}</math> because <math>\frac{91 - 18}{6} = \frac{73}{6}</math>, which is not an integer. === Sample Problem ===15 KB (2,396 words) - 20:24, 21 February 2024
- ...ngruence: <math>x \equiv 5 \pmod{21}</math>, and <math>x \equiv -3 \equiv 18 \pmod{21}</math>. (Note that two values of <math>x</math> that are congrue ...following topics expand on the flexible nature of modular arithmetic as a problem solving tool:14 KB (2,317 words) - 19:01, 29 October 2021
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 10A Problems/Problem 1]]2 KB (180 words) - 18:06, 6 October 2014
- == Problem 1 == ...<math> \overline{AC} </math> is perpendicular to <math> \overline{CD}, AB=18, BC=21, </math> and <math> CD=14. </math> Find the perimeter of <math> ABCD7 KB (1,173 words) - 03:31, 4 January 2023
- == Problem == <math>c=18</math>3 KB (439 words) - 18:24, 10 March 2015
- == Problem == ...ne{AC}</math> is [[perpendicular]] to <math>\overline{CD}</math>, <math>AB=18</math>, <math>BC=21</math>, and <math>CD=14</math>. Find the [[perimeter]]2 KB (217 words) - 21:43, 2 February 2014
- ==Problem 1== [[2021 JMC 10 Problems/Problem 1|Solution]]12 KB (1,784 words) - 16:49, 1 April 2021
- == Problem 1 == [[2006 AMC 12B Problems/Problem 1|Solution]]13 KB (2,058 words) - 12:36, 4 July 2023
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 12A Problems/Problem 1]]1 KB (168 words) - 21:51, 6 October 2014
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AMC 12A Problems/Problem 1]]2 KB (186 words) - 17:35, 16 December 2019
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AMC 12B Problems/Problem 1]]2 KB (181 words) - 21:40, 6 October 2014
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2005 AMC 12A Problems/Problem 1|Problem 1]]2 KB (202 words) - 21:30, 6 October 2014
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AMC 12B Problems/Problem 1|Problem 1]]2 KB (206 words) - 23:23, 21 June 2021
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2000 AMC 12 Problems/Problem 1]]1 KB (126 words) - 13:28, 20 February 2020
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2001 AMC 12 Problems/Problem 1]]1 KB (127 words) - 21:36, 6 October 2014
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2002 AMC 12A Problems/Problem 1]]1 KB (158 words) - 21:33, 6 October 2014
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2003 AMC 12A Problems/Problem 1]]1 KB (162 words) - 21:52, 6 October 2014
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2002 AMC 12B Problems/Problem 1]]1 KB (154 words) - 00:32, 7 October 2014
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2003 AMC 12B Problems/Problem 1]]1 KB (160 words) - 20:46, 1 February 2016
- == Problem 1 == [[2006 AMC 12A Problems/Problem 1|Solution]]15 KB (2,223 words) - 13:43, 28 December 2020
- == Problem 1 == [[2005 AMC 12A Problems/Problem 1|Solution]]13 KB (1,971 words) - 13:03, 19 February 2020
- == Problem 1 == [[2004 AMC 12A Problems/Problem 1|Solution]]13 KB (1,953 words) - 00:31, 26 January 2023
- == Problem 1 == [[2003 AMC 12A Problems/Problem 1|Solution]]13 KB (1,955 words) - 21:06, 19 August 2023
- == Problem 1 == [[2002 AMC 12A Problems/Problem 1|Solution]]12 KB (1,792 words) - 13:06, 19 February 2020
- == Problem 1 == [[2000 AMC 12 Problems/Problem 1|Solution]]13 KB (1,948 words) - 12:26, 1 April 2022
- == Problem 1 == [[2001 AMC 12 Problems/Problem 1|Solution]]13 KB (1,957 words) - 12:53, 24 January 2024
- == Problem 1 == [[2002 AMC 12B Problems/Problem 1|Solution]]10 KB (1,547 words) - 04:20, 9 October 2022
- == Problem 1 == <cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath>13 KB (1,987 words) - 18:53, 10 December 2022
- == Problem 1 == [[2004 AMC 12B Problems/Problem 1|Solution]]13 KB (2,049 words) - 13:03, 19 February 2020
- == Problem 1 == [[2005 AMC 12B Problems/Problem 1|Solution]]12 KB (1,781 words) - 12:38, 14 July 2022
- == Problem == // from amc10 problem series3 KB (458 words) - 16:40, 6 October 2019
- == Problem == \mathrm{(B)}\ \frac 181 KB (188 words) - 22:10, 9 June 2016
- == Problem == {{AMC12 box|year=2006|ab=B|num-b=18|num-a=20}}4 KB (696 words) - 09:47, 10 August 2015
- == Problem == \mathrm{(A)}\ \frac 183 KB (485 words) - 14:09, 21 May 2021
- == Problem == &= \pi^2 \left ( \frac14 - \frac1{36} - \frac1{18}\right ) \\3 KB (563 words) - 22:45, 24 October 2021
- ...reworded and incorporated into this article:--[[User:MCrawford|MCrawford]] 18:54, 30 June 2006 (EDT) ...h>. What if we aren't this lucky? Suppose we want to solve the following problem:4 KB (597 words) - 01:41, 19 December 2013
- == Problem == {{AMC12 box|year=2006|ab=A|num-b=16|num-a=18}}6 KB (958 words) - 23:29, 28 September 2023
- == Problem == {{AMC12 box|year=2006|ab=A|num-b=18|num-a=20}}2 KB (253 words) - 22:52, 29 December 2021
- ...ate|[[2006 AMC 12A Problems|2006 AMC 12A #20]] and [[2006 AMC 10A Problems/Problem 25|2006 AMC 10A #25]]}} == Problem ==5 KB (908 words) - 19:23, 22 September 2022
- == Problem == ...zes left, so she can afford to get less than an <math>A</math> on <math>20-18=\boxed{\textbf{(B) }2}</math> of them.1 KB (197 words) - 14:16, 14 December 2021
- == Problem == {{AMC10 box|year=2005|ab=B|num-b=18|num-a=20}}2 KB (280 words) - 15:35, 16 December 2021
- == Problem == {{AMC12 box|year=2005|ab=B|num-b=16|num-a=18}}1 KB (159 words) - 21:18, 21 December 2020
- == Problem == {{AMC12 box|year=2005|ab=B|num-b=18|num-a=20}}2 KB (283 words) - 20:02, 24 December 2020
- == Problem == ...ad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}} </math>4 KB (761 words) - 09:10, 1 August 2023
- ==Problem 1== [[2006 AMC 10A Problems/Problem 1|Solution]]13 KB (2,028 words) - 16:32, 22 March 2022
- ...ate|[[2006 AMC 12A Problems|2006 AMC 12A #12]] and [[2006 AMC 10A Problems/Problem 14|2006 AMC 10A #14]]}} == Problem ==2 KB (292 words) - 11:56, 17 December 2021
- == Problem == {{AMC10 box|year=2006|num-b=16|num-a=18|ab=A}}6 KB (1,066 words) - 00:21, 2 February 2023
- == Problem == {{AMC10 box|year=2006|ab=A|num-b=18|num-a=20}}2 KB (259 words) - 03:10, 22 June 2023
- ==Problem 1== [[1991 AJHSME Problems/Problem 1|Solution]]17 KB (2,246 words) - 13:37, 19 February 2020
- ==Problem== ...metric sequence to be <math>\{ g, gr, gr^2, \dots \}</math>. Rewriting the problem based on our new terminology, we want to find all positive integers <math>m4 KB (792 words) - 00:29, 13 April 2024
- == Problem == ...sitive integer]]s that are divisors of at least one of <math> 10^{10},15^7,18^{11}. </math>3 KB (377 words) - 18:36, 1 January 2024
- == Problem 1 == [[2005 AIME II Problems/Problem 1|Solution]]7 KB (1,119 words) - 21:12, 28 February 2020
- == Problem == 2 & 18 & no\\ \hline8 KB (1,248 words) - 11:43, 16 August 2022
- == Problem == ...^{34}\cdot 3^{18}}</math> and so the answer is <math>2 + 3 + 5 + 12 + 34 + 18 = \boxed{074}</math>.4 KB (600 words) - 21:44, 20 November 2023
- == Problem == ...next, giving <math>2</math> ways. This totals <math>6 + 3\cdot 2\cdot 2 = 18</math> ways.5 KB (897 words) - 00:21, 29 July 2022
- == Problem == ...eft with <math>c = 2</math>, so our triangle has area <math>\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}</math> and so the answer is <math>24 + 14 =5 KB (906 words) - 23:15, 6 January 2024
- == Problem == ...ac{24}{5}</math>. Consequently, from Pythagorean theorem, <math>SC = \frac{18}{5}</math> and <math>AS = 14-SC = \frac{52}{5}</math>. We can also use the13 KB (2,129 words) - 18:56, 1 January 2024
- == Problem == ...ions in a string format, starting with the operation that sends <math>f(x_{18}) = x_{19}</math> and so forth downwards. There are <math>2^9</math> ways t9 KB (1,491 words) - 01:23, 26 December 2022
- == Problem == P(x) &= \left(\frac{x^{18} - 1}{x - 1}\right)^2 - x^{17} = \frac{x^{36} - 2x^{18} + 1}{x^2 - 2x + 1} - x^{17}\\ &= \frac{x^{36} - x^{19} - x^{17} + 1}{(x -2 KB (298 words) - 20:02, 4 July 2013
- == Problem == Now, consider the strip of length <math>1024</math>. The problem asks for <math>s_{941, 10}</math>. We can derive some useful recurrences f6 KB (899 words) - 20:58, 12 May 2022
- == Problem == ...n't need to be nearly as rigorous). A more natural manner of attacking the problem is to think of the process in reverse, namely seeing that <math>n \equiv 111 KB (1,857 words) - 21:55, 19 June 2023
- == Problem == <math>= 7 + 6 \cdot 3 \cdot 7 = 7 + 18 \cdot 7 = 19 \cdot 7 = 133</math>.8 KB (1,283 words) - 19:19, 8 May 2024
- ==Problem== From the initial problem statement, we have <math>1000w\cdot\frac{1}{4}t=\frac{1}{4}</math>.4 KB (592 words) - 19:02, 26 September 2020
- == Problem == ...wo red candies after Terry chooses two red candies is <math>\frac{7\cdot8}{18\cdot17} = \frac{28}{153}</math>. So the probability that they both pick tw2 KB (330 words) - 13:42, 1 January 2015
- ==Problem== ...possible combinations of <math>B</math> and <math>J</math>, and <math>1 + 18 \cdot 2 + 1 = 38</math> ones that form <math>P</math>, so <math>P = \frac{35 KB (830 words) - 22:15, 28 December 2023
- == Problem 1 == [[1989 AIME Problems/Problem 1|Solution]]7 KB (1,045 words) - 20:47, 14 December 2023
- == Problem 1 == [[1990 AIME Problems/Problem 1|Solution]]6 KB (870 words) - 10:14, 19 June 2021
- == Problem 1 == [[Image:AIME 1995 Problem 1.png]]6 KB (1,000 words) - 00:25, 27 March 2024
- == Problem 1 == [[2002 AIME I Problems/Problem 1|Solution]]8 KB (1,374 words) - 21:09, 27 July 2023
- == Problem 1 == [[2000 AIME II Problems/Problem 1|Solution]]6 KB (947 words) - 21:11, 19 February 2019
- == Problem == ...th of the roots of this equation are real, since its discriminant is <math>18^2 - 4 \cdot 1 \cdot 20 = 244</math>, which is positive. Thus by [[Vieta's f3 KB (532 words) - 05:18, 21 July 2022
- == Problem == ...h>9z^2 - yz + 4 = 0</math> to see that <math>z = \frac{y\pm\sqrt{y^2-144}}{18}</math>. We know that <math>y</math> must be an integer and as small as it4 KB (722 words) - 20:25, 14 January 2023
- == Problem == ...th> and <math>y</math> are different digits, <math>1+0=1 \leq x+y \leq 9+9=18</math>, so the only possible multiple of <math>11</math> is <math>11</math>2 KB (412 words) - 18:23, 1 January 2024
- == Problem == ...hat circle bisects the chord, so <math>QM = MP = PN = NR</math>, since the problem told us <math>QP = PR</math>.13 KB (2,151 words) - 17:48, 27 May 2024
- == Problem == ...achieved with TWO or MORE methods. (Note: This is actually the exact same problem as the original, just reworded differently and now we are adding the score.7 KB (1,163 words) - 23:53, 28 March 2022
- == Problem == If <math>x \ge 18</math> and is <math>0 \bmod{6}</math>, <math>x</math> can be expressed as <8 KB (1,346 words) - 01:16, 9 January 2024
- == Problem == [[Image:AIME 1985 Problem 15.png]]2 KB (245 words) - 22:44, 4 March 2024
- == Problem == ...= Y</math>). Finding the optimal location for <math>X</math> is a classic problem: for any path from <math>F_1</math> to <math>X</math> and then back to <mat5 KB (932 words) - 17:00, 1 September 2020
- == Problem == *<math>1</math>: Easily possible, for example try plugging in <math>x =\frac 18</math>.12 KB (1,859 words) - 18:16, 28 March 2022
- == Problem == D(anglemark(A2,O,A1,5)); D(anglemark(A3,O,A2,5)); D(anglemark(A2,A3,A1,18));5 KB (763 words) - 16:20, 28 September 2019
- == Problem == ...1</math>). By the [[Binomial Theorem]], this is <math>(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}</math>.6 KB (872 words) - 16:51, 9 June 2023
- == Problem == ...of the multiples of <math>222</math> from <math>15\cdot222</math> to <math>18\cdot222</math> by subtracting <math>N</math> from each <math>222(a+b+c)</ma3 KB (565 words) - 16:51, 1 October 2023
- == Problem == ...10}\cdot450=\frac{15}{17}d</math> and <math>EC=\frac{d}{425}\cdot450=\frac{18}{17}d</math>. Since <math>FD'=BC-EE'</math>, we have <math>900-\frac{33}{1711 KB (1,850 words) - 18:07, 11 October 2023
- == Problem == x_3-x_1&=18\\1 KB (212 words) - 16:25, 17 November 2019
- == Problem == &= (x(x-6) + 18)(x(x+6)+18),7 KB (965 words) - 10:42, 12 April 2024
- == Problem == ...than 1, so <math>3n^2 > 999</math> and <math>n > \sqrt{333}</math>. <math>18^2 = 324 < 333 < 361 = 19^2</math>, so we must have <math>n \geq 19</math>.4 KB (673 words) - 19:48, 28 December 2023
- == Problem == ...to 3, it would overlap with case 1). Thus, there are <math>2(3 \cdot 3) = 18</math> cases.3 KB (547 words) - 22:54, 4 April 2016
- == Problem == ...s: <math>\sqrt{(12 - (-2))^2 + (8 - (-10))^2 + (-16 - 5)^2} = \sqrt{14^2 + 18^2 + 21^2} = 31</math>. The largest possible distance would be the sum of th697 bytes (99 words) - 18:46, 14 February 2014
- == Problem == <cmath>3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27</cmath>4 KB (727 words) - 23:37, 7 March 2024
- == Problem == <math>(-3+i)^1=-3+i ; (-3+i)^2=8-6i ; (-3+i)^3=-18+26i</math>2 KB (408 words) - 17:28, 16 September 2023
- == Problem == ...be a [[tetrahedron]] with <math>AB=41</math>, <math>AC=7</math>, <math>AD=18</math>, <math>BC=36</math>, <math>BD=27</math>, and <math>CD=13</math>, as2 KB (376 words) - 13:49, 1 August 2022
- == Problem == ...e [[Pythagorean Theorem]]. Thus <math>A = \frac 12(12\sqrt{3})\sqrt{399} = 18\sqrt{133}</math>.7 KB (1,086 words) - 08:16, 29 July 2023
- == Problem == The sets <math>A = \{z : z^{18} = 1\}</math> and <math>B = \{w : w^{48} = 1\}</math> are both sets of comp3 KB (564 words) - 04:47, 4 August 2023
- == Problem == <!-- replaced: Image:1990 AIME Problem 7.png by I_like_pie -->8 KB (1,319 words) - 11:34, 22 November 2023
- == Problem == The area of a 2x3 rectangle and a 3x4 rectangle combined is 18, so a 4x4 square is impossible without overlapping. Thus, the next smallest1 KB (242 words) - 18:35, 15 August 2023
- == Problem == ...=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18));2 KB (284 words) - 03:56, 23 January 2023
- == Problem == {{AMC10 box|year=2006|ab=B|num-b=16|num-a=18}}1 KB (211 words) - 04:32, 4 November 2022
- == Problem == ...2, \pm 72),( \pm 3, \pm 48),( \pm 4, \pm 36),( \pm 6, \pm 24),( \pm 8, \pm 18),( \pm 9, \pm 16),( \pm 12, \pm 12)</math>; since <math>c</math> is the [[m2 KB (310 words) - 11:25, 13 June 2023
- == Problem == ...s take a value of 7. So, <math>\lfloor r\rfloor \le ...\le \lfloor r+\frac{18}{100}\rfloor \le 7</math> and <math>\lfloor r+\frac{92}{100}\rfloor \ge ...3 KB (447 words) - 17:02, 24 November 2023
- == Problem == <math>a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153</math>5 KB (778 words) - 21:36, 3 December 2022
- == Problem == {{AMC10 box|year=2006|ab=B|num-b=18|num-a=20}}5 KB (873 words) - 15:39, 29 May 2023
- == Problem == ...mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} </math>5 KB (861 words) - 00:53, 25 November 2023
- == Problem == <math> \mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \ma4 KB (558 words) - 14:38, 6 April 2024
- == Problem == Another way to do the problem is by the process of elimination. The only possible correct choices are the5 KB (878 words) - 14:39, 3 December 2023
- == Problem == ...19)\\ &= P_{17}(((x - 20) - 19) - 18)\\ &= \cdots\\ &= P_0(x - (20 + 19 + 18 + \ldots + 2 + 1)).\end{align*}</cmath>2 KB (355 words) - 13:25, 31 December 2018
- == Problem == ...[circumcenter]]s of <math>\triangle PAB, PBC, PCA</math>. According to the problem statement, the circumcenters of the triangles cannot lie within the interio4 KB (717 words) - 22:20, 3 June 2021
- == Problem == ...argest multiple of <math>6</math> that is <math>\le 19</math> is <math>n = 18</math>, which we can easily verify works, and the answer is <math>\frac{13}3 KB (473 words) - 17:06, 1 January 2024
- == Problem == ...ath>78</math> intersect at a point whose distance from the center is <math>18</math>. The two chords divide the interior of the circle into four regions3 KB (484 words) - 13:11, 14 January 2023
- == Problem == Hence, the answer is <math>\frac{18\pi}{15}+\frac{5\pi}{15}=\frac{23\pi}{15}\cdot\frac{180}{\pi}=\boxed{276}</m6 KB (1,022 words) - 20:23, 17 April 2021
- == Problem == ...math> yields <math>x \equiv 141 \cdot 19 \equiv (140+1)(18+1) \equiv 0+140+18+1 \equiv 159 \pmod{180}</math>.4 KB (503 words) - 15:46, 3 August 2022
- == Problem == \frac{5}{18}t^2 - 110t + 110^2 - 51^2 &\le& 0\\4 KB (617 words) - 18:47, 17 July 2022
- ...bsp;by<br/>}}}'''{{{before|[[{{{year}}} AHSME Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]]}}}''' ...nbsp;by<br/>}}}'''{{{after|[[{{{year}}} AHSME Problems/Problem {{{num-a}}}|Problem {{{num-a}}}]]}}}'''4 KB (457 words) - 21:28, 15 August 2014
- == Problem == ...\frac{F_{7}}{F_{8}} \cdot 1000 = \frac{13}{21} \cdot 1000 = 618.\overline{18}</math>2 KB (354 words) - 19:37, 24 September 2023
- == Problem == Though the problem may appear to be quite daunting, it is actually not that difficult. <math>\1 KB (225 words) - 02:20, 16 September 2017
- == Problem == ...itude from <math>P</math> to <math>\triangle ABC</math>. The crux of this problem is the following lemma.7 KB (1,169 words) - 15:28, 13 May 2024
- == Problem == ...s <math>n=1, 9, 10,</math> or <math>18</math>, and the sum is <math>1+9+10+18=\boxed{38}</math>.2 KB (296 words) - 01:18, 29 January 2021
- == Problem == ...h congruent area is <math>621.</math> Therefore, since the height is <math>18,</math> the sum of the bases of each trapezoid must be <math>69.</math>3 KB (423 words) - 11:06, 27 April 2023
- == Problem == This problem just requires a good diagram and strong 3D visualization.3 KB (445 words) - 19:40, 4 July 2013
- == Problem == If we work with the problem for a little bit, we quickly see that there is no direct combinatorics way7 KB (1,011 words) - 20:09, 4 January 2024
- == Problem == ...ive mail and <math>1</math> represent a house that does receive mail. This problem is now asking for the number of <math>19</math>-digit strings of <math>0</m13 KB (2,298 words) - 19:46, 9 July 2020
- == Problem == ...th>-degree arcs can be represented as <math>x + 20</math>, as given in the problem.3 KB (561 words) - 19:25, 27 November 2022
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[2005 AMC 10B Problems/Problem 1]]1 KB (165 words) - 12:40, 14 August 2020
- == Problem == <cmath>18 c_2 \equiv 2 \pmod{31}</cmath>3 KB (493 words) - 13:51, 22 July 2020
- == Problem == <math>0,2,2</math> 188 KB (1,187 words) - 02:40, 28 November 2020
- == Problem == ...ath>\overline{AD}</math> and <math>\overline{CE}</math> have lengths <math>18</math> and <math>27</math>, respectively, and <math>AB=24</math>. Extend <m6 KB (974 words) - 13:01, 29 September 2023
- == Problem == ...ath>19 - n</math> choices for both pairs. This gives <math>\sum_{n = 10}^{18} (19 - n)^2 = \sum_{n = 1}^9 n^2 = 285</math> balanced numbers. Thus, ther4 KB (696 words) - 11:55, 10 September 2023
- == Problem 8== ...d</math>, so only <math>9</math> may work. Hence, the four terms are <math>18,\ 27,\ 36,\ 48</math>, which indeed fits the given conditions. Their sum is5 KB (921 words) - 23:21, 22 January 2023
- == Problem == ...other cases yield non-convex and/or degenerate hexagons, which violate the problem statement.9 KB (1,461 words) - 15:09, 18 August 2023
- == Problem == <cmath>\frac{1}{m_1^2}a^2-[(18/m_1)+12]a+117=0</cmath>7 KB (1,182 words) - 09:56, 7 February 2022
- == Problem == ...23, 14, 5, 21, 12, 3, 19, 10, 1, 17, 8, 24, 15, 6, 22, 13, 4, 20, 11, 27, 18, 9, 0,</math> and then it loops.3 KB (403 words) - 12:10, 9 September 2023
- == Problem == <math>18 > 8 \cdot 3^m</math> which is true for m=0 but fails for higher integers <m3 KB (515 words) - 14:46, 14 February 2021
- == Problem == [[Image:AIME 2002 II Problem 4.gif]]2 KB (268 words) - 07:28, 13 September 2020
- == Problem == ...ath>. The total volume added here is then <math>\Delta P_1 = 4 \cdot \frac 18 = \frac 12</math>.2 KB (380 words) - 00:28, 5 June 2020
- == Problem == ...14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10!}=\frac N{1!18!}</math></center> find the [[floor function|greatest integer]] that is less2 KB (281 words) - 12:09, 5 April 2024
- == Problem == ...is <math>(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)</math>. If a number has <math>18 = 2 \cdot 3 \cdot 3</math> factors, then it can have at most <math>3</math>2 KB (397 words) - 15:55, 11 May 2022
- == Problem == {{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}}2 KB (324 words) - 15:30, 16 December 2021
- ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #3]] and [[2006 AMC 10A Problems/Problem 3|2006 AMC 10A #3]]}} == Problem ==1 KB (155 words) - 17:30, 16 December 2021
- ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}} == Problem ==3 KB (528 words) - 18:29, 7 May 2024
- == Problem == ...rm{(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
- **[[2007 iTest Problems/Problem 1|Problem 1]] **[[2007 iTest Problems/Problem 2|Problem 2]]3 KB (305 words) - 15:10, 5 November 2023
- == Problem 1 == [[2006 AMC 10B Problems/Problem 1|Solution]]14 KB (2,059 words) - 01:17, 30 January 2024
- == Problem 1 == [[2005 AMC 10B Problems/Problem 1|Solution]]12 KB (1,874 words) - 21:20, 23 December 2020
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AMC 10A Problems/Problem 1]]2 KB (182 words) - 18:09, 6 October 2014
- ==Problem== draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);8 KB (1,202 words) - 16:17, 10 May 2024
- ==Problem== Two different prime numbers between <math>4</math> and <math>18</math> are chosen. When their sum is subtracted from their product, which o1 KB (228 words) - 19:31, 29 April 2024
- == Problem == ...th>a=7,</math> <math>b^2+b+1=7^3.</math> Solving this quadratic, <math>b = 18 </math>.713 bytes (114 words) - 01:45, 19 August 2012
- == Problem 1 == [[University of South Carolina High School Math Contest/1993 Exam/Problem 1|Solution]]14 KB (2,102 words) - 22:03, 26 October 2018
- ...>}}}'''{{{before|[[{{{year}}} AMC 12{{{ab|}}} Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]]}}}''' .../>}}}'''{{{after|[[{{{year}}} AMC 12{{{ab|}}} Problems/Problem {{{num-a}}}|Problem {{{num-a}}}]]}}}'''3 KB (429 words) - 09:15, 23 November 2021
- == Problem 13 == [[1987 AIME Problems/Problem 13|Solution]]22 KB (3,778 words) - 19:29, 2 June 2020
- ...an 9 balls. There are <math>{12 + 7 - 1 \choose 7 - 1} = {18 \choose 6} = 18,564</math> ways to place 12 objects into 7 boxes. Of these, 7 place all 12 *[[Mock AIME 1 2006-2007 Problems/Problem 1 | Previous Problem]]1 KB (188 words) - 15:53, 3 April 2012
- == Problem == .../math>. The [[divisor | factors]] of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72 and 144 itself. As both <math>x_1</math> and <math>x_2</ma3 KB (470 words) - 00:33, 10 August 2019
- == Problem == Thus <math>[ABCD]=\frac{25\sqrt{143}}{18}\rightarrow\boxed{186}</math>2 KB (311 words) - 10:53, 4 April 2012
- ==Problem== {{AMC10 box|year=2005|ab=A|num-b=16|num-a=18}}2 KB (266 words) - 03:36, 16 January 2023
- * [[2006_AMC_12A_Problems/Problem_18 | 2006 AMC 12A Problem 18]]363 bytes (53 words) - 14:47, 20 July 2016
- ===Problem 1=== [[2007 iTest Problems/Problem 1|Solution]]30 KB (4,794 words) - 23:00, 8 May 2024
- *[[2006 iTest Problems/Problem 1|Problem 1]] *[[2006 iTest Problems/Problem 2|Problem 2]]3 KB (320 words) - 09:56, 23 April 2024
- == Problem 1 == [[2005 AMC 10A Problems/Problem 1|Solution]]14 KB (2,026 words) - 11:45, 12 July 2021
- ==Problem== ...op of the <math>2^{\text{nd}}</math>, it will be a layer of <math>3\times6=18</math> oranges, etc.1 KB (180 words) - 01:14, 12 April 2022
- ==Problem== ...h>\left(\frac12\right)^3{3\choose2}\left(\frac12\right)^4{4\choose2}=\frac{18}{128}</math>2 KB (221 words) - 21:49, 15 April 2024
- ==Problem== ...\qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ } 24 </math>1 KB (167 words) - 12:10, 17 August 2021
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2003 AMC 10A Problems/Problem 1]]1 KB (165 words) - 18:48, 6 October 2014
- ==Problem 1== [[2003 AMC 10A Problems/Problem 1|Solution]]13 KB (1,900 words) - 22:27, 6 January 2021
- == Problem == <math> \mathrm{(A) \ } 12\qquad \mathrm{(B) \ } 15\qquad \mathrm{(C) \ } 18\qquad \mathrm{(D) \ } 21\qquad \mathrm{(E) \ } 24 </math>2 KB (336 words) - 15:49, 19 August 2023
- == Problem == ...ac{17}{50}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{18}{25} </math>2 KB (261 words) - 14:34, 17 August 2023
- == Problem == {{AMC10 box|year=2003|ab=A|num-b=16|num-a=18}}2 KB (414 words) - 13:48, 4 April 2024
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AMC 10A Problems/Problem 1]]2 KB (182 words) - 01:29, 7 October 2014
- == Problem == When there are 5 equilateral triangles in the base, you need <math>18</math> toothpicks in all.5 KB (686 words) - 18:01, 28 January 2021
- ==Problem 1 == [[2004 AMC 10A Problems/Problem 1|Solution]]15 KB (2,092 words) - 20:32, 15 April 2024
- ...ate|[[2004 AMC 12A Problems|2004 AMC 12A #15]] and [[2004 AMC 10A Problems/Problem 17|2004 AMC 10A #17]]}} ==Problem==2 KB (309 words) - 22:27, 15 August 2023
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2001 AMC 8 Problems/Problem 1]]1 KB (138 words) - 10:26, 22 August 2013
- ==Problem 1== [[2001 AMC 8 Problems/Problem 1 | Solution]]13 KB (1,994 words) - 13:04, 18 February 2024
- == Problem == ...qquad \mathrm{(B) \ } 9\%\qquad \mathrm{(C) \ } 12\%\qquad \mathrm{(D) \ } 18\%\qquad \mathrm{(E) \ } 24\% </math>1 KB (183 words) - 15:36, 19 August 2023
- == Problem == ...{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}</math>. We have thus solved the problem.3 KB (380 words) - 21:53, 19 March 2022
- == Problem == We expand to get <math>2x^2-8x+3x-12+2x^2-12x+3x-18=0</math> which is <math>4x^2-14x-30=0</math> after combining like terms. Us979 bytes (148 words) - 13:06, 8 November 2021
- ==Problem== ...+ \sqrt{4R^2 - 4\cdot24^2}}{2} = \frac{43}{\sqrt 3} + \frac{11}{\sqrt3} = 18\sqrt{3}</math>. (We only take the positive sign because [[angle]] <math>B<3 KB (532 words) - 20:29, 31 August 2020
- ==Problem 1== [[2020 AMC 10A Problems/Problem 1|Solution]]13 KB (1,968 words) - 18:32, 29 February 2024
- ...>}}}'''{{{before|[[{{{year}}} AMC 10{{{ab|}}} Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]]}}}''' .../>}}}'''{{{after|[[{{{year}}} AMC 10{{{ab|}}} Problems/Problem {{{num-a}}}|Problem {{{num-a}}}]]}}}'''3 KB (429 words) - 17:50, 22 November 2021
- **[[2021 Fall AMC 10B Problems/Problem 1|Problem 1]] **[[2021 Fall AMC 10B Problems/Problem 2|Problem 2]]2 KB (205 words) - 10:53, 1 December 2021
- ==Problem== ...r_1=\frac{|AD| \times |DP| \times |AP|}{4A_1}=\frac{(3)(4)(6)}{4A_1}=\frac{18}{A_1}</math>3 KB (563 words) - 02:05, 25 November 2023
- == Problem == ...he result, we find that <math>(81 + 18bi - b^2)(9 + bi) = \ldots + (9\cdot 18 + 81)bi - b^3i</math>.1,003 bytes (163 words) - 15:34, 18 February 2017
- == Problem == ...9</math>. Therefore, it must round to <math>c</math> because <math>\frac 5{18}<\frac 12</math> so <math>c</math> is the closest integer. Therefore there7 KB (1,076 words) - 00:10, 29 November 2023
- == Problem == ...Exclusion]] based on the stipulation that <math>j\ne k</math> to solve the problem:13 KB (2,328 words) - 00:12, 29 November 2023
- == Problem == ..., <math>(15,18,21,26)</math>, <math>(15,18,21,24,26)</math>, and <math>(15,18,21,24)</math>. That is a total of 6.10 KB (1,519 words) - 00:11, 29 November 2023
- == Problem == ...sqrt{28}</math>. The area of <math>\triangle{PHI}=\frac{1}{2}\cdot\sqrt{28-18}\cdot6\sqrt{2}=6\sqrt{5}</math>7 KB (1,034 words) - 21:56, 22 September 2022
- == Problem == ...math> if <math>x + y = n</math>. Thus, there are <math>0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = \boxed{200}</math> solutions of <math>(a,b,c)</math>.1 KB (228 words) - 08:41, 4 November 2022
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2007 AMC 12A Problems/Problem 1]]1 KB (168 words) - 21:50, 6 October 2014
- ==Problem 1== [[2007 AMC 12A Problems/Problem 1 | Solution]]11 KB (1,750 words) - 13:35, 15 April 2022
- a summer camp, a problem-solving strategy, etc.) will be selected. :''September 12 - September 18''11 KB (1,427 words) - 16:25, 9 May 2021
- ...Azjps|Azjps]] ([[User talk:Azjps|<font color="green">talk</font>]])</font> 18:19, 11 April 2007 (EDT) ...ps|Azjps]] ([[User talk:Azjps|<font color="green">talk</font>]])</font> 15:18, 13 April 2007 (EDT)13 KB (2,027 words) - 10:07, 24 August 2008
- == Problem == ..., and 6 apples cost as much as 4 oranges. How many oranges cost as much as 18 bananas?597 bytes (89 words) - 16:33, 15 February 2021
- == Problem == <math> pq + pr + ps = p(9-p) \ge 3(9-3) = 18 </math>.3 KB (542 words) - 17:09, 19 December 2018
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1951 AHSME Problems/Problem 1|Problem 1]]3 KB (258 words) - 14:25, 20 February 2020
- ...n McNugget Theorem''' (or '''Postage Stamp Problem''' or '''Frobenius Coin Problem''') states that for any two [[relatively prime]] [[positive integer]]s <mat ...t Theorem has also been called the Frobenius Coin Problem or the Frobenius Problem, after German mathematician Ferdinand Frobenius inquired about the largest17 KB (2,748 words) - 19:22, 24 February 2024
- ...ate|[[2007 AMC 12A Problems|2007 AMC 12A #10]] and [[2007 AMC 10A Problems/Problem 14|2007 AMC 10A #14]]}} ==Problem==2 KB (231 words) - 14:02, 3 June 2021
- ...ate|[[2007 AMC 12A Problems|2007 AMC 12A #12]] and [[2007 AMC 10A Problems/Problem 16|2007 AMC 10A #16]]}} ==Problem==3 KB (445 words) - 08:59, 24 March 2023
- <b>Problem: </b> ...e add an even and odd. We can use complementary counting to help solve the problem. There are a total of <math>90</math> possibilities since Jack can choose <4 KB (694 words) - 22:00, 12 January 2024
- ==Problem== ...mouse is at <math>(4,-2)</math> and is running up the [[line]] <math>y=-5x+18</math>. At the point <math>(a,b)</math> the mouse starts getting farther fr2 KB (387 words) - 18:20, 27 November 2023
- ...ate|[[2007 AMC 12A Problems|2007 AMC 12A #22]] and [[2007 AMC 10A Problems/Problem 25|2007 AMC 10A #25]]}} == Problem ==15 KB (2,558 words) - 19:33, 4 February 2024
- == Problem == {{AMC12 box|year=2007|ab=A|num-b=16|num-a=18}}1,022 bytes (153 words) - 14:56, 7 August 2017
- == Problem == ...t 2}{223} = 18</math>. Thus <math>A</math> lies on the lines <math>y = \pm 18 \quad \mathrm{(1)}</math>.4 KB (565 words) - 17:01, 2 April 2023
- == Problem == We can simply apply casework to this problem.4 KB (536 words) - 21:18, 22 May 2023
- == Problem == {{AMC12 box|year=2005|num-b=16|num-a=18|ab=A}}2 KB (215 words) - 13:56, 19 January 2021
- == Problem == ...t\rfloor +\left\lfloor\frac{200}{9}\right\rfloor - \left\lfloor \frac{200}{18}\right\rfloor +\left\lfloor \frac{200}{27}\right\rfloor - \left\lfloor \fra4 KB (562 words) - 18:37, 30 October 2020
- == Problem == ...es. Thus, thus total possibilities for <math>(a,b)</math> is <math>576+144+18=738</math>.7 KB (1,114 words) - 03:41, 12 September 2021
- == Problem I1 == [[2005 PMWC Problems/Problem I1|Solution]]9 KB (1,449 words) - 20:49, 2 October 2020
- == Problem I1 == [[1998 PMWC Problems/Problem I1|Solution]]11 KB (1,738 words) - 19:25, 10 March 2015
- == Problem I1 == [[1999 PMWC Problems/Problem I1|Solution]]6 KB (703 words) - 21:21, 21 April 2014
- ==Problem== <math>4</math> | <math>18</math>1 KB (203 words) - 19:14, 7 April 2016
- ==Problem== {{CYMO box|year=2006|l=Lyceum|num-b=16|num-a=18}}789 bytes (123 words) - 22:00, 30 November 2015
- ==Problem== {{CYMO box|year=2006|l=Lyceum|num-b=18|num-a=20}}1 KB (214 words) - 23:44, 22 December 2016
- ==Problem== fill((-30,0)..(-24,18)--(0,0)--(-24,-18)..cycle,gray(0.7));3 KB (476 words) - 03:50, 23 January 2023
- ...ate|[[2004 AMC 12A Problems|2004 AMC 12A #18]] and [[2004 AMC 10A Problems/Problem 22|2004 AMC 10A #22]]}} == Problem ==5 KB (738 words) - 13:11, 27 March 2023
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2007 AMC 12B Problems/Problem 1]]1 KB (168 words) - 21:22, 6 October 2014
- ==Problem 1== [[2007 AMC 12B Problems/Problem 1 | Solution]]12 KB (1,814 words) - 12:58, 19 February 2020
- == Problem 1 == [[1999 AHSME Problems/Problem 1|Solution]]13 KB (1,945 words) - 18:28, 19 June 2023
- ...ms|2004 AMC 12A #14]] and [[2004 AMC 10A Problems/Problem 18|2004 AMC 10A #18]]}} == Problem ==4 KB (689 words) - 03:35, 16 January 2023
- ...ate|[[2004 AMC 12A Problems|2004 AMC 12A #17]] and [[2004 AMC 10A Problems/Problem 24|2004 AMC 10A #24]]}} == Problem ==2 KB (233 words) - 08:14, 6 September 2021
- == Problem == ...>c < \frac 12</math>. Thus the chance is <math>\frac{\frac{1}{4}}2 = \frac 18</math>.3 KB (552 words) - 23:26, 28 December 2020
- *[[2005 iTest Problems/Problem 1|Problem 1]] *[[2005 iTest Problems/Problem 2|Problem 2]]3 KB (283 words) - 02:37, 24 January 2024
- ... by:<br/>'''{{{before|[[{{{year}}} iTest Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]]}}}''' ... by:<br/>'''{{{after|[[{{{year}}} iTest Problems/Problem {{{num-a}}}|Problem {{{num-a}}}]]}}}'''4 KB (470 words) - 19:57, 5 November 2023
- ...bsp;by<br/>}}}'''{{{before|[[{{{year}}} AMC 8 Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]]}}}''' ...nbsp;by<br/>}}}'''{{{after|[[{{{year}}} AMC 8 Problems/Problem {{{num-a}}}|Problem {{{num-a}}}]]}}}'''2 KB (243 words) - 18:01, 29 May 2011
- == Problem == ...2^1</math>, <math>(6-2)2! = 8 > 4 = 2^2</math>, and <math>(6-3) \cdot 3! = 18 > 8 = 2^3</math>. For the other integers less less than or equal to six, w5 KB (919 words) - 23:29, 20 January 2016
- == Problem == ...the first row is numbered <math>1,2,\ldots,17</math>, the second row <math>18,19,\ldots,34</math>, and so on down the board. If the board is renumbered s2 KB (310 words) - 11:28, 3 August 2021
- == Problem == ...trig to guess and check: the only trig facts we need to know to finish the problem is:6 KB (979 words) - 12:50, 17 July 2022
- {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #18]] and [[2000 AMC 10 Problems|2000 AMC 10 #25]]}} == Problem ==2 KB (409 words) - 19:33, 25 May 2024
- == Problem == {{AMC12 box|year=2000|num-b=18|num-a=20}}3 KB (547 words) - 17:37, 17 February 2024
- == Problem == {{AMC10 box|year=2000|num-b=18|num-a=20}}5 KB (804 words) - 01:22, 13 May 2024
- == Problem == ...s <math>36/{\pi}-r</math>. By the Pythagorean Theorem, you get <math>r^2+(18/{\pi})^2=(36/{\pi}-r)^2 \implies r=27/2{\pi}</math>. Finally, we see that t2 KB (263 words) - 19:59, 18 April 2024
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2007 AMC 10A Problems/Problem 1]]2 KB (182 words) - 03:21, 31 December 2019
- == Problem == {{AMC10 box|year=2007|ab=A|num-b=16|num-a=18}}1 KB (201 words) - 08:04, 11 February 2023
- == Problem 1 == [[2007 AMC 10A Problems/Problem 1|Solution]]13 KB (2,058 words) - 17:54, 29 March 2024
- == Problem == <math>\text{(A)}\ 14 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 24</math>1 KB (195 words) - 20:44, 21 January 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1995 AHSME Problems/Problem 1|Problem 1]]2 KB (174 words) - 00:30, 2 October 2014
- == Problem 1 == [[1995 AHSME Problems/Problem 1|Solution]]17 KB (2,387 words) - 22:44, 26 May 2021
- == Problem == ...of five positive integers has [[mean]] <math>12</math> and [[range]] <math>18</math>. The [[mode]] and [[median]] are both <math>8</math>. How many diffe1 KB (180 words) - 20:59, 10 February 2019
- ==Problem== <math> \mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 18 } \qquad \mathrm{(C) \ 24 } \qquad \mathrm{(D) \ 36 } \qquad \mathrm{(E) \2 KB (233 words) - 10:37, 30 March 2023
- == Problem I1 == [[1997 PMWC Problems/Problem I1|Solution]]15 KB (2,057 words) - 19:13, 10 March 2015
- == Problem == ...ath>70 \text{ km/h}</math>. It traveled <math>3</math> rounds within <math>18</math> hours. What is the distance between <math>A</math> and <math>B</math857 bytes (149 words) - 13:13, 21 January 2019
- == Problem 1 == [[1951 AHSME Problems/Problem 1|Solution]]23 KB (3,641 words) - 22:23, 3 November 2023
- == Problem == ...})\ 2x^2 \qquad (\mathrm{C})\ \frac{2x^2}9 \qquad (\mathrm{D})\ \frac{x^2}{18} \qquad (\mathrm{E})\ \frac{x^2}{72}</math>743 bytes (121 words) - 12:19, 5 July 2013
- x^2 + 18 &= 43 \\ ...e right side to maintain equality. The right hand side becomes <math>43 - 18 = 25</math>.4 KB (562 words) - 18:49, 8 November 2020
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1959 AHSME Problems/Problem 1|Problem 1]]3 KB (257 words) - 14:19, 20 February 2020
- == Problem 1== [[1959 AHSME Problems/Problem 1|Solution]]22 KB (3,345 words) - 20:12, 15 February 2023
- == Problem 1 == <math>\text{(A)} \ \frac 18 \qquad \text{(B)} \ \frac 73 \qquad \text{(C)} \ \frac78 \qquad \text{(D)}19 KB (3,159 words) - 22:10, 11 March 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1966 AHSME Problems/Problem 1|Problem 1]]2 KB (217 words) - 14:15, 20 February 2020
- == Problem == <math>\text{(A)} \ \frac 18 \qquad \text{(B)} \ \frac 73 \qquad \text{(C)} \ \frac78 \qquad \text{(D)}712 bytes (99 words) - 12:39, 5 July 2013
- == Problem == {{AMC12 box|year=2002|ab=B|num-b=18|num-a=20}}852 bytes (119 words) - 10:22, 4 July 2013
- ...12B Problems|2002 AMC 12B #14]] and [[2002 AMC 10B Problems|2002 AMC 10B #18]]}} == Problem ==2 KB (282 words) - 14:04, 12 July 2021
- == Problem == [[Image:2002_12B_AMC-18.png]]3 KB (376 words) - 19:16, 20 August 2019
- == Problem == The sum of <math>18</math> consecutive positive integers is a [[perfect square]]. The smallest2 KB (261 words) - 23:34, 18 March 2023
- == Problem == ...thrm{(D)}\ 4}</math> solutions (respectively yielding <math>n = 0, 10, 16, 18</math>).4 KB (579 words) - 05:54, 17 October 2023
- ==Problem 1== [[2007 AMC 8 Problems/Problem 1|Solution]]12 KB (1,800 words) - 20:01, 8 May 2023
- == Problem == We rewrite the logarithms in the problem. <cmath>\log(x) + 3\log(y) = 1</cmath> <cmath>2\log(x) + \log(y) = 1</cmath2 KB (329 words) - 13:49, 4 April 2024
- ==Problem== <cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath>970 bytes (134 words) - 00:09, 14 September 2015
- ==Problem== {{AMC10 box|year=2004|ab=A|num-b=18|num-a=20}}2 KB (220 words) - 14:19, 21 April 2021
- == Problem == {{AMC10 box|year=2007|ab=A|num-b=18|num-a=20}}4 KB (560 words) - 14:57, 3 June 2021
- == Problem == ...elements. Hence the answer is <math>\frac{3 \cdot 3!}{4 \cdot 4!} = \frac {18}{96} = \frac{3}{16}</math>, and <math>a+b=19 \Rightarrow \mathrm{(E)}</math2 KB (320 words) - 22:59, 5 May 2024
- == Problem == {{AHSME box|year=1998|num-b=16|num-a=18}}656 bytes (94 words) - 22:16, 28 March 2024
- == Problem == \qquad\mathrm{(E)}\ 18</math>2 KB (302 words) - 19:59, 3 July 2013
- == Problem == A truncated cone has horizontal bases with radii <math>18</math> and <math>2</math>. A sphere is tangent to the top, bottom, and late3 KB (520 words) - 19:12, 20 November 2023
- == Problem == {{AMC12 box|year=2004|ab=B|num-b=16|num-a=18}}936 bytes (144 words) - 02:47, 13 July 2021
- ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2008 AMC 12A Problems/Problem 1|Problem 1]]2 KB (193 words) - 21:49, 6 October 2014
- ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2008 AMC 12B Problems/Problem 1|Problem 1]]2 KB (193 words) - 00:11, 7 October 2014
- ...The second link contains the answer key. The rest contain each individual problem and its solution. **[[2008 AMC 10A Problems/Problem 1|Problem 1]]2 KB (195 words) - 18:08, 28 June 2021
- ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2008 AMC 10B Problems/Problem 1|Problem 1]]2 KB (188 words) - 18:44, 6 October 2014
- == Problem == <math>\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20</math>3 KB (430 words) - 23:13, 13 September 2023
- ==Problem 1== [[2008 AMC 10A Problems/Problem 1|Solution]]14 KB (2,138 words) - 15:08, 18 February 2023
- ==Problem 1== [[2008 AMC 12A Problems/Problem 1|Solution]]13 KB (2,025 words) - 13:56, 2 February 2021
- ==Problem== {{AMC12 box|year=2008|ab=A|num-b=18|num-a=20}}5 KB (895 words) - 22:54, 9 January 2021
- ==Problem== {{AMC12 box|year=2008|ab=A|num-b=16|num-a=18}}2 KB (386 words) - 13:52, 21 December 2020
- == Problem == {{AMC12 box|year=2002|ab=B|num-b=16|num-a=18}}2 KB (380 words) - 15:00, 17 August 2023
- ==Problem== ...KB}{KD} = \frac{3}{4}</math>, so <math>[\triangle AKB] = \frac{3}{4}(24) = 18</math>. Similarly, we find <math>[\triangle DKC] = \frac{4}{3}(24) = 32</ma2 KB (294 words) - 21:53, 17 October 2023
- ==Problem== <cmath>[ACD] = [ABC] - [BCD] = \frac 12 (3)(4) - \frac{24}{7} = \frac{18}{7}</cmath>-->6 KB (951 words) - 16:31, 2 August 2019
- ...dex.php/2007_AMC_10B_Problems/Problem_25#Solution| 2007 AMC 10B #25] (same problem) == Problem ==9 KB (1,522 words) - 22:46, 12 May 2022
- == Problem == What is the area of the region defined by the [[inequality]] <math>|3x-18|+|2y+7|\le3</math>?902 bytes (125 words) - 10:28, 13 March 2016
- ==Problem== ...}\ 10 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 21</math>5 KB (758 words) - 16:35, 15 February 2021
- ==Problem== ...three smaller triangles. The altitudes of these triangles are given in the problem as <math>PQ</math>, <math>PR</math>, and <math>PS</math>.3 KB (401 words) - 22:58, 8 May 2023
- ==Problem== {{AMC12 box|year=2007|ab=B|num-b=16|num-a=18}}1 KB (242 words) - 03:05, 19 May 2024
- ==Problem== \textbf{(C)}\ \frac{\sqrt3}{18} \qquad2 KB (328 words) - 10:54, 4 July 2013
- ==Problem== {{AMC12 box|year=2007|ab=B|num-b=18|num-a=20}}1 KB (166 words) - 16:35, 15 February 2021
- ==Problem== <math>(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).</math>4 KB (725 words) - 19:59, 4 January 2024
- ==Problem== <math>\mathrm{(A)}\ 15 \qquad \mathrm{(B)}\ 18 \qquad \mathrm{(C)}\ 27 \qquad \mathrm{(D)}\ 54 \qquad \mathrm{(E)}\ 81</ma3 KB (463 words) - 16:35, 15 February 2021
- == Problem 19 == {{AMC12 box|year=2004|ab=A|num-b=18|num-a=20}}5 KB (785 words) - 00:29, 31 July 2023
- ==Problem 1== [[2008 AMC 10B Problems/Problem 1|Solution]]12 KB (1,838 words) - 16:52, 7 October 2022
- ==Problem 1== [[2008 AMC 12B Problems/Problem 1|Solution]]14 KB (2,199 words) - 13:43, 28 August 2020
- ...cate|[[2008 AMC 12B Problems|2008 AMC 12B #2]] and [[2008 AMC 10B Problems/Problem 2|2008 AMC 10B #2]]}} ==Problem==2 KB (219 words) - 23:53, 31 July 2023
- ==Problem== ...ac {2}{9} \qquad \textbf{(B)}\ \frac {1}{4} \qquad \textbf{(C)}\ \frac {5}{18} \qquad \textbf{(D)}\ \frac {7}{24} \qquad \textbf{(E)}\ \frac {3}{10}</mat1 KB (183 words) - 22:35, 10 June 2017
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2008 AIME I Problems/Problem 1]]1 KB (140 words) - 16:18, 17 October 2020
- ==Problem== {{AMC12 box|year=2008|ab=B|num-b=18|num-a=20}}2 KB (307 words) - 19:28, 26 September 2023
- == Problem 1 == [[1985 AJHSME Problem 1 | Solution]]12 KB (1,670 words) - 17:42, 24 November 2021
- ...he second link contains the answer key. The rest contain each individual problem and its solution. ** [[1985 AJHSME Problems/Problem 1|Problem 1]]2 KB (175 words) - 16:47, 12 April 2012
- ...bsp;by<br/>}}}'''{{{before|[[{{{year}}} AMC 8 Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]]}}}''' ...nbsp;by<br/>}}}'''{{{after|[[{{{year}}} AMC 8 Problems/Problem {{{num-a}}}|Problem {{{num-a}}}]]}}}'''3 KB (384 words) - 21:09, 15 August 2014
- == Problem 1 == [[2008 AIME I Problems/Problem 1|Solution]]9 KB (1,536 words) - 00:46, 26 August 2023
- == Problem == Indeed, by solving, we find <math>(x,y) = (18,62)</math> is the unique solution.4 KB (732 words) - 22:17, 28 November 2023
- == Problem == ...from <math>A</math> to line <math>CT</math>. Suppose <math>\overline{AB} = 18</math>, and let <math>m</math> denote the maximum possible length of segmen8 KB (1,333 words) - 00:18, 1 February 2024
- == Problem 1 == [[Mock AIME 1 2007-2008 Problems/Problem 1|Solution]]6 KB (992 words) - 14:15, 13 February 2018
- == Problem== *[http://usamts.org/Solutions/Solution2_2_18.pdf USAMTS Year 18 Problem 2]8 KB (1,338 words) - 23:15, 28 November 2023
- == Problem == <cmath>\sqrt{18},\ \sqrt{13},\ \sqrt{10},\ \sqrt{9},\ \sqrt{8},\ \sqrt{5},\ \sqrt{4},\ \sqr4 KB (569 words) - 09:44, 25 November 2019
- == Problem == The well known problem of ordering <math>x</math> elements of a string of <math>y</math> elements10 KB (1,550 words) - 12:58, 15 July 2023
- ==Problem== {{AHSME box|year=1995|num-b=18|num-a=20}}1 KB (200 words) - 20:58, 10 February 2019
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1952 AHSME Problems/Problem 1|Problem 1]]3 KB (257 words) - 14:25, 20 February 2020
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1953 AHSME Problems/Problem 1|Problem 1]]3 KB (257 words) - 14:24, 20 February 2020
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1954 AHSME Problems/Problem 1|Problem 1]]3 KB (257 words) - 14:23, 20 February 2020
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1955 AHSME Problems/Problem 1|Problem 1]]3 KB (257 words) - 14:23, 20 February 2020
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1956 AHSME Problems/Problem 1|Problem 1]]3 KB (257 words) - 14:22, 20 February 2020
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1957 AHSME Problems/Problem 1|Problem 1]]3 KB (257 words) - 14:21, 20 February 2020
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1958 AHSME Problems/Problem 1|Problem 1]]3 KB (257 words) - 14:20, 20 February 2020
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1960 AHSME Problems/Problem 1|Problem 1]]2 KB (217 words) - 14:19, 20 February 2020
- ==Problem== ...quation by 2 and substituting into the first equation gives <math>128=5u_4+18 \Longrightarrow u_4=22</math>, therefore <math>u_5=\frac{128-22}{2}=53 \lon913 bytes (135 words) - 19:42, 24 October 2022
- ==Problem== In more concise terms, this problem is an extension of the binomial distribution. We find the number of ways on1 KB (202 words) - 12:42, 30 January 2024
- ==Problem== {{AMC10 box|year=2008|ab=B|num-b=18|num-a=20}}2 KB (373 words) - 13:21, 7 June 2021
- ==Problem== <math>\mathrm{(A)}\ 5/18\qquad\mathrm{(B)}\ 7/18\qquad\mathrm{(C)}\ 11/18\qquad\mathrm{(D)}\ 3/4\qquad\mathrm{(E)}\ 8/9</math>4 KB (572 words) - 17:44, 14 June 2023
- ==Problem== ...athrm{(A)}\ 36+24\sqrt{3}\qquad\mathrm{(B)}\ 54+9\pi\qquad\mathrm{(C)}\ 54+18\sqrt{3}+6\pi\qquad\mathrm{(D)}\ \left(2\sqrt{3}+3\right)^2\pi\\\mathrm{(E)}2 KB (348 words) - 19:59, 4 June 2021
- ==Problem== {{AMC10 box|year=2008|ab=A|num-b=18|num-a=20}}2 KB (321 words) - 17:52, 7 November 2021
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1961 AHSME Problems/Problem 1|Problem 1]]2 KB (218 words) - 14:18, 20 February 2020
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1962 AHSME Problems/Problem 1|Problem 1]]2 KB (218 words) - 14:17, 20 February 2020
- == Problem == {{AMC12 box|year=2003|ab=A|num-b=16|num-a=18}}6 KB (1,026 words) - 22:35, 29 March 2023
- ==Problem== ...} \qquad \mathrm{(C) \ 14 } \qquad \mathrm{(D) \ 16 } \qquad \mathrm{(E) \ 18 } </math>760 bytes (100 words) - 13:59, 5 July 2013
- ...e also many books and online handouts/lectures you can use to improve your problem-solving skills. Depending on your current abilities, you will want to star The '''Art of Problem Solving books''' are an excellent resource to help prepare for math contest13 KB (1,926 words) - 11:22, 30 November 2023
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1967 AHSME Problems/Problem 1|Problem 1]]2 KB (216 words) - 14:14, 20 February 2020
- == Problem == {{AHSME box|year=1999|num-b=18|num-a=20}}1 KB (173 words) - 14:35, 5 July 2013
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1999 AHSME Problems/Problem 1|Problem 1]]2 KB (178 words) - 13:28, 20 February 2020
- == Problem 1 == [[1998 AHSME Problems/Problem 1|Solution]]15 KB (2,222 words) - 10:40, 11 August 2020
- == Problem == <math> \mathrm{(A) \ }10 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }22 \qquad \mathrm{(E) \ } 26</math>4 KB (662 words) - 00:51, 3 October 2023
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1998 AHSME Problems/Problem 1|Problem 1]]2 KB (173 words) - 03:44, 29 September 2014
- == Problem == ...\left(\frac {1}{3} + \frac {8a + 6}{6}\right) = 1 + \frac {8(a + b + c) + 18}{6} = 8</math>, as desired.6 KB (1,063 words) - 02:36, 9 August 2023
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1963 AHSME Problems/Problem 1|Problem 1]]2 KB (217 words) - 14:17, 20 February 2020
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1989 AHSME Problems/Problem 1|Problem 1]]2 KB (174 words) - 00:32, 2 October 2014
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1990 AHSME Problems/Problem 1|Problem 1]]2 KB (174 words) - 00:32, 2 October 2014
- == Problem == <cmath>\begin{array}{ccccc}1&3&9&27&81\\2&6&18&54\\4&12&36\\5&15&45\\7&21&63\\8&24&72\\10&30&90\\11&33&99\\13&39\\\vdots&\2 KB (285 words) - 19:25, 25 September 2020
- ==Problem== a_8 + a_{10} & = 18, &&(4) \\3 KB (402 words) - 23:17, 23 September 2023
- == Problem == 4 & 9 & 14 & 17 & 17 & 17 & 18 & 19 \\5 KB (841 words) - 17:19, 5 May 2022
- *[[2008 iTest Problems/Problem 1|Problem 1]] *[[2008 iTest Problems/Problem 2|Problem 2]]5 KB (424 words) - 15:18, 24 January 2020
- ==Problem 1== [[2008 iTest Problems/Problem 1|Solution]]71 KB (11,749 words) - 01:31, 2 November 2023
- == Problem == Then <math>xy = 2^{15} \cdot 3^{17} \cdot 7</math> has <math>16 \cdot 18 \cdot 2 = \boxed{576}</math> factors.4 KB (685 words) - 14:39, 7 October 2017
- ==Problem== draw(rotate(18)*polygon(5));1 KB (201 words) - 16:29, 12 March 2024
- ==Problem== <math> \mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 } </math>5 KB (828 words) - 05:52, 30 September 2023
- ==Problem== ...y=ax+d</math>, <math>y=bx+c</math>, and <math>y=bx+d</math> has area <math>18</math>. The parallelogram bounded by the lines <math>y=ax+c</math>, <math>y7 KB (1,143 words) - 21:25, 20 December 2020
- ==Problem== <math>8+10=18</math>2 KB (242 words) - 19:53, 31 October 2016
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2007 AMC 10B Problems/Problem 1]]2 KB (182 words) - 18:52, 6 October 2014
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2007 AMC 8 Problems/Problem 1]]1 KB (127 words) - 21:46, 25 November 2013
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1980 AHSME Problems/Problem 1|Problem 1]]2 KB (174 words) - 22:29, 1 October 2014
- ==Problem 1== [[2005 AMC 8 Problems/Problem 1|Solution]]13 KB (1,821 words) - 22:18, 5 December 2023
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2005 AMC 8 Problems/Problem 1]]1 KB (127 words) - 21:47, 25 November 2013
- == Problem == ...rac{1}{6}\qquad \mathrm{(C) \ } \frac{2}{9}\qquad \mathrm{(D) \ } \frac{5}{18}\qquad \mathrm{(E) \ } \frac{5}{12} </math>1 KB (213 words) - 10:16, 4 July 2013
- == Problem== ...1{80} \qquad \mathrm{(B) \ } \frac 1{40} \qquad \mathrm{(C) \ } \frac 1{18} \qquad \mathrm{(D) \ } \frac 1{9} \qquad \mathrm{(E) \ } \frac 9{80} </m1 KB (201 words) - 14:36, 13 February 2019
- ...ink contains the answers to each problem. The rest contain each individual problem and its solution. * [[2002 AMC 10A Problems/Problem 1]]1 KB (169 words) - 19:05, 26 November 2019
- ==Problem 1== [[2002 AMC 10A Problems/Problem 1|Solution]]11 KB (1,733 words) - 11:04, 12 October 2021
- ==Problem 1== [[2002 AMC 10B Problems/Problem 1|Solution]]10 KB (1,540 words) - 22:53, 19 December 2023
- ==Problem== {{AMC10 box|ab=A|year=2002|num-b=18|num-a=20}}1 KB (204 words) - 20:45, 28 May 2023
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2002 AMC 10B Problems/Problem 1]]1 KB (157 words) - 00:51, 3 February 2015
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AMC 10B Problems/Problem 1]]2 KB (180 words) - 01:30, 7 October 2014
- ==Problem 1== [[2004 AMC 10B Problems/Problem 1|Solution]]13 KB (1,988 words) - 20:19, 15 May 2024
- ==Problem== ...the positive integers <math>a,b,\ldots, g</math>. For example, <math>(3,6,18)=3</math> and <math>[6,15]=30</math>. Prove that5 KB (1,018 words) - 11:14, 6 October 2023
- ==Problem 1== ...the positive integers <math>a,b,\ldots, g</math>. For example, <math>(3,6,18)=3</math> and <math>[6,15]=30</math>. Prove that2 KB (350 words) - 15:28, 2 June 2018
- ==Problem 1== [[2001 AMC 10 Problems/Problem 1|Solution]]14 KB (1,983 words) - 16:25, 2 June 2022
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2000 AMC 10 Problems/Problem 1]]1 KB (125 words) - 14:27, 20 February 2020
- ==Problem 1== [[2000 AMC 10 Problems/Problem 1|Solution]]14 KB (2,035 words) - 21:57, 2 May 2024
- ==Problem 1== [[2021 GMC 10B Problems/Problem 1|Solution]]11 KB (1,695 words) - 14:33, 7 March 2022
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2003 AMC 10B Problems/Problem 1]]1 KB (165 words) - 18:53, 6 October 2014
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2001 AMC 10 Problems/Problem 1]]2 KB (153 words) - 18:59, 6 October 2014
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1990 AJHSME Problems/Problem 1]]2 KB (141 words) - 23:39, 8 October 2014
- == Problem == ...e digit equal to <math>7</math>. Therefore the probability is <math>\dfrac{18}{90} = \boxed{\dfrac{1}{5}} \Longrightarrow \mathrm{(B)}</math>.1 KB (210 words) - 19:22, 22 October 2022
- == Problem == {{AHSME box|year=1998|num-b=18|num-a=20}}2 KB (368 words) - 11:08, 4 February 2017
- == Problem == ...15 \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad \mathrm{(E) \ }18 </math>1 KB (195 words) - 14:29, 5 July 2013
- ...help improve their problem solving skills. The focus of this program is on problem solving, exploration of mathematical knowledge, and improvement of analytic ...ornia State University, Fullerton (Residential and Day Camp options)- June 18-July 73 KB (428 words) - 16:59, 9 March 2017
- == Problem == According to the problem statement, there are polynomials <math>Q(x)</math> and <math>R(x)</math> su2 KB (361 words) - 05:01, 7 February 2016
- ==Problem== <math>\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20</math>2 KB (313 words) - 23:51, 5 October 2023
- ==Problem== {{AMC10 box|year=2000|num-b=16|num-a=18}}1 KB (211 words) - 03:52, 20 July 2023
- #REDIRECT [[2000 AMC 12 Problems/Problem 18]]45 bytes (4 words) - 00:10, 27 November 2011
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[1983 AHSME Problems/Problem 1|Problem 1]]2 KB (174 words) - 00:34, 2 October 2014
- ==Problem== <math> \mathrm{(A) \ }12 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24</math>3 KB (451 words) - 10:19, 23 January 2024
- ==Problem== {{AJHSME box|year=1985|num-b=18|num-a=20}}1 KB (219 words) - 09:04, 22 January 2023
- ==Problem== {{AJHSME box|year=1985|num-b=16|num-a=18}}1 KB (234 words) - 08:14, 13 January 2023
- == Problem 1 == [[1986 AJHSME Problems/Problem 1|Solution]]14 KB (2,054 words) - 15:41, 8 August 2020
- == Problem == ...\qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 18</math>3 KB (538 words) - 12:02, 17 October 2020
- ...The second link contains the answer key. The rest contain each individual problem and its solution. ** [[1986 AJHSME Problems/Problem 1|Problem 1]]2 KB (173 words) - 10:54, 15 April 2012
- ==Problem== We can solve this problem using logic.3 KB (457 words) - 15:02, 4 April 2021
- ==Problem== {{AJHSME box|year=1986|num-b=18|num-a=20}}1 KB (198 words) - 11:43, 24 February 2023
- ==Problem== <math> \mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 19 \qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 301 KB (208 words) - 19:10, 24 January 2015
- ==Problem== <math> \mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } 42 KB (326 words) - 09:47, 17 October 2020
- ...The second link contains the answer key. The rest contain each individual problem and its solution. ** [[1987 AJHSME Problems/Problem 1|Problem 1]]2 KB (173 words) - 06:52, 16 April 2012
- == Problem 1 == [[1987 AJHSME Problems/Problem 1|Solution]]12 KB (1,568 words) - 09:35, 31 October 2021
- ==Problem== When in the situation <math>H</math>, we have probability <math>\frac 18</math> of winning the game right away, by throwing three more heads in a ro2 KB (374 words) - 11:48, 9 December 2022
- == Problem == {{AMC10 box|year=2002|ab=B|num-b=16|num-a=18}}2 KB (273 words) - 13:27, 21 May 2021
- == Problem == ...> and <math>\sin(\angle E) = \frac{12}{20}</math>, so <math>96 = 18 + 18 + 18 + x</math>.6 KB (904 words) - 12:54, 22 October 2023
- == Problem == {{AMC10 box|year=2002|ab=A|num-b=16|num-a=18}}3 KB (425 words) - 00:37, 1 February 2024
- == Problem == {{AMC12 box|year=2001|num-b=18|num-a=20}}1 KB (179 words) - 22:48, 18 August 2023
- == Problem == ...ad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5</math>1 KB (202 words) - 00:33, 30 December 2023
- == Problem 10 == ...ext{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18</math>976 bytes (145 words) - 00:07, 5 July 2013
- ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2009 AMC 12A Problems/Problem 1|Problem 1]]2 KB (203 words) - 21:48, 6 October 2014
- ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2009 AMC 10A Problems/Problem 1|Problem 1]]2 KB (184 words) - 01:50, 4 March 2020
- ==Problem== From the values given in the problem statement we see that <math>a_3=a_1+2</math>.2 KB (364 words) - 11:41, 13 October 2021
- == Problem 1 == Kim's flight took off from Newark at 10:34 AM and landed in Miami at 1:18 PM. Both cities are in the same time zone. If her flight took <math>h</math13 KB (2,105 words) - 13:13, 12 August 2020
- == Problem 1 == [[2009 AMC 10A Problems/Problem 1|Solution]]14 KB (2,130 words) - 11:32, 7 November 2021
- == Problem == <cmath>\theta_{18}=165</cmath>7 KB (990 words) - 07:23, 24 October 2022
- {{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #18]] and [[2009 AMC 10A Problems|2009 AMC 10A #25]]}} == Problem ==6 KB (1,012 words) - 19:16, 14 September 2022
- == Problem == For simplicity, we solve the same problem with triangle and square instead of pentagon and heptagon. For the triangle4 KB (630 words) - 21:27, 30 December 2023
- == Problem == ...check that <math>r_1-r_2\ne 0</math>, as you cannot divide by zero. As the problem states that the series are different, <math>r_1 \ne r_2</math>, and so ther3 KB (453 words) - 10:22, 6 October 2023
- == Problem == ...4</math> congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to <math>4</m5 KB (759 words) - 16:48, 19 April 2022
- == Problem == ...sum is <math>9+9=18</math>. Hence the sum of digits will be at most <math>18</math>.3 KB (481 words) - 20:06, 17 December 2017
- == Problem == Hence the answer is <math>\frac{36}{70}=\frac{18}{35}</math>. We know this is a little bit larger than <math>\frac 12</math>3 KB (402 words) - 10:29, 2 August 2021
- == Problem == \mathrm{(D)}\ 183 KB (524 words) - 16:26, 23 June 2023
- == Problem == Kim's flight took off from Newark at 10:34 AM and landed in Miami at 1:18 PM. Both cities are in the same time zone. If her flight took <math>h</math834 bytes (126 words) - 21:41, 3 July 2013
- == Problem == ...en be a factor of <math>100</math>, excluding <math>100</math> because the problem says that <math>r<100</math>. <math>100\: =\: 2^2\; \cdot \; 5^2</math>. Th2 KB (276 words) - 09:57, 8 June 2021
- == Problem == {{AMC10 box|year=2009|ab=A|num-b=16|num-a=18}}4 KB (684 words) - 21:14, 23 October 2023
- == Problem == {{AMC10 box|year=2001|num-b=16|num-a=18}}2 KB (279 words) - 00:32, 30 December 2023
- ...AMC 12 Problems|2001 AMC 12 #10]] and [[2001 AMC 10 Problems|2001 AMC 10 #18]]}} == Problem ==2 KB (252 words) - 00:11, 15 August 2022
- == Problem == ...precisely <math>30-6=24</math> numbers that satisfy all criteria from the problem statement.7 KB (1,110 words) - 15:20, 30 May 2022
- ==Problem== What [[fraction]] of the large <math>12</math> by <math>18</math> [[rectangle|rectangular]] region is shaded?996 bytes (130 words) - 23:53, 4 July 2013
- == Problem == {{AMC12 box|year=2001|num-b=16|num-a=18}}5 KB (792 words) - 15:23, 30 November 2021
- == Problem == {{AMC12 box|year=2002|ab=A|num-b=16|num-a=18}}1 KB (161 words) - 15:01, 8 September 2022
- ==Problem== <math>\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mat2 KB (231 words) - 15:59, 9 February 2023
- == Problem == {{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}2 KB (350 words) - 11:09, 18 July 2023
- == Problem == \text{(B) }183 KB (485 words) - 03:13, 1 September 2023
- ==Problem== <math>9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}</math>8 KB (1,425 words) - 23:05, 21 June 2023
- ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2009 AMC 10B Problems/Problem 1|Problem 1]]2 KB (189 words) - 17:22, 28 June 2021
- ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2009 AMC 12B Problems/Problem 1|Problem 1]]2 KB (197 words) - 21:14, 6 October 2014
- == Problem == \mathrm{(D)}\ 18\qquad791 bytes (119 words) - 15:58, 9 June 2021
- ...10B Problems|2009 AMC 10B #23]] and [[2009 AMC 12B Problems|2009 AMC 12B #18]]}} == Problem ==2 KB (304 words) - 13:23, 2 July 2021
- == Problem == <math>\mathrm{(A)}\frac 18\qquad4 KB (637 words) - 04:52, 2 July 2022
- == Problem 1 == [[2009 AMC 12B Problems/Problem 1|Solution]]13 KB (2,030 words) - 03:04, 5 September 2021
- == Problem == \mathrm{(D)}\ 18\qquad1 KB (159 words) - 08:13, 4 November 2022
- == Problem == {{AMC10 box|year=2009|ab=B|num-b=16|num-a=18}}4 KB (695 words) - 21:33, 7 October 2023
- == Problem == ...math> becomes negative for <math>n</math> between <math>2</math> and <math>18</math>, and then <math>f(19)=761</math> is again a prime number. And as <ma4 KB (696 words) - 13:27, 23 December 2020
- == Problem == {{AMC10 box|year=2009|ab=B|num-b=18|num-a=20}}2 KB (372 words) - 17:36, 28 June 2021
- == Problem == ...sum_{a=1}^{4} a(5-a)^2 = 1\cdot 16 + 2\cdot 9 + 3 \cdot 4 + 4 \cdot 1 = 16+18+12+4=50</cmath>15 KB (2,229 words) - 03:36, 4 September 2021
- == Problem == \mathrm{(D)}\ 18\qquad883 bytes (137 words) - 18:42, 23 February 2017
- == Problem 1 == [[2009 AMC 10B Problems/Problem 1|Solution]]15 KB (2,262 words) - 00:53, 18 June 2021
- == Problem == ...ath>\frac{180*(18-2)}{18} = 160^\circ</math> so each interior angle of the 18-gon is <math>160^\circ</math>. Let <math>x</math> be the degree measure of3 KB (546 words) - 04:26, 16 January 2023
- ==Problem== ~Shadow-182 KB (282 words) - 14:47, 8 June 2021
- ==Problem== {{AJHSME box|year=1987|num-b=18|num-a=20}}980 bytes (136 words) - 11:36, 20 May 2017
- ==Problem== {{AJHSME box|year=1987|num-b=16|num-a=18}}934 bytes (156 words) - 23:53, 4 July 2013
- ...The second link contains the answer key. The rest contain each individual problem and its solution. ** [[1988 AJHSME Problems/Problem 1|Problem 1]]2 KB (173 words) - 17:28, 6 January 2024
- == Problem 1 == [[1988 AJHSME Problems/Problem 1|Solution]]14 KB (1,872 words) - 15:23, 17 January 2023
- == Problem == ...egers, the two solutions to this are <math>(y,z) = (39,0)</math> or <math>(18,5)</math>.2 KB (326 words) - 14:12, 16 August 2020
- == Problem == Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem:8 KB (1,224 words) - 19:52, 7 March 2024
- == Problem == (If it's not then the answer to the problem would be irrational which can't be in the form of <math>\frac {m}{n}</math>12 KB (1,970 words) - 22:53, 22 January 2024
- == Problem 1 == [[1989 AJHSME Problems/Problem 1|Solution]]13 KB (1,765 words) - 11:55, 22 November 2023
- == Problem 1 == [[2000 AMC 8 Problems/Problem 1|Solution]]15 KB (2,165 words) - 03:32, 13 April 2024
- ==Problem== ...75688772935274463415059)--(18,1.7320508075688772935274463415059)--(17,0)--(18,-1.7320508075688772935274463415059)--(20,-1.73205080756887729352744634150596 KB (470 words) - 23:56, 4 July 2013
- ==Problem== ...\dfrac{\dfrac{12}{13}}{\sqrt{\dfrac{8}{26}}\cdot\dfrac{5}{13}+\sqrt{\dfrac{18}{26}}\cdot\dfrac{12}{13}}\\9 KB (1,415 words) - 13:56, 18 December 2022
- ==Problem== 18 & 2 & 20 \\ \hline884 bytes (110 words) - 19:37, 10 March 2015
- ==Problem== {{AJHSME box|year=1988|num-b=16|num-a=18}}1 KB (202 words) - 23:56, 4 July 2013
- ==Problem== {{AJHSME box|year=1988|num-b=18|num-a=20}}946 bytes (133 words) - 10:51, 28 June 2023
- ...The second link contains the answer key. The rest contain each individual problem and its solution. ** [[1989 AJHSME Problems/Problem 1|Problem 1]]2 KB (173 words) - 17:29, 6 January 2024
- ...sp;by<br/>}}}'''{{{before|[[{{{year}}} AJHSME Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]]}}}''' ...bsp;by<br/>}}}'''{{{after|[[{{{year}}} AJHSME Problems/Problem {{{num-a}}}|Problem {{{num-a}}}]]}}}'''3 KB (370 words) - 21:07, 15 August 2014
- ===Problem 1=== ===Problem 2===4 KB (668 words) - 14:52, 17 August 2020
- ==Problem== &= -15+18 \\476 bytes (64 words) - 23:57, 4 July 2013
- ==Problem 1== ...x digits <math>4,5,6,7,8,9</math> in one of the six boxes in this addition problem?15 KB (2,059 words) - 15:03, 6 October 2021
- ==Problem 17== Note that we can also use coordinates to solve this problem. WLOG, set the side length of square <math>ABCD</math> equal to <math>6</ma2 KB (321 words) - 16:54, 27 November 2023
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1991 AJHSME Problems/Problem 1]]2 KB (141 words) - 23:39, 8 October 2014
- ==Problem== unitsize(18);2 KB (255 words) - 03:28, 28 November 2019
- ==Problem== ...ext{(A)}\ 6 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 18 \qquad \text{(E)}\ 24</math>1 KB (177 words) - 16:03, 19 April 2021
- ==Problem== ...the cut rectangles to the whole uncutted rectangle will be <math>\frac{15}{18} = \frac{5}{6}.</math>2 KB (333 words) - 20:05, 12 December 2021
- ==Problem== With this knowledge, we can break down the problem into smaller problems, first, the probability that the first spinner lands3 KB (437 words) - 12:57, 16 October 2023
- ==Problem== <math>\text{(A)}\ 12 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 36</math>833 bytes (136 words) - 20:58, 15 September 2019
- ==Problem== {{AJHSME box|year=1990|num-b=18|num-a=20}}727 bytes (119 words) - 01:59, 25 November 2020
- ==Problem== fill((18,1)--(19,1)--(19,3)--(18,3)--cycle,gray);2 KB (306 words) - 22:17, 11 July 2009
- ...lso, due to variances within a contest, ranges shown may overlap. A sample problem is provided with each entry, with a link to a solution. ...ection Test.] When considering problem difficulty, '''put more emphasis on problem-solving aspects and less so on technical skill requirements'''.41 KB (6,751 words) - 18:21, 31 May 2024
- ==Problem== <math>\text{(A)}\ 18 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 25 \qquad \t1 KB (149 words) - 00:07, 5 July 2013
- ==Problem== <cmath>\begin{align*}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end4 KB (704 words) - 19:25, 28 March 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1992 AJHSME Problems/Problem 1]]2 KB (141 words) - 23:40, 8 October 2014
- ==Problem== unitsize(18);2 KB (314 words) - 00:07, 5 July 2013
- ==Problem== {{AJHSME box|year=1991|num-b=16|num-a=18}}2 KB (266 words) - 00:07, 5 July 2013
- ==Problem== {{AJHSME box|year=1991|num-b=18|num-a=20}}895 bytes (135 words) - 12:16, 8 October 2015
- ==Problem== In the addition problem, each digit has been replaced by a letter. If different letters represent1 KB (203 words) - 13:06, 16 August 2020
- ==Problem 1== [[1992 AJHSME Problems/Problem 1|Solution]]17 KB (2,346 words) - 13:36, 19 February 2020
- == Problem== <math>\textbf{(A)}\ \sqrt{18} \qquad1 KB (176 words) - 23:27, 26 May 2018
- ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1993 AJHSME Problems/Problem 1]]2 KB (141 words) - 23:41, 8 October 2014
- ==Problem 1== [[1993 AJHSME Problems/Problem 1|Solution]]14 KB (1,797 words) - 11:13, 28 December 2022
- ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2006 AMC 8 Problems/Problem 1|Problem 1]]2 KB (154 words) - 21:46, 25 November 2013
- == Problem == \qquad\mathrm{(C)}\ \dfrac{1}{18}4 KB (635 words) - 11:46, 1 September 2022
- == Problem == {{AMC10 box|year=2005|ab=A|num-b=18|num-a=20}}3 KB (315 words) - 12:46, 14 December 2021
- ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2010 AMC 12A Problems/Problem 1|Problem 1]]2 KB (203 words) - 21:47, 6 October 2014
- **[[2010 AMC 10A Problems/Problem 1|Problem 1]] **[[2010 AMC 10A Problems/Problem 2|Problem 2]]2 KB (175 words) - 21:24, 15 November 2022
- == Problem 1 == [[2010 AMC 12A Problems/Problem 1|Solution]]12 KB (1,817 words) - 15:00, 12 August 2020
- == Problem == {{AMC12 box|year=2010|num-b=18|num-a=20|ab=A}}2 KB (379 words) - 18:30, 10 July 2022
- == Problem == {{AMC10 box|year=2010|ab=A|num-b=16|num-a=18}}2 KB (383 words) - 17:42, 28 June 2021
- ...12A Problems|2010 AMC 12A #16]] and [[2010 AMC 10A Problems|2010 AMC 10A #18]]}} == Problem ==5 KB (796 words) - 22:51, 26 November 2023
- == Problem == {{AMC10 box|year=2010|ab=A|num-b=18|num-a=20}}4 KB (690 words) - 10:13, 14 October 2022
- == Problem == There might be a problem when you cancel out the <math>10</math>s from <math>90!</math>. One method10 KB (1,525 words) - 09:44, 24 April 2024
- == Problem == <math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36</math>9 KB (1,434 words) - 17:54, 17 August 2022
- == Problem == <math>\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 18 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 36</math>679 bytes (109 words) - 12:22, 16 August 2021
- == Problem == ...age of <math>5</math> people in a room is <math>30</math> years. An <math>18</math>-year-old person leaves1 KB (192 words) - 12:22, 16 August 2021