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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 (382 words) - 19:20, 12 May 2023
- == 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
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 (1,107 words) - 07:35, 26 March 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 (2,675 words) - 10:57, 7 March 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
