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- == Problem == == Solution 1 ==1 KB (174 words) - 16:10, 3 February 2025
- ==Problem== ==Solution 1 (Linear Polynomials)==4 KB (670 words) - 13:03, 13 November 2023
- == Problem == ...given by the [[binomial coefficient]] <math>{n \choose 6} = \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}<1 KB (239 words) - 11:54, 31 July 2023
- == Problem == [[Image:2005 AIME I Problem 1.png]]1 KB (213 words) - 13:17, 22 July 2017
- == Problem == == Solution 1==2 KB (250 words) - 23:56, 2 December 2024
- == Problem == [[Image:2004 AIME II Problem 1.png]]2 KB (329 words) - 23:20, 4 July 2013
- == Problem == Let <math>x</math>, <math>y</math> and <math>z</math> all exceed <math>1</math> and let <math>w</math> be a positive number such that <math>\log_x w4 KB (641 words) - 10:19, 6 January 2025
- == Problem == ...h>a_3\ldots</math> is an [[arithmetic progression]] with common difference 1, and <math>a_1+a_2+a_3+\ldots+a_{98}=137</math>.4 KB (576 words) - 15:33, 24 July 2024
- == Problem == {{AIME box|year=1986|before=First Question|num-a=2}}688 bytes (104 words) - 13:34, 22 July 2020
- == Problem == ...s, the number of [[ordered pair]]s will be <math>(1 + 1)(4 + 1)(9 + 1)(2 + 1) = 2\cdot 5\cdot 10\cdot 3 = \boxed{300}</math>.1 KB (191 words) - 14:42, 17 September 2016
- == Problem == ...any order -- the correct five buttons. The sample shown below has <math>\{1,2,3,6,9\}</math> as its [[combination]]. Suppose that these locks are redes1 KB (181 words) - 18:23, 26 August 2019
- == Problem == Compute <math>\sqrt{(31)(30)(29)(28)+1}</math>.4 KB (523 words) - 00:12, 8 October 2021
- == Problem == == Solution 1==2 KB (283 words) - 23:11, 25 June 2023
- == Problem == == Solution 1 ==4 KB (628 words) - 22:05, 7 June 2021
- == Problem == ===Solution 1===1 KB (190 words) - 20:02, 23 February 2022
- == Problem == == Solution 1 ==3 KB (440 words) - 21:20, 22 July 2021
- == Problem == ...= 4 + (997-1) \cdot 3 = 2992</math>. The value of <math>n^2 - 1 = 2992^2 - 1 \pmod{1000}</math> is <math>\boxed{063}</math>.946 bytes (139 words) - 21:05, 1 September 2023
- == Problem == ...<math>S_{i+2}.</math> The total area enclosed by at least one of <math>S_{1}, S_{2}, S_{3}, S_{4}, S_{5}</math> can be written in the form <math>m/n,</2 KB (302 words) - 19:29, 4 July 2013
- == Problem == 1&a&b\\hline2 KB (332 words) - 11:28, 4 August 2021
- == Problem == How many of the integers between 1 and 1000, inclusive, can be expressed as the [[difference of squares|differ801 bytes (115 words) - 15:52, 2 March 2020
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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 === Proof 1 ===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 ...Sets [//competifyhub.com/resources/ Free Competition Resources for Grades 1-12]14 KB (1,913 words) - 23:52, 6 March 2025
- \1-\frac{13}{x+5}+4&\ge 3 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
- ...th Jam''' is a free online class or information session hosted by [[Art of Problem Solving]] in the [[AoPS Schoolhouse|classroom]]. ...discussion of the problems from each year's [[AMC 10]], [[AMC 12]], and [[AIME]] exams.891 bytes (116 words) - 18:36, 31 January 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 == ...lutions <math>x,y,z</math> of the equation <math>x^2+5y^2+10z^2=4xy+6yz+2z-1</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
- ...as a great introductory video to combinations, permutations, and counting problems in general! [https://bit.ly/CombinationsAndPermutations Permutations & Comb * <math>\binom{n-1}{r-1}+\binom{n-1}{r}=\binom{n}{r}</math>4 KB (638 words) - 21:55, 5 January 2025
- ...[recursion|recursive definition]] for the factorial is <math>n!=n \cdot (n-1)!</math>. * <math>0! = 1</math> (remember! this is 1, not 0! (the '!' was an exclamation mark, not a factorial sign))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
- ...ot [[divisibility|divisible]] by <math>p</math>, then <math>a^{p-1} \equiv 1 \pmod{p}</math>. ...denotes [[Euler's totient function]]. In particular, <math>\varphi(p) = p-1</math> for prime numbers <math>p</math>. In turn, this is a special case of15 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
- ...er [[relatively prime]] to <math>a</math>, then <math>{a}^{\phi (m)}\equiv 1 \pmod {m}</math>. ...hi(m)} \pmod{m} </math> <math> \implies </math> <math> a^{\phi (m)} \equiv 1 \pmod{m}</math> as desired. Note that dividing by <math> n_1 n_2 ... n_{\ph4 KB (569 words) - 22:34, 30 December 2024
- Substituting <math>\sin^2B=1-\cos^2B</math> results in <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
