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- == Problem == ...\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, </math> where <math> a, b, </math> and <math> c </math> are [[positive]]3 KB (439 words) - 18:24, 10 March 2015
- #REDIRECT [[2006 AIME I Problems/Problem 5]]44 bytes (5 words) - 18:41, 26 June 2009
- == Problem == ...<math> (a,b) </math> of [[integer]]s such that <math> \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </math>3 KB (547 words) - 19:15, 4 April 2024
- == Problem == There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation.5 KB (830 words) - 01:51, 1 March 2023
- == Problem == ...ss than Alpha's on that day. Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could achieve is <mat3 KB (436 words) - 18:31, 9 January 2024
- ==Problem== From the initial problem statement, we have <math>1000w\cdot\frac{1}{4}t=\frac{1}{4}</math>.4 KB (592 words) - 19:02, 26 September 2020
- == Problem == One way to solve this problem is by [[substitution]]. We have4 KB (672 words) - 10:17, 17 March 2023
- == Problem == Determine the value of <math>ab</math> if <math>\log_8a+\log_4b^2=5</math> and <math>\log_8b+\log_4a^2=7</math>.8 KB (1,319 words) - 22:42, 6 March 2025
- == Problem == The problem gives us a sequence defined by a [[recursion]], so let's calculate a few va2 KB (406 words) - 11:20, 15 February 2025
- == Problem == The key to this problem is to realize that <math>n+10 \mid n^3 +1000</math> for all <math>n</math>.2 KB (338 words) - 19:56, 15 October 2023
- == Problem == {{AIME box|year=1987|num-b=4|num-a=6}}1 KB (160 words) - 04:44, 21 January 2023
- == Problem == ...o draw a [[bijection]] to the number of factors that <math>10^{11} = 2^{11}5^{11}</math> has, which is <math>(11 + 1)(11 + 1) = 144</math>. Our probabil822 bytes (108 words) - 22:21, 6 November 2016
- == Problem == ...bility that the coin comes up heads in exactly <math>3</math> out of <math>5</math> flips. Find <math>i+j</math>.2 KB (258 words) - 00:07, 25 June 2023
- == Problem == ...ath>n = 2^43^45^2</math> and <math>\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}</math>.1 KB (175 words) - 03:45, 21 January 2023
- == Problem == ...d(a,b) = 1</math>. There are 8 [[prime number]]s less than 20 (<math>2, 3, 5, 7, 11, 13, 17, 19</math>), and each can only be a factor of one of <math>a919 bytes (141 words) - 20:00, 4 July 2022
- == Problem == {{AIME box|year=1992|num-b=4|num-a=6}}2 KB (293 words) - 17:12, 2 January 2025
- == Problem == <cmath>\implies C_1(X) = \frac{-77(1-X)^4 + 3X(X^2+4X+1) - 626X(1-X)^2}{(1-X)^5}</cmath>5 KB (923 words) - 13:17, 16 September 2024
- == Problem == Suppose we write each number in the form of a three-digit number (so <math>5 \equiv 005</math>), and since our <math>p(n)</math> ignores all of the zero2 KB (275 words) - 10:27, 14 June 2024
- == Problem == ...ive <math>3+4i</math>. This gets three equations necessary for solving the problem. <cmath>p+r = 3</cmath> <cmath>pr-qs = 13</cmath> <cmath>-q-s = 4</cmath> S3 KB (457 words) - 14:08, 4 July 2024
- == Problem == {{AIME box|year=1996|num-b=4|num-a=6}}3 KB (585 words) - 19:39, 8 September 2024
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>3-4-5</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
- 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
- ...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
- ...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. ...ficulty=7-9|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
- * 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
- ...ypically contain a squared term such as <math>(x-3)^2</math>. However, the problem may be posed as to convert from an expanded form to a factored perfect squa All kinds of exotic factoring techniques are used on the [[AIME]], including completing the square.2 KB (422 words) - 16:20, 5 March 2023
- ...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>5! = 120</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 real r = 0.3, row1 = 3.5, row2 = 0, row3 = -3.5;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
- ...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
- ...ve results. Examples include the [[Monty Hall paradox]] and the [[birthday problem]]. Probability can be loosely defined as the chance that an event will hap <math>\mathit{P}(\{H\})=0.5</math>,4 KB (590 words) - 11:52, 28 September 2024
- 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
- ..., 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
