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- ==Problem==900 bytes (126 words) - 18:16, 15 October 2023
- == Problem ==5 KB (871 words) - 18:59, 10 May 2023
- == Problem ==2 KB (324 words) - 20:45, 2 January 2018
- #REDIRECT [[2003 AMC 10B Problems/Problem 8]]45 bytes (5 words) - 00:17, 5 January 2014
- ==Problem==1 KB (190 words) - 11:26, 13 June 2022
- ==Problem == ...extbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 10 </math>518 bytes (78 words) - 18:23, 24 September 2016
- ==Problem==2 KB (277 words) - 13:08, 1 July 2023
- == Problem== Since the problem doesn't specify the number of 3-point shots she attempted, it can be assume3 KB (419 words) - 11:39, 10 March 2024
- #REDIRECT [[2013 AMC 12B Problems/Problem 5]]45 bytes (5 words) - 12:10, 7 April 2013
- {{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #6]] and [[2013 AMC 10B Problems|2013 AMC 10B #11]]}} ==Problem==2 KB (291 words) - 18:04, 14 July 2021
- ==Problem== ...is equivalent to <math>\frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6}{144}=12\cdot11\cdot10\cdot7\cdot3</math>. If all of the math textbooks a5 KB (831 words) - 17:20, 9 January 2024
- == Problem ==886 bytes (130 words) - 18:13, 5 September 2021
- ==Problem 6== ...is small, so <math>10=6x+9</math>. <math>x=1/6\implies \sqrt{10}\approx 19/6</math>. This is 3.16.5 KB (842 words) - 19:42, 15 June 2024
- ==Problem==4 KB (691 words) - 18:29, 10 May 2023
- ==Problem== ...more than 2/5 of the contestants. Moreover, no contestant solved all the 6 problems. Show that there are at least 2 contestants who solved exactly 5 p394 bytes (56 words) - 01:00, 19 November 2023
- ==Problem== ...ls a six twice when using the fair die is <math>\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}</math>. The probability that he rolls a six twice using the2 KB (356 words) - 09:03, 14 June 2021
- ==Problem 6== Again, we experiment. If x = 2, y = 3, and z = 3, then <math>18 > 7 + 4\sqrt{6}</math>.3 KB (517 words) - 20:02, 30 April 2014
- == Problem ==530 bytes (88 words) - 01:35, 16 August 2023
- 780 bytes (136 words) - 10:02, 3 June 2021
- ==Problem== ...equilateral triangle (it is well-known that this is possible; in fact, the problem here is the <math>3</math>-dimensional version of this), and then complete7 KB (1,370 words) - 15:42, 29 January 2021
Page text matches
- == Problem == real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD;4 KB (729 words) - 01:00, 27 November 2022
- == Problem == ...19}+ \frac3{19}+ \frac 1{17}+ \frac2{17}= \frac6{19} + \frac 3{17} = \frac{6\cdot17 + 3\cdot19}{17\cdot19} = \frac{159}{323}</math> and so the answer is2 KB (298 words) - 20:02, 4 July 2013
- == Problem == ...left(\frac{\frac{4}{5}}{1-\frac{1}{25}}\right)= \frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9},</cmath> and the answer is <math>m+n = 5 + 9 = \boxed{014}</2 KB (303 words) - 22:28, 11 September 2020
- == Problem == ...th>r_{ABC} = \frac{[ABC]}{s_{ABC}} = \frac{15 \cdot 36 /2}{(15+36+39)/2} = 6</math>. Thus <math>r_{A'B'C'} = r_{ABC} - 1 = 5</math>, and since the ratio5 KB (836 words) - 07:53, 15 October 2023
- == Problem == ...th> be a [[triangle]] with sides 3, 4, and 5, and <math> DEFG </math> be a 6-by-7 [[rectangle]]. A segment is drawn to divide triangle <math> ABC </math4 KB (618 words) - 20:01, 4 July 2013
- == Problem == ...n + 1) + n = 3210 + 1111n</math>, for <math>n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace</math>.2 KB (374 words) - 14:53, 27 December 2019
- == Problem == The thing about this problem is, you have some "choices" that you can make freely when you get to a cert8 KB (1,437 words) - 21:53, 19 May 2023
- == Problem == Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted quest3 KB (436 words) - 18:31, 9 January 2024
- == Problem == ...\le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{align*}</cmath>6 KB (941 words) - 11:37, 27 May 2024
- == Problem == There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars are similar, but there are two n4 KB (620 words) - 21:26, 5 June 2021
- == Problem 1 == [[2004 AIME I Problems/Problem 1|Solution]]9 KB (1,434 words) - 13:34, 29 December 2021
- == Problem == Now, consider the strip of length <math>1024</math>. The problem asks for <math>s_{941, 10}</math>. We can derive some useful recurrences f6 KB (899 words) - 20:58, 12 May 2022
- == Problem == ...n't need to be nearly as rigorous). A more natural manner of attacking the problem is to think of the process in reverse, namely seeing that <math>n \equiv 111 KB (1,857 words) - 21:55, 19 June 2023
- == Problem == ...CD </math> be an [[isosceles trapezoid]], whose dimensions are <math> AB = 6, BC=5=DA, </math>and <math> CD=4. </math> Draw [[circle]]s of [[radius]] 33 KB (431 words) - 23:21, 4 July 2013
- == Problem == ...}</math>. For example, with the binary string 0001001000 <math>y</math> is 6 and <math>x</math> is 3 (note that it is zero indexed).8 KB (1,283 words) - 19:19, 8 May 2024
- == Problem == ...ays and likewise we can partition the 167 in three ways. So we have <math>6\cdot 3\cdot 3 = \boxed{54}</math> as our answer.2 KB (353 words) - 18:08, 25 November 2023
- == Problem == It is clear from the problem setup that <math>0<\theta<\frac\pi2</math>, so the correct value is <math>\9 KB (1,501 words) - 05:34, 30 October 2023
- == Problem == ...al. Also note some other conditions we have picked up in the course of the problem, namely that <math>b_1</math> is divisible by <math>8</math>, <math>b_2</ma6 KB (950 words) - 14:18, 15 January 2024
- ==Problem== From the initial problem statement, we have <math>1000w\cdot\frac{1}{4}t=\frac{1}{4}</math>.4 KB (592 words) - 19:02, 26 September 2020
- == Problem 1 == [[2004 AIME II Problems/Problem 1|Solution]]9 KB (1,410 words) - 05:05, 20 February 2019
