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- == Problem == ...at the right end of the decimal representation of the product <math> 1!2!3!4!\cdots99!100!. </math> Find the remainder when <math> N </math> is divided2 KB (353 words) - 22:56, 27 September 2024
- #REDIRECT [[2006 AIME I Problems/Problem 4]]44 bytes (5 words) - 12:03, 28 June 2009
- == Problem == {{AIME box|year=2005|n=II|num-b=3|num-a=5}}3 KB (377 words) - 18:36, 1 January 2024
- == Problem == ...erfect square]] strictly between <math>(n + 3)^2 + 5</math> and <math>(n + 4)^2 + 5</math>. Thus, if the number of columns is <math>n</math>, the numbe8 KB (1,249 words) - 21:25, 20 November 2024
- == Problem == ...h> and <math>(0,y)</math>. Because the segment has length 2, <math>x^2+y^2=4</math>. Using the midpoint formula, we find that the midpoint of the segmen4 KB (647 words) - 21:51, 12 January 2025
- == Problem == ...>-digit number, for a total of <math>(2^1 - 2) + (2^2 - 2) + (2^3 -2) + (2^4 - 2) = 22</math> such numbers (or we can list them: <math>AB, BA, AAB, ABA,3 KB (508 words) - 01:16, 19 January 2024
- ==Problem== <cmath> AC = \sqrt{AB^2 + BC^2} = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10}. </cmath>11 KB (1,747 words) - 20:54, 31 December 2024
- == Problem == {{AIME box|year=1984|num-b=3|num-a=5}}2 KB (319 words) - 03:38, 16 January 2023
- == Problem == [[File:Aime.png]]3 KB (484 words) - 21:40, 2 March 2020
- == Problem == ...ve [[equation]]s gives us <math>6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)</math> so <math>x_1 + x_2 + x_3 + x_4 + x_5 = 31</math>. Subtrac1 KB (212 words) - 16:25, 17 November 2019
- == Problem == ...of the region enclosed by the [[graph]] of <math>|x-60|+|y|=\left|\frac{x}{4}\right|.</math>2 KB (371 words) - 17:25, 13 February 2024
- == Problem == ...S is <math>1.</math> Similarly testing <math>1,-1,-1,1</math> yields <math>4</math> on the LHS and <math>0</math> on the RHS. It seems for every negativ2 KB (394 words) - 10:21, 27 January 2024
- == Problem == ...h> equal <math>a+1</math>, <math>a+2</math>, <math>a+3</math>, and <math>a+4</math>, respectively. Call the square and cube <math>k^2</math> and <math>m3 KB (552 words) - 12:41, 3 March 2024
- == Problem == {{AIME box|year=1990|num-b=3|num-a=5}}1 KB (156 words) - 07:35, 4 November 2022
- == Problem == ...<math>-1 \le y \le 1</math>. It is [[periodic function|periodic]] (in this problem) with a period of <math>\frac{2}{5}</math>.2 KB (300 words) - 16:01, 26 November 2019
- == Problem == \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\vspace{4pt}4 KB (488 words) - 07:07, 26 February 2025
- == Problem == ===Solution 4===8 KB (1,343 words) - 22:42, 19 February 2025
- == Problem == {{AIME box|year=1994|num-b=3|num-a=5}}2 KB (264 words) - 13:33, 11 August 2018
- == Problem == <cmath>PQ^2 = 4(A_9P)^2 = 4[(O_9P)^2-(O_9A_9)^2] = 4[9^2-5^2] = \boxed{224}</cmath>3 KB (605 words) - 11:30, 5 May 2024
- == Problem == {{AIME box|year=1996|num-b=3|num-a=5}}2 KB (257 words) - 17:50, 4 January 2016
Page text matches
- ...e many tools in a good geometer's arsenal. A very large number of geometry problems can be solved by building right triangles and applying the Pythagorean Theo B = (4, 3);6 KB (978 words) - 12:02, 6 March 2025
- **[[AIME]] ....mathcon.org MathCON] hosts annual math competition for students in grades 4-12, with more than 200,000 participants since 2008.6 KB (625 words) - 23:27, 13 January 2025
- ...idual articles often have sample problems and solutions for many levels of problem solvers. Many also have links to books, websites, and other resources rele * CompetifyHub's Problem Sets [//competifyhub.com/resources/ Free Competition Resources for Grades 114 KB (1,913 words) - 23:52, 6 March 2025
- A more complex example is <math>\frac{x-8}{x+5}+4\ge 3</math>. \frac{x-8}{x+5}+4&\ge 312 KB (1,806 words) - 06:07, 19 June 2024
- ...ke the more challenging [[American Invitational Mathematics Examination]] (AIME). ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC.5 KB (646 words) - 04:52, 3 February 2025
- ...ed States at the [[International Mathematics Olympiad]] (IMO). While most AIME participants are high school students, some bright middle school students a High scoring AIME students are invited to take the prestigious [[United States of America Mat5 KB (669 words) - 17:19, 11 March 2025
- ...ticipate in the [[Math Olympiad Summer Program]] (MOP), a challenging 3 to 4 week math program for the brightest students in the country. ...dministered by the [[Mathematical Association of America]] (MAA). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the MAA.6 KB (836 words) - 04:57, 3 February 2025
- * Performance on problems at practice sessions. * Performance on AMC and AIME, including current and previous years.22 KB (3,532 words) - 11:25, 27 September 2024
- == Problems == *Show that <math>x^2+y^4\geq 2x+4y^2-5</math> for all real <math>x</math> and <math>y</math>.3 KB (583 words) - 21:20, 2 August 2024
- ...nd make the equation factorable. It can be used to solve more than algebra problems, sometimes going into other topics such as number theory. == Fun Practice Problems ==4 KB (682 words) - 10:25, 18 February 2025
- ...d only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. ==== Problem ====9 KB (1,703 words) - 01:20, 7 December 2024
- * <math>4! = 24</math> ==Problems==10 KB (809 words) - 16:40, 17 March 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME II Problems]]1 KB (133 words) - 12:32, 22 March 2011
- In contest problems, Fermat's Little Theorem is often used in conjunction with the [[Chinese Re === Proof 4 (Geometry) ===15 KB (2,618 words) - 12:03, 19 February 2025
- ==Problems== ...style="text-align:right;">([[2006 AMC 10A Problems/Problem 8|2006 AMC 10A, Problem 8]])</div>3 KB (551 words) - 16:22, 13 September 2023
- ==Problems== ...ast two digits of <math> 7^{81}-3^{81} </math>. ([[Euler's Totient Theorem Problem 1 Solution|Solution]])4 KB (569 words) - 22:34, 30 December 2024
- <cmath>4[ABCD]^2=\sin^2 B(ab+cd)^2</cmath> <cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2</cmath>3 KB (543 words) - 19:35, 29 October 2024
- ...uality]]. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures. == Problems ==6 KB (922 words) - 17:34, 13 January 2025
- An example of a classic problem is as follows: ...a coincidence that 67 is close to two-thirds of 100! We can approach this problem in a constructive way, building the set based on the remainders when divide4 KB (635 words) - 12:19, 2 January 2022
- This is a problem where constructive counting is not the simplest way to proceed. This next e ...proceed with the construction. If we were to go like before and break the problem down by each box, we'd get a fairly messy solution.13 KB (2,018 words) - 15:31, 10 January 2025
