Search results
Create the page "AIME Problems/Problem 9" on this wiki! See also the search results found.
Page title matches
- == Problem == {{AIME box|year=2006|n=I|num-b=8|num-a=10}}4 KB (651 words) - 18:27, 22 May 2021
- == Problem == This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</ma6 KB (1,154 words) - 03:30, 11 January 2024
- == Problem == [[Image:2005_I_AIME-9.png]]4 KB (600 words) - 21:44, 20 November 2023
- == Problem == draw((9*4/14,0)--(9*4/14,5*3/14),dashed);4 KB (647 words) - 17:43, 23 November 2024
- == Problem == We can find the value of <math>a_{9}</math> by its bounds using three conditions:3 KB (535 words) - 22:25, 5 March 2025
- == Problem == <math>f'(y)</math> = <math>9 - 4y^{-2}</math>5 KB (824 words) - 19:34, 20 July 2024
- == Problem == ...B \cdot h_{ABD}</math>, we find that <math>h_{ABD} = 8</math>. Because the problem does not specify, we may assume both <math>ABC</math> and <math>ABD</math>6 KB (947 words) - 18:00, 20 February 2025
- == Problem == == Solution 1== <!-- Images obsoleted Image:1985_AIME-9.png, Image:1985_AIME-9a.png by asymptote -->5 KB (784 words) - 21:05, 8 December 2024
- == Problem == </asy></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps -->11 KB (1,879 words) - 21:04, 8 December 2024
- == Problem == {{AIME box|year=1987|num-b=8|num-a=10}}1 KB (212 words) - 13:37, 11 December 2024
- == Problem == ...}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework:6 KB (893 words) - 08:15, 2 February 2023
- == Problem == {{AIME box|year=1989|num-b=8|num-a=10}}6 KB (874 words) - 15:50, 20 January 2024
- == Problem == ...<math>\frac{144}{1024} = \frac{9}{64}</math>. Thus, our solution is <math>9 + 64 = \boxed{073}</math>.3 KB (427 words) - 15:03, 24 June 2024
- == Problem == Note: The problem is much easier computed if we consider what <math>\sec (x)</math> is, then10 KB (1,590 words) - 14:04, 20 January 2023
- == Problem == ...> is irrelevant as long as there still exists a circle as described in the problem.5 KB (896 words) - 21:03, 8 December 2024
- == Problem == ...<math>1994 + n</math> to be a multiple of <math>125</math>, <math>n\equiv 9\pmod{16}</math> and <math>n\equiv 6\pmod{125}</math>. The smallest <math>n<3 KB (488 words) - 02:06, 22 September 2023
- == Problem == ...ously <math>P_1 = 1</math>), we get <math>\frac {3^5}{11*9*7*5*3} = \frac {9}{385}</math>, and <math>p+q=\boxed{394}</math>.5 KB (880 words) - 12:49, 20 July 2024
- == Problem == {{AIME box|year=1995|num-b=8|num-a=10}}7 KB (1,149 words) - 13:36, 26 November 2024
- == Problem == (Note: If you try to do this, first look through all the problems! -Guy)3 KB (525 words) - 23:51, 6 September 2023
- == Problem == <math>a^{12} = a^{11}+a^{10}= 2a^{10} + a^{9} = 3a^9+ 2a^8 = ... = 144a^1+89a^0 = 144a+89</math>4 KB (586 words) - 21:53, 30 December 2023
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 <cmath>9-40-41</cmath>6 KB (978 words) - 12:02, 6 March 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
- ...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. ...down=<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
- == Problems == ...largest area that this triangle can have? ([[1992 AIME Problems/Problem 13|AIME 1992]])3 KB (583 words) - 21:20, 2 August 2024
- ...ded to this, then we would have a [[perfect square]], <math>(x-3)^2=x^2-6x+9</math>. To do this, add <math>7</math> to each side of the equation to get <math>x^2-6x+9=7\Rightarrow(x-3)^2=7\Rightarrow x-3=\pm{\sqrt7}\Rightarrow x=3\pm\sqrt{7}<2 KB (422 words) - 16:20, 5 March 2023
- ...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>9! = 362880</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 == Problems ==15 KB (2,618 words) - 12:03, 19 February 2025
- xaxis(-9,9,Ticks(f, 1.0)); yaxis(-9,9,Ticks(f, 1.0));3 KB (551 words) - 16:22, 13 September 2023
- ...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
- ...e multiply the possibilities for each digit to arrive at our answer: <math>9 \cdot 10 \cdot 10 \cdot 10 = 9000</math>. <math>\square</math> This is a problem where constructive counting is not the simplest way to proceed. This next e13 KB (2,018 words) - 15:31, 10 January 2025
- ..., then adding together the totals of each part. Casework is a very general problem-solving approach, and as such has wide applicability. ...emonstrate casework in action. Unlike the selections in this article, most problems cannot be completely solved through casework. However, it is crucial as an5 KB (709 words) - 17:40, 24 September 2024
- We have <math>93 \equiv 9 \pmod{42}</math>, so <math>\gcd(93,42) = \gcd(42,9)</math>. <br/> Similarly, <math>42 \equiv 6 \pmod{9}</math>, so <math>\gcd(42,9) = \gcd(9,6)</math>. <br/>6 KB (923 words) - 17:39, 30 September 2024
- ...ue in the second. For instance, one function may map 1 to 1, 2 to 4, 3 to 9, 4 to 16, and so on. This function has the rule that it takes its input va ...t function. For an example, consider the function: <math>f(x) = \sqrt{x^2-9}</math>. The domain of the function is the set <math>\{x:|x| \geq 3\}</mat10 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
- * Because the sum of digits of <math>9513</math> is a multiple of 9, we can rewrite <math>38052</math> as <math>2^2 \cdot 3^2 \cdot 1057</math> ==Problems==3 KB (496 words) - 22:14, 5 January 2024
- D(P--P-(9/5,4)); ==Problems==5 KB (892 words) - 21:52, 1 May 2021
- Base-10 uses digits 0-9. Usually, the base, or '''radix''', of a number is denoted as a subscript == Example Problems ==4 KB (547 words) - 17:23, 30 December 2020
