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- == Problem == *Person 2: <math>\frac{6 \cdot 4 \cdot 2}{6 \cdot 5 \cdot 4} = \frac 25</math>4 KB (628 words) - 11:28, 14 April 2024
- == Problem == ...values of <math> k </math> does <math> S_k </math> contain the term <math>2005</math>?2 KB (303 words) - 01:31, 5 December 2022
- ==Problem== ...ets of <math>7</math> digits, consider <math>9</math> urns labeled <math>1,2,\cdots,9</math> (note that <math>0</math> is not a permissible digit); then950 bytes (137 words) - 10:16, 29 November 2019
- == Problem == ...y tangent to a larger circle <math>\omega_2</math> of radius <math>12\sqrt{2}</math> such that the center of <math>\omega_2</math> lies on <math>\omega_752 bytes (117 words) - 21:16, 8 October 2014
- == Problem == ...math>. We focus on the large prime <math>167</math> as the powers of <math>2</math> and <math>3</math> in the prime factorization of <math>2004!</math>643 bytes (89 words) - 16:01, 4 August 2019
- == Problem == ...ath>x^4 + \dfrac{1}{x^4} = \left(x^2 + \dfrac{1}{x^2}\right)^2 - 2 = 7^2 - 2 = 47.</cmath> Finally, <cmath>x^7 + \dfrac{1}{x^7} = \left(x^3 + \dfrac{1}{883 bytes (128 words) - 16:14, 4 August 2019
- == Problem == ...ox, there are <math>4</math> green balls, <math>4</math> blue balls, <math>2</math> red balls, a brown ball, a white ball, and a black ball. These balls1 KB (170 words) - 17:15, 4 August 2019
- == Problem == ...th> numbers with leading digit <math>1</math> among the set <math>\{5^1, 5^2, 5^3, \cdots 5^{2003}\}.</math> However, <math>5^0</math> also starts with817 bytes (114 words) - 17:16, 4 August 2019
- == Problem == ...3^3 \cdot 37,</cmath> the number <math>10^{12} -1</math> has <math>4 \cdot 2^6 = 256</math> divisors and our answer is <math>256 - 1 = \boxed{255}.</mat1 KB (171 words) - 17:38, 4 August 2019
- ==Problem== ...utations of <math>(4, 5)</math>, so there are a total of <math>2 \cdot 1 = 2</math> ways for this case.4 KB (690 words) - 14:32, 6 January 2021
- ...ro; modulo five, the remainder is <math>2^{2001\pmod{5}} \equiv 2^1 \equiv 2\pmod{5}</math>, so we have <math>2002^{2001} \equiv 12\pmod{20}</math>. {{Mock AIME box|year=Pre 2005|n=2|num-b=7|num-a=9}}1 KB (188 words) - 12:01, 10 August 2020
- ...a</math> are the roots of <math>x(x-200)(x+1/4)-1/4 = x^3 - \frac{799}{4}x^2 - 50x - \frac{1}{4}</math>. By Vieta's formulas, we have: {{Mock AIME box|year=Pre 2005|n=2|num-b=10|num-a=12}}1 KB (192 words) - 18:03, 13 September 2020
- == Problem == Let <cmath>(1+x^3)\left(1+2x^{3^2}\right)\cdots \left(1+kx^{3^k}\right) \cdots \left(1+1997x^{3^{1997}}\right2 KB (232 words) - 00:22, 1 January 2021
- ==Problem== &\text {i. In form of abcde and fghij such that fghij = 2(abcde)}\2 KB (428 words) - 17:27, 23 February 2025
- The number of ways to choose 2 suits: \(\binom{4}{2} = 6\)543 bytes (64 words) - 16:34, 24 May 2024
Page text matches
- ...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>\sum_{x=0}^{n} \binom{n}{x} = 2^n</math>4 KB (638 words) - 21:55, 5 January 2025
- ...th>, <math>d</math> are the four side lengths and <math>s = \frac{a+b+c+d}{2}</math>. .../math>. Hence, <math>[ABCD]=\frac{\sin B(ab+cd)}{2}</math>. Multiplying by 2 and squaring, we get:3 KB (543 words) - 19:35, 29 October 2024
- This is a problem where constructive counting is not the simplest way to proceed. This next e === Example 2 ===13 KB (2,018 words) - 15:31, 10 January 2025
- ...llipse is the origin this simplifies to <math>\frac{x^2}{a^2}+\frac{y^2}{b^2}=1</math>. ...<math>P</math>. Then it is true that the bisector of angle <math>F_{1}PF_{2}</math> is perpendicular to <math>\ell</math>.5 KB (892 words) - 21:52, 1 May 2021
- ...mplex numbers''' arise when we try to solve [[equation]]s such as <math> x^2 = -1 </math>. ...from this addition, we are not only able to find the solutions of <math> x^2 = -1 </math> but we can now find ''all'' solutions to ''every'' polynomial.5 KB (860 words) - 15:36, 10 December 2023
- ...0, AMC 12, and AIME competitions it is very common to use WLOG in geometry problems by assuming one undefined side length to be equal to 1. == Problems using WLOG ==2 KB (318 words) - 23:35, 13 September 2024
- ..., -50, 25, -25/2</math> is a geometric sequence with common ratio <math>-1/2</math>; however, <math>1, 3, 9, -27</math> and <math>-3, 1, 5, 9, \ldots</m ...onsecutive terms <math>a_{n-1}, a_n, a_{n+1}</math>. In symbols, <math>a_n^2 = a_{n-1}a_{n+1}</math>. This is mostly used to perform substitutions, thou4 KB (649 words) - 21:09, 19 July 2024
- For example, <math>1, 2, 3, 4</math> is an arithmetic sequence with common difference <math>1</math ...-1}, a_n, a_{n+1}</math>. In symbols, <math>a_n = \frac{a_{n-1} + a_{n+1}}{2}</math>. This is mostly used to perform substitutions, though it occasional4 KB (736 words) - 02:00, 7 March 2024
- ...ree [[integer]]s that satisfy the [[Pythagorean Theorem]], <math>a^2+b^2=c^2</math>. There are three main methods of finding Pythagorean triples: ...th> is an [[odd]] number, then <math>m, \frac {m^2 -1}{2}, \frac {m^2 + 1}{2}</math> is a Pythagorean triple.9 KB (1,434 words) - 01:15, 4 July 2024
- ...s a [[secant line|secant]] (middle figure). In this case, we have <math>AB^2 = BC\cdot BD</math>. <cmath>PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i</cmath>5 KB (948 words) - 17:04, 21 February 2025
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME II Problems]]1 KB (135 words) - 12:24, 22 March 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME I Problems]]1 KB (154 words) - 12:30, 22 March 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME I Problems]]1 KB (135 words) - 12:31, 22 March 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME II Problems]]1 KB (135 words) - 12:30, 22 March 2011
- ...ake. Sometimes, the administrator may ask other people to sign up to write problems for the contest. * Look at past [[AMC]]/[[AHSME]] tests to get a feel for what kind of problems you should write and what difficulty level they should be.51 KB (6,175 words) - 21:41, 27 November 2024
- ...in the AoPSWiki, please consider adding them. Also, if you notice that a problem in the Wiki differs from the original wording, feel free to correct it. Fi | 2025 || [[2025 AIME I | AIME I]] || [[2025 AIME II | AIME II]]3 KB (400 words) - 20:21, 13 February 2025
- {{AIME Problems|year=2006|n=I}} == Problem 1 ==7 KB (1,173 words) - 03:31, 4 January 2023
- ...ple indistinct elements, such as the following: <math>\{1,4,5,3,24,4,4,5,6,2\}</math> Such an entity is actually called a multiset. ...describe sets should be used with extreme caution. One way to avoid this problem is as follows: given a property <math>P</math>, choose a known set <math>T<11 KB (2,019 words) - 17:20, 7 July 2024
