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- ==Problem==1 KB (171 words) - 18:51, 28 July 2018
- ==Problem== ...f a palace is in a shape of regular hexagon, where the sidelength is <math>6 \text{ m}</math>. The floor of the hall is covered with an equilateral tri2 KB (255 words) - 00:13, 11 August 2018
- ==Problem==2 KB (286 words) - 01:33, 13 August 2018
- ==Problem==1 KB (176 words) - 12:40, 31 August 2018
- ==Problem== ...He drives [mathjax]50[/mathjax] miles, so he drives for [mathjax]\dfrac{5}{6}[/mathjax] hours, which is equal to [mathjax]50[/mathjax] minutes (Note tha2 KB (291 words) - 01:21, 4 April 2024
- == Problem == ...<math>i,j</math>; in other words, this sequence simultaneously solves the problem for all <math>a\geq 1</math> simultaneously.6 KB (1,068 words) - 03:20, 24 January 2024
- ==Problem== ...''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}1 KB (167 words) - 23:31, 3 November 2023
- 3 bytes (0 words) - 22:27, 31 March 2019
- ==Problem== {{USAMO newbox|year=2019|num-b=5|aftertext=|after=Last Problem}}3 KB (648 words) - 17:10, 20 February 2020
- 1 KB (164 words) - 19:46, 4 October 2023
- == Problem ==503 bytes (75 words) - 03:52, 12 January 2019
- == Problem == <math>a_2 = 2+0+1+6+2+0+1+6 = 18</math>659 bytes (94 words) - 20:02, 6 August 2023
- == Problem ==1 KB (167 words) - 04:18, 19 January 2019
- == Problem ==634 bytes (108 words) - 04:47, 20 January 2019
- ==Problem ==332 bytes (39 words) - 04:21, 22 January 2019
- ==Problem==2 KB (301 words) - 12:51, 16 July 2024
- 44 bytes (5 words) - 17:28, 9 February 2019
- {{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #6]] and [[2019 AMC 12B Problems|2019 AMC 12B #4]]}} ==Problem==3 KB (488 words) - 10:24, 24 June 2023
- 45 bytes (5 words) - 15:42, 14 February 2019
- ==Problem== ...own above. Note that <math>m\angle KPL = 90^{\circ}</math> as given in the problem. Since <math>\angle KPL \cong \angle KLN</math> and <math>\angle PKL \cong11 KB (1,717 words) - 20:11, 19 January 2024
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- ...k contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2002 Pan African MO Problems/Problem 1]]581 bytes (73 words) - 13:47, 4 December 2019
- == Problem == <math> \textbf{(A) } \frac{1}{10}\qquad \textbf{(B) } \frac{1}{6}\qquad \textbf{(C) } \frac{1}{5}\qquad \textbf{(D) } \frac{1}{3}\qquad \tex1 KB (211 words) - 04:32, 4 November 2022
- == Problem == Note: The problem is much easier computed if we consider what <math>\sec (x)</math> is, then10 KB (1,590 words) - 14:04, 20 January 2023
- == Problem == Clearly, the sequence repeats every 6 terms.1 KB (158 words) - 01:33, 29 May 2023
- == Problem == ...ath>(\pm1,\pm144),( \pm 2, \pm 72),( \pm 3, \pm 48),( \pm 4, \pm 36),( \pm 6, \pm 24),( \pm 8, \pm 18),( \pm 9, \pm 16),( \pm 12, \pm 12)</math>; since2 KB (310 words) - 11:25, 13 June 2023
- == Problem == {{AIME box|year=1991|num-b=6|num-a=8}}2 KB (285 words) - 05:15, 13 June 2022
- == Problem == {{AIME box|year=1991|num-b=4|num-a=6}}919 bytes (141 words) - 20:00, 4 July 2022
- == Problem == ...<math>-1 \le y \le 1</math>. It is [[periodic function|periodic]] (in this problem) with a period of <math>\frac{2}{5}</math>.2 KB (300 words) - 16:01, 26 November 2019
- == Problem == ...st going to be equivalent to multiplying this fraction by <math>\frac{995}{6}</math>. Notice that this fraction's numerator plus denominator is equal to5 KB (865 words) - 12:13, 21 May 2020
- == Problem == ...g this inequality may be found by Stars and Bars to be <math>\binom{7+6-1}{6-1} = \boxed{792}</math>.2 KB (443 words) - 22:41, 22 December 2021
- == Problem == ...> is irrelevant as long as there still exists a circle as described in the problem.5 KB (874 words) - 10:27, 22 August 2021
- == Problem == ...means <math>A</math> looks like <math>(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)</math>. More specifically, <math>A_n=a_1+k(n-1)+\frac{(n-1)(n-2)}{2}</5 KB (778 words) - 21:36, 3 December 2022
- == Problem == {{AIME box|year=1992|num-b=6|num-a=8}}800 bytes (114 words) - 17:40, 14 March 2017
- == Problem == ...math>h+d\ge 10</math> or <math>c+g\ge 10</math> or <math>b+f\ge 10</math>. 6. Consider <math>c \in \{0, 1, 2, 3, 4\}</math>. <math>1abc + 1ab(c+1)</mat3 KB (455 words) - 02:03, 10 July 2021
- == Problem == {{AIME box|year=1992|num-b=4|num-a=6}}2 KB (277 words) - 20:45, 4 March 2024
- == Problem == \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt}3 KB (476 words) - 14:13, 20 April 2024
- == Problem == ...math>1000n+3000>1006n+2012</math>, so <math>n<\frac{988}{6}=164 \dfrac {4}{6}=164 \dfrac{2}{3}</math>. Thus, the answer is <math>\boxed{164}</math>.2 KB (251 words) - 08:05, 2 January 2024
- == Problem == ...ntial ascending numbers, one for each [[subset]] of <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9\}</math>.2 KB (336 words) - 05:18, 4 November 2022
- == Problem == \qquad \mathrm{(C) \ } \pi(2-\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}+\frac{\sqrt{3}+1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}-1+\sqrt{3} </math5 KB (873 words) - 15:39, 29 May 2023
- == Problem == In rectangle <math>ABCD</math>, we have <math>A=(6,-22)</math>, <math>B=(2006,178)</math>, <math>D=(8,y)</math>, for some inte4 KB (594 words) - 15:45, 30 July 2023
