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2006 AMC 12B Problems/Problem 18
== Problem ==

2 KB (354 words) - 16:57, 28 December 2020
2006 AMC 12A Problems/Problem 18
== 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> which

2 KB (334 words) - 18:34, 18 September 2020
2005 AMC 12B Problems/Problem 18
== Problem ==

2 KB (262 words) - 21:20, 21 December 2020
2006 AMC 10A Problems/Problem 18
== Problem ==

2 KB (254 words) - 14:39, 5 April 2024
2006 AMC 10B Problems/Problem 18
== Problem ==

1 KB (158 words) - 01:33, 29 May 2023
2005 AMC 10B Problems/Problem 18
== Problem ==

1 KB (195 words) - 15:33, 16 December 2021
University of South Carolina High School Math Contest/1993 Exam/Problem 18
== Problem == ...y of South Carolina High School Math Contest/1993 Exam/Problem 17|Previous Problem]]

2 KB (331 words) - 00:37, 26 January 2023
2005 AMC 10A Problems/Problem 18
==Problem==

2 KB (260 words) - 17:42, 7 July 2023
2003 AMC 10A Problems/Problem 18
== 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
2003 AMC 12A Problems/Problem 18
== Problem ==

3 KB (426 words) - 18:20, 18 July 2022
2007 AMC 12A Problems/Problem 18
== Problem ==

2 KB (319 words) - 00:37, 25 March 2024
2005 AMC 12A Problems/Problem 18
== Problem ==

2 KB (277 words) - 18:15, 25 November 2020
2004 AMC 12A Problems/Problem 18
...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
2000 AMC 12 Problems/Problem 18
{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #18]] and [[2000 AMC 10 Problems|2000 AMC 10 #25]]}} == Problem ==

2 KB (382 words) - 19:20, 12 May 2023
2002 AMC 12B Problems/Problem 18
== Problem == [[Image:2002_12B_AMC-18.png]]

3 KB (376 words) - 19:16, 20 August 2019
2007 AMC 10A Problems/Problem 18
==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/Prob

6 KB (867 words) - 00:17, 20 May 2023
2008 AMC 12A Problems/Problem 18
==Problem==

2 KB (302 words) - 04:51, 16 January 2023
2007 AMC 12B Problems/Problem 18
==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
2007 AMC 10B Problems/Problem 18
==Problem== You can solve this problem by setting up a simple equation with the Pythagorean Theorem. The hypotenus

2 KB (385 words) - 14:17, 4 June 2021
2008 AMC 10B Problems/Problem 18
==Problem==

875 bytes (139 words) - 20:19, 23 March 2023
2008 AMC 10A Problems/Problem 18
== Problem == From the problem, we know that

8 KB (1,339 words) - 14:15, 1 August 2022
2003 AMC 12B Problems/Problem 18
== 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
1995 AHSME Problems/Problem 18
==Problem==

1 KB (167 words) - 13:59, 5 July 2013
2008 AMC 12B Problems/Problem 18
==Problem==

1 KB (214 words) - 12:01, 2 February 2015
2002 AMC 10A Problems/Problem 18
== Problem ==

999 bytes (153 words) - 20:43, 28 May 2023
1998 AHSME Problems/Problem 18
== Problem ==

1 KB (187 words) - 14:29, 5 July 2013
1999 AHSME Problems/Problem 18
== Problem ==

2 KB (270 words) - 14:35, 5 July 2013
2000 AMC 10 Problems/Problem 18
==Problem==

2 KB (274 words) - 10:26, 8 November 2021
1985 AJHSME Problems/Problem 18
==Problem==

1 KB (235 words) - 09:03, 22 January 2023
1986 AJHSME Problems/Problem 18
==Problem==

2 KB (270 words) - 18:54, 28 December 2023
2004 AMC 12B Problems/Problem 18
== Problem ==

2 KB (461 words) - 16:29, 27 August 2016
2002 AMC 10B Problems/Problem 18
#REDIRECT[[2002 AMC 12B Problems/Problem 14]]

45 bytes (5 words) - 16:15, 29 July 2011
2004 AMC 10B Problems/Problem 18
== 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
2009 AMC 12A Problems/Problem 18
{{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
2009 AMC 10A Problems/Problem 18
== 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
2001 AMC 12 Problems/Problem 18
== Problem ==

5 KB (822 words) - 01:35, 7 February 2024
2002 AMC 12A Problems/Problem 18
== Problem == \text{(B) }18

3 KB (485 words) - 03:13, 1 September 2023
2009 AMC 10B Problems/Problem 18
== Problem ==

6 KB (930 words) - 22:14, 18 January 2024
1987 AJHSME Problems/Problem 18
==Problem==

1 KB (185 words) - 19:11, 26 August 2016
1988 AJHSME Problems/Problem 18
==Problem==

1 KB (170 words) - 23:56, 4 July 2013
1972 AHSME Problems/Problem 18
==Problem==

2 KB (267 words) - 04:54, 23 June 2022
1989 AJHSME Problems/Problem 18
==Problem==

1 KB (195 words) - 13:25, 28 December 2021
1991 AJHSME Problems/Problem 18
==Problem==

2 KB (266 words) - 00:07, 5 July 2013
2010 AMC 12A Problems/Problem 18
== 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 multip

5 KB (910 words) - 01:40, 2 February 2021
2010 AMC 10A Problems/Problem 18

46 bytes (5 words) - 13:27, 26 May 2020
2010 AMC 12B Problems/Problem 18
== 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 calculus

6 KB (1,105 words) - 13:39, 9 January 2024
2007 AMC 8 Problems/Problem 18
==Problem==

1,000 bytes (149 words) - 05:43, 31 December 2022
2010 AMC 10B Problems/Problem 18
#redirect [[2010 AMC 12B Problems/Problem 16]]

46 bytes (5 words) - 20:45, 26 May 2020
2011 AMC 12A Problems/Problem 18
== Problem ==

2 KB (306 words) - 21:50, 2 November 2021
2011 AMC 10A Problems/Problem 18
#redirect [[2011 AMC 12A Problems/Problem 11]]

46 bytes (5 words) - 19:19, 27 June 2020

Page text matches

2005 AMC 12A Problems/Problem 16
== Problem == draw((0,0)--(18,0));

2 KB (307 words) - 15:30, 30 March 2024
2016 AMC 10A Problems/Problem 7
== Problem == ...[[User:Integralarefun|Integralarefun]] ([[User talk:Integralarefun|talk]]) 18:19, 27 September 2023 (EDT)

2 KB (268 words) - 18:19, 27 September 2023
1953 AHSME Problems/Problem 13
== 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
Simon's Favorite Factoring Trick
...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 (1,107 words) - 07:35, 26 March 2024
Pigeonhole Principle
...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
Principle of Inclusion-Exclusion
...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
Factorial
* <math>18! = 6402373705728000</math> ([[2007 iTest Problems/Problem 6|Source]])

10 KB (809 words) - 16:40, 17 March 2024
Fermat's Little Theorem
...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
Induction
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/Fo

5 KB (768 words) - 20:45, 1 September 2022
Complex number
*[[2007 AMC 12A Problems/Problem 18]] *[[1984 AIME Problems/Problem 8|1984 AIME Problem 8]]

5 KB (860 words) - 15:36, 10 December 2023
Without loss of generality
* [[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
2000 AMC 12 Problems/Problem 1
==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 * 3

2 KB (276 words) - 05:25, 9 December 2023
Functional equation
...ns can significantly help in solving functional identities. Consider this problem: === Problem Examples ===

2 KB (361 words) - 14:40, 24 August 2021
2006 AMC 10B
...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
2006 AMC 12B
...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
Modular arithmetic/Introduction
* <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
Modular arithmetic/Intermediate
...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
2006 AMC 10A
...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
2006 AIME I Problems
== 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> ABCD

7 KB (1,173 words) - 03:31, 4 January 2023
2006 AIME I Problems/Problem 5
== Problem == <math>c=18</math>

3 KB (439 words) - 18:24, 10 March 2015
2006 AIME I Problems/Problem 1
== 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
2021 JMC 10
==Problem 1== [[2021 JMC 10 Problems/Problem 1|Solution]]

12 KB (1,784 words) - 16:49, 1 April 2021
2006 AMC 12B Problems
== Problem 1 == [[2006 AMC 12B Problems/Problem 1|Solution]]

13 KB (2,058 words) - 12:36, 4 July 2023
2006 AMC 12A
...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
2004 AMC 12A
...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
2004 AMC 12B
...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
2005 AMC 12A
...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
2005 AMC 12B
...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
2000 AMC 12
...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
2001 AMC 12
...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
2002 AMC 12A
...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
2003 AMC 12A
...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
2002 AMC 12B
...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
2003 AMC 12B
...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
2006 AMC 12A Problems
== Problem 1 == [[2006 AMC 12A Problems/Problem 1|Solution]]

15 KB (2,223 words) - 13:43, 28 December 2020
2005 AMC 12A Problems
== Problem 1 == [[2005 AMC 12A Problems/Problem 1|Solution]]

13 KB (1,971 words) - 13:03, 19 February 2020
2004 AMC 12A Problems
== Problem 1 == [[2004 AMC 12A Problems/Problem 1|Solution]]

13 KB (1,953 words) - 00:31, 26 January 2023
2003 AMC 12A Problems
== Problem 1 == [[2003 AMC 12A Problems/Problem 1|Solution]]

13 KB (1,955 words) - 21:06, 19 August 2023
2002 AMC 12A Problems
== Problem 1 == [[2002 AMC 12A Problems/Problem 1|Solution]]

12 KB (1,792 words) - 13:06, 19 February 2020
2000 AMC 12 Problems
== Problem 1 == [[2000 AMC 12 Problems/Problem 1|Solution]]

13 KB (1,948 words) - 12:26, 1 April 2022
2001 AMC 12 Problems
== Problem 1 == [[2001 AMC 12 Problems/Problem 1|Solution]]

13 KB (1,957 words) - 12:53, 24 January 2024
2002 AMC 12B Problems
== Problem 1 == [[2002 AMC 12B Problems/Problem 1|Solution]]

10 KB (1,547 words) - 04:20, 9 October 2022
2003 AMC 12B Problems
== 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
2004 AMC 12B Problems
== Problem 1 == [[2004 AMC 12B Problems/Problem 1|Solution]]

13 KB (2,049 words) - 13:03, 19 February 2020
2005 AMC 12B Problems
== Problem 1 == [[2005 AMC 12B Problems/Problem 1|Solution]]

12 KB (1,781 words) - 12:38, 14 July 2022
2006 AMC 12B Problems/Problem 15
== Problem == // from amc10 problem series

3 KB (458 words) - 16:40, 6 October 2019
2006 AMC 12B Problems/Problem 17
== Problem == \mathrm{(B)}\ \frac 18

1 KB (188 words) - 22:10, 9 June 2016
2006 AMC 12B Problems/Problem 19
== Problem == {{AMC12 box|year=2006|ab=B|num-b=18|num-a=20}}

4 KB (696 words) - 09:47, 10 August 2015
2006 AMC 12B Problems/Problem 20
== Problem == \mathrm{(A)}\ \frac 18

3 KB (485 words) - 14:09, 21 May 2021
2006 AMC 12B Problems/Problem 24
== Problem == &= \pi^2 \left ( \frac14 - \frac1{36} - \frac1{18}\right ) \\

3 KB (563 words) - 22:45, 24 October 2021

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