Search results
Create the page "Problem 6" on this wiki! See also the search results found.
Page title matches
- ==Problem==705 bytes (108 words) - 16:52, 19 January 2021
- ==Problem==3 KB (485 words) - 16:50, 5 August 2022
- ==Problem== ...use the <math>19</math> possible integer rod lengths that fall into <math>[6, 24]</math>. However, she has already used the rods of length <math>7</math1 KB (183 words) - 13:58, 19 December 2020
- ==Problem== ...extbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7</math>989 bytes (157 words) - 20:04, 1 May 2021
- ==Problem== The circle having <math>(0,0)</math> and <math>(8,6)</math> as the endpoints of a diameter intersects the <math>x</math>-axis a1 KB (217 words) - 14:19, 26 July 2022
- ==Problem 6==732 bytes (126 words) - 09:22, 23 October 2023
- ==Problem 6== ...not need to spend time factoring <math>x^2 - 120x + 3024</math>. Since the problem asks for <math>|x_1 - x_2|</math>, where <math>x_1</math> and <math>x_2</ma4 KB (726 words) - 16:18, 5 January 2024
- ==Problem== ==Solution 6==7 KB (1,076 words) - 14:13, 12 June 2024
- == Problem == ...uence starts as <math>-4, -6, -9</math>. The common ratio is <math>\frac{-6}{-4} = \frac{3}{2}</math>. The next term is <math>\frac{3}{2} \cdot -9 = \1 KB (223 words) - 04:12, 23 July 2019
- ==Problem== {{USAJMO newbox|year=2017|num-b=5|aftertext=|after=Last Problem}}8 KB (1,465 words) - 15:30, 12 June 2020
- ==Problem==399 bytes (66 words) - 20:57, 26 December 2017
- 3 KB (622 words) - 07:40, 22 May 2024
- == Problem ==1 KB (239 words) - 17:53, 16 January 2023
- <math>\emph{Problem:}</math> An international society has its members from six different countr ...Then there exists a partition of <math>\{1,2,3,\ldots, 1978\}</math> into 6 difference-free subsets <math>A,B,C,D,E,F</math>. A set S is difference-fre3 KB (515 words) - 19:45, 7 July 2023
- ==Problem==4 KB (715 words) - 04:21, 19 November 2023
- ==Problem==1 KB (202 words) - 15:27, 26 May 2024
- ==Problem==647 bytes (115 words) - 15:55, 28 January 2021
- == Problem 6 ==991 bytes (150 words) - 12:01, 13 February 2021
- ==Problem== {{IMO box|year=2017|num-b=5|after=Last Problem}}582 bytes (105 words) - 03:09, 19 November 2023
- ===Problem 6===476 bytes (82 words) - 21:44, 7 January 2018
Page text matches
- == Problem == {{AIME box|year=1987|num-b=4|num-a=6}}1 KB (160 words) - 04:44, 21 January 2023
- == Problem == Thus, listing out the first ten numbers to fit this form, <math>2 \cdot 3 = 6,\ 2^3 = 8,\ 2 \cdot 5 = 10,</math> <math>\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,3 KB (511 words) - 09:29, 9 January 2023
- == Problem == ...ring the day, and the boss delivers them in the order <math>1, 2, 3, 4, 5, 6, 7, 8, 9</math>.7 KB (1,186 words) - 10:16, 4 June 2023
- == Problem == ...+ 1</math> into the polynomial with a higher degree, as shown in Solution 6.10 KB (1,585 words) - 03:58, 1 May 2023
- == Problem == ...[[face]]s 12 [[square]]s, 8 [[regular polygon|regular]] [[hexagon]]s, and 6 regular [[octagon]]s. At each [[vertex]] of the polyhedron one square, one6 KB (902 words) - 17:40, 19 May 2024
- == Problem == ...this in to <math>5a^2 + 12a \equiv 1 \pmod{25}</math> gives <math>10a_1 + 6 \equiv 1 \pmod{25}</math>. Obviously, <math>a_1 \equiv 2 \pmod 5</math>, so6 KB (893 words) - 08:15, 2 February 2023
- == Problem == ...8}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot f(4,6)\\4 KB (538 words) - 13:24, 12 October 2021
- == Problem == {{AIME box|year=1988|num-b=6|num-a=8}}1 KB (178 words) - 23:25, 20 November 2023
- == Problem == [[Image:1988_AIME-6.png]]5 KB (878 words) - 23:06, 20 November 2023
- == Problem == {{AIME box|year=1988|num-b=4|num-a=6}}822 bytes (108 words) - 22:21, 6 November 2016
- == Problem == ...rder -- the correct five buttons. The sample shown below has <math>\{1,2,3,6,9\}</math> as its [[combination]]. Suppose that these locks are redesigned1 KB (181 words) - 18:23, 26 August 2019
- == Problem == ...e figure below). Given that <math>AP=6</math>, <math>BP=9</math>, <math>PD=6</math>, <math>PE=3</math>, and <math>CF=20</math>, find the area of <math>\13 KB (2,091 words) - 00:20, 26 October 2023
- == Problem == ...pairs: <math>[2,9]</math>, <math>[3,7]</math>, <math>[4,11]</math>, <math>[6,10]</math>. Now, <math>1989 = 180\cdot 11 + 9</math>. Because this isn't an2 KB (274 words) - 04:07, 17 December 2023
- == Problem == === Solution 6===8 KB (1,401 words) - 21:41, 20 January 2024
- == Problem == &>1.6\cdot1.6\cdot1.3 \\6 KB (874 words) - 15:50, 20 January 2024
- == Problem == ...math>f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7</cmath> for some <math>k\in\{1,2,3\}.</math>8 KB (1,146 words) - 04:15, 20 November 2023
- == Problem == {{AIME box|year=1989|num-b=6|num-a=8}}2 KB (320 words) - 15:21, 14 May 2024
- == Problem == pair A=(0,0),B=(10,0),C=6*expi(pi/3);6 KB (980 words) - 15:08, 14 May 2024
- == Problem == ...g a heads in one flip of the biased coin as <math>h</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3</math>.2 KB (258 words) - 00:07, 25 June 2023
- == Problem == ...> in which plugging into the other expression we know <math>3(25k^3 - 2) + 6 = 75k^3</math> is a perfect square. We know <math>75 = 5^2 \cdot 3</math> s3 KB (552 words) - 12:41, 3 March 2024
