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Create the page "Problem 6" on this wiki! See also the search results found.
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- ==Problem==538 bytes (85 words) - 06:09, 3 October 2014
- ...C 12B Problems|2003 AMC 12B #5]] and [[2003 AMC 10B Problems|2003 AMC 10B #6]]}} ==Problem==1 KB (220 words) - 14:53, 20 October 2020
- == Problem == ...s is <math>28 + 5x - {x \choose 2}</math>, which is maximized at x=5 and x=6, and the maximum value is <math>43</math>. Choosing the first 5 numbers as3 KB (477 words) - 17:52, 15 January 2022
- == Problem ==1 KB (157 words) - 14:28, 5 July 2013
- ==Problem== <math>\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math>2 KB (294 words) - 17:52, 26 October 2020
- ==Problem==1 KB (184 words) - 12:49, 5 July 2013
- ==Problem== Use logic to solve this problem. You don't actually need to use any equations.903 bytes (171 words) - 21:17, 30 October 2016
- ==Problem==1,012 bytes (143 words) - 00:26, 5 July 2013
- == Problem ==6 KB (1,107 words) - 14:12, 12 April 2023
- {{IMO box|year=2011|num-b=5|after=Last Problem}}2 KB (317 words) - 01:22, 19 November 2023
- == Problem == {{IMO box|year=1966|num-b=5|after=Last Problem}}2 KB (397 words) - 00:37, 17 May 2015
- ==Problem== label("$144$",(6,-7.5),N);845 bytes (116 words) - 00:47, 5 July 2013
- ==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
Page text matches
- == Problem == ...h> can have is <math>(1+2+3+4)=10</math>, and the greatest value is <math>(6+7+8+9)=30</math>. Therefore, <math>(e+f+g+h)</math> must equal <math>11</ma2 KB (411 words) - 21:02, 21 December 2020
- == Problem == f.p=fontsize(6);2 KB (262 words) - 21:20, 21 December 2020
- == Problem == ...h>. Now we have <math>(a+b)+(a-b)=11+1</math>, <math>2a=12</math>, <math>a=6</math>, then <math>b=5</math>, <math>x=65</math>, <math>y=56</math>, <math>2 KB (283 words) - 20:02, 24 December 2020
- == Problem == ...and <math>h</math> be distinct elements in the set <math>\{-7,-5,-3,-2,2,4,6,13\}.</math>3 KB (463 words) - 19:28, 6 November 2022
- == Problem == Our sum is simply <math>2 - 2\cdot\frac{8}{11} = \frac{6}{11}</math>, and thus we can divide by <math>2</math> to obtain <math>\frac4 KB (761 words) - 09:10, 1 August 2023
- == Problem == We approach this problem by counting the number of ways ants can do their desired migration, and the10 KB (1,840 words) - 21:35, 7 September 2023
- == Problem == Applying the Power of a Point Theorem gives <math> 6\cdot x = 4\cdot 1 </math>, so <math> x = \frac 23 </math>289 bytes (45 words) - 13:14, 16 July 2017
- ==Problem 1== [[2006 AMC 10A Problems/Problem 1|Solution]]13 KB (2,028 words) - 16:32, 22 March 2022
- == Problem == <math>b=-6</math>2 KB (348 words) - 23:10, 16 December 2021
- ...k contains the full set of test problems. The rest contain each individual problem and its solution. **[[2008 USAMO Problems/Problem 1]]471 bytes (52 words) - 21:46, 12 August 2014
- == 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 (254 words) - 14:39, 5 April 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 (2,021 words) - 00:00, 17 July 2011
- ...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
