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Create the page "Problem 6" on this wiki! See also the search results found.
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- == Problem == Let <math>\triangle ABC</math> and <math>K, L, M</math> be as in the problem.5 KB (872 words) - 16:06, 22 September 2024
- ==Problem== label("$144$",(6,-7.5),N);1 KB (148 words) - 23:02, 29 August 2024
- ==Problem==1 KB (201 words) - 17:51, 19 December 2023
- == Problem == <math>\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10</math>664 bytes (103 words) - 00:11, 5 July 2013
- ==Problem==1 KB (144 words) - 13:55, 23 October 2016
- ==Problem== draw((0,0)--(15,0)--(15,6)--(12,6)--(12,9)--(0,9)--cycle);2 KB (277 words) - 11:37, 27 June 2023
- ==Problem== <math> 1,-2,3,-4,5,-6,\ldots, </math>888 bytes (149 words) - 14:12, 5 July 2013
- == Problem == Dividing the gray square into four smaller squares, there are <math>6</math> gray tiles and <math>10</math> white tiles, giving a ratio of <math>913 bytes (136 words) - 19:21, 8 August 2021
- ==Problem 6== ...= 6,000</math> gallons of water. At the rate it goes at it will take <math>6,000/2.5 = 2400</math> minutes, or <math>\boxed{\textbf{(A)}\ 40}</math> hou1 KB (158 words) - 08:23, 30 May 2023
- ==Problem== ...\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math>781 bytes (118 words) - 14:25, 12 January 2014
- ==Problem==1 KB (169 words) - 14:07, 5 July 2013
- == Problem == draw((0,6)--(4,6)--(4,4)--(3,4)--(3,0)--(1,0)--(1,4)--(0,4)--cycle, linewidth(1));</asy>809 bytes (117 words) - 21:45, 2 January 2023
- ==Problem==1 KB (185 words) - 18:59, 19 March 2024
- == Problem== <cmath>2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1=0</cmath>976 bytes (151 words) - 11:57, 5 July 2013
- ==Problem==1 KB (194 words) - 00:34, 27 December 2022
- ==Problem== ...solution was posted and copyrighted by DAFR. The original thread for this problem can be found here: [https://aops.com/community/p2751173]9 KB (1,788 words) - 00:02, 30 January 2021
- == Problem == <math>0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,</math>2 KB (325 words) - 20:05, 12 June 2024
- == Problem ==2 KB (293 words) - 21:24, 21 December 2011
- == Problem == ...equal <math>x</math> and <math>y</math>. From the information given in the problem, two equations can be written:1 KB (161 words) - 13:55, 1 July 2023
- #REDIRECT [[2012 AMC 10A Problems/Problem 8]]45 bytes (5 words) - 14:32, 12 February 2012
Page text matches
- == Problem == size(150); pathpen = linewidth(0.6); defaultpen(fontsize(10));3 KB (424 words) - 10:14, 17 December 2021
- == Problem == ...}{6}</math>. The probability of winning is <math>\frac{1}{2}\cdot \frac{1}{6} =\frac{1}{12}</math>. If the game is to be fair, the amount paid, <math>5<1 KB (207 words) - 09:39, 25 July 2023
- == Problem == draw((6,0){up}..{left}(0,6),blue);3 KB (532 words) - 17:49, 13 August 2023
- == Problem == D((2*t,2)--(2*t,4)); D((2*t,5)--(2*t,6));5 KB (732 words) - 23:19, 19 September 2023
- == Problem == So, there are <math>6 - 1 = 5</math> choices for the position of the letters.2 KB (252 words) - 15:33, 30 July 2024
- == Problem == ...5</math> possible values in <math>\bmod{5}</math> and we are picking <math>6</math> numbers, by the [[Pigeonhole Principle]], two of the numbers must be1 KB (187 words) - 08:21, 17 March 2023
- == Problem == ...gements for the non-<math>2</math> or <math>3</math> digits. We have <math>6 \cdot 56</math> = <math>336</math> arrangements for this case. We have <mat3 KB (525 words) - 20:25, 30 April 2024
- == Problem == <math>\textbf{(A) } \frac{1}{8}\qquad\textbf{(B) } \frac{1}{6}\qquad\textbf{(C) } \frac{1}{4}\qquad\textbf{(D) } \frac{1}{3}\qquad\textbf2 KB (292 words) - 10:19, 19 December 2021
- ...tiple 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
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. **[[2006 USAMO Problems/Problem 1]]467 bytes (51 words) - 09:25, 6 August 2014
- === Problem 1 === [[2006 USAMO Problems/Problem 1 | Solution]]3 KB (520 words) - 09:24, 14 May 2021
- ==Problem 1== [[1991 AJHSME Problems/Problem 1|Solution]]17 KB (2,246 words) - 13:37, 19 February 2020
- #REDIRECT [[2006 AIME I Problems/Problem 6]]44 bytes (5 words) - 12:05, 28 June 2009
- ==Problem== ...x digits <math>4,5,6,7,8,9</math> in one of the six boxes in this addition problem?1 KB (191 words) - 17:12, 29 October 2016
- ==Problem== ...metric sequence to be <math>\{ g, gr, gr^2, \dots \}</math>. Rewriting the problem based on our new terminology, we want to find all positive integers <math>m5 KB (883 words) - 01:05, 2 June 2024
- == Problem == ...The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find <math>1 KB (239 words) - 11:54, 31 July 2023
- == Problem == *Person 1: <math>\frac{9 \cdot 6 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{9}{28}</math>4 KB (628 words) - 11:28, 14 April 2024
- == Problem == ...d pair]]s <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. </mat3 KB (547 words) - 19:15, 4 April 2024
- == Problem == ...xample, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical st2 KB (384 words) - 00:31, 26 July 2018
- == Problem 1 == ...The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find <math>7 KB (1,119 words) - 21:12, 28 February 2020
