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- ==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) - 09:16, 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) - 20:16, 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) - 15:01, 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) - 15:14, 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) - 16:15, 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) - 16:16, 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) - 16:38, 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) - 13:32, 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) - 11:01, 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) - 17:03, 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) - 23:22, 31 December 2020
- If <math>a=2</math> and <math>b=0</math>, then <math>g=0=b</math>, a contradiction. If <math>a=2</math> and <math>b=1</math>, then <math>f=1=b</math>, a contradiction.2 KB (335 words) - 05:41, 5 September 2023
- The number of ways to choose 2 suits: \(\binom{4}{2} = 6\)543 bytes (64 words) - 15:34, 24 May 2024
Page text matches
- ...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 Mat8 KB (1,062 words) - 18:04, 17 January 2025
- ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC and of the recent expansion ...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 (874 words) - 22:02, 10 November 2024
- ...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) - 18:35, 29 October 2024
- ...A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member ...AMC]] competition. There is no guarantee that community members will make Mock AMCs in any given year, but there probably will be one.51 KB (6,175 words) - 20:41, 27 November 2024
- The '''Mock AIME 2 Pre 2005''' was written by [[Art of Problem Solving]] community member Mildorf. * [[Mock AIME 2 Pre 2005 Problems|Entire Exam]]2 KB (181 words) - 09:58, 18 March 2015
- The '''Mock AIME 7 Pre 2005''' was written by [[Art of Problem Solving]] community member Mildorf. * [[Mock AIME 7 Pre 2005 Problems|Entire Exam]]1 KB (146 words) - 15:33, 14 October 2022
- The '''Mock AIME 1 2005-2006''' was written by [[Art of Problem Solving]] community member paladin8. * [[Mock AIME 1 2005-2006/Answer Key|Answer Key]]1 KB (135 words) - 16:41, 21 January 2017
- == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,100 words) - 21:35, 9 January 2016
- ==Problem 1== [[Mock AIME 3 Pre 2005/Problem 1|Solution]]7 KB (1,135 words) - 22:53, 24 March 2019
- ==Problem== {{Mock AIME box|year=Pre 2005|n=3|before=First Question|num-a=2}}795 bytes (129 words) - 09:22, 4 April 2012
- ==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) - 09:16, 29 November 2019
- ==Problem== 0 &= x^2 - \frac{3 \times 2004 - 4}{10}x + \frac 52\end{align*}</cmath>1 KB (191 words) - 09:22, 4 April 2012
- ==Problem== <cmath>\zeta_1^2+\zeta_2^2+\zeta_3^2=3</cmath>2 KB (221 words) - 01:49, 19 March 2015
- ==Problem== ...VC</tt> - the only other combination, two vowels, is impossible due to the problem statement). Then, note that:5 KB (795 words) - 15:03, 17 October 2021
- ==Problem== <cmath>\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}</cmath>3 KB (501 words) - 13:48, 29 November 2019
- ==Problem== ...algebraically or by inspection) to get that <math>BC=3</math> and <math>AD=2</math>.2 KB (330 words) - 09:23, 4 April 2012
- ==Problem== <math>\left|a_1^{2} - a_2^{2}\right| = 10</math>3 KB (520 words) - 11:55, 11 January 2019
- ==Problem== .../math> such that <math>\angle{BAE}</math> is right. If <math>BD = 15, DE = 2,</math> and <math>BC = 16</math>, then <math>CD</math> can be expressed as2 KB (278 words) - 15:32, 27 December 2019
- ==Problem== <math>a_{n} = 2a_{n-1} + n^2</math>2 KB (306 words) - 09:36, 4 April 2012
- ==Problem== ...second equation and rearranging variables gives <math>45x=8\cdot 15+bx=15^2\implies x=5</math>. Back-substitution yields <math>AC=21</math> and consequ2 KB (379 words) - 00:27, 6 December 2024