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- ==Problem 6==6 KB (962 words) - 00:39, 1 September 2021
- == Problem == ...math>y = <math>x</math>. Plugging that in to the original equation, <math> 6/7 y</math> = <math>90</math> and <math> y =105</math>3 KB (457 words) - 14:38, 3 July 2023
- ==Problem== The mean and median are <cmath>\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,</cmath>so <math>3m+17=2n</math> and <math>m+11=n</math2 KB (349 words) - 20:56, 28 October 2022
- == Problem == ...\frac{1}{15} \qquad \textbf{(B)} \frac{1}{10} \qquad \textbf{(C)} \frac{1}{6} \qquad \textbf{(D)} \frac{1}{5} \qquad \textbf{(E)} \frac{1}{4}</math>2 KB (337 words) - 00:15, 11 November 2023
- ==Problem==2 KB (297 words) - 22:05, 27 May 2023
- ==Problem== .../math>. So <math>m\leq 6</math>. If <math>n\geq4</math> then <math>m-n\leq 6-n\leq 2</math>. Now assume that <math>n=3</math>. In this case if <math>m\l3 KB (560 words) - 16:02, 29 January 2021
- ==Problem==1 KB (216 words) - 10:58, 30 September 2022
- ==Problem 6== {{USAJMO newbox|year=2018|num-b=5|aftertext=|after=Last Problem}}2 KB (333 words) - 18:30, 7 April 2023
- ==Problem 6==4 KB (700 words) - 04:36, 5 March 2023
- ==Problem==1 KB (178 words) - 18:57, 17 May 2018
- 0 bytes (0 words) - 16:47, 11 October 2021
- == Problem ==1 KB (153 words) - 07:39, 5 June 2018
- == Problem 6==1 KB (242 words) - 22:16, 25 February 2018
- ==Problem== ...rs <math>z</math> with the properties that <math>|z|=1</math> and <math>z^{6!}-z^{5!}</math> is a real number. Find the remainder when <math>N</math> is7 KB (1,211 words) - 00:23, 20 January 2024
- ==Problem==2 KB (286 words) - 12:20, 27 March 2022
- ==Problem== {{IMO box|year=2012|num-b=5|after=Last Problem}}392 bytes (53 words) - 01:29, 19 November 2023
- ==Problem== {{IMO box|year=2013|num-b=5|after=Last Problem}}1,007 bytes (176 words) - 01:32, 19 November 2023
- ==Problem== <math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\text885 bytes (128 words) - 13:57, 20 February 2020
- ==Problem== [[File:2018 IMO 6.png|470px|right]]8 KB (1,407 words) - 01:47, 19 November 2023
- ==Problem== ...equal <math>5</math> is when <math>p = 1</math> or <math>p = \tfrac{\sqrt{6}}{3}</math>, which are not prime numbers. Thus, the rest of the primes can2 KB (241 words) - 00:10, 4 August 2018
Page text matches
- == Problem == ...28)+1=755161.</math> Since the alternating sum of the digits <math>7-5+5-1+6-1=11</math> is divisible by <math>11,</math> we conclude that <math>755161<4 KB (523 words) - 00:12, 8 October 2021
- == Problem == pair D=origin, A=(13,0), B=(13,12), C=(0,12), P=(6.5, 6);7 KB (1,086 words) - 08:16, 29 July 2023
- == Problem == ...gon can be written in the form <math>a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},</math> where <math>a^{}_{}</math>, <math>b^{}_{}</math>, <math>c^{}_{}</m6 KB (906 words) - 13:23, 5 September 2021
- == Problem == Someone observed that <math>6! = 8 \cdot 9 \cdot 10</math>. Find the largest [[positive]] [[integer]] <m3 KB (519 words) - 09:28, 28 June 2022
- == Problem == ...^{16}, \ldots n^{144}\}</math> and of set <math>B</math> as <math>\{n^3, n^6, \ldots n^{144}\}</math>. <math>n^x</math> can yield at most <math>144</mat3 KB (564 words) - 04:47, 4 August 2023
- == Problem == ...pattern, we find that there are <math>\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}3 KB (427 words) - 15:03, 24 June 2024
- == Problem == <!-- replaced: Image:1990 AIME Problem 7.png by I_like_pie -->8 KB (1,319 words) - 11:34, 22 November 2023
- == Problem == {{AIME box|year=1990|num-b=4|num-a=6}}1 KB (175 words) - 03:45, 21 January 2023
- == Problem == Find the value of <math>(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}</math>.5 KB (765 words) - 23:00, 26 August 2023
- == Problem == The [[increasing sequence]] <math>2,3,5,6,7,10,11,\ldots</math> consists of all [[positive integer]]s that are neithe2 KB (283 words) - 23:11, 25 June 2023
- == Problem == <math> \textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 6\qquad \textbf{(D) } 8\qquad \textbf{(E) } 9 </math>1 KB (172 words) - 10:47, 19 December 2021
- == Problem == {{AMC10 box|year=2006|ab=B|num-b=4|num-a=6}}1 KB (242 words) - 18:35, 15 August 2023
- == Problem == {{AMC10 box|year=2006|ab=B|num-b=6|num-a=8}}1 KB (179 words) - 10:33, 19 August 2022
- == Problem == ...bf{(A) } 1\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 6\qquad \textbf{(E) } 9 </math>1 KB (170 words) - 14:00, 26 January 2022
- == Problem == ...f{(C) }1 \qquad \textbf{(D) \ } \frac{14}{13}\qquad \textbf{(E) } \frac{7}{6} </math>1 KB (182 words) - 14:11, 26 January 2022
- == Problem == ...inds that <math>n=44</math> is the largest possible integer satisfying the problem conditions.7 KB (1,328 words) - 20:24, 5 February 2024
- == Problem == ...4\sqrt{3}\qquad \textbf{(C) } 8\qquad \textbf{(D) } 9\qquad \textbf{(E) } 6\sqrt{3} </math>3 KB (445 words) - 22:01, 20 August 2022
- == Problem == ...o more similar triangles, we find that point <math>E</math> is at <math>(9.6, 7.2)</math>.8 KB (1,270 words) - 23:36, 27 August 2023
- == Problem == ...multiples of <math>4</math> (with a few exceptions that don't affect this problem).2 KB (336 words) - 10:51, 11 May 2024
- == Problem == dot((cos(i*pi/6), sin(i*pi/6)));4 KB (740 words) - 17:46, 24 May 2024
