Search results
Create the page "Problem 6" on this wiki! See also the search results found.
Page title matches
- == Problem == [[Image:AIME_1998-6.png|350px]]2 KB (254 words) - 19:38, 4 July 2013
- == Problem == [[Image:1999_AIME-6.png]]2 KB (354 words) - 16:42, 20 July 2021
- == Problem == ...]s <math>(x,y)</math> of [[integer]]s is it true that <math>0 < x < y < 10^6</math> and that the [[arithmetic mean]] of <math>x</math> and <math>y</math6 KB (966 words) - 11:31, 24 June 2024
- == Problem == Recast the problem entirely as a block-walking problem. Call the respective dice <math>a, b, c, d</math>. In the diagram below,11 KB (1,729 words) - 20:50, 28 November 2023
- == Problem == ...{225}(x_2)=(3+\sqrt{5})+(3-\sqrt{5})=6</math> <math>\Rightarrow x_1x_2=225^6=15^{12}</math>.1 KB (194 words) - 19:55, 23 April 2016
- == Problem == ...h>{4\choose 3} = 4</math> triangles of the first type, and there are <math>6</math> faces, so there are <math>24</math> triangles of the first type. Eac3 KB (477 words) - 18:35, 27 December 2021
- == Problem == G=(6.3333,4);5 KB (787 words) - 17:38, 30 July 2022
- == Problem == ...ve terms are <math>1,2,..,9998</math>, while the negative ones are <math>5,6,...,10002</math>. Hence we are left with <math>1000 \cdot \frac{1}{4} (1 +2 KB (330 words) - 05:56, 23 August 2022
- == Problem == ...a = b</math> would imply <math>m = n</math>, and <math>m < n</math> in the problem, we must use the other factor. We get <math>b = 2/5a</math>, meaning the ra4 KB (772 words) - 19:31, 6 December 2023
- == Problem ==3 KB (433 words) - 19:42, 20 December 2021
- #REDIRECT[[2005 AMC 12B Problems/Problem 4]]44 bytes (5 words) - 10:51, 29 June 2011
- == Problem ==5 KB (986 words) - 22:46, 18 May 2015
- ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}} == Problem ==3 KB (528 words) - 18:29, 7 May 2024
- {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #6]] and [[2000 AMC 10 Problems|2000 AMC 10 #11]]}} ==Problem==1 KB (228 words) - 19:31, 29 April 2024
- == Problem == ...ty of South Carolina High School Math Contest/1993 Exam/Problem 5|Previous Problem]]2 KB (299 words) - 21:06, 5 July 2017
- ==Problem== *[[Mock AIME 1 2006-2007 Problems/Problem 5 | Previous Problem]]3 KB (460 words) - 15:52, 3 April 2012
- == Problem ==1 KB (157 words) - 10:51, 4 April 2012
- == Problem ==2 KB (383 words) - 05:58, 11 February 2024
- == Problem ==2 KB (275 words) - 20:33, 27 November 2023
- ==Problem==909 bytes (130 words) - 19:09, 25 December 2022
Page text matches
- * <math>3! = 6</math> * <math>6! = 720</math>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/Problem 1]]1 KB (133 words) - 12:32, 22 March 2011
- int n = 6; ...),red+linewidth(2));match(2,3,1); </asy>For <math>p=2,3</math> and <math>a=6,4</math>, respectively.</center>16 KB (2,658 words) - 16:02, 8 May 2024
- f.p=fontsize(6); f.p=fontsize(6);3 KB (551 words) - 16:22, 13 September 2023
- ...ast two digits of <math> 7^{81}-3^{81} </math>. ([[Euler's Totient Theorem Problem 1 Solution|Solution]]) ...ast two digits of <math> 3^{3^{3^{3}}} </math>. ([[Euler's Totient Theorem Problem 2 Solution|Solution]])3 KB (542 words) - 17:45, 21 March 2023
- ...uited to studying large-scale properties of prime numbers. The most famous problem in analytic number theory is the [[Riemann Hypothesis]]. ...es <math>G_4</math> and <math>G_6</math> are modular forms of weight 4 and 6 respectively.5 KB (849 words) - 16:14, 18 May 2021
- ...instance, if we tried to take the harmonic mean of the set <math>\{-2, 3, 6\}</math> we would be trying to calculate <math>\frac 3{\frac 13 + \frac 16 * [[2002 AMC 12A Problems/Problem 11]]1 KB (196 words) - 00:49, 6 January 2021
- ...if its units digit is divisible by 2, i.e. if the number ends in 0, 2, 4, 6 or 8. ...s at the end of it to be divisible by 1,000,000 because <math>1,000,000=10^6</math>.8 KB (1,298 words) - 15:07, 23 May 2024
- Other, odder inductions are possible. If a problem asks you to prove something for all integers greater than 3, you can use <m ...ernational Mathematics Olympiad | IMO]]. A good example of an upper-level problem that can be solved with induction is [http://www.artofproblemsolving.com/Fo5 KB (768 words) - 20:45, 1 September 2022
- An example of a classic problem is as follows: ...hem twice. A number that is divisible by both 2 and 3 must be divisible by 6, and there are 16 such numbers. Thus, there are <math>50+33-16=\boxed{67}</4 KB (635 words) - 12:19, 2 January 2022
- This is a problem where constructive counting is not the simplest way to proceed. This next e ...wer is <math>8 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 = 8 \cdot 7^6 = 941,192</math>, as desired. <math>\square</math>12 KB (1,896 words) - 23:55, 27 December 2023
- Similarly, <math>42 \equiv 6 \pmod{9}</math>, so <math>\gcd(42,9) = \gcd(9,6)</math>. <br/> Continuing, <math>9 \equiv 3 \pmod{6}</math>, so <math>\gcd(9,6) = \gcd(6,3)</math>. <br/>6 KB (924 words) - 21:50, 8 May 2022
- ...multiply the functions together, getting <math>1+3x+6x^2+8x^3+8x^4+6x^5+3x^6+x^7</math>. We want the number of ways to choose 4 eggs, so we just need to4 KB (659 words) - 12:54, 7 March 2022
- ...th>\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots</cmath> for all <math>x</math>. ...[[polynomial]]s with [[binomial coefficient]]s. For example, if a contest problem involved the polynomial <math>x^5+4x^4+6x^3+4x^2+x</math>, one could factor5 KB (935 words) - 13:11, 20 February 2024
- ...function: <math>f(x) = x^2 + 6</math>. The function <math>g(x) = \sqrt{x-6}</math> has the property that <math>f(g(x)) = x</math>. Therefore, <math>g ([[2006 AMC 10A Problems/Problem 2|Source]])10 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
- [[2008 AMC 12B Problems/Problem 22]] [[2001 AIME I Problems/Problem 6]]1,016 bytes (141 words) - 03:39, 29 November 2021
- .../www.artofproblemsolving.com/Forum/viewtopic.php?p=394407#394407 1986 AIME Problem 11] ...lving.com/Forum/resources.php?c=182&cid=45&year=2000&p=385891 2000 AIME II Problem 7]12 KB (1,993 words) - 20:06, 23 June 2024
- ...- 3)^2 + (y + 6)^2 = 25</math> represents the circle with center <math>(3,-6)</math> and radius 5 units. ([[2006 AMC 12A Problems/Problem 16|Source]])9 KB (1,581 words) - 18:59, 9 May 2024
- ...umber can be rewritten as <math>2746_{10}=2\cdot10^3+7\cdot10^2+4\cdot10^1+6\cdot10^0.</math> ...of the decimal place (recall that the decimal place is to the right of the 6, i.e. 2746.0) tells us that there are six <math>10^0</math>'s, the second d4 KB (547 words) - 17:23, 30 December 2020
