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- == Problem == 2^2+3^2+6^2 &= 7^2 \\1 KB (202 words) - 22:13, 19 August 2021
- == Problem == ...l of the positive integer divisors of <math>6^2=36</math> is <math>1+2+3+4+6+9+12+18+36=91</math>549 bytes (81 words) - 02:33, 13 January 2019
- == Problem ==834 bytes (136 words) - 01:45, 23 October 2014
- ==Problem==1 KB (194 words) - 14:31, 31 August 2015
- ==Problem== label("x", (2.25,6));1 KB (175 words) - 12:32, 1 March 2018
- ==Problem== path table = origin--(1,0)--(1,6)--(6,6)--(6,0)--(7,0)--(7,7)--(0,7)--cycle;2 KB (221 words) - 18:11, 1 April 2018
- == Problem ==871 bytes (143 words) - 20:57, 25 July 2016
- ==Problem==830 bytes (120 words) - 11:15, 2 July 2023
- ==Problem== 2xy + 2yz + 2xz &= 2x^2 + 2y^2 + 2z^2 - 6 \\2 KB (286 words) - 11:56, 17 March 2020
- ==Problem== draw((6,1+4*sqrt(3))--(4,1));3 KB (539 words) - 03:41, 25 December 2022
- == Problem ==926 bytes (128 words) - 09:43, 6 March 2024
- ...C 12A Problems|2015 AMC 12A #4]] and [[2015 AMC 10A Problems|2015 AMC 10A #6]]}} ==Problem==1 KB (203 words) - 23:05, 26 June 2023
- #REDIRECT[[2015 AMC 10A Problems/Problem 8]]44 bytes (5 words) - 20:44, 4 February 2015
- ==Problem== ...3, 5, 7, 9, 11,</math> hence the number of odd products is <math>6 \times 6 = 36</math>. Therefore the answer is <math>36/169 \doteq \boxed{\textbf{(A)1 KB (200 words) - 20:51, 18 January 2019
- ==Problem==1 KB (199 words) - 17:11, 2 August 2022
- ==Problem== [[File:2015 AIME I 6.png|500px|right]]5 KB (782 words) - 16:04, 21 July 2023
- ==Problem==417 bytes (65 words) - 23:54, 18 November 2023
- ==Problem== We get <math>(1, 4, 8)</math>, <math>(3, 6, 6)</math>, and <math>(4, 4, 7)</math>.5 KB (946 words) - 14:06, 14 February 2023
- ==Problem==767 bytes (132 words) - 16:33, 15 April 2015
- ==Problem == {{UMO box|year=2015|num-b=5|after=Last Problem}}642 bytes (99 words) - 03:03, 6 November 2015
Page text matches
- == Problem == ...in which the trio sits together (as a single entity). There are <math>3! = 6</math> ways to determine their order, and there are <math>(23-1)! = 22!</ma9 KB (1,392 words) - 20:37, 19 January 2024
- == Problem == There are <math>6</math> possible sequences.5 KB (855 words) - 20:26, 14 January 2023
- == Problem == ...>2s</math>. All other edges have length <math>s</math>. Given that <math>s=6\sqrt{2}</math>, what is the volume of the solid?5 KB (865 words) - 21:11, 6 February 2023
- == Problem == draw((-2.6,0)--(-2.6,0.6));2 KB (412 words) - 18:23, 1 January 2024
- == Problem == ...example, the alternating sum for <math>\{1, 2, 3, 6,9\}</math> is <math>9-6+3-2+1=5</math> and for <math>\{5\}</math> it is simply <math>5</math>. Find5 KB (894 words) - 22:02, 5 April 2024
- == Problem == In the adjoining figure, two circles with radii <math>8</math> and <math>6</math> are drawn with their centers <math>12</math> units apart. At <math>P13 KB (2,149 words) - 18:44, 5 February 2024
- == Problem == ...h>. Suppose that the radius of the circle is <math>5</math>, that <math>BC=6</math>, and that <math>AD</math> is bisected by <math>BC</math>. Suppose fu19 KB (3,221 words) - 01:05, 7 February 2023
- == Problem == D(A--B--C--cycle); D(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12);4 KB (726 words) - 13:39, 13 August 2023
- == Problem == == Solution 6 ==6 KB (863 words) - 16:10, 16 May 2024
- == Problem == [[Image:1984_AIME-6.png]]6 KB (1,022 words) - 19:29, 22 January 2024
- == Problem == Open up a coding IDE and use recursion to do this problem. The idea is to define a function (I called it <math>f</math>, you can call4 KB (617 words) - 22:09, 15 May 2024
- == Problem == The equation <math>z^6+z^3+1=0</math> has complex roots with argument <math>\theta</math> between3 KB (430 words) - 19:05, 7 February 2023
- == Problem == draw(surface(A--B--C--cycle),rgb(0.6,0.7,0.6),nolight);6 KB (947 words) - 20:44, 26 November 2021
- == Problem == ...>s, we are placing the extra <math>3</math> <math>n</math>s into the <math>6</math> intervals beside the <math>b</math>s.7 KB (1,115 words) - 00:52, 7 September 2023
- == Problem == ...either <math>0 \bmod{6}</math>, <math>2 \bmod{6}</math>, or <math>4 \bmod{6}</math>. Notice that the numbers <math>9</math>, <math>15</math>, <math>21<8 KB (1,346 words) - 01:16, 9 January 2024
- == Problem == ...math> \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 </math><br /><math> \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^6 KB (1,051 words) - 04:52, 8 May 2024
- == Problem == === Solution 6 ===6 KB (1,122 words) - 12:23, 6 January 2022
- == Problem == [[Image:AIME 1985 Problem 15.png]]2 KB (245 words) - 22:44, 4 March 2024
- == Problem == ...0 = n^2 + 19n + 90</math> and <math>n^2 -21n + 90 = 0</math> and <math>n = 6</math> or <math>n = 15</math>. Now, note that the top <math>n</math> playe5 KB (772 words) - 22:14, 18 June 2020
- == Problem == <cmath>\begin{alignat*}{6}17 KB (2,837 words) - 13:34, 4 April 2024
