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- == Problem == {{AIME box|year=2006|n=I|num-b=5|num-a=7}}2 KB (237 words) - 19:14, 20 November 2023
- #REDIRECT [[2006 AIME I Problems/Problem 6]]44 bytes (5 words) - 12:05, 28 June 2009
- == Problem == ...xample, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical st2 KB (384 words) - 00:31, 26 July 2018
- == Problem == ==Solution 6 (De Moivre's Theorem)==4 KB (686 words) - 12:52, 13 June 2024
- == Problem == We divide the problem into two cases: one in which zero is one of the digits and one in which it3 KB (562 words) - 18:12, 4 March 2022
- == Problem == ...al. Also note some other conditions we have picked up in the course of the problem, namely that <math>b_1</math> is divisible by <math>8</math>, <math>b_2</ma6 KB (950 words) - 14:18, 15 January 2024
- == Problem == Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math3 KB (361 words) - 20:20, 14 January 2023
- == Problem == [[Image:1984_AIME-6.png]]6 KB (1,022 words) - 19:29, 22 January 2024
- == Problem == [[Image:AIME 1985 Problem 6.png]]5 KB (814 words) - 18:12, 29 January 2025
- == Problem == {{AIME box|year=1986|num-b=5|num-a=7}}2 KB (336 words) - 14:13, 6 September 2020
- == Problem == {{AIME box|year=1987|num-b=5|num-a=7}}3 KB (530 words) - 07:46, 1 June 2018
- ==Problem== [[Image:1988_AIME-6.png]]5 KB (877 words) - 20:57, 27 December 2024
- == Problem == pair A=(0,0),B=(10,0),C=6*expi(pi/3);6 KB (980 words) - 15:08, 14 May 2024
- == Problem == ...<math>n</math> were not there on May 1, we have the equality <cmath>\frac{6}{10}\cdot n = \frac{3}{4} \cdot k</cmath> which reduces to <cmath>n=\frac{53 KB (457 words) - 12:48, 27 November 2024
- == Problem == {{AIME box|year=1991|num-b=5|num-a=7}}3 KB (447 words) - 17:02, 24 November 2023
- == Problem == ...math>h+d\ge 10</math> or <math>c+g\ge 10</math> or <math>b+f\ge 10</math>. 6. Consider <math>c \in \{0, 1, 2, 3, 4\}</math>. <math>1abc + 1ab(c+1)</mat5 KB (680 words) - 12:27, 16 January 2025
- == Problem == ...be divisible by 495. The only multiples of 495 that are small enough to be AIME answers are 495 and 990. From the second of the three expressions above, we4 KB (676 words) - 18:07, 12 July 2024
- == Problem == ...}\right)^2 = 100</math>. Thus, the total number of unit triangles is <math>6 \times 100 = 600</math>.4 KB (721 words) - 16:14, 8 March 2021
- == Problem == Incorporating this into our problem gives <math>19\times31=\boxed{589}</math>.2 KB (407 words) - 08:14, 4 November 2022
- == Problem == ...rical to the first one. Therefore, there are <math>5 \cdot 2^6 + 5 \cdot 2^6</math> ways for an undefeated or winless team.3 KB (461 words) - 00:33, 16 May 2024
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- ...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 <cmath>6-8-10 = (3-4-5)*2</cmath>6 KB (978 words) - 12:02, 6 March 2025
- ...parentheses. For instance, <math>x \in [3,6)</math> means <math>3 \le x < 6</math>. The problem here is that we multiplied by <math>x+5</math> as one of the last steps. W12 KB (1,806 words) - 06:07, 19 June 2024
- ...are invited to take the [[American Invitational Mathematics Examination]] (AIME). ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC.4 KB (596 words) - 04:53, 3 February 2025
- ...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
- ...dministered by the [[Mathematical Association of America]] (MAA). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the MAA. ...9|breakdown=<u>Problem 1/4</u>: 7<br><u>Problem 2/5</u>: 8<br><u>Problem 3/6</u>: 9}}6 KB (836 words) - 04:57, 3 February 2025
- ...tices held during the Spring Semester to determine the team each year. The 6 practices include 3 individual tests to help determine the team and some le * Performance on problems at practice sessions.22 KB (3,532 words) - 11:25, 27 September 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>3! = 6</math> * <math>6! = 720</math>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 int n = 6;15 KB (2,618 words) - 12:03, 19 February 2025
- f.p=fontsize(6); f.p=fontsize(6);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
- An example of a classic problem is as follows: ...hem twice. A number that is divisible by both 2 and 3 must be divisible by 6, and there are 16 such numbers. Thus, there are <math>50+33-16=\boxed{67}</4 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 ...wer is <math>8 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 = 8 \cdot 7^6 = 941,192</math>, as desired. <math>\square</math>13 KB (2,018 words) - 15:31, 10 January 2025
- Similarly, <math>42 \equiv 6 \pmod{9}</math>, so <math>\gcd(42,9) = \gcd(9,6)</math>. <br/> Continuing, <math>9 \equiv 3 \pmod{6}</math>, so <math>\gcd(9,6) = \gcd(6,3)</math>. <br/>6 KB (923 words) - 17:39, 30 September 2024
- ...function: <math>f(x) = x^2 + 6</math>. The function <math>g(x) = \sqrt{x-6}</math> has the property that <math>f(g(x)) = x</math>. Therefore, <math>g ==Problems==10 KB (1,761 words) - 03:16, 12 May 2023
- ...ng may lead to a quick solution is the phrase "not" or "at least" within a problem statement. ''[[2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive integers have at least one digit that is8 KB (1,192 words) - 17:20, 16 June 2023
- [[2008 AMC 12B Problems/Problem 22]] [[2001 AIME I Problems/Problem 6]]1,016 bytes (141 words) - 03:39, 29 November 2021
